Answer:
True
Explanation:
In 1909 Ernest Rutherford proposed to Geiger and Marsden to conduct an experiment in which they would have to launch alpha particles from a radioactive source against a gold foil a few atoms thick.
The aim of the experiment was to corroborate Rutherford's idea that the particles would pass through the metal foil with little deviation.
Rutherford's interpretation of the experimental results gave rise to a new atomic model, and concluded that the mass of the atom was concentrated in a small region of positive charges that impeded the passage of alpha particles. He suggested a new model in which the atom had a nucleus or center in which the mass and the positive charge are concentrated, and that in the extra nuclear zone are the negatively charged electrons.
A seed crystal of diameter D (mm) is placed in a solution of dissolved salt, and new crystals are observed to nucleate (form) at a constant rate r (crystals/min). Experiments with seed crystals of different sizes show that the rate of nucleation varies with the seed crystal diameter as r(crystals/min)=200D−10D2(D in mm)
a. What are the units of the constants 200 and 10? (Assume the given equation is valid and therefore dimensionally homogeneous.)
b. Calculate the crystal nucleation rate in crystals/s corresponding to a crystal diameter of 0.050 inch.
c. Derive a formula for r(crystals/s) in terms of D(inches). (See Example 2.6-1.) Check the formula using the result of Part b.
Answer:
Explanation:
a) the units of the constants 200 and 10 are as follows:
unit of 200 = unit of r / unit of D
= crystals/min× mm
= crystals / (min×mm)
unit of 10 = unit of r / unit of D^2
= crystals/min × mm²
= crystals / (min×mm²)
b) The objective here is to determine the crystal nucleation rate in crystals/s corresponding to a crystal diameter of 0.050 inch; T o do that ; let's first convert the inch to mm
We all know that
1 inch = 25.4 mm
0.050 inch = 0.050 ×25.4 mm
= 1.27 mm
nucleation rate = 200×D - 10×D²
= 200×1.27 - 10×(1.27)²
=237.9 Crystals/min
=237.9/60 crystals/sec
= 3.96 crystals/sec
c) Derive a formula for r(crystals/s) in terms of D(inches). (See Example 2.6-1.) Check the formula using the result of Part b.
r(crystals/sec)=A D−B D² (D in inch)
unit of 200= crystals / (min×mm)
unit of 10=crystals / (min×mm² )
A = 200 crystals / (min×mm) × 1/60 min/sec ×25.4 mm/inch
= 84.7 crystals/(sec-inch)
B = 10 crystals / (min×mm² ) × 1/60 min/sec ×25.4 mm/inch×25.4 mm/inch
=107.5 crystals/(sec-inch)
What is the molarity of a solution when 3.58 moles of NaCl are dissolved in 2.12 L of solution?
Answer:
1.69 M
Explanation:
Molarity is the ratio of moles to liters:
(3.58 moles)/(2.12 liters) = 1.6887 moles/liter ≈ 1.69 M
Tellurium is a period 5 chalcogen. Selenium is a period 4 chalcogen. If the only factor affecting ionization energies was the nuclear charge, then electrons would be easier to remove (ionize) from Se than Te. Experimentally the opposite is true. It takes 941.0 kJ/mol of energy to ionize the outermost electron from Se while it only takes 869.3 kJ/mol to ionize from Te. A good model should account for this. Quantum mechanical calculations do predict this but require access to sophisticated software, large amounts of computing power and technical expertise. Slater suggested that some simple empirical rules that take into account electron electron repulsion (or shielding) could give a good estimate of the effective nuclear charge (Zeff). The Zeff for the outermost electron in Te is . The Zeff for the outermost electrons in Se is . According to Slater's calculation of effective nuclear charge (does or does not) predict the correct ordering of ionization energies for Se and Te. A better means of rationalizing ionization energies is to include the atomic as follows: [Z subscript e f f end subscript over r squared] . For Te, r = 136 pm and for Se r = 117 pm. This new model (does or does not) predict the correct ordering of first ionization energies for Se and Te.
Answer:
yes
Explanation: took quiz
How does a mixture of benzoic acid and benzaldehyde can be isolated separately by extractions?
Answer:
Dissolve benxioc acid and benzaldehyde in organic solvent. The two compounds are not miscible.with water. Put the two in separating funnel. Then use aqueous sodium bicarbonate to extract. Benzioc acid will be in aqueous layer as benzioate ion. Benzaldehyde remain insoluble and can be isolated.
Explanation:
Extractions are techniques use to separate desired compounds when mixed together. The mixture is brought in contact with solvent in which the sites substance is soluble and other is insoluble. Extractions use imissicible stages to separate substance from another.
Can you please help me ?
Answer:
6.82 moles of Fe2O3
Explanation:
Step 1:
Determination of the number of mole of in 450g of CO2.
