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3 years ago

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Atomic number and atomic mass

1 answer:

Alexeev081 [22]3 years ago

5 0

Answer:

Together, the number of protons and the number of neutrons determine an element's mass number: mass number = protons + neutrons. If you want to calculate how many neutrons an atom has, you can simply subtract the number of protons, or atomic number, from the mass number.

Explanation:

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A 1.00 L sample of a gas at a pressure of 1.000atm is compressed to 0.25 L at a constanttemperature. What is the new pressure of

lord [1]

Answer:

<h3>The answer is 4 atm </h3>

Explanation:

The new pressure can be found by using the formula for Boyle's law which is

$P_1V_1 = P_2V_2$

Since we are finding the new pressure

$P_2 = \frac{P_1V_1}{V_2} \\$

From the question we have

$P_2 = \frac{1 \times 1}{0.25} = \frac{1}{0.25} \\$

We have the final answer as

<h3>4 atm </h3>

Hope this helps you

5 0

3 years ago

Rubidium has two naturally occurring isotopes, rubidium -85 ( atomic mass = 84.9118 amu; abundance = 72.15%) and rubidium -87 (a

son4ous [18]

Answer:

The answer to your question is: 85.458 amu

Explanation:

data

Rubidium-85 A = 84.9118 amu abundance = 72.15%

Rubidium - 87 A = 86.9092 amu abundance = 27.85%

Atomic weight = ?

Atomic weight = 84.9118(0.7215) + 86.9092(0.2785)

Atomic weight = 61.2538 + 24.2042

Atomic weight = 85.458 amu

6 0

4 years ago

1. Explain how and why stars appear to move through the sky throughout the night. Then, explain why Polaris does not move.

Bas_tet [7]

Polaris the north star is very bright and it doesn’t move

3 0

3 years ago

If the heating curve is reversed, what describes the melting point?

olasank [31]

The point of Condensation

6 0

3 years ago

The compound trimethylamine, (CH3)3N, is a weak base when dissolved in water. Write the Kb expression for the weak base equilibr

GREYUIT [131]

Answer:

The $K_b$ expression for the weak base equilibrium is:

$K_b=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N]}$

Explanation:

$(CH_3)_3N(aq)+H_2O(l)\rightlefharpoons (CH_3)_3NH^++OH^-(aq)$

The expression of the equilibrium constant of base $K_c$ can be given as:

$K_c=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N][H_2O]}$

] $K_b=K_c\times [H_2O]=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N]}$

As we know, water is pure solvent, we can put $[H_2O]=1$

$K_b=K_c\times 1=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N]}$

So, the the $K_b$ expression for the weak base equilibrium is:

$K_b=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N]}$

6 0

3 years ago