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Korvikt [17]
3 years ago
8

3. What does this experiment reveal about the impact of ocean acidification on shelled organisms? How

Chemistry
1 answer:
Snezhnost [94]3 years ago
6 0

Answer:

Ocean acidification is sometimes called “climate change’s equally evil twin,” and for good reason: it's a significant and harmful consequence of excess carbon dioxide in the atmosphere that we don't see or feel because its effects are happening underwater. At least one-quarter of the carbon dioxide (CO2) released by burning coal, oil and gas doesn't stay in the air, but instead dissolves into the ocean. Since the beginning of the industrial era, the ocean has absorbed some 525 billion tons of CO2 from the atmosphere, presently around 22 million tons per day.

At first, scientists thought that this might be a good thing because it leaves less carbon dioxide in the air to warm the planet. But in the past decade, they’ve realized that this slowed warming has come at the cost of changing the ocean’s chemistry. When carbon dioxide dissolves in seawater, the water becomes more acidic and the ocean’s pH (a measure of how acidic or basic the ocean is) drops. Even though the ocean is immense, enough carbon dioxide can have a major impact. In the past 200 years alone, ocean water has become 30 percent more acidic—faster than any known change in ocean chemistry in the last 50 million years.

Scientists formerly didn’t worry about this process because they always assumed that rivers carried enough dissolved chemicals from rocks to the ocean to keep the ocean’s pH stable. (Scientists call this stabilizing effect “buffering.”) But so much carbon dioxide is dissolving into the ocean so quickly that this natural buffering hasn’t been able to keep up, resulting in relatively rapidly dropping pH in surface waters. As those surface layers gradually mix into deep water, the entire ocean is affected.

Such a relatively quick change in ocean chemistry doesn’t give marine life, which evolved over millions of years in an ocean with a generally stable pH, much time to adapt. In fact, the shells of some animals are already dissolving in the more acidic seawater, and that’s just one way that acidification may affect ocean life. Overall, it's expected to have dramatic and mostly negative impacts on ocean ecosystems—although some species (especially those that live in estuaries) are finding ways to adapt to the changing conditions.

However, while the chemistry is predictable, the details of the biological impacts are not. Although scientists have been tracking ocean pH for more than 30 years, biological studies really only started in 2003, when the rapid shift caught their attention and the term "ocean acidification" was first coined. What we do know is that things are going to look different, and we can't predict in any detail how they will look. Some organisms will survive or even thrive under the more acidic conditions while others will struggle to adapt, and may even go extinct. Beyond lost biodiversity, acidification will affect fisheries and aquaculture, threatening food security for millions of people, as well as tourism and other sea-related economies

there is more lines go to https://ocean.si.edu/ocean-life/invertebrates/ocean-acidification

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explain the relationship between the rate of effusion of a gas and its molar mass. methane gas (ch4) effuses 3.4 times faster th
Musya8 [376]

The molar mass of the unknown gas is 184.96 g/mol

<h3>Graham's law of diffusion </h3>

This states that the rate of diffusion of a gas is inversely proportional to the square root of the molar mass i.e

R ∝ 1/ √M

R₁/R₂ = √(M₂/M₁)

<h3>How to determine the molar mass of the unknown gas </h3>

The following data were obtained from the question:

  • Rate of unknown gas (R₁) = R
  • Rate of CH₄ (R₂) = 3.4R
  • Molar mass of CH₄ (M₂) = 16 g/mol
  • Molar mass of unknown gas (M₁) =?

The molar mass of the unknown gas can be obtained as follow:

R₁/R₂ = √(M₂/M₁)

R / 3.4R = √(16 / M₁)

1 / 3.4 = √(16 / M₁)

Square both side

(1 / 3.4)² = 16 / M₁

Cross multiply

(1 / 3.4)² × M₁ = 16

Divide both side by (1 / 3.4)²

M₁ = 16 / (1 / 3.4)²

M₁ = 184.96 g/mol

Learn more about Graham's law of diffusion:

brainly.com/question/14004529

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3 0
2 years ago
Each reation releases or absorbs a very large amount of energy per atom.
mario62 [17]

Answer:

chemical and nuclear reaction

Explanation:

4 0
3 years ago
Read 2 more answers
2,3,5-trimethylhexane
dedylja [7]

2,3,5-trimethylhexane

C9H20

Molecular weight= 128.5g/mol

CH3-CH(CH3)-CH(CH3)-CH2-CH(CH3)-CH3

5 0
3 years ago
A 8.96-L sample of gas has a pressure of 1.86 atm and a temperature of 94 °C. The sample is allowed to expand to a volume of 11.
Varvara68 [4.7K]

Answer:

Explanation:

Explanation:

All you have to do here is use the ideal gas law equation, which looks like this

P

V

=

n

R

T

−−−−−−−−−−

Here

P

is the pressure of the gas

V

is the volume it occupies

n

is the number of moles of gas present in the sample

R

is the universal gas constant, equal to

0.0821

atm L

mol K

T

is the absolute temperature of the gas

Rearrange the equation to solve for

T

P

V

=

n

R

T

⇒

T

=

P

V

n

R

Before plugging in your values, make sure that the units given to you match those used in the expression of the universal gas constant.

In this case, the volume is given in liters and the pressure in atmospheres, so you're good to go.

Plug in your values to find

T

=

3.10

atm

⋅

64.51

L

9.69

moles

⋅

0.0821

atm

⋅

L

mol

⋅

K

T

=

251 K

−−−−−−−−−

The answer is rounded to three

8 0
3 years ago
Calculate the second volumes. 51.7 L at 27 C and 90.9 KPa to STP
Arturiano [62]

The second volume : 42.2 L

<h3>Further explanation</h3>

Given

51.7 L at 27 C and 90.9 KPa

Required

The second volume

Solution

STP = P₂=1 atm, T₂=273 K

T₁ = 27 + 273 = 300 K

P₁ = 90.9 kPa = 0,897 atm

Use combine gas law :

P₁V₁/T₁ = P₂V₂/T₂

Input the value :

0.897 x 51.7/300 = 1 x V₂/273

V₂= 42.20 L

6 0
3 years ago
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