Answer:
The Density Of Ar (g) At -11°C And 675 MmHg (R =0.08206 L·atm/mol·K, 1 Atm = 760mmHg).
determine the mass in grams of 3.75 x 10^21 atoms of zinc. (the mass of one mole of zinc is 65.39 g)
Answer: The mass in [tex]3.75 \times 10^{21}[/tex] atoms of zinc is 0.405 g.
Explanation:
Given: Atoms of zinc = [tex]3.75 \times 10^{21}[/tex]
It is known that 1 mole of every substance contains [tex]6.022 \times 10^{23}[/tex] atoms. So, the number of moles in given number of atoms is as follows.
[tex]Moles = \frac{3.75 \times 10^{21}}{6.022 \times 10^{23}}\\= 0.622 \times 10^{-2}\\= 0.0062 mol[/tex]
As moles is the mass of a substance divided by its molar mass. So, mass of zinc (molar mass = 65.39 g/mol) is calculated as follows.
[tex]Moles = \frac{mass}{molar mass}\\0.0062 mol = \frac{mass}{65.39 g}\\mass = 0.405 g[/tex]
Thus, we can conclude that the mass in [tex]3.75 \times 10^{21}[/tex] atoms of zinc is 0.405 g.
What is the first step in the curved arrow mechanism for the chlorination of benzene in the presence of FeCl3
Answer:
See explanation
Explanation:
The chlorination of benzene occurs in the presence of a Lewis acid. A Lewis acid is a compound that can accept a lone pair of electrons.
The first step in the chlorination of benzene is the formation of the ion Cl^+ which attacks the benzene ring.
This ion is formed when the Cl2 molecule undergoes heterolytic fission assisted by FeCl3 to yield FeCl4^- and Cl^+.
The Cl^+ electrophile now attacks the benzene ring to yield chlorobenzene.
determine the budget for tge fitness event.consider the attendance (should tou ask for registration fee.give shirts etc)
Draw bond-line formulas of all dichloro derivatives that might be formed when 1-chloro-2,2,3,3,-tetramethylpentane is allowed to react with Cl 2 under UV irradiation. For each structure, indicate, with an asterisk, any stereocenters that might be present. Predcit the percentage of each product using the relative reactivities 3 0 = 5.3, 2 0 = 3.6, 1 0 = 1
Answer:
Explanation:
This is a halogenation reaction i.e substitution or replacement of a single or more than a single hydrogen atom in the organic alkane compound with the halogen(here it is chlorine).
The chlorination of 1-chloro-2,2,3,3-tetramethylpentane under UV light resulted in the formation of five (5) dichloro derivatives which are shown in the image attached below.
Also, the compounds containing a stereocenter (i.e a location within the compound composing of various substituents in which the interchangeability of these substituents has the tendency of resulting into a stereoisomer) are indicated with an asterisk in the image below.
From the image below:
compound 1 ⇒ 1,1-dichloro-2,2,3,3-tetramethylpentane = 2° C
∴
The given relative reactivity rate for 2° = 3.6x
For compound 2 ⇒ 1,4-dichloro-2,2,3,3-tetramethylpentane = 2° = 3.6x
For compound 3 ⇒ 1,5-dichloro-2,2,3,3-tetramethylpentane = 1° = 1x
For compound 4 ⇒ 1-chloro-2-chloromethyl-2,3,3-trimethylpentane
= 1° = 1x
For compound 5 ⇒ 1-chloro-3-chloromethyl-2,2,3-trimethylpentane
= 1° = 1x
As such, we have:
2(3.6x) + 3(1x) = 100
7.2x + 3x = 100
10.2x = 100
x = 100/10.2
x = 9.803°
∴
For compound (1) = 3.6(9.803) = 35.3%
For compound (2) = 3.6(9.803) = 35.3%
For compound (3) = 1(9.803) = 9.803°%
For compound (4) = 1(9.803) = 9.803°%
For compound (5) = 1(9.803) = 9.803°%
Which atomic model states that it is impossible to know the exact location of electrons around the nucleus?
