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Kisachek [45]
3 years ago
6

Drag the tiles to the boxes to form correct pairs.

Chemistry
2 answers:
strojnjashka [21]3 years ago
8 0

Answer:

1) Proportions decreased as it reacted with other elements to form nitrogen. -AMMONIA

2) Proportions decreased as this lighter gas was blown away by solar winds.- HYDROGEN

3) Proportions increased as this gas was released through volcanic eruptions. -WATER VAPOR

4) Proportions increased as plants released this gas into the atmosphere.- OXYGEN

5)Proportions decreased as plants used this gas during photosynthesis.- CARBON DIOXIDE

Explanation:

IT'S RIGHT 100 %

77julia77 [94]3 years ago
5 0

Answer:

Oxygen - Proportions increased as plants released this gas into the atmosphere.

Water Vapor -  Proportions decreased as this lighter gas was blown away by solar winds.

Hydrogen - Proportions increased as this gas was released through volcanic eruptions.

Ammonia - Proportions decreased as it reacted with other elements to form nitrogen.

Carbon Dioxide - Proportions decreases as plants used this gas during photosynthesis.

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The combustion of 1.5011.501 g of fructose, C6H12O6(s)C6H12O6(s) , in a bomb calorimeter with a heat capacity of 5.205.20 kJ/°C
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Explanation :

First we have to calculate the heat gained by the calorimeter.

q=c\times (T_{final}-T_{initial})

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q = heat gained = ?

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T_{initial} = initial temperature = 22.93^oC

Now put all the given values in the above formula, we get:

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Now we have to calculate the enthalpy change during the reaction.

\Delta H=-\frac{q}{n}

where,

\Delta H = enthalpy change = ?

q = heat gained = 23.4 kJ

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\Delta H=-\frac{23.4kJ}{0.00834mole}=-2805.8kJ/mole

Therefore, the enthalpy change during the reaction is -2805.8 kJ/mole

Now we have to calculate the internal energy change for the combustion of 1.501 g of fructose.

Formula used :

\Delta H=\Delta U+\Delta n_gRT

or,

\Delta U=\Delta H-\Delta n_gRT

where,

\Delta H = change in enthalpy = -2805.8kJ/mol

\Delta U = change in internal energy = ?

\Delta n_g = change in moles = 0   (from the reaction)

R = gas constant = 8.314 J/mol.K

T = temperature = 27.43^oC=273+27.43=300.43K

Now put all the given values in the above formula, we get:

\Delta U=\Delta H-\Delta n_gRT

\Delta U=(-2805.8kJ/mol)-[0mol\times 8.314J/mol.K\times 300.43K

\Delta U=-2805.8kJ/mol-0

\Delta U=-2805.8kJ/mol

Therefore, the internal energy change is -2805.8 kJ/mol

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