This is illustrated below:
Molar Mass of CO2 = 12 + (2x16) = 44g/mol
Mass of CO2 = 450g
Number of mole of CO2 =.?
Number of mole = Mass/Molar Mass
Number of mole of CO2 = 450/44 = 10.23 moles
Step 2:
Determination of the number of mole of Fe2O3 needed for the reaction. This is illustrated below:
2Fe2O3 + 3C—> 4Fe + 3CO2
From the balanced equation above,
2 moles of Fe2O3 reacted to produce 3 moles of CO2.
Therefore, Xmol of Fe2O3 will react to produce 10.23 moles of CO2 i.e
Xmol of Fe2O3 = (2x10.23)/3
Xmol of Fe2O3 = 6.82 moles
Therefore, 6.82 moles of Fe2O3 is required.
In the news recently, an individual poured an unknown liquid substance over a female victim and fled on
foot. The victim sustained chemical burns to her face, neck, shoulder, and back (source: NYPD). What
could the unknown substance have been?
water
a base, only
either an acid or a base
an acid, only
Answer:
either an acid or a base
Explanation:
Especially, concentrated acids and bases tend to do enormous amounts of chemical burns. People usually think its only acids but the strongest of bases can easily melt our skin.
Watch the video to determine which of the following relationships are correct according to Boyle’s law.
1. PV∝P
2. P∝V
3. PV∝V
4. P∝1V
5. V∝1P
6. V∝P
Part B. Standard temperature and pressure (STP) are considered to be 273 K and 1.0 atm. Predict which of the following changes will cause the volume of the balloon to increase or decrease assuming that the temperature and the gas filling the balloon remain unchanged. Drag the appropriate items to their respective bins.
A. Volume increases
B. Volume decreases
C. Volume is unchanged
Part C. If a certain gas occupies a volume of 20 L when the applied pressure is 10 atm, find the pressure when the gas occupies a volume of 5.0 L. Express your answer to two significant figures, and include the appropriate units.
ki, kf =
Part D. If a certain gas occupies a volume of 18 L when the applied pressure is 9.0 atm , find the pressure when the gas occupies a volume of 4.5 L .Express your answer to two significant figures and include the appropriate units.
Answer:
Part A
Boyle's Law is given mathematically as
P ∝(1/V) or V ∝(1/P)
Options 4 and 5, if they are properly written.
Part B
At constant temperature, and according to the Boyle's law for an ideal gas,
A. What can cause a Volume increase is a corresponding decrease in pressure.
B. What can cause a Volume decrease is a corresponding increase in pressure.
C. The Volume is unchanged if the pressure of the gas is unchanged too.
Part C
The pressure when the gas occupies a volume of 5.0 L = 40 atm
Part D
The pressure when the gas occupies a volume of 4.5 L = 36 atm
Explanation:
Part A
Boyle's Law states that at constant temperature, the pressure of an ideal gas is inversely proportional to the volume occupied by the gas.
So, mathematically, Boyle's Law is given as
P ∝(1/V) or V ∝(1/P)
Part B
Inverse relationship between two quantities means that the higher the value of one of the quantities go, the lower the value of the other quantity goes and vice versa.
So, at constant temperature, and according to the Boyle's law for an ideal gas.
A. What can cause a Volume increase is a corresponding decrease in pressure.
B. What can cause a Volume decrease is a corresponding increase in pressure.
C. The Volume is unchanged if the pressure of the gas is unchanged too.
Part C
A certain gas occupies a volume of 20 L when the applied pressure is 10 atm, find the pressure when the gas occupies a volume of 5.0 L.
According to Boyle's Law for an ideal gas,
P ∝(1/V)
P = (k/V)
where k is the constant of proportionality
PV = k
Therefore,
P₁V₁ = P₂V₂ = k
P₁ = 10 atm
V₁ = 20 L
P₂ = ?
V₂ = 5.0 L
10 × 20 = P₂ × 5
P₂ = 40 atm
Part D
If a certain gas occupies a volume of 18 L when the applied pressure is 9.0 atm , find the pressure when the gas occupies a volume of 4.5 L
P₁V₁ = P₂V₂ = k
P₁ = 9.0 atm
V₁ = 18 L
P₂ = ?
V₂ = 4.5 L
9 × 18 = P₂ × 4.5
P₂ = 36 atm
Hope this Helps!!!
Which type of reaction is the Haber process: N2(g) + 3 H2(g) → 2 NH3(g) + heat? *
exothermic, with an increase in entropy
exothermic, with a decrease in entropy
endothermic, with an increase in entropy
endothermic, with a decrease in entropy
Answer:
exothermic, with a decrease in entropy
Explanation:
Whenever you produce heat as a product in a reaction, the reaction is exothermic. To determine entropy, we know we have 4 moles of gas on reactant (1 from N2 and 3 from H2) and in produce side we only have two moles (2 from NH3) thus since we are decreasing the number of gas molecules, there is going to be less disorder, hence decrease in entropy.
