All categories

2 years ago

How many moles of sodium sulfate will be formed if you start with 2.00 moles of sodium

Chemistry

1 answer:

dimaraw [331]2 years ago

3 0

Answer:

The molecular weight of sodium sulfate is 142.1 g/mol. To calculate the mass of sodium sulfate, substitute the values in the formula. Therefore, the mass of sodium sulfate formed when we start with 200.0 grams of sodium hydroxide and you have excess sulphuric acid is 355.25g.

You might be interested in

Science, please help and thank you!!

lapo4ka [179]

Swings hope this helps

8 0

3 years ago

One of the products of a fermentation reaction is

Alex73 [517]

There are two types of fermentation aerobic and anaerobic. Aerobic are those that need oxygen to ferment while anaerobic are those who do not need oxygen. Products of fermentation are usually water, ethanol, lactic acid and carbon dioxide. Other products are acetone and butyric acid.

5 0

3 years ago

During a titration the following data were collected. A 50.0 mL portion of an HCl solution was titrated with 0.500 M NaOH; 200.

Travka [436]

<u>Answer:</u> The mass of HCl present in 500 mL of acid solution is 36.5 grams

<u>Explanation:</u>

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is HCl

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=1\\M_1=?M\\V_1=50.00mL\\n_2=1\\M_2=0.500M\\V_2=200.mL$

Putting values in above equation, we get:

$1\times M_1\times 50.00=1\times 0.500\times 200\\\\M_1=\frac{1\times 0.500\times 200}{1\times 50.00}=2M$

To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Molar mass of HCl = 36.5 g/mol

Molarity of solution = 2 M

Volume of solution = 500 mL

Putting values in above equation, we get:

$2mol/L=\frac{\text{Mass of solute}\times 1000}{36.5g/mol\times 500}\\\\\text{Mass of solute}=\frac{2\times 36.5\times 500}{1000}=36.5g$

Hence, the mass of HCl present in 500 mL of acid solution is 36.5 grams

4 0

2 years ago

Quiz time!

kykrilka [37]

potassium belongs to group IA of the periodic table.

7 0

2 years ago

Read 2 more answers

Acetaldehyde and butanoic acid must have the same

ANTONII [103]

Answer:

dsvcsdc

Explanation:

7 0

3 years ago