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OLEGan [10]
2 years ago
11

G in your own words, summarize the issue of ocean acidification: what is it and what's causing it?

Chemistry
1 answer:
AleksandrR [38]2 years ago
8 0

Ocean acidification refers to a reduction in the pH of the ocean over an extended period of time, caused primarily by the uptake of carbon dioxide (CO2) from the atmosphere.

Because of human-driven increased levels of carbon dioxide in the atmosphere, there is more CO2 dissolving into the ocean. The ocean's average pH is now around 8.1, which is basic (or alkaline), but as the sea continues to absorb more CO2, the pH decreases and the ocean becomes more acidic.

How does ocean acidification affect us?

Ocean acidification can modify the abundance and chemical composition of harmful algal blooms in such a way that shellfish toxicity increases and, therefore, human health are negatively affected.

How we can reduce ocean acidification?

The most effective way to limit ocean acidification is to act on climate change, implementing solutions to dramatically reduce the use of fossil fuels. If we dramatically cut our global warming emissions, and we limit future warming, we can significantly reduce the harm to marine ecosystems.

To know more about ocean acidification:

brainly.com/question/7604502

#SPJ4

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How many moles are in 128.9 grams of Cr2(SO3)2?
Dominik [7]

Answer: 0.5 moles

Explanation:

Cr2(SO3)2 is the chemical formula for chromium sulphate.

Given that,

Amount of moles of Cr2(SO3)2 (n) = ?

Mass of Cr2(SO3)2 in grams = 128.9g

For molar mass of Cr2(SO3)2, use the atomic masses:

Chromium, Cr = 52g;

Sulphur, S = 32g;

Oxygen, O = 16g

Cr2(SO3)2 =

(52g x 2) + [(32g + 16g x 3) x 2]

= 104g + [(32g + 48g) x 2]

= 104g + [80g x 2]

= 104g + 160g

= 264g/mol

Since, n = mass in grams / molar mass

n = 128.9g / 264g/mol

n = 0.488 mole [Round the value of n to the nearest tenth which is 0.5

Thus, there are 0.5 moles in 128.9 grams of Cr2(SO3)2

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A chemist mixes 75.0 g of an unknown substance at 96.5°C with 1,150 g of water at 25.0°C. If the final temperature of the system
AleksAgata [21]
To do this problem it is necessary to take into account that the heat given by the unknown substance is equal to the heat absorbed by the water, but considering the correct sign:

-m\cdot c_e\cdot \Delta T = m_w\cdot c_e_w\cdot \Delta T_w

Clearing the specific heat of the unknown substance:

c_e = \frac{m_w\cdot c_e_w\cdot \Delta T_w}{m\cdot \Delta T} = -\frac{1\ 150\ g\cdot 4.184\frac{J}{g\cdot ^\circ C}\cdot (37.1 - 25.0)^\circ C}{75\ g\cdot (37.1 - 96.5)^\circ C}

c_e = -\frac{1\ 150\ g\cdot 4.184\frac{J}{g\cdot ^\circ C}\cdot 12.1^\circ C}{75\ g\cdot (-59.4)^\circ C} = \bf 13.07\frac{J}{g\cdot ^\circ C}
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