For the equation 3Cu + 8HNO3 → 3Cu(NO3)2 + 2NO + 4H2O, how many units of NO3 are represented on the products side?
A 2
B 3
C 6
D 8
Answer:
C
Explanation:
Following are the calculation to units of [tex]\bold{NO_{3}}[/tex]:
Given equation:
[tex]\bold{3Cu + 8HNO_3 \longrightarrow 3Cu(NO_3)2 + 2NO + 4H_2O}[/tex]
To find:
units of [tex]\bold{NO_{3}}[/tex]=?
Solution:
[tex]\bold{3Cu + 8HNO_3 \longrightarrow 3Cu(NO_3)2 + 2NO + 4H_2O}[/tex]
The above-given equation when the 3 mol copper reacts with Nitric acid so, it will give 3 mol Cupric nitrate, 2 mol nitric oxide, and 4 mol of water.In the above-given equation[tex]\bold{3Cu(NO_3)_2= 3Cu +3(NO_3)_2 =3Cu +6(NO_3) }[/tex] there are 6 mols of[tex]\bold{NO_3}[/tex].Therefore, the final answer is "Option C".
Learn more:
brainly.com/question/18492732
If I have a 200 L container filled with nitrogen at a pressure of 1.0 atm, how many moles of nitrogen are present at 25 C?
0
Select one:
O a. 0,085 moles
O b. 81.8 moles
O C. 19.3 moles
O d. 8.18 moles
3. A thin lead apron is used to protect patients from harmful X rays. If the sheet measures 75.0 cm by 55.0 cm by 0.10 cm, and the density of lead is 11.3 g/cm3, what is the mass of the apron in grams?
Answer:
4.67 kg
Explanation:
Given data
Dimensions of the lead sheet: 75.0 cm by 55.0 cm by 0.10 cmDensity of lead: 11.3 g/cm³Step 1: Calculate the volume of the sheet
The volume of the sheet is equal to the product of its dimensions.
[tex]V = 75.0 cm \times 55.0 cm \times 0.10 cm = 413 cm^{3}[/tex]
Step 2: Calculate the mass of the sheet
The density (ρ) is equal to the mass divided by the volume.
[tex]\rho = \frac{m}{V} \\m = \rho \times V = \frac{11.3g}{cm^{3} } \times 413cm^{3} = 4.67 \times 10^{3} g = 4.67 kg[/tex]
Convection refers to the movement of heat through a fluid, such as water. Which best describes one way that convection
influences the circulation of ocean water?
Cold water at the surface flows toward the poles and sinks as it cools.
Deep, warm water flows toward the equator and rises as it cools.
Warm water at the surface flows toward the poles and sinks as it cools.
Deep, cold water flows toward the equator and rises as it cools,
help me please
Answer:
the second answer where it says deep warm water flows, etc.
Explanation:
i hope that helps you
Answer:
C) Warm water at the surface flows toward the poles and sinks as it cools.
Explanation:
have a nice day.
I need help solving this problem:
How many moles is 1.5x10^23 atoms of Carbon
To find out how many moles is 1.5x10^23 atoms of Carbon, we will use Avogadro's number, which is represented by 6.022x10^23 atoms or molecules per mol.
To convert from atoms to moles, divide the atoms by Avogadro's number.
So, now we can convert the numbers into proper form:
1.5x10^23 ---> 1.5E23
6.022x10^23 ---> 6.022E23
Then, we divide them.
1.5E23/6.022E23 = 0.24908
Round to nearest hundredth.
0.24908 --> 0.25
Therefore, there are 0.25 moles of carbon in 1.5E23 atoms of carbon.
The original fluid mosaic model stated that membrane proteins move freely in the plane of the lipid bilayer via lateral diffusion "like protein icebergs in a sea of lipids." Since its initial description in 1927, the fluid mosaic model has been refined.
Complete the statements that describe some of the refinements of the fluid mosaic model as it relates to protein mobility.
Choose the answer that best completes each sentence:
Membrane (lipid) rafts, Vesicles, Cytoskeletal elements, Protein-protein interactions, Bilayer leaflets.
1) ______ may form a locally rigid, non-lipid matrix surrounded by the fluid lipid matrix.
2) ______, membrane domains rich in sphingolipids and cholesterol, are regions with reduced lipid and protein mobility.
3) ______ can be attached to proteins or can "fence in" proteins, reducing protein mobility.
