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2 years ago

How many hydrogen molecules are there in 1 ton of hydrogen?

1 answer:

6 0

Hydrogen gas(H2) has a molar mass of 2 g. Molar mass of a substance is defined as the mass of 1 mole of that substance. And by 1 mole it is meant a collection of 6.022*10^23 particles of that substance.

So number of moles of H2 are 0.5 in this case. And thus it means there are (6.022*10^23)*0.5 particles( here they are molecules) in 1g of H2.

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An unknown gas is contained in a sealed container. Over time, the gas is gradually cooled until it becomes a solid. Determine wh

Alecsey [184]

Answer:D molecular attraction increases as temp decreases

Explanation:A,B would require data not available. C is missing. The question is a mess.

5 0

3 years ago

Read 2 more answers

Calculate the number of O atoms in 0.150 g of CaSO4 · 2H2O.

natulia [17]

Answer:

There are 0.00523 moles of oxygen in 0.150 grams of calcium sulfate crystal.

Explanation:

Mass of calcium sulfate crystal = m = 0.150 g

Molar mass of calcium sulfate crystal = M = 172 g/mol

Moles of magnesium nitrate = n

$n=\frac{m}{M}$

$n=\frac{0.150 g}{172 g/mol}=0.0008721 mol$

1 mole of calcium sulfate crystal has 6 moles of oxygen atoms. Then 0.004446 moles calcium sulfate crystal will have :

$6\times 0.0008721 mol=0.0052326mol\approx 0.00523 mol$

There are 0.00523 moles of oxygen in 0.150 grams of calcium sulfate crystal.

8 0

3 years ago

Determine the number of centiliters in 2.4 x 10^3 milliliters.

AnnZ [28]

2.4(10^3)

=2.4*10^3

=2.4*(10*10*10)

=2400 milliliters

To centiliters is 2400mL= 240.0000cl. 

8 0

3 years ago

Use the given data at 500 K to calculate ΔG°for the reaction

Anton [14]

Answer : The value of $\Delta G^o$ for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

$2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

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

$\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]$

$\Delta H^o=-1035.6kJ=-1035600J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]$