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Anestetic [448]
3 years ago
6

Which of the following statements is true about both nuclear fusion and nuclear fission? They occur in the sun. Heavy atoms are

split. Two light nuclei combine. Some mass changes into energy.
Chemistry
1 answer:
galina1969 [7]3 years ago
6 0
<h2>Answer </h2>

Some mass changes into energy

<u>Explanation </u>

Some mass changes into energy are true about both nuclear fusion and nuclear fission. These both reactions produce large amounts of energy. Nuclear fusion is the process in which two light nuclei combine to form a larger nucleus. On the other hand, nuclear fission is reverse in which a heavy nucleus breaks into two light nuclei. Nuclear decay and transmission are also types of nuclear reactions. The matter is not destroyed in nuclear reactions.

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If a weak acid is 25% deprotonated at ph 4, what is the pka?if a weak acid is 25% deprotonated at ph 4, what is the pka?
Oksanka [162]
4.48
pH=pKa+log([A-/HA])

25% deprotonated tells us that A- is .25 and that the rest (75% is protonated) thats .75.

4 = pKa + log\frac{.25}{.75}
4 - log\frac{.25}{.75} = pKa
4.48=pKa

7 0
3 years ago
Read 2 more answers
18 ) Liquid A and Liquid B are clear liquids. They are placed in open containers and allowed to evaporate. When evaporation is c
arsen [322]

Answer:

Liquid A is a homogeneous mixture and Liquid B is a heterogeneous mixture

Explanation:

5 0
3 years ago
Solve for ‘x’: 12(x-7)+3(2x+2)=50x-2
Alex Ar [27]

Answer:

Explanation:

Hi there!

12(x-7) = 12x -84

3(2x+2) = 6x+6

Total 18x-78

18x-78 = 50x-2

-78= 32x-2

-76= 32x

19/8 =x

Hope you understand

Have a nice day,

Wolfiex.

5 0
3 years ago
How many minutes are in 3.5 years
-Dominant- [34]

Answer: 1,839,600 minutes.

Explanation: 3.5 years to minutes has been calculated by multiplying 3.5 years by 525,600

7 0
3 years ago
A container has a mixture of NO2 gas and N2O4 gas in equilibrium. The chemical reaction between the two gases is described by th
kondaur [170]

Answer: The most likely partial pressures are 98.7MPa for NO₂ and 101.3MPa for N₂O₄

Explanation: To determine the partial pressures of each gas after the increase of pressure, it can be used the equilibrium constant Kp.

For the reaction 2NO₂ ⇄ N₂O₄, the equilibrium constant is:

Kp = \frac{P(N_{2}O_{4} )}{P(NO_{2} ^{2}) }

where:

P(N₂O₄) and P(NO₂) are the partial pressure of each gas.

Calculating constant:

Kp = \frac{38.8}{61.2^{2} }

Kp = 0.0104

After the weights, the total pressure increase to 200 MPa. However, at equilibrium, the constant is the same.

P(N₂O₄) + P(NO₂) = 200

P(N₂O₄) = 200 - P(NO₂)

Kp = \frac{P(N_{2}O_{4} )}{P(NO_{2} ^{2}) }

0.0104 = \frac{200 - P(NO_{2})  }{[P(NO_{2} )]^{2}}

0.0104[P(NO_{2} )]^{2} + P(NO_{2} ) - 200 = 0

Resolving the second degree equation:

P(NO_{2} ) = \frac{-1+\sqrt{9.32} }{0.0208}

P(NO_{2} ) = 98.7

Find partial pressure of N₂O₄:

P(N₂O₄) = 200 - P(NO₂)

P(N₂O₄) = 200 - 98.7

P(N₂O₄) = 101.3

The partial pressures are P(NO_{2} ) = 98.7 MPa and P(N₂O₄) = 101.3 MPa

3 0
3 years ago
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