Answer:
1. Protein-protein interactions
2. Lipid rafts
3. Cytoskeletal elements
Explanation:
The fluid mosaic model was proposed by the Singer and Nicolson in 1972 which stated that the membrane is a mosaic of the lipids, carbohydrates and the proteins.
The fluid mosaic model has been revised many times which are:
1. Protein-protein interactions: the proteins present in the lipid bilayer can easily float in the lipid layer, can be found integrated into the layer and exist on the outside of the layer. These proteins can form a bond with each other which are rigid and are non-lipid matrix.
2. Lipid rafts: the structure formed in the bilayer which is rich in cholesterol and the sphingolipids. These regions have a reduced amount of lipids and reduced protein mobility.
3. Cytoskeletal elements: are attached to the proteins and act as a fence in proteins which reduces the mobility of the proteins.
A student runs two experiments with a constant-volume "bomb" calorimeter containing 1500.g of water.
First, a 7.500g tablet of benzoic acid C6H5CO2H is put into the "bomb" and burned completely in an excess of oxygen. (Benzoic acid is known to have a heat of combustion of 26.454 kJ/g.) The temperature of the water is observed to rise from 10.00°C to 36.99°C over a time of 13.0 minutes.
Next, 5.260g of ethanol C2H5OH are put into the "bomb" and similarly completely burned in an excess of oxygen. This time the temperature of the water rises from 10.00°C to 28.03°C.
Use this information, to answer the questions below about this reaction:
C2H5OH(l)+ 3O2(g)→ 2CO2(g)+ 3H2O(g)
a. Is this reaction exothermic, endothermic, or neither?
b. If you said the reaction was exothermic or endothermic, calculate the amount of heat that was released or absorbed by the reaction in the second experiment.
c. Calculate the reaction enthalpy ΔHrxn per mole of CO2
Answer:
See explaination
Explanation:
Please kindly check attachment for the step by step solution of the given problem
Since there has been a rise in the reaction temperature, there has been an exothermic reaction.
The amount of heat energy released in the second step has been -132.54 kJ.
The reaction enthalpy per mole has been -1160.85 kJ/mol.
(a) To determine whether the reaction has been exothermic or endothermic, the heat absorbed or released has been calculated.
Since there has been a rise in the temperature of the solution with the combustion, the reaction has been termed as the exothermic reaction.
(b) Amount of heat released in second experiment:
In the bomb calorimeter:
[tex]\rm q_r_e_a_c_t_i_o_n\;+\;q_b_o_m_b\;+\;q_w_a_t_e_r\;=\;0[/tex]
[tex]\rm q_w_a_t_e_r[/tex] has been given a:q = mc[tex]\Delta[/tex]T
q = 1500 g × 4.184 J/g.[tex]\rm ^\circ C[/tex] × (36.99 - 10[tex]\rm ^\circ C[/tex] )
[tex]\rm q_w_a_t_e_r[/tex] = 169389.24 J.
[tex]\rm q_b_o_m_b\;[/tex] can be given as:q = C [tex]\Delta[/tex]T
q = c (36.99 - 10[tex]\rm ^\circ C[/tex] )
[tex]\rm q_b_o_m_b\;[/tex] = 26.99 [tex]\rm ^\circ C[/tex] × c
[tex]\rm q_r_e_a_c_t_i_o_n[/tex] can be given by:q = mass × heat of combustion of benzoic acid
q = 7.5 g × 26.454 kJ/g
[tex]\rm q_r_e_a_c_t_i_o_n[/tex] = 198405 J
[tex]\rm q_r_e_a_c_t_i_o_n\;+\;q_b_o_m_b\;+\;q_w_a_t_e_r\;=\;0[/tex]
[tex]\rm q_b_o_m_b\;[/tex] = - ([tex]\rm q_r_e_a_c_t_i_o_n\;+\;q_w_a_t_e_r[/tex])
[tex]\rm q_b_o_m_b\;[/tex] = - (198405 J + 169389.24 J )
[tex]\rm q_b_o_m_b\;[/tex] = 29015.76 J.
[tex]\rm q_b_o_m_b\;[/tex] = 26.99 [tex]\rm ^\circ C[/tex] × c
29015.76 J = 26.99 [tex]\rm ^\circ C[/tex] × c
c of bomb = 1075.05 J/[tex]\rm ^\circ C[/tex].
For the second reaction of combustion of ethanol:
[tex]\rm q_r_e_a_c_t_i_o_n\;+\;q_b_o_m_b\;+\;q_w_a_t_e_r\;=\;0[/tex]
[tex]\rm q_w_a_t_e_r[/tex] has been given as:q = mc[tex]\Delta[/tex]T
q = 1500 g × 4.184 J/g.[tex]\rm ^\circ C[/tex] × (28.03 - 10[tex]\rm ^\circ C[/tex] )
[tex]\rm q_w_a_t_e_r[/tex] = 113156.28 J.
