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2 years ago

How do SMRs provide more flexibility than ordinary nuclear reactors?

Chemistry

1 answer:

Shkiper50 [21]2 years ago

7 0

Answer:

The SMRs can adjust their energy output using coolants and control rods.

Power plants can be built to fit locations with less demand or less space.

Explanation:

I think that it is but I am not sure.

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Which of the following is considered an ignition source?

matrenka [14]

A.

Weldings sparks could start fire on wood.

Lightning can ignite a fire one a tall tree

Open flames can start another fire if made in contact especially with flammable substances

4 0

3 years ago

Which of the following represents the greatest number of atoms?

forsale [732]

The correct option is B. To get the number of atom for each compound, each element in the compound will be counted as an atom. For instance, for Fe[ClO4]2, there are 1 atom of Fe, 2 atoms of Cl, and 8 atoms of O, making a total of 11 atoms [1 + 2 + 8= 11]. The other options have less than 11 atoms.

7 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

What is the relationship between reaction rate and reaction time

Natali [406]

Answer:

The reaction rate is inversely proportional to the reaction time.

Explanation:

The reaction rate is the change of the concentration of reactants and products with the time.

<em>∵ Reaction rate = - Δ[reactants]/Δt = Δ[products]/Δt.</em>

<em>∴ The reaction rate is inversely proportional to the time, as the reaction rate increases it will take a lower time.</em>

4 0

3 years ago

A generator makes electricity from _____.

marysya [2.9K]

Answer: kinetic energy

Explanation: searched it up

6 0

3 years ago

Read 2 more answers