Answer:
Bohr Model is the correct answer
Answer:
Electron -Cloud Model
Explanation:
Just took the quiz got 100%
pls help ive been stuck on this question for a while im not good with chemistry lol.
Answer:
The answer would be B, putting thermal energy into something means you're adding heat into it.
I need help with the practice question at the bottom. Thank you.
As the gas expands on the surrounding, work is done by the system.
Therefore, W= -279J
Absorbtion of heat,q= +216J
∆U=q+W = (216-279)J= -63J
Which equation was used by Albert Einstein to explain the photoelectric effect? [E = energy, h= planck's constan, and v = frequency]
Answer:
E = hv
Explanation:
Energy = planck constant × frequency
Which safety feature works to slow down nuclear-fission chain reactions?
A. Proliferation
B. Control rods
C. Cooling rods
D. Fuel rods
Answer:
The answer is
-control rods
Answer:
B. Control Rods.... via A P E X
0.300 moles of sodium nitrite are needed for a reaction. The solution is 0.450 M. How many mL are needed
Answer:
The solution is 0.450 M. How many mL are needed?
- 0.667 mL
Explanation:
Calculate the moles of H3PO4 that reacted (8). Consult the coefficients in the balance chemical reaction to obtain the mole ratio. Show your calculation here.
Answer:
The number of moles of H₃PO₄ that reacted is 0.000343 moles
Note: Some data is missing. Data from the attachment is used for the calculationsinnthe explanation below.
Explanation:
The reaction is a neutralization reaction between NaOH and H₃PO₄. The equation of the reaction is given as follows:
3 NaOH + H₃PO₄ ---> Na₃PO₄ + 3 H₂O
The molarity of the NaOH solution is 0.238 mol/L.
Average volume of NaOH used during the titration to arrive to endpoint = (4.6 + 3.9 + 4.5) mL / 3 = 4.33 mL
Molarity is defined ratio of the number of moles of solute to the volume of solution. Mathematically, molarity = number of moles/volume in Litres
Number of moles of NaOH reacted = 0.238 mol/L × (4.33mL × 1 L/1000 mL)
Number of moles of NaOH = 0.00103 moles
From the equation of the reaction, 3 moles of NaOH reacts with 1 mole of H₃PO₄
0.00103 moles of NaOH will react with 0.00103 x 1/3 moles of H₃PO₄ = 0.000343 moles of H₃PO₄.
Therefore, number of moles of H₃PO₄ that reacted is 0.000343 moles
Pl hep help help me
Gaseous methane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 9.6 g of methane is mixed with 64.9 g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to significant digits.
Answer:
21.6 g
Explanation:
The reaction that takes place is:
CH₄ + 2O₂ → CO₂ + 2H₂OFirst we convert the given masses of both reactants into moles, using their respective molar masses:
9.6 g CH₄ ÷ 16 g/mol = 0.6 mol CH₄64.9 g O₂ ÷ 32 g/mol = 2.03 mol O₂0.6 moles of CH₄ would react completely with (2 * 0.6) 1.2 moles of O₂. As there are more O₂ moles than required, O₂ is the reactant in excess and CH₄ is the limiting reactant.
Now we calculate how many moles of water are produced, using the number of moles of the limiting reactant:
0.6 mol CH₄ * [tex]\frac{2molH_2O}{1molCH_4}[/tex] = 1.2 mol H₂OFinally we convert 1.2 moles of water into grams, using its molar mass:
1.2 mol * 18 g/mol = 21.6 gEthylene glycol flows at 0.01 kg/s through a 3-mm diameter, thin-walled tube. The tube is coiled and submerged in well-stirred water bath maintained at 25°C. If the fluid enters the tube at 85°C, what heat rate and tube length are required for the fluid to leave at 35°C?