[tex]\rm q_b_o_m_b\;[/tex] can be given as:q = C [tex]\Delta[/tex]T
q = 1075.05 J/[tex]\rm ^\circ C[/tex] × (28.03 - 10[tex]\rm ^\circ C[/tex] )
[tex]\rm q_b_o_m_b\;[/tex] = 19383.15 J
Moles of ethanol = [tex]\rm \dfrac{weight}{molecular\;weight}[/tex]
Moles of ethanol = [tex]\rm \dfrac{5.26\;g}{46.07\;g/mol}[/tex]
Moles of ethanol = 0.11417 mol.
[tex]\rm q_r_e_a_c_t_i_o_n[/tex] for ethanol combustion can be given by:q = moles of ethanol × [tex]\Delta[/tex]H of reaction
[tex]\rm q_r_e_a_c_t_i_o_n[/tex] = 0.11417 × [tex]\Delta[/tex]H of reaction
[tex]\rm q_r_e_a_c_t_i_o_n\;+\;q_b_o_m_b\;+\;q_w_a_t_e_r\;=\;0[/tex]
[tex]\rm q_r_e_a_c_t_i_o_n[/tex] = - ([tex]\rm q_b_o_m_b\;[/tex] + [tex]\rm q_w_a_t_e_r[/tex])
[tex]\rm q_r_e_a_c_t_i_o_n[/tex] = - (19383.15 J + 113156.28 J)
[tex]\rm q_r_e_a_c_t_i_o_n[/tex] = -132539.43 J
[tex]\rm q_r_e_a_c_t_i_o_n[/tex] = -132.54 kJ.
The amount of heat energy released in the second step has been -132.54 kJ.
(c) The reaction enthalpy per mole can be given as:
[tex]\rm q_r_e_a_c_t_i_o_n[/tex] = 0.11417 mol × [tex]\Delta[/tex]H of reaction
-132.54 kJ = 0.11417 mol × [tex]\Delta[/tex]H of reaction
[tex]\Delta[/tex]H of reaction = -1160.85 kJ/mol.
The reaction enthalpy per mole has been -1160.85 kJ/mol.
Since there has been a rise in the reaction temperature, there has been an exothermic reaction.
The amount of heat energy released in the second step has been -132.54 kJ.
The reaction enthalpy per mole has been -1160.85 kJ/mol.
For more information about the reaction in the bomb calorimeter, refer to the link:
https://brainly.com/question/14989357
Which of the following acts as a bronsted Lowry acid, but not as a bronsted Lowry base
Answer:
D. (HCIO4(AQ)
Explanation:
briefly describe how a potentiometric ph meter works. [Hint describe how the pH meter measures the amount of H+ or OH- ions in a sample]
Answer:
pH meter measures the degree of acidity or alkalinity in a substance. The H+ ions measures its acidity and the OH- measures the alkalinity.It usually has a voltmeter which is connected to a pH-responsive electrode and a standard electrode which has no degree of variation.
The potentiometric ph meter works functions by measuring the voltage between two electrodes and the result are usually displayed after conversion into the corresponding pH value takes place.
With 21 g of Zinc, and 7 g of CuCl2, how much ZnCl2 is made in grams?
Answer: 7.07 grams
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}\times{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} zinc=\frac{21g}{65g/mol}=0.32moles[/tex]
[tex]\text{Moles of} CuCl_2=\frac{7g}{134g/mol}=0.052moles[/tex]
[tex]Zn+CuCl_2\rightarrow Cu+ZnCl_2[/tex]
According to stoichiometry :
1 mole of [tex]CuCl_2[/tex] require 1 mole of [tex]Zn[/tex]
Thus 0.052 moles of [tex]CuCl_2[/tex] will require=[tex]\frac{1}{1}\times 0.052=0.052moles[/tex] of [tex]Zn[/tex]
Thus [tex]CuCl_2[/tex] is the limiting reagent as it limits the formation of product and [tex]Zn[/tex] is the excess reagent.
As 1 mole of [tex]CuCl_2[/tex] give = 1 mole of [tex]ZnCl_2[/tex]
Thus 0.052 moles of [tex]CuCl_2[/tex] give =[tex]\frac{1}{1}\times 0.052=0.052moles[/tex] of [tex]ZnCl_2[/tex]
Mass of [tex]ZnCl_2=moles\times {\text {Molar mass}}=0.052moles\times 136g/mol=7.07g[/tex]
Thus 7.07 g of [tex]ZnCl_2[/tex] will be produced from the given masses of both reactants.
Which step is this ?prophase I, Anaphase I, prophase II, Anaphase II
Answer:
Prophase 1
Explanation:
Answer:
The correct answer will be prophase I
Explanation:
During this stage, the genetic code is together and hasn't split yet therefore, the answer is prophase I hopefully this helps!
