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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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For the equilibrium reaction 2NH3(g) + 22 kJ ↔ N2(g) + 3H2(g), which of the following changes would result in the formation of m
daser333 [38]
Hello! Your question seems incomplete, still I will try to answer it in a general way. As shown in in equation 22 KJ heat is absorbed during the reaction so it is endothermic reaction.

Effect of Temperature:
According to Le Chatelier's principle, by increasing temperature the equilibrium will shift in the forward direction hence more nitrogen and hydrogen will produce.

Effect of Pressure:
It is evident from the reaction that there is an increase in volume in the formation of N2 and H2 (4 moles in whole), Hence, decrease in pressure on this system in equilibrium state will adjust itself in a direction in which the volume is increased i.e. formation of N2 and H2.

Effect of Concentration:
Increasing concentration of NH3, removal of N2 or removal of H2 will shift the equilibrium in forward direction, Hence more product is formed.
3 0
3 years ago
Compare the electronegativity of H and O in the polar covalent bonds of a water
ExtremeBDS [4]

Answer:

The O atom will tend to attract the electrons.

Explanation:

The electronegativity of O (3.5) is much higher than H (2.1), which means it is more likely to attract electrons. The higher the electronegativity, the more attractive.

7 0
3 years ago
Calculate the volume of the acid solution and the volume of the conjugate base solution that would be needed to prepare a buffer
bogdanovich [222]

Answer:

Explanation:

This can be contradictory, depending on whether the 0.1 M

is the total species concentration or the concentration of each of the two components. I'll consider this to be the former...

VA− = 9.125 mL

VHA = 15.875 mL

The Henderson-Hasselbalch equation is:

pH = pKa + log [A−][HA]

We have a pH 4.5

solution of acetic acid and acetate, so from there we can get the ratio of weak acid to conjugate base:

[A−][HA]=10

pH − pKa = 104.5 − 4.74 = 0.5754

Now, if the total concentration is

0.10 M , then:

[HA] + [A−] 0.5754

[HA] = 0.10 M

⇒[HA] = 0.10 M 1.0000 +0.5754

= 0.0635 M

−−−−−−−−

⇒[A−] = 0.0365 M

−−−−−−−−

and these concentrations are AFTER mixing. Since the total volume is 50 mL , or 0.050 L, the mols of each component (which are constant!) are:

nA − = 0.0365 molL × 0.050L =

0.001825 mols

−−−−−−−−−−−−

nHA = 0.0635 molL × 0.050L =

0.003175 mols

−−−−−−−−−−−−

So, if both of the starting concentrations were

0.20 M, we can find the volume they each start with:

VA − = 1 L0.20mols

A− × 0.001825mols A− = 0.009125 L = 9.125 mL

−−−−−−−−

VHA = 1 L 0.20 mols HA × 0.003175

mols HA = 0.015875 L = 15.875 mL

−−−−−−−−−

And this should make sense, because the total starting volume is

25.000 mL , the total ending volume is twice as large; the total species concentration is half the concentration that both species started with.

6 0
3 years ago
Draw the structure of bromous acid, hbro2. optimize formal charges.
ArbitrLikvidat [17]

STRUCTURE OF BROMOUS ACID: H–O–Br=O

<span>In this structure, all the elements have a formal charge of zero. The formal charge of each element is calculated below: </span><span>

H: 1 – 1/2(2) – 0 = 0
O: 6 – 1/2(4) – 4 = 0
Br: 7 – 1/2(6) – 4 = 0
<span>O: 6 – 1/2(4) – 4 = 0</span></span>

8 0
3 years ago
What is the pressure in a 245 L tank that contains 5.21 kg of helium at 27 Celsius ​
Kitty [74]
<h2><u>Answer:</u></h2>

(These are not rounded to the correct decimal)

130.94 atm

13,266.6 kPa

99,571.4 mmHg

<h2><u>Explanation:</u></h2>

<u></u>

PV = nRT

V = 245L

P = ?

R = 0.08206 (atm) , 8.314 (kPa) , 62.4 (mmHg)

T = 273.15 + 27 = 300.15K

n = 1302.5 moles

How I found (n).

5.21kg x 1000g/1kg x 1 mole/4.0g = 1302.5 moles

Now, plug all the numbers into the equation.

Pressure in atm = (1302.5)(0.08206)(300.15) / 245 = 130.94 atm (not rounded to the correct decimal)

Pressure in kPa = (1302.5)(8.314)(300.15) / 245 = 13,266.6 kPa (not rounded to the correct decimal)

Pressure in mmHg = (1302.5)(62.4)(300.15) / 245 = 99,571.4 mmHg (not rounded to the correct decimal)

5 0
3 years ago
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