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2 years ago

79. Explain why the production of a gas does

1 answer:

4 0

Is evaporation of water considered a chemical reaction? No because it’s a physical change. Some reactions have gases as a product and some do not. Just because there’s a gas doesn’t mean a chemical reaction has occurred.

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How many grams of sodium metal must be introduced to water to produce 3.3 grams of hydrogen gas?

Cloud [144]

Answer:

The mass of sodium metal that must be introduced to water to produce 3.3 grams of hydrogen gas, H₂, is approximately 18.82 grams of sodium metal

Explanation:

The given mass of hydrogen gas produced = 3.3 grams

The molar mass of hydrogen gas, H₂ = 2.016 g/mol

The number of moles of hydrogen gas in 3.3 grams of H₂, 'n', is given as follows;

n = Mass/(Molar mass)

n = 3.3 g/(2.016 g/mol) = 1.63690476 moles of H₂

The reaction of sodium and water can be written as follows;

2Na + 2H₂O → 2NaOH + H₂ (g)

2 moles of sodium produces 1 mole of hydrogen gas, H₂

Therefore;

1.63690476/2 moles of sodium will produce 1.63690476 moles of hydrogen gas, H₂

The molar mass of sodium, Na ≈ 22.989 g/mol

The mass of 1.63690476/2 moles of sodium, 'm', is given as follows;

m = 1.63690476/2 moles × 22.989 g/mol ≈ 18.8154018 grams ≈ 18.82 grams

The mass of sodium that will produce 3.3 grams of hydrogen, m ≈ 18.82 grams of sodium metal.

8 0

3 years ago

How many half-lives have passed when there are 3 times as much daughter isotope as parent isotope?

Citrus2011 [14]

Answer:

2 half lives.

Explanation:

Suppose there are 100g of parent isotope at the start.

After 1 half-life there will be 50g of parent and 50g of daughter isotope.

After another half life there is 25 g of parent and 75g of daughter isotope.

3 0

3 years ago

A solution is made from ethanol, C2H5OH and water is 0.75 m. How many grams of ethanol are contained per 250. g of water?

kumpel [21]

Answer:

$m_{solute}=8.6g$

Explanation:

Hello,

In this case, the unit of concentration is molality which is defined by:

$m=\frac{n_{solute}}{m_{solvent}}$

Whereas the mass of the solvent is measured in kilograms. In such a way, with the given data, we first compute the kilograms of water:

$m_{solvent}=250g*\frac{1kg}{1000g} =0.25kg$

Then, we solve for the moles of the solute:

$n_{solute}=0.75\frac{mol}{kg}*0.25kg =0.19mol$

Finally since the molar mass of ethanol is 46 g/mol, we compute the grams for the given solution:

$m_{solute}=0.19*\frac{46g}{1mol} \\\\m_{solute}=8.6g$

Best regards.

8 0

3 years ago

Instant cold packs, often used to ice athletic injuries on the field, contain ammonium nitrate and water separated by a thin pla

RSB [31]

Answer: The enthalpy change of the reaction is -27. kJ/mol

Explanation:

To calculate the mass of water, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of water = 1 g/mL

Volume of water = 25.0 mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{25.0mL}\\\\\text{Mass of water}=(1g/mL\times 25.0mL)=25g$

To calculate the heat released by the reaction, we use the equation:

$q=mc\Delta T$

where,

q = heat released

m = Total mass = [1.25 + 25] = 26.25 g

c = heat capacity of water = 4.18 J/g°C

$\Delta T$ = change in temperature = $T_2-T_1=(21.9-25.8)^oC=-3.9^oC$

Putting values in above equation, we get:

$q=26.25g\tiimes 4.18J/g^oC\times (-3.9^oC)=-427.9J=-0.428kJ$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of ammonium nitrate = 1.25 g

Molar mass of ammonium nitrate = 80 g/mol

Putting values in above equation, we get:

$\text{Moles of ammonium nitrate}=\frac{1.25g}{80g/mol}=0.0156mol$

To calculate the enthalpy change of the reaction, we use the equation:

$\Delta H_{rxn}=\frac{q}{n}$

where,

q = amount of heat released = -0.428 kJ

n = number of moles = 0.0156 moles

$\Delta H_{rxn}$ = enthalpy change of the reaction

Putting values in above equation, we get:

$\Delta H_{rxn}=\frac{-0.428kJ}{0.0156mol}=-27.44kJ/mol$

Hence, the enthalpy change of the reaction is -27. kJ/mol

4 0

3 years ago

Calculate kb the base dissociation constant for [c2h3o2-], acetate anion, for each of your trials from the concentrations of spe

Pani-rosa [81]

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
Ka (mol/L ) = 0.00002340

5 0

3 years ago