Answers only please 100pts and brainliest. you will be reported for wrong answers.
Answer:
1. 42.75 grams KOH
2. Add 150ml of water to the 250ml containing 9.8 grams 42.75 grams KOH
Explanation
1. moles KOH in 500ml in 1.5M solution = 0.500L x 1.5 mole/Liter = 0.75 mol KOH
grams KOH in 0.75 mole = 0.75 mole x 57 grams/mole = 42.75 grams KOH
2. formula weight H₃PO₄ = 98 grams/mole
9.8 grams H₃PO₄ in 250ml H₂O => 9.8g/98g·mole⁻¹/0.250Liters = 0.40M in H₃PO₄
Final volume of 0.25M H₃PO₄ solution = (0.40M)(250ml)/(0.25M) = 400ml
∴ Add 400ml - 250ml = 150ml of water to the 250ml of 0.40M H₃PO₄ solution => 400ml of 0.25M H₃PO₄ solution.
Answer:
1. 42.75 grams koh
2. add 150 ml of water to the 250 ml containing 9.8 grams 42.75 grams koh
Explanation:
The outer surface of a steel gear is to be hardened by increasing its carbon content; the carbon is to be supplied from an external carbon-rich atmosphere, which is maintained at an elevated temperature. A diffusion heat treatment at 850 °C (1123 K) for 10 min increases the carbon concentration to 0.90 wt% at a position 1.0 mm below the surface. Estimate the diffusion time required at 650 °C (923 K) to achieve this same concentration also at a 1.0-mm position. Assume that the surface carbon content is the same for both heat treatments, which is maintained constant. D = 1.1 x 10-6 m2 /s and Qd = 87,400 J/mol C diffusion in a-Fe. (8 points)
Answer:
[tex]\mathbf{t_2 = 75.696 \ min}[/tex]
Explanation:
From the question:
The outer surface of a steel gear is to be hardened by increasing its carbon content
Given that :
Diffusion of heat temperature at [tex]T_1[/tex] 850 °C = 1123 K
Diffusion time [tex]t_1[/tex] = 10 min
diffusion after the carbon concentration at a position [tex]x_1[/tex] ( 1.0 mm) below the surface = 0.90 wt%
Preexponential = 1.1 × 10⁻⁶ m²/s
Activation Energy [tex]Q_d[/tex] = 87400 J/mol
We are to determine the time [tex]t_2[/tex] at 650 °C (923 K) to achieve the same diffusion result as at 850 °C (1123 K) for [tex]t_1[/tex] = 10 min
Considering Fick's second law for the condition of Constant surface concentration; we have:
[tex]\frac{Cx-C_0}{C_s-C_0} = 1-erf(\frac{x}{2\sqrt{Dt} } )[/tex] ------ equation (1)
where;
[tex]C_0 =[/tex] concentration of the diffusing solute atom before diffusion
[tex]C_s[/tex] = Constant surface concentration
[tex]C_x[/tex] = Concentration at depth x after time t
[tex]erf(\frac{x}{2\sqrt{Dt} } )[/tex] = Gaussian error function
At some desired specific concentration of solute [tex]C_1[/tex] in an alloy ; the left side of the above equation (1) thus becomes constant ;
i.e [tex]\frac{Cx-C_0}{C_s-C_0} = \mathbf{ constant}[/tex]
Then ; [tex]\frac{x}{2\sqrt{Dt} }[/tex] = constant
[tex]\frac{x^2}{Dt}[/tex] = constant
Dt = constant
Thus; [tex]D_1t_1 = D_2t_2[/tex]
Therefore, the time [tex]t_2[/tex] at 650°C([tex]T_2[/tex] = 923 K) required to produce the same diffusion on result as at 850°C ([tex]T_1[/tex] = 1123 K) for [tex]t_1[/tex] = 10 min is [tex]t_2 = \frac{D_1t_1}{D_2}[/tex]
We need to first determine the Diffusion coefficient at 1123 K and 923 K ( i.e [tex]D_1[/tex] and [tex]D_2[/tex])
At [tex]T_1[/tex] = 1123 K , Diffusion coefficient [tex]D_1[/tex] is calculated by the equation [tex]D_1 = D_0 exp ( - \frac{Q_d}{RT_1})[/tex] (equation from temperature dependence of the diffusion coefficient)
[tex]D_1 = 1.1 * 10^{-6} \ exp ( - \frac{87,400}{8.314*1123} )[/tex]
[tex]D_1 = 9.462*10^{-11} \ m^2/s[/tex]
[tex]D_2 = D_0 exp ( - \frac{Q_d}{RT_2})[/tex]
[tex]D_2 = 1.1 * 10^{-6} \ exp ( - \frac{87,400}{8.314*923} )[/tex]
[tex]D_2 = 1.25*10^{-11} m^2/s[/tex]
[tex]t_2 = \frac{ 9.462*10^{-11}*10}{ 1.25*10^{-11} }[/tex]
[tex]\mathbf{t_2 = 75.696 \ min}[/tex]
When wave-particle duality was applied to the electron, it explained why the energy of the electron is quantized because:
a. The electron is a standing wave that can only have an integer number of wavelengths
b. An electron can only exhibit specific velocities
c. The electron must have a specific mass
d. An electron will emit and absorb light quanta
Answer:
a. The electron is a standing wave that can only have an integer number of wavelengths.
