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1 year ago

Please help! it’s due date is in a few minutes

Chemistry

2 answers:

Musya8 [376]1 year ago

4 0

Answer:

The answer is A

Explanation:

Because the x would be smalle than |-40|

Anastasy [175]1 year ago

3 0

Answer:

Answer is a

Explanation:

It is a put it quick

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2 years ago

450g of chromium (III) sulfate reacts with excess potassium phosphate. How many grams of potassium sulfate will be produced? (AN

Oduvanchick [21]

Answer:

599.26 grams of potassium sulfate will be produced.

Explanation:

$Cr_2(SO_4)_3(aq)+2K_3PO_4(aq)\rightarrow 2CrPO_4(s)+3K_2SO_4(aq)$

Moles of chromium (III) sulfate = $\frac{450 g}{392 g/mol}=1.1480 mol$

According to reaction, 1 mole of chromium (III) sulfate gives 3 moles of potassium sulfate.

Then 1.1480 moles of chromium (III) sulfate will give:

$\frac{3}{1}\times 1.1480 mol=3.4440 mol$

Mass of 3.4440 moles of potassium sulfate:

= 3.4440 mol × 174 g/mol = 599.26 g

599.26 grams of potassium sulfate will be produced.

8 0

3 years ago

How much of 0.5 M HNO3 is necessary to titrate 25.0 mL of 0.05 M KOH

Ivan

Answer: 2.5 ml of 0.5 M HNO3 is necessary to titrate 25.0 mL of 0.05 M KOH

solution to the endpoint

Explanation:

To calculate the volume 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 $HNO_3$

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

We are given:

$n_1=1\\M_1=0.5M\\V_1=?mL\\n_2=1\\M_2=0.05M\\V_2=25.0mL$

Putting values in above equation, we get:

$1\times 0.5\times V_1=1\times 0.05\times 25.0\\\\V_1=2.5mL$

Thus 2.5 ml of 0.5 M HNO3 is necessary to titrate 25.0 mL of 0.05 M KOH

solution to the endpoint

6 0

2 years ago