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3 years ago

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You are given a cube with the length of 2.5 cm , a width of 2.5 cm , and a height of 2.5 cm. You place it on the scale, and its

mass is 295 g. Calculate the density. Please show your work.

1 answer:

kakasveta [241]3 years ago

8 0

Density is equal to the mass divided by the voulme.
So: 2.5 multiplied by 2.5 by 2.5.
Then: 295 divided by 16.625=18.88g/cm to the power 3

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Air containing 20.0 mol% water vapor at an initial pressure of 1 atm absolute is cooled in a 1- liter sealed vessel from 200°C t

chubhunter [2.5K]

Answer:

This solution is quite lengthy

Total system = nRT

n was solved to be 0.02575

nH20 = 0.2x0.02575

= 0.00515

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Pair = 1-0.19999

= 0.80001

At 15⁰c

Pair = 0.4786atm

I used antoine's equation to get pressure

The pressure = 0.50

2. Moles of water vapor = 0.0007084

Moles of condensed water = 0.0044416

Grams of condensed water = 0.07994

Please refer to attachment. All solution is in there.

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VashaNatasha [74]

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3 years ago

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miv72 [106K]

Answer:

29.42 Litres

Explanation:

The general/ideal gas equation is used to solve this question as follows:

PV = nRT

Where;

P = pressure (atm)

V = volume (L)

n = number of moles (mol)

R = gas law constant (0.0821 Latm/molK)

T = temperature (K

According to the information provided in this question;

mass of nitrogen gas (N2) = 25g

Pressure = 0.785 atm

Temperature = 315K

Volume = ?

To calculate the number of moles (n) of N2, we use:

mole = mass/molar mass

Molar mass of N2 = 14(2) = 28g/mol

mole = 25/28

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3 years ago

Carbonyl fluoride, COF2, is an important intermediate used in the production of fluorine-containing compounds. For instance, it

lions [1.4K]

<u>Answer:</u> The equilibrium concentration of $COF_2$ is 0.332 M

<u>Explanation:</u>

We are given:

Initial concentration of $COF_2$ = 2.00 M

The given chemical equation follows:

$2COF_2(g)\rightleftharpoons CO_2(g)+CF_4(g)$

<u>Initial:</u> 2.00

<u>At eqllm:</u> 2.00-2x x x

The expression of $K_c$ for above equation follows:

$K_c=\frac{[CO_2][CF_4]}{[COF_2]^2}$

We are given:

$K_c=6.30$

Putting values in above expression, we get:

$6.30=\frac{x\times x}{(2.00-2x)^2}\\\\x=0.834,1.25$

Neglecting the value of x = 1.25 because equilibrium concentration of the reactant will becomes negative, which is not possible

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Hence, the equilibrium concentration of $COF_2$ is 0.332 M

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3 years ago