Answer:
heat rate= 1281W
length = 15.8m
Explanation:
we have this data to answer this question with
Tmi = 85 degrees
Tmo = 35 degrees
Ts = 25 dgrees
flow rate = 25 degrees
using engine oil property from table a-5
Tm = Tmo - TMi/2 = 333k
u =0.522x10⁻²
k = 0.26
pr = 51.3
cp = 2562 J/kg.k
mcp(Tmo-Tmi) =
0.01 x 2562(35-85)
= 1281 W
we find the change in Tim
= [(35-25)-(85-25)]/ln[(35-25)/(85-25)]
= -50/ln0.167
= -50/-1.78976
= 27.9°c
we finf the required reynold number
4x0.01/πx0.003x0.522x10⁻²
= 0.04/0.00004921
= 812.8
= 813
we find approximate correlation
NuD = hd/k
NuD = 3.66
3.66 = 0.003D/0.26
cross multiply
0.003D = 3.66x0.26
D = 3.66x0.26/0.003
= 317.2
As = 1281/317x27.9
= 0.145
As = πDL
L = As/πD
= 0.145/π0.003
= 0.145/0.009429
L = 15.378
Explain the sample prep for IR
Answer:
) Making a sandwich To prepare a liquid sample to IR analysis, firstly place a drop of the liquid on the face of a highly polished salt plate (such as NaCl, AgCl or KBr), then place a second plate on top of the first plate so as to spread the liquid in a thin layer between the plates, and clamps the plates together.
Question 16(Multiple Choice Worth 5 points)
(04.01 LC) Which statement is true about the total mass of the reactants during a chemical change?
O It is destroyed during chemical reaction.
O It is less than the total mass of the products. O It is equal to the total mass of the products.
O It is greater than the total mass of the products.
Answer:
It is equal to the total mass of the products.
Explanation:
Hope this helps :)
For the molecular compound dioxygen difluoride , what would you multiply "grams of O2F2 " by to get the units "molecules of O2F2 "
Answer:
Molecules of O₂F₂ = mass of O₂F₂ × (1 mole O₂F₂ / 70 g O₂F₂) × (6.02 × 10²³ molecules / one mole of O₂F₂)
Explanation:
The Avogadros constant gives the the number of specified entities in one mole of a substance. One mole of any substance contains 6.02 × 10²³ particles. Therefore, one mole of O₂F₂ contains 6.02 × 10²³ molecules.
Also, the molar mass of a substance is the mass in grams of one mole of that substance. It is obtained by summing the relative atomic masses of all the atoms of the elements in the substance. For O₂F₂, the molar mass = (2 × 16 + 2 × 19) g/mol = 70 g/mol
Converting to molecules of O₂F₂;
To convert from grams of a substance to molecules of that substance, multiply by the ratio of one mole and mass of one mole, and then by the number of molecules per mole.
Molecules of A = mass of A × (1 mole / mass of one mole) × (6.02 × 10²³ molecules / 1 mole)
Therefore,Molecules of O₂F₂ = mass of O₂F₂ × (1 mole O₂F₂ / 70 g O₂F₂) × (6.02 × 10²³ molecules /one mole of O₂F₂)
can anybody pls help me with chemistry how to balance an equation of class 7 ??? pls
really I don't like chemistry..
I need only the step(s)..
Answer:
[tex] This\:may\: help[/tex]
Answer:
(1) Write down the chemical reaction in the form of word equation,keeping reactants on left hand side and products on right hand side.
(2) Write symbol and formula of all reactants and products in word equation. (3) Balance the equation by multiplying the symbols and formula by smallest possible figures.
o Calculate the pH of 500 ml of buffer solution containing .2M of ammonium sulphate and 0.3M ammonium hydroxide which 5 2.1% ionized in dilute solution. (kb of ammonium hydroxide is 1.8×10^-5)
Answer is 9
pKb=−logK
b=−log10^-5=5
A mixture of ammonium hydroxide and ammonium chloride forms a basic buffer solution.Henderson's equation for the basic buffer solution is as given below-pOH=pKb+log[acid]*[salt]
Substitute values in the above expression.
pOH=5+log0.1*0.1=5
Hence, the pH of the solution is pH=14−pOH=14−5=9
What do you mean by pH ?potential of hydrogena measure of the acidity or alkalinity of a solution equal to the common logarithm of the reciprocal of the concentration of hydrogen ions in moles per cubic decimetre of solution. Pure water has a pH of 7, acid solutions have a pH of less than 7, and alkaline solution with a pH greater than 7.
