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Ludmilka [50]
3 years ago
13

Climatology is a subspecialty of

Chemistry
2 answers:
stellarik [79]3 years ago
8 0

Climatology is a subspecialty of meterology in which these are considered atmospheric sciences which study weather patterns and their effects on humans and the environment.

andrew-mc [135]3 years ago
6 0
Climatology is a subspecialty of Meterology.
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Which of the following is a change from a more condensed to a less condensed state of matter? Freezing Condensing Evaporating So
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Evaporating.

This is because evaporation occurs for things such as water coming in contact with a heat source that cause molecules to run crazy and begin to go farther and farther apart, becoming steam and evaporation in the air.

Therefore, during the process of evaporation, molecules become less condensed.

Hope this helps!
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3 years ago
Water can act as both an acid and a base in an ionization reaction to form (H) and OH ions.
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True or Fase. Every substance has to be classified as either acid or base?
Sever21 [200]

Answer:

False

Explanation:

Some substances don't have to be

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2 years ago
Which term is described as a long, narrow depression in the ocean floor?
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2 years ago
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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
3 years ago
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