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choli [55]
2 years ago
10

Question 8 (5 points)

Chemistry
1 answer:
Gnom [1K]2 years ago
7 0

Answer:

d . The volume of the solvent used was less than 5 liters.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to calculate the volume of the stock (initial) solution by using the following equation:

M_1V_1=M_2V_2

Thus, we solve for, V1, which stands for the aforementioned volume of stock solution:

V_1=\frac{M_2V_2}{M_1}

Then, we plug in to obtain:

V_1=\frac{5L*1M}{10M}\\\\V_1=0.5L

Now, since the final volume was 5 L, we can infer that the volume of solvent is 4.5 L and that of the stock solution 0.5 L for a total of 5 L of diluted solution; therefore, the correct reasoning is d . The volume of the solvent used was less than 5 liters.

Regards!

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Draw the structures of all monobromo derivatives of pentane, c5h11br, which contain a 4-carbon chain.
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5 0
3 years ago
How many grams of NO can be produced if 204 g of NO2 is mixed with 58.1 g of H2O?
Goshia [24]

Answer:

44.4 grams of NO can be produced

Explanation:

Step 1: Data given

Mass of NO2 = 204 grams

Molar mass NO2 = 46.0 g/mol

Mass of H2O = 58.1 grams

Molar mass H2O = 18.02 g/mol

Step 2: The balanced equation

3NO2 + H2O→ 2HNO3 + NO

Step 3: Calculate moles NO2

Moles NO2 = 204 grams / 46.0 g/mol

Moles NO2 = 4.43 moles

Step 4: Calculate moles H2O

Moles H2O = 58.1 grams / 18.02 g/mol

Moles H2O = 3.22 moles

Step 5: Calculate limiting reactant

For 3 moles NO2 we need 1 mol H2O to produce 2 moles HNO3 and 1 mol NO

NO2 is the limiting reactant. It will completely be consumed (4.43 moles). H2O is in excess. there will react 4.43 /3 = 1.48 moles. There will remain 3.22 - 1.48 = 1.74 moles

Step 6: Calculate moles NO

For 3 moles NO2 we need 1 mol H2O to produce 2 moles HNO3 and 1 mol NO

For 4.43 moles NO2 we'll have 4.43/3 = 1.48 moles NO

Step 7: Calculate mass NO

Mass NO = 1.48 moles * 30.01 g/mol

Mass NO = 44.4 grams

44.4 grams of NO can be produced

3 0
3 years ago
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