Learn more about pH here
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Silver has two naturally occurring isotopes with the following isotopic masses: 10747Ag 10947Ag 106.90509 108.9047 The average atomic mass of silver is 107.8682 amu. The fractional abundance of the lighter of the two isotopes is ________.
Answer: The fractional abundance of lighter isotope is 0.518
Explanation:
Average atomic weight is the sum of the masses of the individual isotopes each multiplied by its fractional abundance. The equation used is:[tex]\text{Average atomic weight}=\sum_{i=1}^{n}\text{(Atomic mass of isotope)}_i\times \text{(Fractional abundance)}_i[/tex] ......(1)
Let the fractional abundance of Ag-107 isotope be 'x'
For Ag-107 isotope:Atomic mass = 106.90509 amu
Fractional abundance = x
For Ag-109 isotope:Atomic mass = 108.9047 amu
Fractional abundance = (1 - x)
Average atomic mass of silver = 107.8682 amu
Plugging values in equation 1:
[tex]107.8682=(106.90509 \times x) + (108.9047 \times (1-x))\\\\107.8682=106.90509x+108.9047-108.9047x\\\\1.99961x=1.0365\\\\x=0.518[/tex]
Fractional abundance of Ag-107 isotope (lighter) = x = 0.518
Hence, the fractional abundance of lighter isotope is 0.518
what is the mass of insoluble calcium phosphate produced from .555 grams of calcium chloride
Answer:
0.518 g
Explanation:
Step 1: Write the balanced equation
3 CaCl₂ + 2 H₃PO₄ ⇒ Ca₃(PO₄)₂ + 6 HCl
Step 2: Calculate the moles corresponding to 0.555 g of CaCl₂
The molar mass of CaCl₂ is 110.98 g/mol.
0.555 g × 1 mol/110.98 g = 5.00 × 10⁻³ mol
Step 3: Calculate the moles of Ca₃(PO₄)₂ produced
5.00 × 10⁻³ mol CaCl₂ × 1 mol Ca₃(PO₄)₂/3 mol CaCl₂ = 1.67 × 10⁻³ mol Ca₃(PO₄)₂
Step 4: Calculate the mass corresponding to 1.67 × 10⁻³ moles of Ca₃(PO₄)₂
The molar mass of Ca₃(PO₄)₂ is 310.18 g/mol.
1.67 × 10⁻³ mol × 310.18 g/mol = 0.518 g
what does LPG stand for? mention one important source of LPG give sort answer
Answer:
liquefied petroleum gas
LPG is prepared by refining natural gas. it is made by refining crude oil or from extracted natural gas streams as they emerge from the ground.
For an ideal gas, classify the pairs of properties as directly or inversely proportional. You are currently in a sorting module. a. P and n b. V and n c. P and T d. T and V e. P and V1. directly proportional2. inversely proportional
Answer:
the result for the following are (a) P is directly proportional to n
(b) V is directly proportional to T (c) P is directly proportional to T (d) T is inversly proportional to V (e) P is inversely proportional to V
Which of the following statements correctly explains why bromination reactions are more selective than chlorination reactions.
a. bromine radical is less stable than chlorine radical, so it is more reactive and less choosy
b. bromine radical is more stable than chlorine radical, so it is more reactive and less choosy
c. bromine radical is more stable than chlorine radical, so it is less reactive and more choosy
d. bromine radical is less stable than chlorine radical, so it is less reactive and more choosy
e. relative radical stability is 3' radicals > 2" radicals> 1 radicals when bromine radicals snatch hydrogens from alkanes, but when chlorine radicals snatch hydrogens the resulting alkyl radical stability is 3 radicals < 2 radicals< 1' radicals
Answer: A bromine radical is more stable than chlorine radical, so it is less reactive and more choosy.
Explanation:
A chlorine atom being more electronegative in nature is able to attract a hydrogen atom more readily towards itself as compared to a bromine atom.
Since bromine is less electronegative in nature so bromine will be more selective as a hydrogen abstracting agent. As a result, bromine radical is more stable in nature than chlorine radical.
