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3 years ago

Question 7 of 15

Chemistry

1 answer:

Crazy boy [7]3 years ago

3 0

Answer: 0.4 moles

Explanation:

Given that:

Volume of gas V = 11L

(since 1 liter = 1dm3

11L = 11dm3)

Temperature T = 25°C

Convert Celsius to Kelvin

(25°C + 273 = 298K)

Pressure P = 0.868 atm

Number of moles N = ?

Note that Molar gas constant R is a constant with a value of 0.00821 atm dm3 K-1 mol-1

Then, apply ideal gas equation

pV = nRT

0.868atm x 11dm3 = n x (0.00821 atm dm3 K-1 mol-1 x 298K)

9.548 atm dm3 = n x 24.47atm dm3mol-1

n = (9.548 atm dm3 / 24.47atm dm3 mol-1)

n = 0.4 moles

Thus, there are 0.4 moles of the gas.

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How many grams of H2O will be formed when 32.0 g H2 is mixed with 84.0 g of O2 and allowed to react to form water

Zarrin [17]

Answer:

94.58 g of $H_2O$

Explanation:

For this question we have to start with the reaction:

$H_2~+~O_2~->~H_2O$

Now, we can balance the reaction:

$2H_2~+~O_2~->~2H_2O$

We have the amount of $H_2$ and the amount of $O_2$ . Therefore we have to find the limiting reactive, for this, we have to follow a few steps.

1) Find the moles of each reactive, using the molar mass of each compound ( $H_2~=~2~g/mol~~O_2=~32~g/mol$ ).

2) Divide by the coefficient of each compound in the balanced reaction ("2" for $H_2$ and "1" for $O_2$ ).

Find the moles of each reactive

$32.0~g~H_2\frac{1~mol~H_2}{2~g~of~H_2}=15.87~mol~H_2$

$84.0~g~of~O_2\frac{1~mol~of~O_2}{32~g~of~O_2}=2.62~mol~of~O_2$

Divide by the coefficient

$\frac{15.87~mol~H_2}{2}=7.94$

$\frac{2.62~mol~of~O_2}{1}=2.62$

The smallest values are for $H_2$ , so hydrogen is the limiting reagent. Now, we can do the calculation for the amount of water:

$32.0~g~H_2\frac{1~mol~H_2}{2~g~of~H_2}\frac{2~mol~H_2O}{2~mol~H_2}\frac{18~g~H_2O}{1~mol~H_2O}=94.58~g~H_2O$

We have to remember that the molar ratio between $H_2O$ and $H_2$ is 2:2 and the molar mass of $H_2O$ is 18 g/mol.

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3 years ago

Sand dunes protect beaches against erosion and flooding due to storms.

kogti [31]

Answer:

B. Building houses near the beach

Explanation:

it was right on edge

7 0

4 years ago

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Which reactions are oxidation-reduction reactions? Check all that apply

tia_tia [17]

2 Na + Cl₂ → 2 NaCl

Cu + 2 AgNO₃ → Cu(NO₃)₂ + 2 Ag

CH₄ + 2 O₂ → CO₂ + 2 H₂O

are oxidation-reduction reactions

Explanation:

In the oxidation-reduction reaction there is a component which accept electrons and it is reduced and another component which donates electrons and it is oxidized.

Now in the case of the reaction of copper (Cu) with silver nitrate (AgNO₃):

Cu + 2 AgNO₃ → Cu(NO₃)₂ + 2 Ag

Cu⁰ with a oxidation number 0 donates 2 electrons⁻ and is transformed in Cu⁺² with the oxidation number 2+.

Ag⁺ with a oxidation number 1+ accepts 1 electron⁻ and is transformed in Ag with the oxidation number 0.

In the case of CH₄ + 2 O₂ → CO₂ + 2 H₂O reaction we have:

C⁴⁻ with a oxidation number +4 donates 8 electrons⁻ and is transformed in C⁴⁺ with the oxidation number 4+.

O⁰ with a oxidation number 0 accepts 2 electrons⁻ and is transformed in O²⁻ with the oxidation number 2-.

And in the case of 2 Na + Cl₂ → 2 NaCl reaction we have:

Na⁰ with a oxidation number 0 donates 1 electron⁻ and is transformed in Na⁺ with the oxidation number 1+.

Cl⁰ with a oxidation number 0 accepts 1 electron⁻ and is transformed in Clwith the oxidation number 1-.

Learn more about:

oxidation-reduction reactions

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4 0

3 years ago

Calculate your experimentally determined percent mass of water in Manganese(II) sulfate monohydrate. Report your result to 2 or

antoniya [11.8K]

Answer:

10.6%

Explanation:

The determined percent mass of water can be calculated from the formula of the hydrate by

dividing the mass of water in one mole of the hydrate by the molar mass of the hydrate and

multiplying this fraction by 100.

Manganese(ii) sulphate monohydrate is MnSO4 . H2O

1. Calculate the formula mass. When determining the formula mass for a hydrate, the waters of

hydration must be included.

1 Manganes 52.94 g = 63.55 g

1 Sulphur 32.07 g =

32.07 g 2 Hydrogen is = 2.02 g

4 Oygen =

64.00 g 1 Oxygen 16.00 = 16.00 g

151.01 g/mol 18.02 g/mol

Formula Mass = 151.01 + (18.02) = 169.03 g/mol

2. Divide the mass of water in one mole of the hydrate by the molar mass of the hydrate and

multiply this fraction by 100.

Percent hydration = (18.02 g /169.03 g) x (100) = 10.6%

The final result is 10.6% after the two steps calculations

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3 years ago

The discovery that uranium is a radioactive element was made by Marie Curie. True or False

navik [9.2K]

Your answer is true.

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3 years ago

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