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Sergeeva-Olga [200]
3 years ago
6

What is the volume of 2 moles of He at STP?

Chemistry
1 answer:
larisa86 [58]3 years ago
6 0

Answer:

STP conditions imply a temperature of 273.15 K and a pressure of 100 kPa. When tose conditions are met, 1 mole of any ideal gas will have a volume of 22.7 L. So, if 1 mole occupies a volume of 22.7 L, 2 moles will occupy a volume twice as big. Likewise, 0.5 moles will occupy half the volume 1 mole occupies. Assuming that the gas is at standard temperature and pressure (STP), one mole of any gas occupies 22.4 L . This means the number of moles of O2 is 222.4=0.089 mol .

Explanation:

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damaskus [11]
<span>Copper II Sulfate- Cu 2+ and a sulfate SO4 2- add together nicely to make CuSO4. </span>
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Select all that apply.
wariber [46]

Answer:

The reaction is endothermic

Explanation:

we know that the reactants had to get energy from its surroundings which means that the reaction is endothermic

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For the following balanced equation: 3 Cu(s) + 8 HNO3(aq) → 3 Cu(NO3)2(aq) + 2 NO(g) + 4 H2O(l) a) How many moles of HNO3 will r
s344n2d4d5 [400]

Answer:

a) <u>26.67 moles HNO3 </u>

b) <u>0.33 moles NO</u>

c) <u>0.40 moles NO is produced</u>

d)<u>.157 moles Cu</u>

e) <u>0.105 moles NO</u>

f) <u>26.4 grams HNO3</u>

g) <u>Cu is in excess</u>

h) <u>2.41 grams Cu remain</u>

i) <u>2.37 grams NO</u>

Explanation:

Step 1: Data given

Molar mass of Cu = 63.55 g/mol

Molar mass of HNO3 = 63.01 g/mol

Molar mass of Cu(NO3)2 = 187.56 g/mol

Molar mass of NO = 30.01 g/mol

Molar mass of H2O = 18.02 g/mol

Step 2: The balanced equation

3 Cu(s) + 8 HNO3(aq) → 3 Cu(NO3)2(aq) + 2 NO(g) + 4 H2O(l)

a) How many moles of HNO3 will react with 10 moles of Cu?

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

For 10 moles Cu we need 8/3 *10 = <u>26.67 moles HNO3 </u>

b) How many moles of NO will form if 0.50 moles of Cu reacts?

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

For 0.50 moles Cu we'll have 2/3 *0.50 = <u>0.33 moles NO</u>

c) If 0.80 moles of H2O forms, how much NO must also form?

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

If 0.80 moles H2O is produced, 0.80/2 = <u>0.40 moles NO is produced</u>

d) How many moles of Cu are in 10.0 grams of Cu?

Moles Cu = 10.0 grams / 63.55 g/mol = 0.157 moles

In 10.0 grams Cu we have <u>0.157 moles Cu</u>

e) If 10.0 g of Cu reacts, how many moles of NO will form?

10.0 grams Cu = 0.157 moles

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

For 0.157 moles Cu we'll have 2/3 * 0.157 = <u>0.105 moles NO</u>

f) If 10.0 g of Cu reacts, how many grams of HNO3 are required?

10.0 grams Cu = 0.157 moles

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

For 0.157 moles Cu we'll need 0.419 moles HNO3

This is 0.419 moles * 63.01 g/mol = <u>26.4 grams HNO3</u>

g) If 10.0 g of Cu and 20.0 g of HNO3 are put together in a reaction vessel, which one will be in excess?

Moles Cu = 0.157 moles

Moles HNO3 = 20.0 grams / 63.01 g/mo = 0.317 moles

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

The limiting reactant is HNO3. It will completely be consumed (0.317 moles). <u>Cu is in excess.</u> There will react 3/8 * 0.317 = 0.119 moles Cu

There will remain 0.157 - 0.119 = 0.038 moles

h) How many grams of the excess substance will be left over?

There will react 3/8 * 0.317 = 0.119 moles Cu

There will remain 0.157 - 0.119 = 0.038 moles

This is 0.038 moles * 63.55 g/mol = 2.41 grams

i) How many grams of NO will form in the reaction described in part g?

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

For 0.317 moles HNO3 we'll have 0.317/4 = 0.0793 moles NO

This is 0.079 mol * 30.01 g/mol =<u> 2.37 grams NO</u>

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What the sample will look like after the temperature is increased from 300 K to 450 KK at constant pressure?
sleet_krkn [62]

Answer:

The sample will look expanded and occupy more space.

Explanation:

Since, the pressure is constant here, but the temperature is changed. Therefore, according to Charles' Law Volume is directly proportional to Temperature, provided the pressure is kept constant. Mathematically:

V1/T1 = V2/T2

V1 = (T1/T2)(V2)

V1 = (300 k/450 k)(V2)

<u>V1 = (0.67)V2</u>

The equation indicates that The fina volume of the gas V2 will be greater than the initial volume V1. <u>Thus, sample will look expanded and occupy more space than the previous state.</u>

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Answer:

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Explanation:

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