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ivanzaharov [21]
3 years ago
8

Number 7 plz!!!! I have no clue and it’s due by midnight!!!!!!!

Chemistry
1 answer:
IrinaVladis [17]3 years ago
5 0
The answer would be high pressure
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If the temperature of a gas is raised from 30°C to 60°C, what happens to the pressure?
Fofino [41]
<h3>Answer:</h3>

The pressure increases by 10% of the original pressure

Thus the new pressure is 1.1 times the original pressure.

<h3>Explanation:</h3>

We are given;

  • Initial temperature as 30°C, but K = °C + 273.15
  • Thus, Initial temperature, T1 =303.15 K
  • Final temperature, T2 is 333.15 K

We are required to state what happens to the pressure;

  • We are going to base our arguments to Pressure law;
  • According to pressure law, the pressure of a gas and its temperature are directly proportional at a constant volume
  • That is; P α T
  • Therefore, at varying pressure and temperature

\frac{P1}{T1}=\frac{P2}{T2}

Assuming the initial pressure, P1 is P

Rearranging the formula;

[tex]P2=\frac{P1T2}{T1}[/tex]

P2=\frac{(P)(333.15K)}{303.15K}

     P2 = 1.099P

                 = 1.10 P

The new pressure becomes 1.10P

This means the pressure has increased by 10%

We can conclude that, the new pressure will be 1.1 times the original pressure.

5 0
3 years ago
At high pressures how does the volume of a real gas compare with the volume of an ideal gas under the same conditions and why
malfutka [58]
Answer: No, a<span>t high pressures, volume of a real gas does not  compare with the volume of an ideal gas under the same conditions.

Reason: 
For an ideal gas, there should not be any intermolecular forces of interaction. However, for real gases there are intermolecular forces of interaction like dipole-dipole and dipole-induced dipole. Further, at high pressures, molecules are close by. Hence, extend of these intermolecular forces is expected to be high. This results in decreases in volume of real gas. Thus, </span>volume of a real gas does not  compare with the volume of an ideal gas under the same conditions.

3 0
3 years ago
Read 2 more answers
How much solar radiation is reflected back by clouds? 5 percent
Ipatiy [6.2K]
I think five percent.
5 0
3 years ago
Read 2 more answers
The atomic mass of helium-4 is 4.0026 amu. How many of each
umka21 [38]

Answer:

Protons: 2.

Electrons: 2.

Neutrons: 2.

Explanation:

Hello,

In this case, since an atom's atomic number is equal to the number of electrons, considering the electron configurations, taking into account that helium-4 is neither positively nor negatively charged, we can infer that the number of electrons equal the number of protons, which in this case are 2, due to the fact that is atomic number is 2.

Moreover, as helium-4's atomic number is 4 as a whole number, we compute the number of neutrons by using the shown below equation:

Neutrons=mass\ number-atomic\ number\\\\Neutrons=4-2\\\\Neutrons=2

Regards.

7 0
3 years ago
A solution of 0.0470 M HCl is used to titrate 26.0 mL of an ammonia solution of unknown concentration. The equivalence point is
DanielleElmas [232]

The pH at equivalence point is 12.46

At equivalence point, number of moles of acid, n equals number of moles of base, n'

So, n = n'

CV = C'V' where

  • C = concentration of acid (HCl) = 0.0470 M,
  • V = volume of acid = 16.0 mL,
  • C' = concentration of base (ammonia solution) and
  • V' = volume of base = 26.0 mL.
<h3>Concentration of ammonia solution</h3>

Making C' subject of the formula, we have

C' = CV/V'

Substituting the values of the variables into the equation, we have

C' = CV/V'

C' = 0.0470 M × 16.0 mL/26.0 mL

C' = 0.752 MmL/26.0 mL

C' = 0.0289 M

<h3>The concentration of acid at equivalence point</h3>

We know that the ion-product of water Kw is

Kw = [H⁺][OH⁻] =  where

  • [H⁺] = concentration of HCl at equivalence point,
  • [OH⁻] = C' = concentration of ammonia solution = 0.0289 M and
  • Kw = 1.01 × 10⁻¹⁴

Making [H⁺] subject of the formula, we have

[H⁺} = Kw/[OH⁻]

[H⁺] = 1.01 × 10⁻¹⁴/0.0289

[H⁺] = 34.95 × 10⁻¹⁴

[H⁺] = 3.495 × 10⁻¹³

<h3>pH at equivalence point</h3>

Since pH = -㏒[H⁺]

pH = -㏒[3.495 × 10⁻¹³]

pH = -㏒[3.495] + (-㏒10⁻¹³)

pH = -㏒[3.495] + [-13(-㏒10)]

pH = 13 - 0.5434

pH = 12.4566

pH ≅ 12.46

So, the pH at equivalence point is 12.46

Learn more about pH at equivalence point here:

brainly.com/question/25487920

3 0
2 years ago
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