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2 years ago

An ion has 56 protons, 54 electrons, and 81 neutrons. What is its atomic number ?

Chemistry

2 answers:

-BARSIC- [3]2 years ago

8 0

The Atomic Number indicates the number of protons = 56

EastWind [94]2 years ago

5 0

Answer: The barium – 137 atom would have 56 protons, 56 electrons and 81 neutrons. The barium – 137 cation would have 56 protons, 54 electrons and 81 neutrons.

Explanation: Hope this is helpful

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A solution contains 42.0 g of heptane (C7H16) and 50.5 g of octane (C8H18) at 25 ∘C. The vapor pressures of pure heptane and pur

Kruka [31]

Answer:

(a) 22.3 torr; 5.6 torr; (b) 27.9 torr; (c) 77.7 % heptane; 23.3 % octane

(d) Heptane is more volatile than octane

Explanation:

We can use Raoult's Law to solve this problem.

It states that the partial pressure of each component of an ideal mixture of liquids is equal to the vapour pressure of the pure component multiplied by its mole fraction. In symbols,

$p_{i} = \chi_{i} p_{i}^{\circ}$

(a) Vapour pressure of each component

Let heptane be Component 1 and octane be Component 2.

(i) Moles of each component

$n_{1} = \text{42.0 g} \times \dfrac{\text{1 mol}}{\text{100.20 g}} = \text{0.4192 mol}\\n_{2} = \text{50.5 g} \times \dfrac{\text{1 mol}}{\text{114.23 g}} = \text{0.4421 mol}$

(ii) Total moles

$n_{\text{tot}} = 0.4192 + 0.4421 = \text{0.8613 mol}$

(iiii) Mole fractions of each component

$p_{1} = 0.4867 \times 45.8 = \textbf{22.3 torr}\\p_{2} = 0.5133 \times 10.9 = \ \textbf{5.6 torr}$

(iv) Partial vapour pressures of each component

$p_{1} = 0.4867 \times 45.8 = \textbf{22.3 torr}\\p_{2} = 0.5133 \times 10.9 = \textbf{5.6 torr}$

(b) Total pressure

$p_{\text{tot}} = p_{1} + p_{2} = 22.3 + 5.6 = \text{27.9 torr}$

(c) Mass percent of each component in vapour

$\chi_{1} = \dfrac{p_{1}}{p_{\text{Tot}}} = \dfrac{22.3}{27.9} =0.799\\\chi_{2} = \dfrac{p_{2}}{p_{\text{Tot}} }= \dfrac{5.6}{27. 9} =0.201$

The ratio of the mole fractions is the same as the ratio of the moles.

$\dfrac{n_{1}}{n_{2}} = \dfrac{0.799}{0.201}$

If we have 1 mol of vapour, we have 0.799 mol of heptane and 0.201 mol of octane

$m_{1} = 0.799 \times 100.20 = \text{80.1 g}\\m_{2} = 0.201\times 114.23 = \text{23.0 g}\\m_{\text{tot}} = 80.1 + 23.0 = \text{103.1 g}\\\\\text{ mass percent heptane} = \dfrac{80.1}{103.1} \times 100 \, \% = \mathbf{77.7\, \%}\\\\\text{ mass percent octane} = \dfrac{23.0}{103.1} \times 100 \, \% = \mathbf{22.3\, \%}}$

(d) Enrichment of vapour

The vapour is enriched in heptane because heptane is more volatile than octane.

5 0

4 years ago

There is a container on the lab bench containing some unknown chemical. what should you do?

iragen [17]

Ask your lab manager for instructions. If you happen to be the lab manager, take out the container (making a note of where it is), find the culprit, and try to figure out what might be in the beaker so you can know how to properly dispose of it.

Proper safety measures should be followed.

<h2>What are the safety measures in chemistry lab?</h2>

In the lab, always use the proper eye protection, such as chemical splash goggles. When handling hazardous items, put on the disposable gloves that the laboratory has given. Before leaving the lab, take the gloves off. Put on a full-length, long-sleeved lab coat (apron) or apron that can withstand chemicals.

To know more about Safety measures, go to the URL

brainly.com/question/8430576

#SPJ4

6 0

2 years ago

The mass of 100cm3 of water is 100 g. Use the formula to calculate its density.

yanalaym [24]

Answer:

It should be 1 g/cm3. Hope this helpd!

4 0

3 years ago

Read 2 more answers