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3 years ago

What is the delta H fusion used to calculate volume of a liquid frozen that produces 1 kg of energy?

Chemistry

2 answers:

gulaghasi [49]3 years ago

5 0

Answer:

The correct option is C. 1 kJ x 1/ΔHfus x g/mol x ml/g liquid

Explanation:

The change in enthalpy of fusion (ΔHfus) has units of energy divided into moles of substance:

ΔHfus = energy/mol of substance = J/mol or KJ/mol

Thus, we multiply the kJ of energy produced by the inverse of ΔHfus (in kJ/mol) we obtain the moles of substance:

energy (kJ) x 1/ΔHfus (mol/kJ) = moles of substance(mol)

Then, if we multiply the moles of substance by the molecular weight of the substance (in g/mol) we obtain the grams:

moles of substance(mol) x molecular weight (g/mol) = grams of substance (g)

Finally, if we multiply the grams of substance (in g) by the density of the substance (in g/ml) we obtain the volume (in ml):

grams of substance (g) x 1/density (ml/g) = volume (ml)

We combine all these parts to obtain the equation for the volume required to produce 1 kJ of energy:

energy (kJ) x 1/ΔHfus (mol/kJ) x molecular weight (g/mol) x 1/density (ml/g)

Therefore, the correct option is:

C → 1 kJ x 1/ΔHfus x g/mol x ml/g liquid

natita [175]3 years ago

4 0

1 kJ x 1/ΔHfus x g/mol x ml/g liquid

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2 years ago

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What's the balanced equation for cellular respiration?

Alik [6]

<span>Correct answer is:

$C_{6}H_{12}O_6+6O_{2}\Rightarrow 6CO_{2}+6H_{2}O+energy$

But how to get there?

Let's start with simple explanation of what exactly is cellular respiration.

Cellular respiration is a multistage biochemical oxidation process of organic substances when prime product is energy (ATP - adenosine triphosphate) and other are released waste products. Cellular respiration takes place even if other metabolic processes are stopped, but cellular respiration may differ in particular organism groups.Some reactions during whole process of cellular respiration are similar in all types of living organisms.

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In cellular respiration all substrates which are in the cell might be organic, but mostly we are using sugar oxidation - glucose in the presence of oxygen. Chemical formula of sugar looks like this:

$C_{6}H_{12}O_6$

Oxygen is just

$O_2$

so for now we have just part of the equation:

$C_{6}H_{12}O_6+O_{2}\Rightarrow$

But what would be on the right hand side?

It's quite simple, remember equation of full combustion? If we want to burn something we need oxygen like in the equation, so the product of this equation would be carbon dioxide, water and of course energy (ATP).Carbon dioxide formula looks like this:

$CO_{2}$

As a reminder water formula:

$H_{2}O$

Full formula would look like that:

$C_{6}H_{12}O_6+O_{2}\Rightarrow\ CO_{2}+H_{2}O+energy$

But still as you see this equation is unbalanced, after balancing it would like that:

$C_{6}H_{12}O_6+6O_{2}\Rightarrow 6CO_{2}+6H_{2}O+energy$

At the end I would like to explain one more thing. Energy which has been released during this process is part of high-energy connection which might be used to perform chemical reactions in the cell or to move organism for example in muscles. We need to remember that production of ATP is not happening with 100% efficiency and part of this energy is released as heat.</span>

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3 years ago

Hurry up as fast as possible Ill mark you brainlist

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2 years ago

Which is the best practice recommended in the safety video to mix and acid or a base with a solvent?

Vesna [10]

Answer:

Never pour water into acid but acid into water

Explanation:

If water is poured into extremely concentrated acid/bases, the rate of volatility and exothermic reaction is too rapid and might cause a chemical eruption, leading to acid burns.

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I believe this is a clear answer.

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3 years ago

30cm^3 of a dilute solution of Ca(OH)2 required 11 cm^3 of 0.06 mol/dm^. Hcl for complete neutralization. Calculate the concentr

Alenkasestr [34]

Answer: Thus concentration of $Ca(OH)_2$ in $mol/dm^3$ is 0.011 and in $g/dm^3$ is 0.814

Explanation:

To calculate the concentration of $Ca(OH)_2$ , we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $HCl$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is $Ca(OH)_2$

We are given:

$n_1=1\\M_1=0.06mol/dm^3\\V_1=11cm^3=0.011dm^3\\n_2=2\\M_2=?\\V_2=30cm^3=0.030dm^3$ $1cm^3=0.001dm^3$

Putting values in above equation, we get:

$1\times 0.06mol/dm^3\times 0.011dm^3=2\times M_2\times 0.030dm^3\\\\M_2=0.011mol/dm^3$

The concentration in $g/dm^3$ is $0.011mol/dm^3\times 74g/mol=0.814g/dm^3$

Thus concentration of $Ca(OH)_2$ is $0.011mol/dm^3$ and $0.814g/dm^3$

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3 years ago