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3 years ago

12

What properties applies to solids but not to liquids or gases?

1 answer:

qaws [65]3 years ago

4 0

A higher density (besides H2O) as they have a more rigid structure, they have better conductive abilities than their liquid or gas forms. there are a couple more that i cannot think of at this moment in time.

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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

avanturin [10]

Answer : The internal energy change is -2805.8 kJ/mol

Explanation :

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

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

where,

q = heat gained = ?

c = specific heat = $5.20kJ/^oC$

$T_{final}$ = final temperature = $27.43^oC$

$T_{initial}$ = initial temperature = $22.93^oC$

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

$q=5.20kJ/^oC\times (27.43-22.93)^oC$

$q=23.4kJ$

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

n = number of moles fructose = $\frac{\text{Mass of fructose}}{\text{Molar mass of fructose}}=\frac{1.501g}{180g/mol}=0.00834mole$

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

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

5 0

3 years ago

Help! giving brainliest

kirill [66]

It’s the 3d one

Good luckkkkkkkk

8 0

2 years ago

If you start with 0.020 g of Mg, how many moles of H2 will you make if the reaction is complete?

irina [24]

If one starts with 0.020 g of Mg, 0.0008 moles of H2 would be made if the reaction is complete.

Going by the balanced equation of reaction in the image, 1 mole of Mg will produce 1 mole of H2 in a complete reaction.

If 0.020 g of Mg is started with:

mole of Mg = mass/molar mass

= 0.020/24.3

= 0.0008 moles

Since the mole of Mg to H2 is 1:1, thus, 0.0008 moles of H2 will also be made from the reaction.

More on stoichiometry can be found here: brainly.com/question/9743981

6 0

2 years ago

Q2.Which is true about potassium?

Olenka [21]

$\bf \LARGE\rightarrow{Very \: reactive}$

5 0

3 years ago

Read 2 more answers

What is the mass of oxygen in 10.0 grams of water?

skelet666 [1.2K]

Answer:

The answer is 16.00 amu.

7 0

3 years ago