Thus, we can conclude that bromine radical is more stable than chlorine radical, so it is less reactive and more choosy.
9. Which property is not important when selecting a material to use as a light bulb filament?
d
high melting point
high vapor pressure
high ductility
Answer:
high melting point
Explanation:
The filament of a bulb is often heated to very high temperatures as the bulb is in operation.
Many times, electric bulbs may have to be on for a whole day and they may reach temperatures that are outrageously high in the process.
The material of the filament must have a very high melting point so that it doesn't melt while the bulb is still in operation.
Answer:
Actually the answer is High Vapor pressure
Explanation:
Question 1
1 pts
How many mols of bromine are present in 35.7g of
Tin(IV) bromate?
Answer:
n = 0.0814 mol
Explanation:
Given mass, m = 35.7g
The molar mass of Tin(IV) bromate, M = 438.33 g/mol
We need to find the number of moles of bromine. We know that,
No. of moles = given mass/molar mass
So,
[tex]n=\dfrac{35.7}{438.33}\\\\n=0.0814\ mol[/tex]
So, there are 0.0814 moles of bromine in 35.7g of Tin(IV) bromate.
In an exothermic reaction, reactants are __ products.
a more stable than
b equal to
c equally stable as
d less stable than
Answer:
Less stable
Explanation:
When an exothermic reaction occurs, the reactants have a greater energy level than the outputs. The products, in other respects, are more stable than that of the reactants.
The outcomes of an exothermic reaction have a lower energy to react. The enthalpy of a process is the difference between some of the activation energy and the power of the products.
Energy Changes in Reactions
Independent Practice
In a synthesis reaction, one reactant contains 256 J of chemical energy, and one reactant contains 328 J of
chemical energy. The product contains 407 J of chemical energy.
Because energy is conserved, what energy change occurs during the reaction?
A. 177 J of energy is released.
B. 177 J of energy is absorbed.
C. 335 J of energy is released,
D. 335 J of energy is absorbed.
SAVE
SAVE AND CLOSE
What should be done if a spectrophotometer reports an absorbance that is too high? Select one: Reposition the cuvette in the spectrophotometer. Pour out half the volume of the sample. Restart the spectrophotometer and try again. Dilute the sample. g
Answer:
The sample should be diluted
Explanation:
According to Beer Lambert's law, the absorbance of a sample depends on the concentration of the sample.
Hence, if the concentration of the sample is very high, the spectrophotometer will also report a very high value of absorbance.
When this is the case, the sample should simply be diluted and the readings are taken again using the spectrophotometer.
A 1.5-m 3 insulated rigid tank contains 2.7 kg of carbon dioxide at 100 kPa. Now paddle-wheel work is done on the system until the pressure in the tank rises to 150 kPa. Determine the entropy change of carbon dioxide during this process. Assume constant specific heats
Answer:
The entropy change of carbon dioxide = 0.719 kJ/k
Explanation:
Given:
1.5 m - 3 insulated rigid tank contains 2.7 kg of carbon dioxide at 100 kPa
The objective is to determine the entropy change of carbon dioxide
Formula used:
ΔS=
Solution:
On considering,
[tex]C_{P} =0.846 kJ/kg K\\C_V=0.657 kJ/kg k\\[/tex]
ΔS=[tex]mc_{v} lu\frac{p_{2} }{P_{1} }[/tex]
On substituting the values,
ΔS=[tex]2.7*0.657lu\frac{150}{100}[/tex]
ΔS=0.719 kJ/k
The entropy change is "0.719 kJ/K".
Given values are:
Mass of tank,
m = 2.7 kgPressure,
P₁ = 100 kPaRised pressure,
P₂ = 150 kPaAssumption of constant specific heat is,
[tex]C_v = 0.657 \ kJ/kgK[/tex]As we know the formula,
→ [tex]\Delta S = mC_v \ ln(\frac{P_2}{P_1} )[/tex]
[tex]= (2.7)(0.657) \ ln (\frac{150}{100} )[/tex]
[tex]= 1.7739\times 0.4055[/tex]
[tex]= 0.7193 \ kJ/K[/tex]
Thus above answer is right.
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