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nirvana33 [79]
3 years ago
13

Hola!

Chemistry
1 answer:
Lisa [10]3 years ago
3 0

Answer:

Hi! I'm feeling great and happy! Life's been good! How are you? :) Hope life's good for you and that you are staying safe.

Explanation:

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What will happen to the chemical equilibrium if MgCl2 is added? PLEASE HURRY
kari74 [83]

Answer:

a

Explanation:

5 0
3 years ago
Read 2 more answers
What is the mass of an 80 mL aliquot of a solution with a density of 5.80<br> g/mL?
Setler [38]

Answer:

The answer is

<h2>464 g</h2>

Explanation:

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

<h3>mass = Density × volume</h3>

From the question

volume of solution = 80 mL

density = 5.80 g/mL

The mass is

mass = 80 × 5.8

We have the final answer as

<h3>464 g</h3>

Hope this helps you

4 0
3 years ago
This organism breaks down nonliving matter into energy-rich compounds giving it back to the soil is called a​
salantis [7]

Answer:

A decomposer

Explanation:

An organism like a fungus or a worm breaks down dead, nonliving matter which ends up giving nutrients back into the soil.

7 0
3 years ago
Why is particle physics related to the model of the universe
Nataly_w [17]
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5 0
4 years ago
How many grams of nitric acid HNO₃, are required to neutralize (completely react with) 4.30 grams of Ca(OH)2 according to the ac
Brrunno [24]

Answer:

7.32g of HNO3 are required.

Explanation:

1st) From the balanced reaction we know that 2 moles of HNO3 react with 1 mole of Ca(OH)2 to produce 2 moles of H2O and 1 mole of Ca(NO3)2.

From this, we find that the relation between HNO3 and Ca(OH)2 is that 2 moles of HNO3 react with 1 mole of Ca(OH)2.

2nd) This is the order of the relations that we have to use in the equation to calculate the grams of nitric acid:

• starting with the 4.30 grams of Ca(OH)2.

,

• using the molar mass of Ca(OH)2 (74g/mol).

,

• relation of the 2 moles of HNO3 that react with 1 mole of Ca(OH)2 .

,

• using the molar mass of HNO3 (63.02g/mol).

4.30g\text{ Ca\lparen OH\rparen}_2*\frac{1\text{ mol Ca\lparen OH\rparen}_2}{74g\text{ Ca\lparen OH\rparen}_2}*\frac{2\text{ moles HNO}_3}{1\text{ mole Ca\lparen OH\rparen}_2}*\frac{63.02g\text{ HNO}_3}{1\text{ mole HNO}_3}=7.32g\text{ HNO}_3

So, 7.32g of HNO3 are required.

4 0
2 years ago
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