Explanation:
As per quantum physics, the theory of wave-particle duality refers to the notion that matter and light show the characteristics of both waves and particles, based on the case of the experiment. Much like light, the matter appeared to possess both wave and particle properties. Large objects display very low wavelengths, but for small particles, the wavelength may be detected and important, as shown by a double-slit experiment with electrons.
A chemical engineer must calculate the maximum safe operating temperature of a high-pressure gas reaction vessel. The vessel is a stainless-steel cylinder that measures wide and high. The maximum safe pressure inside the vessel has been measured to be .
For a certain reaction the vessel may contain up to of carbon monoxide gas. Calculate the maximum safe operating temperature the engineer should recommend for this reaction. Write your answer in degrees Celsius. Be sure your answer has the correct number of significant digits.
Answer:
T = 643K = 643-273.15 = 369.85°C = 370°C = maximum safe operating temperature
Explanation:
The actual question has following data which the question given has missing
The vessel is a stainless-steel cylinder that measures wide 25.0 cm and 30.0 cm high.
The maximum safe pressure inside the vessel is 6.60 MPa.
The vessel may contain up to 0.800 kg of carbon monoxide gas.
Volume of cylinder = pi x r^2 x h = 3.14 x (25/2)cm^2 x 30cm = 14725.78 cm^3 or 0.01472578 m^3
Now, using ideal gas law, PV = nRT
P = 6.60×10^6 Pa
V = 0.01472578 m^3
n= no. of moles = mas staken/ molar mass of CO2 = 800g/44g/mol = 18.18
moles
R = 8.314 J/mol.K
So, (6.60x10^6 ) x 0.01472578 = 18.18×8.314×T
T = 643 K = 643-273.15 = 369.85°C = 370°C = maximum safe operating temperature
A solution is prepared by dissolving 0.26 mol of hydrazoic acid and 0.26 mol of sodium azide in water sufficient to yield 1.00 L of solution. The addition of 0.05 mol of HCl to this buffer solution causes the pH to drop slightly. The pH does not decrease drastically because the HCl reacts with the ________ present in the buffer solution. The Ka of hydrazoic acid is 2.5 x 10^-5.
Answer:
The pH does not decrease drastically because the HCl reacts with the sodium azide (NaN₃) present in the buffer solution.
Explanation:
The buffer solution is formed by 0.26 moles of the weak acid, hydrazoic acid (HN₃), and by 0.26 moles of sodium azide (NaN₃). The equilibrium reaction of this buffer solution is the following:
HN₃(aq) + H₂O(l) ⇄ N₃⁻(aq) + H₃O⁺(aq)
The pH of this solution is:
[tex] pH = pka + log(\frac{[N_{3}^{-}]}{[HN_{3}]}) = -log(2.5 \cdot 10^{-5}) + log(\frac{0.26 mol/1 L}{0.26 mol/1 L}) = 4.60 [/tex]
When 0.05 moles of HCl is added to the buffer solution, the following reaction takes place:
H₃O⁺(aq) + N₃⁻(aq) ⇄ HN₃(aq) + H₂O(l)
The number of moles of NaN₃ after the reaction with HCl is:
[tex] \eta_{NaN_{3}} = \eta_{i} - \eta_{HCl} = 0.26 moles - 0.05 moles = 0.21 moles [/tex]
Now, the number of moles of HN₃ is:
[tex] \eta_{HN_{3}} = \eta_{i} + \eta_{HCl} = 0.26 moles + 0.05 moles = 0.31 moles [/tex]
Then, the pH of the buffer solution after the addition of HCl is:
[tex] pH = pka + log(\frac{[N_{3}^{-}]}{[HN_{3}]}) = -log(2.5 \cdot 10^{-5}) + log(\frac{0.21 mol/V_{T}}{0.31 mol/V_{T}}) = 4.43 [/tex]
The pH of the buffer solution does not decrease drastically, it is 4.60 before the addition of HCl and 4.43 after the addition of HCl.
Therefore, the pH does not decrease drastically because the HCl reacts with the sodium azide (NaN₃) present in the buffer solution.
