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Na + O2 → Na2O2<br> Can you guys help me !

vfiekz [6]

Answer:

Explanation:

Assuming that you are asking for the full-balanced equations...

2Na + O2 = Na2O2

H2 + O2 = H2O2; This equation is already balanced, there is no need for change.

Please let me know if this is not the answer you are looking for and I will amend the answer.

6 0

3 years ago

Enter your answer in the box provided. How many grams of helium must be added to a balloon containing 6.24 g helium gas to doubl

Archy [21]

Answer : The mass of helium gas added must be 12.48 grams.

Explanation : Given,

Mass of helium (He) gas = 6.24 g

Molar mass of helium = 4 g/mole

First we have to calculate the moles of helium gas.

$\text{Moles of }He=\frac{\text{Mass of }He}{\text{Molar mass of }He}=\frac{6.24g}{4g/mole}=1.56moles$

Now we have to calculate the moles of helium gas at doubled volume.

According to the Avogadro's law, the volume of gas is directly proportional to the number of moles of gas at same pressure and temperature. That means,

$V\propto n$

or,

$\frac{V_1}{V_2}=\frac{n_1}{n_2}$

where,

$V_1$ = initial volume of gas = V

$V_2$ = final volume of gas = 2V

$n_1$ = initial moles of gas = 1.56 mole

$n_2$ = final moles of gas = ?

Now we put all the given values in this formula, we get

$\frac{V}{2V}=\frac{1.56mole}{n_2}$

$n_2=3.12mole$

Now we have to calculate the mass of helium gas at doubled volume.

$\text{Mass of }He=\text{Moles of }He\times \text{Molar mass of }He$

$\text{Mass of }He=3.12mole\times 4g/mole=12.48g$

Therefore, the mass of helium gas added must be 12.48 grams.

4 0

3 years ago

An irregularly shaped piece of magnesium with a mass of 11.81 g was dropped into a graduated cylinder partially filled with wate

dlinn [17]

Answer:

The answer is

Explanation:

The density of a substance can be found by using the formula

$density = \frac{mass}{volume} \\$

From the question

mass of magnesium = 11.81 g

volume = 6.80 mL

The density of the metal is

$density = \frac{11.81}{6.80} \\ = 1.73676470...$

We have the final answer as

Hope this helps you

7 0

4 years ago

Points

ExtremeBDS [4]

The answer is D. A compound

6 0

3 years ago

What is the boiling point of a 1.5 m aqueous solution of fructose? the boiling point elevation constant of water is 0.515Â°c/m.

Ronch [10]

Answer is: <span>he boiling point of a 1.5 m aqueous solution of fructose is </span>100.7725°C.
The boiling point elevation is directly proportional to the molality of the solution according to the equation: ΔTb = Kb · b.<span>
ΔTb - the boiling point elevation.
Kb - the ebullioscopic constant. of water.
b - molality of the solution.
Kb = 0.515</span>°C/m.
b = 1.5 m.
ΔTb = 0.515°C/m · 1.5 m.
ΔTb = 0.7725°C.
Tb(solution) = Tb(water) + ΔTb.
Tb(solution) = 100°C + 0.7725°C = 100.7725°C.

4 0

3 years ago

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