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3 years ago

As of January 2009, the USA has produced 60,000 metric tons of nuclear waste in 60 yesrs of operating 104 nuclear power plants.

Chemistry

2 answers:

marta [7]3 years ago

8 0

I beleieve its b honey i hope i dont get u wrong :P but sorry if i do

Ann [662]3 years ago

3 0

B. 83.3 tons p. month

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What phrase from the Preamble justifies a law like the Affordable Care Act?

Alexandra [31]

<span>The Affordable Care Act which was instated during Barack ObamaÂ´s presidency and is sometimes known as Obamacare is justified by the phrase Â¨for the general welfareÂ¨ in the Preamble. This concept can be defined as holding the US government responsable for providing care for all people, not just a few elites. Therefore the Affordable Care Act, which aims to increase access to healthcare to all Americans is adhering to this phrase.</span>

4 0

3 years ago

Describe some conditions in Antarctica that could lead to chemical weathering

harkovskaia [24]

There's a heck ton of snow! What does that mean? It means metals that may not be water-soluble can rust. <span />

5 0

4 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 $(\Delta S^o)$ .

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

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

where,

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

n = number of moles

$\Delta S_f^0$ = standard entropy of formation

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

$\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]$

$\Delta S^o=-153J/K$

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 500 K.

$\Delta G^o=(-1035600J)-(500K\times -153J/K)$

$\Delta G^o=-959100J=-959.1kJ$

Therefore, the value of $\Delta G^o$ for the reaction is -959.1 kJ

3 0

3 years ago

Hydrogenotrophy is Choose one: A. an award given to fuel companies that have the most fuel-efficient vehicles. B. the oxidation

Mrrafil [7]

Answer:

C. the use of hydrogen gas as an electron donor.

Explanation:

Hydrogenotrophy is the convertion of hydrogen gas to other compounds as part of its metabolism.

8 0

3 years ago

How many milligrams of a 20mg sample of cesium-137 remain after 60 years

andrey2020 [161]

Cs -137 has a half life of about 30 years. If 60 years pass, there is two half lives passed so 1/2 * 1/2= 1/4. Take 1/4 and multiply it with the mass given:
(1/4)*20mg=5mg left

5 0

3 years ago