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djyliett [7]
2 years ago
5

8

Chemistry
1 answer:
mr_godi [17]2 years ago
7 0

Answer:8

in the respiratory system

a.

Explanation:

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PLEASE HELP ASAP
Mekhanik [1.2K]

The mass of magnesium, which has a density of 1.74 g/cm is 504.6 g.

<h3>What is mass?</h3>

Mass is the quantity of matter. Mass can be calculated by multiplying density by volume.

Magnesium is a chemical element with the atomic number 12. It is needed in the body in trace amounts. It can cause malnutrition in the body.

Mass = Density x volume

We know the density and the volume of magnesium.

Density = 1.74

Volume = 290

Density x volume

Putting the value in the equation

1.74 x 290 = 504.6 g

Thus, the mass of magnesium is 504.6 g.

To learn more about mass, refer to the below link:

brainly.com/question/22795877

#SPJ1

4 0
1 year ago
.asdfafERf EfgAsdf as
Bogdan [553]

Can't say i can answer this. :/

8 0
3 years ago
Do you think that all foods containing carbohydrates are healthful? Explain plz:(
deff fn [24]

Answer:

Yes

Explanation:

because some foods have carbohydrates but are healthy

8 0
3 years ago
A solution of 100.0 mL of 0.200 M KOH is mixed with a solution of 200.0 mL of 0.150 M NiSO4. (a) Write the balanced chemical equ
NISA [10]

Answer:

a) 2KOH + NiSO₄ → K₂SO₄ + Ni(OH)₂

b) Ni(OH)₂

c) KOH

d) 0.927 g

e) K⁺=0.067 M, SO₄²⁻=0.1 M, Ni²⁺=0.067 M

Explanation:

a) The equation is:

2KOH + NiSO₄ → K₂SO₄ + Ni(OH)₂   (1)        

b) The precipitate formed is Ni(OH)₂  

 

c) The limiting reactant is:

n_{KOH} = V*M = 100.0 \cdot 10^{-3} L*0.200 mol/L = 0.020 moles

n_{NiSO_{4}} = V*M = 200.0 \cdot 10^{-3} L*0.150 mol/L = 0.030 moles

From equation (1) we have that 2 moles of KOH react with 1 mol of NiSO₄, so the number of moles of KOH is:

n = \frac{2}{1}*0.030 moles = 0.060 moles                  

Hence, the limiting reactant is KOH.  

d) The mass of the precipitate formed is:

n_{Ni(OH)_{2}} = \frac{1}{2}*n_{KOH} = \frac{1}{2}*0.020 moles = 0.010 moles

m = n*M = 0.010 moles*92.72 g/mol = 0.927 g  

e) The concentration of the SO₄²⁻, K⁺, and Ni²⁺ ions are:

C_{K^{+}} = \frac{2*\frac{1}{2}*n_{KOH}}{V} = \frac{0.020 moles}{0.300 L} = 0.067 M  

C_{SO_{4}^{2-}} = \frac{\frac{1}{2}*n_{KOH + (0.03 - 0.01)}}{V} = \frac{0.030 moles}{0.300 L} = 0.1 M

C_{Ni^{2+}} = \frac{0.020 moles}{0.300 L} = 0.067 M

I hope it helps you!                                                                        

5 0
4 years ago
Question is in picture! Due at 3:45!!!
olga nikolaevna [1]
The answer is D.2HCl
6 0
3 years ago
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