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olasank [31]
3 years ago
14

Calculate the mass of O2 produced after 20.0 grams of KClO3 have reacted?

Chemistry
1 answer:
IrinaVladis [17]3 years ago
5 0

Answer:

RFM \: of \: potassium \: chlorate = (39 + 35.5 + (16 \times 3)) = 122.5 \: g \\ 122.5 \: g \: of \: potassium \: chlorate \: produce \: (16 \times 3) \: g \: of \: oxygen \\ 20.0 \: g \: of \: potassium \: chlorate \: will \: produce \:  (\frac{(20.0 \times 16 \times 3)}{122.5} ) \: g \\  = 7.84 \: g \: of \: oxygen

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<h3>What is a Food web?</h3>

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The food web also helps to determine the various ways in which energy flows and it usually starts from the producers which are the plants as a result of their photosynthetic ability which involves the conversion of solar energy to chemical energy.

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4 0
1 year 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]

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

<u>Explanation:</u>

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

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4 years ago
Please match the example with the classification of matter in which it belongs. Column A 1. soil: soil 2. chocolate milk: chocol
bekas [8.4K]

Answer:

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Explanation:

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3 years ago
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