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3 years ago

13

The nation on average uses around 10 Million tons of salt per year to de-ice roads. How many kilograms of salt is this? (please

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1 answer:

Westkost [7]3 years ago

3 0

45359237 kilograms :))))))))))

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Find concentration of solution of 45g of glucose is dissolved in water to prepare 500g solution?

ivolga24 [154]

Answer:

9% solution by mass

Explanation:

If there is 500 gm of solution and 45 g of it is glucose then:

45/500 * 100% = 9 % solution

3 0

2 years ago

The range in size of most atomic radii is approximately what?

Nastasia [14]

It is approximately 10 ^ -10

3 0

3 years ago

Using any data you can find in the ALEKS Data resource, calculate the equilibrium constant K at 30.0 °C for the following reacti

gayaneshka [121]

Answer : The value of $K$ for this reaction is, $2.6\times 10^{15}$

Explanation :

The given chemical reaction is:

$CH_3OH(g)+CO(g)\rightarrow HCH_3CO_2(g)$

Now we have to calculate value of $(\Delta G^o)$ .

$\Delta G^o=G_f_{product}-G_f_{reactant}$

$\Delta G^o=[n_{HCH_3CO_2(g)}\times \Delta G^0_{(HCH_3CO_2(g))}]-[n_{CH_3OH(g)}\times \Delta G^0_{(CH_3OH(g))}+n_{CO(g)}\times \Delta G^0_{(CO(g))}]$

where,

$\Delta G^o$ = Gibbs free energy of reaction = ?

n = number of moles

$\Delta G^0_{(HCH_3CO_2(g))}$ = -389.8 kJ/mol

$\Delta G^0_{(CH_3OH(g))}$ = -161.96 kJ/mol

$\Delta G^0_{(CO(g))}$ = -137.2 kJ/mol

Now put all the given values in this expression, we get:

$\Delta G^o=[1mole\times (-389.8kJ/mol)]-[1mole\times (-163.2kJ/mol)+1mole\times (-137.2kJ/mol)]$

$\Delta G^o=-89.4kJ/mol$

The relation between the equilibrium constant and standard Gibbs, free energy is:

$\Delta G^o=-RT\times \ln K$

where,

$\Delta G^o$ = standard Gibbs, free energy = -89.4 kJ/mol = -89400 J/mol

R = gas constant = 8.314 J/L.atm

T = temperature = $30.0^oC=273+30.0=303K$

$K$ = equilibrium constant = ?

Now put all the given values in this expression, we get:

$-89400J/mol=-(8.314J/L.atm)\times (303K)\times \ln K$

$K=2.6\times 10^{15}$

Thus, the value of $K$ for this reaction is, $2.6\times 10^{15}$

4 0

4 years ago

A mixture of two miscible liquids with a widely different boiling point is distilled. The temperature of distilled liquid is obs

svetoff [14.1K]

Answer:

The difference in temperature is significant means that the lower-boiling liquid finishes distilling at a temperature that is too low for the higher-boiling liquid to be in vapor form yet.

Explanation:

The temperature will rise as the vapor of lower-boiling liquid rushes into the distillation head. However once the lower-boiling liquid is done distilling, there is a temperature drop because while the lower temperature liquid is done distilling, the temperature is still too low for the higher-boiling liquid to be rushing in as a vapour, so the temperature drops.

8 0

3 years ago

How many grams of lead(II) sulfate (303 g/mol) are needed to react with sodium chromate (162 g/mol) in order to produce 0.162 kg

Afina-wow [57]

Answer : The mass of $PbSO_4$ needed are, 1.515 grams.

Explanation :

First we have to calculate the mole of $PbCrO_4$ .

$\text{Moles of }PbCrO_4=\frac{\text{Mass of }PbCrO_4}{\text{Molar mass of }PbCrO_4}=\frac{0.162g}{323g/mole}=0.005mole$

Now we have to calculate the moles of $PbSO_4$ .

The balanced chemical reaction will be,

$PbSO_4+Na_2CrO_4\rightarrow PbCrO_4+Na_2SO_4[tex]From the balanced chemical reaction, we conclude thatAs, 1 mole of [tex]PbCrO_4$ produced from 1 mole of $PbSO_4$

So, 0.005 mole of $PbCrO_4$ produced from 0.005 mole of $PbSO_4$

Now we have to calculate the mass of $PbSO_4$

$\text{Mass of }PbSO_4=\text{Moles of }PbSO_4\times \text{Molar mass of }PbSO_4$

$\text{Mass of }PbSO_4=0.005mole\times 303g/mole=1.515g$

Therefore, the mass of $PbSO_4$ needed are, 1.515 grams.

6 0

3 years ago