I hope it helps you!
Which is not a physical property?
A) Hardness
B) Boiling point
C)ability to conduct electricity
D) ability to combine with hydrogen
Answer:
D
Explanation:
Combing with hydrogen is a chemical property.
Answer:
D. ability to combine with hydrogen
explanation: It is a chemical property.
What is the relationship between the concentration of the hydronium and hydroxide ion and pH in any water solution?
Answer:
The concentration of hydronium ions and the pH value is related by the equation: pH=-Log[H+]
Explanation:
We have two different concepts pH and concentration of hydronium ions. Lets start with the hydronium ion.
An hydronium ion is a species that is produced by an acid:
[tex]HA~->~H^+~+~A^-[/tex]
Aditionally, we can have the production of hydroxide ions. The subtances that have the capacity to produce this ions are called "bases":
[tex]BOH~->~B^+~+~OH^-[/tex]
Now we can continue "pH"
The pH is a scale that indicates if the substance is and acid (higher concentration of [tex]H^+[/tex]), neutral (equal amounts of [tex]H+[/tex] and [tex]OH^-[/tex]) or basic (higher amount of [tex]OH^-[/tex]).
Finally, the "pH" is calculated with the concentration of the hydronium ions (
[tex]H^+[/tex]), the letter "p" is "-Log", therefore:
[tex]pH=-Log[H^+] [/tex]
I hope it helps!
There are certain trends with which you should become very familar (recognizing these trends will save you time!) This part of the question is dedicated to that task.
a) Whenever you see 9 or a multiple of 9 in the integration ratio, which group should you first consider for being responsible for that signal.
b) Whenever you see a quartet and triplet on a spectrum, which group should you first consider for being responsible for those signals?
c) Whenever you see septet and a doublet on a spectrum, which group should you first consider for being responsible for those signals?
d) Whenever you see 6 or a multiple of 6 in the integration ratio, which group should you first consider for being responsible for that signal
e) Whenever you see 3 as the actual number of protons for a given signal, which group should you first consider for being responsible for that signal
Answer:
Explanation:
The objective of this question is all about identifying the phenomena that holds true for the statement being said in each instance. Let; walk through them.
a) Whenever you see 9 or a multiple of 9 in the integration ratio, which group should you first consider for being responsible for that signal.
( C₄H₉ )Tert. Butyl group
b) Whenever you see a quartet and triplet on a spectrum, which group should you first consider for being responsible for those signals?
(CH₃CH₂) ethyl group
c) Whenever you see septet and a doublet on a spectrum, which group should you first consider for being responsible for those signals?
(CH(CH₃)₂) isoproply group
d) Whenever you see 6 or a multiple of 6 in the integration ratio, which group should you first consider for being responsible for that signal
(CH(CH₃)₂) isoproply group
e) Whenever you see 3 as the actual number of protons for a given signal, which group should you first consider for being responsible for that signal.
CH₃- methyl group
A perfect description showing the explanation of each answers chosen is explained with an aid of diagram below.
Types of radioactive decay
PLEASE HELP BEFORE 7 A.M. PACIFIC TIME
SEE ATTACHED.
Answer:
(i) 32.9 mL; (ii) 37.5 mL; (iii) 21.73 mL
Explanation:
You should always try to read a measuring instrument to a tenth of the smallest scale division.
Here, you are measuring liquids, so you take the scale reading from the bottom of the meniscus.
(i) Graduated cylinder
There are 10 divisions between 30 mL and 40 mL, so each division represents 1 mL.
The level of the liquid appears to be between 32 mL and 33 mL. It is much closer to 33 mL (perhaps right on 33 mL).
You should report the volume to the nearest 0.1 mL. I would read the volume as 32.9 mL, but 32.8 and 33.0 are also acceptable.
Note: If you think the level is right on the 33 mark, you report the volume as 33.0 mL (NOT 33 mL).
(ii) Thermometer
The reading is about half-way between 87 ° and 88 °.
I would report the temperature as 87.5 °, but 87.4 ° and 87.6 ° would also be acceptable.
(iii) Buret
There are 10 divisions between 21 mL and 22 mL, so each division represents 0.1 mL.
You should estimate to the nearest 0.01 mL.
The liquid level is about a third of the way from 21.7 mL to 21.8 mL.
I would report the volume as 21.73 mL, but 21.72 mL and 21.74 mL are also acceptable.
A chemistry graduate student is studying the rate of this reaction:
NH4OH(aq)→NH3(aq)+H2O(aq)
She fills a reaction vessel with and measures its concentration as the reaction
proceeds: time (seconds).
1. Write the rate law for this reaction.
2. Calculate the value of the rate constant.
Answer:
See explaination
Explanation:
Please kindly check attachment for the step by step solution of the given problem