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3 years ago

Suggest why sodium wasn’t produced until the early 19th century.

Chemistry

1 answer:

AlexFokin [52]3 years ago

5 0

Answer:

The reason is that sodium attaches itself very strongly to other elements. Its compounds are very difficult to break apartand also because it is so reactive.

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Occurs when energy is added or removed

stepan [7]

Answer:

melting

Explanation:

one change of state happens when you add energy to the substance. This change of state is called melting. By adding energy to the molecules in a solid the molecules begin to move quicker and can break away from the other molecules.

5 0

3 years ago

What is the atomic mass that of magnesium

hram777 [196]

Answer:

Name: Magnesium

Atomic Number: 12

Atomic Mass: 24.305 atomic mass units

Number of Protons: 12

Number of Neutrons: 12

3 0

3 years ago

How many molecules are in 6.20 moles of CaCO3?

DedPeter [7]

Answer:

Moles to Grams caco3

1 mole is equal to 1 moles CaCO3, or 100.0869 grams.

6 0

3 years ago

Read 2 more answers

If the dosage for a medication is 225mg/lb, how many grams should a 25-lb child be given?

Oksanka [162]

Answer:

5.625 grams

Explanation:

Start your equation with what you have been given. Place the units you need in your answer on the right side of the equal sign.

225mg

----------- X ----------- X ------------- = ? g

Now start to fill in your equation and use a conversions to get rid of the units you don't want. Convert mg into grams first. The child's weight (25 lb) is placed over 1 just to get the equation lined up properly so you can see how the units cancel out.

225 mg 1 g 25 lb 5.625 g

--------------- X --------------- X ------------- = ---------------

lb 1000 mg 1 1

The lb on the top and bottom cancel each other out and you are left with just grams. Even though it is over one, that is the same at just 5.625 grams.

3 0

3 years ago

Ethylene (CH2CH2) is the starting point for a wide array of industrial chemical syntheses. For example, worldwide about 8.0 x 10

mylen [45]

Answer: The percent by mass of ethylene in the equilibrium gas mixture is 3.76 %

Explanation:

We are given:

Initial partial pressure or ethane = 24.0 atm

The chemical equation for the dehydration of ethane follows:

$C_2H_6(g)\rightleftharpoons C_2H_4(g)+H_2(g)$

Initial: 24.0

At eqllm: 24-x x x

The expression of $K_p$ for above equation follows:

$K_p=\frac{p_{C_2H_4}\times p_{H_2}}{p_{C_2H_6}}$

We are given:

$K_p=0.040$

Putting values in above expression, we get:

$0.040=\frac{x\times x}{24-x}\\\\x^2+0.04x-0.96=0\\\\x=0.96,-1$

Neglecting the value of x = -1 because partial pressure cannot be negative.

So, partial pressure of hydrogen gas at equilibrium = x = 0.96 atm

Partial pressure of ethylene gas at equilibrium = x = 0.96 atm

Partial pressure of ethane gas at equilibrium = (24-x) = (24 - 0.96) atm = 23.04 atm

To calculate the number of moles, we use the equation given by ideal gas, which follows:

$PV=nRT$ .........(1)

To calculate the mass of a substance, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ..........(2)

For ethane gas:

We are given:

$P=23.04atm\\V=30.0L\\T=800^oC=[800+273]K=1073K\\R=0.0821\text{ L. atm }mol^{-1}K^{-1}$

Putting values in equation 1, we get:

$23.04atm\times 30.0L=n\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 1073K\\\\n=\frac{23.04\times 30.0}{0.0821\times 1073}=7.85mol$

We know that:

Molar mass of ethane gas = 30 g/mol

Putting values in equation 2, we get:

$7.85mol=\frac{\text{Mass of ethane gas}}{30g/mol}\\\\\text{Mass of ethane gas}=(7.85mol\times 30g/mol)=235.5g$

For ethylene gas:

We are given:

$P=0.96atm\\V=30.0L\\T=800^oC=[800+273]K=1073K\\R=0.0821\text{ L. atm }mol^{-1}K^{-1}$

Putting values in equation 1, we get:

$0.96atm\times 30.0L=n\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 1073K\\\\n=\frac{0.96\times 30.0}{0.0821\times 1073}=0.33mol$

We know that:

Molar mass of ethylene gas = 28 g/mol

Putting values in equation 2, we get:

$0.33mol=\frac{\text{Mass of ethylene gas}}{28g/mol}\\\\\text{Mass of ethylene gas}=(0.33mol\times 28g/mol)=9.24g$

For hydrogen gas:

We are given:

$P=0.96atm\\V=30.0L\\T=800^oC=[800+273]K=1073K\\R=0.0821\text{ L. atm }mol^{-1}K^{-1}$

Putting values in equation 1, we get:

$0.96atm\times 30.0L=n\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 1073K\\\\n=\frac{0.96\times 30.0}{0.0821\times 1073}=0.33mol$

We know that:

Molar mass of hydrogen gas = 2 g/mol

Putting values in equation 2, we get:

$0.33mol=\frac{\text{Mass of hydrogen gas}}{2g/mol}\\\\\text{Mass of hydrogen gas}=(0.33mol\times 2g/mol)=0.66g$

To calculate the mass percentage of ethylene in equilibrium gas mixture, we use the equation:

$\text{Mass percent of ethylene gas}=\frac{\text{Mass of ethylene gas}}{\text{Mass of equilibrium gas mixture}}\times 100$

Mass of equilibrium gas mixture = [235.5 + 9.24 + 0.66] = 245.4 g

Mass of ethylene gas = 9.24 g

Putting values in above equation, we get:

$\text{Mass percent of ethylene gas}=\frac{9.24g}{245.5g}\times 100=3.76\%$

Hence, the percent by mass of ethylene in the equilibrium gas mixture is 3.76 %

5 0

3 years ago