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Gwar [14]
3 years ago
8

What is the answer to this question

Chemistry
1 answer:
Stels [109]3 years ago
4 0

Answer:

38 kg/m³

0.038 g/mL

Explanation:

Volume of a cube is the side length cubed.

V = s³

Given s = 0.65 m:

V = (0.65 m)³

V ≈ 0.275 m³

The mass is 10.5 kg.  The density is the mass divided by volume:

ρ = (10.5 kg) / (0.275 m³)

ρ ≈ 38 kg/m³

Or:

ρ ≈ 0.038 g/mL

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Calculate the mass of 2.50 mol of CH,OH(1). Show your work. Use the appropriate
Sophie [7]

Answer:

80.1 grams

Explanation:

Find the molar mass of CH3OH first by using the periodic table values.

12.011 g/mol C + (1.008*3 g/mol H) + 15.999g/mol O + 1.008 g/mol H

=32.042 so that is the molar mass

Now that you have 2.50 moles of CH3OH, you can calculate the mass in g

2.50molCH3OH * (32.042g CH3OH / 1 mol CH3OH) = 80.105

32.042g / 1 mol is the same as 32.042 g/mol

Since there are 3 sig figs in the problem (2.50 has 3 sig figs), you round to 80.1 g CH3OH

7 0
2 years ago
Assessment timer and count Assessment items Item 3 What condition would most likely cause a decrease in the salinity of ocean wa
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Answer:

it's a river flowing into the sea

Explanation:

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3 years ago
Which of the following would be appropriate reason(s) for scientists to reject atomic theory?​
leonid [27]
It posed a contradiction to Quantum Theory
3 0
3 years ago
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If 13.0 g of MgSO4⋅7H2O is thoroughly heated, what mass of anhydrous magnesium sulfate will remain?
Paul [167]

6.349 g mass of anhydrous magnesium sulfate will remain.

<h3>What are moles?</h3>

A mole is defined as 6.02214076 × 1023 of some chemical unit, be it atoms, molecules, ions, or others. The mole is a convenient unit to use because of the great number of atoms, molecules, or others in any substance.

Molar mass MgSO₄.7 H₂O = 246.52 g/mol

Moles =\frac{mass}{molar \;mass}

Moles =\frac{13.0 g}{246.52}

0.0527 moles

Molar mass MgSO₄ = 120.4 g/mol

Mass of anhydrous magnesium sulfate :

( 0.0527 x 120.4 ) => 6.349 g

Learn more about moles here:

brainly.com/question/8455949

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4 0
1 year ago
While ethanol is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by reacting e
sp2606 [1]

The question is incomplete, here is the complete question:

While ethanol is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by reacting ethylene with water vapor at elevated temperatures.

A chemical engineer studying this reaction fills a 1.5 L flask at 12°C with 1.8 atm of ethylene gas and 4.7 atm of water vapor. When the mixture has come to equilibrium she determines that it contains 1.16 atm of ethylene gas and 4.06 atm of water vapor.

The engineer then adds another 1.2 atm of ethylene, and allows the mixture to come to equilibrium again. Calculate the pressure of ethanol after equilibrium is reached the second time. Round your answer to 2 significant digits.

<u>Answer:</u> The partial pressure of ethanol after equilibrium is reached the second time is 1.0 atm

<u>Explanation:</u>

We are given:

Initial partial pressure of ethylene gas = 1.8 atm

Initial partial pressure of water vapor = 4.7 atm

Equilibrium partial pressure of ethylene gas = 1.16 atm

Equilibrium partial pressure of water vapor = 4.06 atm

The chemical equation for the reaction of ethylene gas and water vapor follows:

                     CH_2CH_2(g)+H_2O(g)\rightleftharpoons CH_3CH_2OH(g)

<u>Initial:</u>                  1.8                4.7

<u>At eqllm:</u>           1.8-x             4.7-x

Evaluating the value of 'x'

\Rightarrow (1.8-x)=1.16\\\\x=0.64

The expression of K_p for above equation follows:

K_p=\frac{p_{CH_3CH_2OH}}{p_{CH_2CH_2}\times p_{H_2O}}

p_{CH_2CH_2}=1.16atm\\p_{H_2O}=4.06atm\\p_{CH_3CH_2OH}=0.64atm

Putting values in above expression, we get:

K_p=\frac{0.64}{1.16\times 4.06}\\\\K_p=0.136

When more ethylene is added, the equilibrium gets re-established.

Partial pressure of ethylene added = 1.2 atm

                     CH_2CH_2(g)+H_2O(g)\rightleftharpoons CH_3CH_2OH(g)

<u>Initial:</u>                2.36             4.06               0.64

<u>At eqllm:</u>           2.36-x        4.06-x             0.64+x

Putting value in the equilibrium constant expression, we get:

0.136=\frac{(0.64+x)}{(2.36-x)\times (4.06-x)}\\\\x=0.363,13.41

Neglecting the value of x = 13.41 because equilibrium partial pressure of ethylene and water vapor will become negative, which is not possible.

So, equilibrium partial pressure of ethanol = (0.64 + x) = (0.64 + 0.363) = 1.003 atm

Hence, the partial pressure of ethanol after equilibrium is reached the second time is 1.0 atm

3 0
3 years ago
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