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3 years ago

A neutron breaks down to form ?

1 answer:

4 0

A neutron breaks down to form a proton and an electron.
Change in atomic mass when an atom emits gamma radiation. The gamma radiation consists of photons which have no mass so the mass would remain the same.
Re the change in atomic number when an atom emits an alpha particle it loses two protons and two neutrons so the atomic number drops by 2.

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A 500.0-mL buffer solution is 0.100 M in HNO2 and 0.150 M in KNO2. Determine whether each addition would exceed the capacity of

Leviafan [203]

Answer:

None of the additions will exceed the capacity of the buffer.

Explanation:

As we know a buffer has the ability to resist pH changes when small amounts of strong acid or base are added.

The pH of the buffer is given by the Henderson-Hasselbach equation:

pH = pKa + log [A⁻] / [HA]

where A⁻ is the conjugate base of the weak acid HA.

Now we can see that what is important is the ratio [A⁻] / [HA] to resist a pH change brought about by the addition of acid or base.

It follows then that once we have consumed by neutralization reaction either the acid or conjugate base in the buffer, this will lose its ability to act as such and the pH will increase or decrease dramatically by any added acid or base.

Therefore to solve this question we must determine the number of moles of acid HNO₂ and NO₂⁻ we have in the buffer and compare it with the added acid or base to see if it will deplete one of these species.

Volume buffer = 500.0 mL = 0.5 L

# mol HNO₂ = 0.5 L x 0.100 mol/L = 0.05 mol HNO₂

# mol NO₂⁻ = 0.5 L x 0.150 mol/L = 0.075 mol NO₂⁻

a. If we add 250 mg NaOH (0.250 g)

molar mass NaOH =40 g/mol

# mol NaOH =0.250 g/ 40g/mol = 0.0063 mol

0.0063 mol NaOH will be neutralized by 0.0063 mol HNO₂ and we have plenty of it, so it would not exceed the capacity of the buffer.

b. If we add 350 mg KOH (0.350 g)

molar mass KOH =56.10 g

# mol KOH = 0.350 g/56.10 g/mol = 0.0062 mol

Again the capacity of the buffer will not be exceeded since we have 0.05 mol HNO₂ in the buffer.

c. If we add 1.25 g HBr

molar mass HBr = 80.91 g/mol

# mol HBr = 1.25 g / 80.91 g/mol = 0.015 mol

0.015 mol Hbr will neutralize 0.015 mol NO₂⁻ and we have to start with 0.075 mol in the buffer, therefore the capacity will not be exceeded.

d. If we add 1.35 g HI

molar mass HI = 127.91 g/mol

# mol HI = 1.35 g / 127.91 g/mol = 0.011 mol

Again the capacity of the buffer will not be exceed since we have plenty of it in the buffer after the neutralization reaction.

7 0

3 years ago

Which two come together to form an ionic bond?

Lelechka [254]

$\huge \fbox \pink {A}\huge \fbox \green {n}\huge \fbox \blue {s}\huge \fbox \red {w}\huge \fbox \purple {e}\huge \fbox \orange {r}$

b. a positive cation and a negative anion

✏ Anions are negative in nature while cations are positive in nature. Together they come together by an attractive electrostatic force to form an ionic bond.

ʰᵒᵖᵉ ⁱᵗ ʰᵉˡᵖˢ

$\huge\blue{ \mid{ \underline{ \overline{ \tt ꧁❣ ʀᴀɪɴʙᴏᴡˢᵃˡᵗ2²2² ࿐ }} \mid}}$

6 0

2 years ago

Help? The question is down below.

Murrr4er [49]

The first three are T I don’t know about the next two and the last one is T

6 0

2 years ago

Read 2 more answers

How many electrons will a neon (Ne) atom have when it has no charge?

Darya [45]

Answer:

hope it helps you

Explanation:

Once one shell is full, the next electron that is added has to move to the next shell. So... for the element of NEON, you already know that the atomic number tells you the number of electrons. That means there are 10 electrons in a neon atom.

6 0

3 years ago

A sample of nitrogen gas is at a temperature of 50 c and a pressure of 2 atm. If the volume of the sample remains constant and t

Lilit [14]

Answer:

The new temperature of the nitrogen gas is 516.8 K or 243.8 C.

Explanation:

Gay-Lussac's law indicates that, as long as the volume of the container containing the gas is constant, as the temperature increases, the gas molecules move faster. Then the number of collisions with the walls increases, that is, the pressure increases. That is, the pressure of the gas is directly proportional to its temperature.

Gay-Lussac's law can be expressed mathematically as follows:

$\frac{P}{T} =k$

Where P = pressure, T = temperature, K = Constant

You want to study two different states, an initial state and a final state. You have a gas that is at a pressure P1 and at a temperature T1 at the beginning of the experiment. By varying the temperature to a new value T2, then the pressure will change to P2, and the following will be fulfilled:

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