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3 years ago

14

Consider Example 12.10. Suppose the experiment is repeated with 0.032 mol of helium instead with everything else staying the sam

e, what is the amount of heat required now to achieve that process

1 answer:

Lemur [1.5K]3 years ago

5 0

Answer:

120 extracted from the gas

Explanation:

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The atomic mass of chlorine is listed as 35.454 amu on the periodic table. It has two naturally occurring isotopes, Cl-35 and Cl

Fynjy0 [20]

X=percentage of Cl-35
(1-x)=percentage of Cl-35

35(x)+37(1-x)=35.454
35x+37-37x=35.454
-2x=35.454-37
-2x=-1.546
x=-1.546/(-2)
x=0.773

0.773=77.3/100=77.3%

(1-x)=1-0.773=0.227

0.227=22.7/100=22.7%

answer: the percentages are: 77.3% of Cl-35 and 22.7% of Cl-37.

7 0

3 years ago

I need help with my work

elena-14-01-66 [18.8K]

Answer:

2 the mass is greater than 10.0gram

Explanation:

2 the mass is greater than 10.0gram

8 0

2 years ago

Write the net ionic equation for the reaction that takes places between aqueous copper and nitrate.

PIT_PIT [208]

Answer:

Cu+HNO3

Explanation:

7 0

3 years ago

If 5.0 liters H2 (g) at STP is heated to a temperature of 985, pressure remaining constant, the new volume of the gas will be?

vlada-n [284]

Answer:

$V_2=18 \ L \ H_2$

General Formulas and Concepts:

<u>Chemistry - Gas Laws</u>

STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K
Charles' Law: $\frac{V_1}{T_1} =\frac{V_2}{T_2}$

Explanation:

<u>Step 1: Define</u>

Initial Volume: 5.0 L H₂ gas

Initial Temp: 273 K

Final Temp: 985 K

Final Volume: ?

<u>Step 2: Solve for new volume</u>

Substitute: $\frac{5 \ L \ H_2}{273 \ K} =\frac{x \ L \ H_2}{985 \ K}$
Cross-multiply: $(5 \ L \ H_2)(985 \ K) = (x \ L \ H_2)(273 \ K)$
Multiply: $4925 \ L \ H_2 \cdot K = 273x \ L \ H_2 \cdot K$
Isolate <em>x</em>: $18.0403 \ L \ H_2 = x$
Rewrite: $x=18.0403 \ L \ H_2$

<u>Step 3: Check</u>

<em>We are given 2 sig figs as the smallest. Follow sig fig rules and round.</em>

<em /> $18.0403 \ L \ H_2 \approx 18 \ L \ H_2$ <em />

5 0

3 years ago

How many oxygen atoms are represented by the formula Fe(CIO4)3? iron(III) chlorate

zysi [14]

Answer:

12

Explanation:

the 4 by the element symbol O multiplied by the 3 on the outside of the parentheses

5 0

3 years ago

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