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3 years ago

How many molecules are in 2.3 moles of h2o

Chemistry

2 answers:

Naddik [55]3 years ago

7 0

Each mole of substance contains 6.02 x 1023 component parts, in this case water molecules.

If you have 2.3 moles of water you will have 2.3 x 6.02 x 1023 which is 1.3846 x 1024 molecules.

Each molecule contains 2 hydrogen atoms, so the total number of hydrogen atoms in 2.3 moles of water will be 2 x 1.3846 x 1024 = 2.7692 x 1024.

Please mark brainliest, thanks :)

ololo11 [35]3 years ago

5 0

1.385 e 24 molecules or 2.7692 x 10^{24} atoms.

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How much time does it take for light from the center of the galaxy to reach Earth?

Ede4ka [16]

Answer:

Light moves at 300,000 kilometers per second, divide these and you get 500 seconds, or 8 minutes and 20 seconds this is an average number.

Explanation:

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3 years ago

How many MOLECULES are in 50.0 grams of lithium phosphate?

Ray Of Light [21]

Answer: 2.60 x 10^23 molecules

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8 0

3 years ago

Be sure to answer all parts. The standard enthalpy of formation and the standard entropy of gaseous benzene are 82.93 kJ/mol and

skelet666 [1.2K]

Answer : The values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $33.89kJ,95.94J/K\text{ and }5.299kJ/mol$ respectively.

Explanation :

The given balanced chemical reaction is,

$C_6H_6(l)\rightarrow C_6H_6(g)$

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

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

$\Delta H^o=[n_{C_6H_6(g)}\times \Delta H_f^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta H_f^0_{(C_6H_6(l))}]$

where,

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

n = number of moles

$\Delta H_f^0_{(C_6H_6(g))}$ = standard enthalpy of formation of gaseous benzene = 82.93 kJ/mol

$\Delta H_f^0_{(C_6H_6(l))}$ = standard enthalpy of formation of liquid benzene = 49.04 kJ/mol

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

$\Delta H^o=[1mole\times (82.93kJ/mol)]-[1mole\times (49.04J/mol)]$

$\Delta H^o=33.89kJ/mol=33890J/mol$

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

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

$\Delta S^o=[n_{C_6H_6(g)}\times \Delta S^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta S^0_{(C_6H_6(l))}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S^0_{(C_6H_6(g))}$ = standard entropy of formation of gaseous benzene = 269.2 J/K.mol

$\Delta S^0_{(C_6H_6(l))}$ = standard entropy of formation of liquid benzene = 173.26 J/K.mol

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

$\Delta S^o=[1mole\times (269.2J/K.mol)]-[1mole\times (173.26J/K.mol)]$

$\Delta S^o=95.94J/K.mol$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is $25^oC\text{ or }298K$ .

$\Delta G^o=(33890J)-(298K\times 95.94J/K)$

$\Delta G^o=5299.88J/mol=5.299kJ/mol$

Therefore, the values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $33.89kJ,95.94J/K\text{ and }5.299kJ/mol$ respectively.

6 0

3 years ago

Read 2 more answers

Which of the following is the chemical formula for Hydrogen Chloride?

anygoal [31]

Answer:

The answer is (b) HCl

Explanation:

6 0

3 years ago

Read 2 more answers

The molar mass of H2O is 18.01 g/mol. The molar mass of O2 is 32.00 g/mol. What mass of H2O, ins grams, must react to produce 50

Ivan

Answer:

56.28 g

Explanation:

First change the grams of oxygen to moles.

(50.00 g)/(32.00 g/mol) = 1.5625 mol O₂

You have to use stoichiometry for the next part. Looking at the equation, you can see that for every 2 moles of H₂O, 1 mole of O₂ is produced. Convert from moles of O₂ to moles of H₂O using this relation.

(1.5625 mol O₂) × (2 mol H₂O/1 mol O₂) = 3.125 mol H₂O

Now convert moles of H₂O to grams.

(3.125 mol) × (18.01 g/mol) = 56.28125 g

Convert to significant figures.

56.28125 ≈ 56.28

5 0

3 years ago