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2 years ago

How many grams of NaCl can be produced from 4.2 moles of Cl2?

Chemistry

1 answer:

Vikentia [17]2 years ago

6 0

Answer:

491.4 g

Explanation:

This a stoichiometry question, let me repeat the reaction equation here;

2 Na + Cl2 --> 2 NaCl

The balanced reaction equation must serve as a guide in solving any problem on stoichiometry.

From the balanced reaction equation;

1 mole of chlorine gas yields 2 moles of sodium chloride

Therefore, 4.2 moles of chlorine gas yields 4.2 ×2/1 = 8.4 moles of sodium chloride.

Molar mass of sodium chloride= 23 + 35.5= 58.5 gmol-1

Mass of sodium chloride formed= number of moles × molar mass

Mass of sodium chloride formed= 8.4 × 58.5= 491.4 g

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Answer:

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2 years ago

Student Z completes his/her calibration step for part 3 and finds an average massA to be 95.237 g ± 0.005 g. However, in his/her

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How much heat is released as the temperature of 25.2 grams of iron is decreased from 72.1°C to 9.8°C? The specific heat of iron

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Answer:

$Q=-697.06\ J$

Negative sign says that release of heat.

Explanation:

The expression for the calculation of the heat released or absorbed of a process is shown below as:-

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Where,

$\Delta H$ is the heat released or absorbed

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C is the specific heat capacity

$\Delta T$ is the temperature change

Thus, given that:-

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So,

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2 years ago

Consider the following reactions and their respective equilibrium constants: NO(g)+12Br2(g)⇌NOBr(g)Kp=5.3 2NO(g)⇌N2(g)+O2(g)Kp=2

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Answer:

Equilibrium constant of the given reaction is $1.3\times 10^{-29}$

Explanation:

$NO+\frac{1}{2}Br_{2}\rightleftharpoons NOBr$ .... $(K_{p})_{1}=5.3$

$2NO\rightleftharpoons N_{2}+O_{2}$ .... $(K_{p})_{2}=2.1\times 10^{30}$

The given reaction can be written as summation of the following reaction-

$2NO+Br_{2}\rightleftharpoons 2NOBr$

$N_{2}+O_{2}\rightleftharpoons 2NO$

......................................................................................

$N_{2}+O_{2}+Br_{2}\rightleftharpoons 2NOBr$

Equilibrium constant of this reaction is given as-

$\frac{[NOBr]^{2}}{[N_{2}][O_{2}][Br_{2}]}$

$=(\frac{[NOBr]}{[NO][Br_{2}]^{\frac{1}{2}}})^{2}(\frac{[NO]^{2}}{[N_{2}][O_{2}]})$

$=\frac{(K_{p})_{1}^{2}}{(K_{p})_{2}}$

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2 years ago

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This means that the steel bar lost 14900 J of thermal energy. What is the change in temperature of the steel bar? Recall that th

stiv31 [10]

Answer:

ΔT=-747,13°C

Explanation:

Sensible heat is the amount of thermal energy that is required to change the temperature of an object, the equation for calculating the heat change is given by:

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Q, heat that has been absorbed or realeased by the substance [J]
m, mass of the substance [g]
s, specific heat capacity [J/g°C] (
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To solve the problem, we clear ΔT of the equation and then replace our data:

Q=msΔT

ΔT=Q/ms

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(Note that Q=-14900 J because there is a LOST of thermal energy)

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