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Thepotemich [5.8K]
3 years ago
9

WILL GIVE BRAINLEST

Chemistry
2 answers:
Ber [7]3 years ago
8 0
Answer:increased pressure to increase solubility
Bad White [126]3 years ago
7 0

Answer:

A

Explanation:

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Lyrx [107]
The heart is a part of the Circulatory system.
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3 years ago
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A sample of an ideal gas at 1.00 atm and a volume of 1.87 L was placed in a weighted balloon and dropped into the ocean. As the
Ludmilka [50]

Answer:

Volume of sample after droping into the ocean=0.0234L

Explanation:

As given in the question that gas is idealso we can use ideal gas equation to solve this;

Assuming that temperature is constant;

Lets P_1 and V_1 are the initial gas parameter before dropping into the ocean

and P_2 and V_2 are the final gas parameter after dropping into the ocean

according to boyle 's law pressure is inversly proportional to the volume at constant temperature.

hence,

P_1V_1=P_2V_2

P1=1 atm

V1=1.87L

P2=80atm

V2=?

After putting all values we get;

V2=0.0234L

Volume of sample after droping into the ocean=0.0234L

7 0
3 years ago
Determine the pH of a 2.8 ×10−4 M solution<br> of Ca(OH)2.
shepuryov [24]

Answer:

pH = 10.75

Explanation:

To solve this problem, we must find the molarity of [OH⁻]. With the molarity we can find the pOH = -log[OH⁻]

Using the equation:

pH = 14 - pOH

We can find the pH of the solution.

The molarity of Ca(OH)₂ is 2.8x10⁻⁴M, as there are 2 moles of OH⁻ in 1 mole of Ca(OH)₂, the molarity of [OH⁻] is 2*2.8x10⁻⁴M = 5.6x10⁻⁴M

pOH is

pOH = -log 5.6x10⁻⁴M

pOH = 3.25

pH = 14-pOH

<h3>pH = 10.75</h3>
3 0
3 years ago
A rigid tank contains 1.40 moles of an ideal gas. Determine the number of moles of gas that must be withdrawn from the tank to l
sergey [27]

Answer : The final number of moles of gas that withdrawn from the tank to lower the pressure of the gas must be, 0.301 mol.

Explanation :

As we know that:

PV=nRT

At constant volume and temperature of gas, the pressure will be directly proportional to the number of moles of gas.

The relation between  pressure and number of moles of gas will be:

\frac{P_1}{P_2}=\frac{n_1}{n_2}

where,

P_1 = initial pressure of gas = 24.5 atm

P_2 = final pressure of gas = 5.30 atm

n_1 = initial number of moles of gas = 1.40 moles

n_2 = final number of moles of gas = ?

Now put all the given values in the above expression, we get:

\frac{24.5atm}{5.30atm}=\frac{1.40mol}{n_2}

n_2=0.301mol

Therefore, the final number of moles of gas that withdrawn from the tank to lower the pressure of the gas must be, 0.301 mol.

8 0
3 years ago
How do genetic factors influence the formation of "dead zones” around creosote bushes?
Korvikt [17]

Answer:

Genetic factors contributes to the formation of the dead zones is by their ability to expand their abilities that enables them to spread and contribute to the formation of the dead zones. These dead zones are made when the oxygen are low where it is necessarily important for the aquatic life, if the oxygen needed is depleted or too low, instead of supporting aquatic life, dead zones are created and factors contribute to these occurrences with their ability to expand.

Explanation:

5 0
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