Answers
Answer:
Here's what I get
Explanation:
A. Prepare Ca(NO₃)₂ solution
Scoopula — to transfer the solid to an Erlenmeyer flask
Erlenmeyer flask — to hold the solid for weighing and to prepare the solution
Volumetric flask — to hold the liquid
Funnel — to transfer the solution to the volumetric flask
Wash bottle — to fill the volumetric flask to the mark
B. Weigh K₂SO₄
Scoopula — to transfer the solid to a weighing bottle
Weighing bottle — to hold the solid for weighing.
C. Add acetic acid to a volumetric flask
Volumetric flask — to hold the acid
Pipet —to measure out the acid and transfer it to the flask
Pipet bulb — safety feature for the pipet and for control of the acid volume.
D. Add NaOH to an acid
Beaker — to hold the NaOH solution
Ring stand — to hold a buret
Buret clamp — to hold the buret
Funnel — to transfer the solution to a buret
Buret — to hold the NaOH solution for addition to the acid
Related Questions
Can you please help me ?
Answers
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.
Predict what will be observed in experiment below.
Experiment:
Rock candy is formed when excess sugar is dissolved in hot water followed by crystallization. A student wants to make two batches of rock candy. He finds an unopened box of "cane sugar" in the pantry. He starts preparing batch A by dissolving sugar in 500mL of hot water (70°C). He keeps adding sugar until no more sugar dissolves in the hot water. He cools the solution to room temperature.
He prepares batch B by dissolving sugar in 500mL of water at room temperature until no more sugar is dissolved. He lets the solution sit at room temperature.
Predicted observation (Choose one).
O It is likely that more rock candy will be formed in batch A.
O It is likely that less rock candy will be formed in batch A.
O It is likely that no rock candy will be formed in either batch.
O I need more information to predict which batch is more likely to form rock candy.
Answers
Answer:
Option A: It is likely that more rock candy will be formed in batch A.
Explanation:
The difference between batch A and batch B is that batch A uses temperature to dissolve the sugar, while the dissolution of the sugar in batch B is produced at room temperature.
When he use temperature (hot water) to dissolve the sugar he is increasing the solubility of the sugar in the water, so in batch A we will have more quantity of sugar dissolved than in the batch B. The cooling of the solution at room temperature favors the formation of bigger sugar crystals in the process of crystallization.
From all of the above, the correct predicted observation is the option A: It is likely that more rock candy will be formed in batch A, for the increase of the solubility by the use of hot water. Also, you say that Rock candy is formed when excess sugar is dissolved in hot water followed by crystallization.
I hope it helps you!
what kind of air pressure would you find in Bridgeport, Connecticut?
Answers
Answer:
Warm
Explanation: because it would be less hot air population
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
Answers
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
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.
Answers
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.
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?
Answers
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]
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
Answers
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.
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)
Answers
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]
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.
Answers
Answer:
yes
Explanation: took quiz
Which is not a physical property?
A) Hardness
B) Boiling point
C)ability to conduct electricity
D) ability to combine with hydrogen
Answers
Answer:
D
Explanation:
Combing with hydrogen is a chemical property.
Answer:
D. ability to combine with hydrogen
explanation: It is a chemical property.
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
Answers
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.
How does a mixture of benzoic acid and benzaldehyde can be isolated separately by extractions?
Answers
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.
How do you draw structural formulas 2,4-dimethylhexane; 4-methyl-2-pentene; 4-chloro-7-methyl-2-nonyne
Answers
Answer:
Structural formulas are the graphical representation of chemical compunds and shows the chemical bonds between the atoms of a molecule.
structural formulas of following compunds is attached below:
2,4-dimethylhexane - C8H18, it will have single bonds between carbon atom and will have methyl group at position 2 and 4 carbon.
4-methyl-2-pentene - C6H12, it will have double bond at 2nd carbon and methyl group at 4th carbon.
4-chloro-7-methyl-2-nonyne - C10H17Cl, it will have triple bond at 2nd position, chloride group at 4th carbon and a methyl group at 7th carbon.
What distinguishes effusion from diffision? How are these processes similar?
Answers
Answer:
In effusion, a substance escapes through a tiny pinhole, or other hole whereas diffusion is the spreading out of a substance within a dispersing medium.
Effusion and diffusion are similar in terms of rates and speeds.
Explanation:
According to effusion, if there is a small hole in a container, a gas will leak from it. So, in this process a substance escapes through a tiny pinhole, or other hole.
According to diffusion, the two solutions will reach the lowest possible concentration of both, if they are combined and not mixed. So, diffusion is the spreading out of a substance within a dispersing medium.
Effusion and diffusion are similar in terms of rates and speeds.
Answers only please 100pts and brainliest. you will be reported for wrong answers.
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:
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
Answers
I need help solving this problem:
How many moles is 1.5x10^23 atoms of Carbon
Answers
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.
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
Answers
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.
A 2.40 kg block of ice is heated with 5820 J of heat. The specific heat of water is 4.18 J•g^-1•C^-1. By how much will it’s temperature rise, assuming it does not melt?
Answers
Answer: The temperature rise is [tex]0.53^0C[/tex]
Explanation:
The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.
[tex]Q=m\times c\times \Delta T[/tex]
Q = Heat absorbed by ice = 5280 J
m = mass of ice = 2.40 kg = 2400 g (1kg=1000g)
c = heat capacity of water = [tex]4.18J/g^0C[/tex]
Initial temperature = [tex]T_i[/tex]
Final temperature = [tex]T_f[/tex]
Change in temperature ,[tex]\Delta T=T_f-T_i=?[/tex]
Putting in the values, we get:
[tex]5280J=2400g\times 4.18J/g^0C\times \Delta T[/tex]
[tex]\Delta T=0.53^0C[/tex]
Thus the temperature rise is [tex]0.53^0C[/tex]
Answer: 0.580 C
Explanation: On Ck-12 I got it right sooo...
Which of the following acts as a bronsted Lowry acid, but not as a bronsted Lowry base
Answers
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]
Answers
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.
What is the relationship between the concentration of the hydronium and hydroxide ion and pH in any water solution?
Answers
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!
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
Answers
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
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.
Answers
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)
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.
Answers
Answer:
See explaination
Explanation:
Please kindly check attachment for the step by step solution of the given problem
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
Answers
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 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.
Answers
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!
If the volume of a spherical ball is 113.04 cubic inches, what is the radius?
3 inches
9 inches
18 inches
27 inches
Answers
Answer:
r= 3 inches
Explanation:
V= (4/3)*pi* r^3
113.04= (4/3)*3.14*r^3
113.04*(3/4)= 3.14*r^3
84.78 = 3.14*r^3
84.78/3.14 = r^3
27 = r^3 Take the cubed root of both sides.
r = 3 inches
Gneiss rock forms from a great amount of pressure and heat underground. What type of rock is it?
Answers
Answer:
Gneiss is a foliated metamorphic rock
Explanation:
Gneiss is a high grade metamorphic rock, meaning that it has been subjected to higher temperatures and pressures .
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
Answers
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.