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Doss [256]
3 years ago
8

What is the mass of 2.25 moles of manganese(lv) sulfide(MnS2)

Chemistry
2 answers:
sertanlavr [38]3 years ago
8 0
Hi! I'm hoping I understand your question correctly.. To calculate the mass or grams of MnS^2 you will have to find the molar mass. The molar mass is 87.003 g/mol. All you have to do is use stoichiometry and multiply that molar mass by the mols (2.25)! Hope this helped! Remember your sig figs, too!
Alona [7]3 years ago
5 0

Explanation:

Number of moles is defined as mass divided by molar mass of the substance or atom.

Mathematically,     No. of moles = \frac{mass}{molar mass}

It is known that molar mass of MnS_{2} is 119.058 g/mol and number of moles is given as 2.25 mol.

Therefore, calculate mass of MnS_{2} as follows.

                          No. of moles = \frac{mass}{molar mass}

                                   2.25 mol = \frac{mass}{119.058 g/mol}

                                 mass = 267.88 g

Thus, we can conclude that mass of 2.25 moles of manganese(lv) sulfide(MnS2) is 267.88 g.

                 

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Answer:

Fuel experiment

Explanation:

since it involves direct observation of the animals

6 0
2 years ago
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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Answer:

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Explanation:

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2 years ago
Please help me to do this assignment
kirill [66]

Answer:

1. Objective

2. Objective

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5. Opinion

6. Opinion

7. Opinion (I think.)

8. Opinion (I think.)

Explanation:

7 0
2 years ago
If the density of mercury is 13.6g/ml, how many ml of mercury are in 1,000 g of mercury?
Illusion [34]

Answer:

<h2>73.53 mL</h2>

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

volume =  \frac{mass}{density} \\

From the question we have

volume =  \frac{1000}{13.6}  \\  = 73.52941...

We have the final answer as

<h3>73.53 mL</h3>

Hope this helps you

6 0
3 years ago
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