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3 years ago

What particles make up the atomic nucleus

Chemistry

2 answers:

musickatia [10]3 years ago

5 0

The necleus f an atom is make up of protons and neutrons (two types of baryons) joined by the neclear force.

White raven [17]3 years ago

4 0

The particles that make up the atomic nucleus of all atoms are both protons and neutrons.

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What is heat

Nimfa-mama [501]

Answer:

energy that freely moves from one place to another. It is responsible for vibration of particles of matter

8 0

3 years ago

A sample of argon gas has a volume of 73 mL at a pressure of 1.20 atm and a temperature of 112 degrees Celsius. What is the fina

NeX [460]

Answer:

V₂ → 106.6 mL

Explanation:

We apply the Ideal Gases Law to solve the problem. For the two situations:

P . V = n . R . T

Moles are still the same so → P. V / R. T = n

As R is a constant, the formula to solve this is: P . V / T

P₁ . V₁ / T₁ = P₂ .V₂ / T₂ Let's replace data:

(1.20 atm . 73mL) / 112°C = (0.55 atm . V₂) / 75°C

((87.6 mL.atm) / 112°C) . 75°C = 0.55 atm . V₂

58.66 mL.atm = 0.55 atm . V₂

58.66 mL.atm / 0.55 atm = V₂ → 106.6 mL

3 0

3 years ago

What is the pH of a solution in which [H 3O +] = 3.8 × 10 -8 M?

7nadin3 [17]

Answer:

<h3>The answer is 7.42 </h3>

Explanation:

The pH of a solution can be found by using the formula

$pH = - log [ { H_3O}^{+}]$

From the question we have

$pH = - log(3.8 \times {10}^{ - 8} ) \\ = 7.420216...$

We have the final answer as

Hope this helps you

4 0

3 years ago

The radioactivity due to carbon-14 measured in a piece of a wooden casket from an ancient burial site was found to produce 20 co

bekas [8.4K]

Answer:

17202.6 years

Explanation:

Activity of the living sample (Ao) = 160 counts per minute

Activity of the wood sample (A) = 20 counts per minute

Half life of carbon-14 = 5730 years

t= age of the artifact

From;

0.693/t1/2= 2.303/t log Ao/A

Then;

0.693/ 5730= 2.303/t log Ao/A

Substituting values;

0.693/5730= 2.303/t log (160/20)

Then we obtain;

1.209×10^-4 = 2.0798/t

t= 2.0798/1.209×10^-4

Thus;

t= 17202.6 years

Therefore the artifact is 17202.6 years old.

3 0

3 years ago

Where the oxygen comes from the air (21% O2 and 79% N2). If oxygen is fed from air in excess of the stoichiometric amount requir

guajiro [1.7K]

Answer:

$y_{O2} =4.3$ %

Explanation:

The ethanol combustion reaction is:

$C_{2}H_{5} OH+3O_{2}$ → $2CO_{2}+3H_{2}O$

If we had the amount (x moles) of ethanol, we would calculate the oxygen moles required:

$x*1.10(excess)*\frac{3 O_{2}moles }{etOHmole}$

Dividing the previous equation by x:

$1.10(excess)*\frac{3 O_{2}moles}{etOHmole}=3.30\frac{O_{2}moles}{etOHmole}$

We would need 3.30 oxygen moles per ethanol mole.

Then we apply the composition relation between O2 and N2 in the feed air:

$3.30(O_{2} moles)*\frac{0.79(N_{2} moles)}{0.21(O_{2} moles)}=121.414 (N_{2} moles )$

Then calculate the oxygen moles number leaving the reactor, considering that 0.85 ethanol moles react and the stoichiometry of the reaction:

$3.30(O_{2} moles)-0.85(etOHmoles)*\frac{3(O_{2} moles)}{1(etOHmoles)} =0.75O_{2} moles$

Calculate the number of moles of CO2 and water considering the same:

$0.85(etOHmoles)*\frac{3(H_{2}Omoles)}{1(etOHmoles)}=2.55(H_{2}Omoles)$

$0.85(etOHmoles)*\frac{2(CO_{2}moles)}{1(etOHmoles)}=1.7(CO_{2}moles)$

The total number of moles at the reactor output would be:

$N=1.7(CO2)+12.414(N2)+2.55(H2O)+0.75(O2)\\ N=17.414(Dry-air-moles)$

So, the oxygen mole fraction would be:

$y_{O_{2}}=\frac{0.75}{17.414}=0.0430=4.3$ %

6 0

3 years ago