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Stella [2.4K]
3 years ago
8

What would happen to the annual rates of limestone (CaCO3) precipitation in the ocean if the ocean water were to become much war

mer and more acidic?
Chemistry
1 answer:
stiv31 [10]3 years ago
4 0

Answer:

CaCO3 exoskeleton dissolves in acidic water

Explanation:

The increasing CO2 level makes the ocean water acidic and hence reduces the pH. In such acidic environment, marine organism that produce calcium carbonate shells or skeletons are negatively affected. Coral reefs and coralline algae abilities to produce skeleton also reduces.  

Calcium carbonate dissolves in acid. Thus, the more acidic the ocean water is the faster and easier it is to dissolve the exoskeleton and shell of marine organisms made up of calcium carbonate  

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Do electric or magnetic fields affect the x-rays?
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Electric and magnetic fields do not affect xrays as they only affect charged particles and xrays have no charge

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3 years ago
Silicon dioxide (quartz) is usually quite unreactive but reacts readily with hydrogen fluoride according to the following equati
Scrat [10]

Here's your answer, hope this helps!

6 0
3 years ago
Given the following chemical reaction: 2 O2 + CH4 CO2 + 2 H2O What mass of CH4 is required to completely react with 100 grams of
AfilCa [17]

Answer:

The mass methane (CH4) required is, 25 grams  (option C)

Explanation:

Mass of oxygen gas = 100 g

Molar mass of oxygen gas = 32 g/mole

Molar mass of methane gas = 16 g/mole

<u>Step 1:</u> Balance the reaction

2O2 + CH4 → CO2 + 2H2O

<u>Step 2:</u> Calculate the moles of O2:

Moles O2 = mass of O2 / Molar mass of o2  

Moles O2 = 100g / (32g/mole) = 3,125 moles

⇒From the balanced reaction we conclude that  2 moles of O2 react with 1 mole of CH4

⇒ So, 3.125 moles of O2 react with 3.125/2 = 1.563 moles of CH4

<u>Step 3:</u> Calculate the mass of CH4:

Mass of CH4 = moles of CH4 x Molar mass of CH4

Mass of CH4 = 1.563 moles / (16g/ moles) = 25.008 grams CH4

6 0
3 years ago
What is non point solution?
____ [38]
Duy ryrjudjvjfjjfdnndjdjdd
8 0
3 years ago
The wording of these questions are confusing how would you set this up ?
xxMikexx [17]

Answer:

8. 171074.8 mL

9. 3475 mL.

Explanation:

8. Determination of the volume of the diluted solution.

Initial Molarity (M₁) = 14 M

Initial volume (V₁) = 523 mL

Final Molarity (M₂) = 0.0428 M

Final volume (V₂) =?

Using the dilution formula, we can obtain the volume of the diluted solution as follow:

M₁V₁ = M₂V₂

14 × 523 = 0.0428 × V₂

7322 = 0.0428 × V₂

Divide both side by 0.0428

V₂ = 7322 / 0.0428

V₂ = 171074.8 mL

Therefore, the volume of the diluted solution is 171074.8 mL

9. Determination of the volume of water added.

We'll begin by calculating the final volume of the solution. This can be obtained as follow:

Initial Molarity (M₁) = 3.2 M

Initial volume (V₁) = 973 mL

Final Molarity (M₂) = 0.7 M

Final volume (V₂) =?

M₁V₁ = M₂V₂

3.2 × 973 = 0.7 × V₂

3113.6 = 0.7 × V₂

Divide both side by 0.7

V₂ = 3113.6 / 0.7

V₂ = 4448 mL

Thus, the final volume of the solution is 4448 mL

Finally, we shall determine the volume of water added. This can be obtained as follow:

Initial volume (V₁) = 973 mL

Final volume (V₂) = 4448 mL

Volume of water added =?

Volume of water added = V₂ – V₁

Volume of water added = 4448 – 973

Volume of water added = 3475 mL

7 0
2 years ago
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