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2 years ago

What is one benefit associated with nuclear fission

1 answer:

4 0

Nuclear fission share a number of advantages. These include less emission of greenhouse gases, relatively lower operating costs, this process is well understood. Also, since the population is rising, the demand is as well rising and with the aid of this process we can meet that high demand.

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What is the maximum number of grams of ammonia, nh3, which can be obtained from the reaction of 10.0 g of h2 and 80.0 g of n2? n

Lelechka [254]

Answer: The mass of ammonia produced is 28.22 g

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For hydrogen gas:

Given mass of hydrogen gas = 10.0 g

Molar mass of hydrogen gas = 2 g/mol

Putting values in equation 1, we get:

$\text{Moles of hydrogen gas}=\frac{10.0g}{2g/mol}=5mol$

For nitrogen gas:

Given mass of nitrogen gas = 80.0 g

Molar mass of nitrogen gas = 28 g/mol

Putting values in equation 1, we get:

$\text{Moles of nitrogen gas}=\frac{80.0g}{28g/mol}=2.86mol$

The given chemical equation follows:

$N_2+3H_2\rightarrow 2NH_3$

By Stoichiometry of the reaction:

3 moles of hydrogen gas reacts with 1 mole of nitrogen gas

So, 5 moles of hydrogen gas will react with = $\frac{1}{3}\times 5=1.66mol$ of nitrogen gas

As, given amount of nitrogen gas is more than the required amount. So, it is considered as an excess reagent.

Thus, hydrogen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

3 moles of hydrogen gas produces 1 mole of ammonia

So, 5 moles of hydrogen gas will produce = $\frac{1}{3}\times 5=1.66moles$ of ammonia

Now, calculating the mass of ammonia from equation 1, we get:

Molar mass of ammonia = 17 g/mol

Moles of ammonia = 1.66 moles

Putting values in equation 1, we get:

$1.66mol=\frac{\text{Mass of ammonia}}{17g/mol}\\\\\text{Mass of ammonia}=(1.66mol\times 17g/mol)=28.22g$

Hence, the mass of ammonia produced is 28.22 g

4 0

2 years ago

What is the mass of a sample of water containing 3.55×1022 molecules of h2o?

ad-work [718]

6,02×10²³ -------------- 18g
3,55×10²² -------------- x $x=\frac{3,55*10^{22}*18g}{6,02*10^{23}}=10,61*10^{22-23}g=10,61*10^{-1}g=1,061g$

6 0

2 years ago

Read 2 more answers

According to the equation below, how many moles of Ca(OH)2 are required to react with 1.36 mol H3PO4 to produce Ca3(PO4)2? 3Ca(O

allsm [11]

Answer: The amount of calcium hydroxide needed to react is 2.04 moles

Explanation:

We are given:

Moles of phosphoric acid = 1.36 moles

For the given chemical equation:

$3Ca(OH)_2+2H_3PO_4\rightarrow Ca_3(PO_4)_2+6H_2O$

By Stoichiometry of the reaction:

2 moles of phosphoric acid reacts with 3 moles of calcium hydroxide

So, 1.36 moles of phosphoric acid will react with = $\frac{3}{2}\times 1.36=2.04mol$ of calcium hydroxide

Hence, the amount of calcium hydroxide needed to react is 2.04 moles

3 0

2 years ago

What is the overall balanced equation for the precipitation reaction occurring between silver nitrate and calcium bromide? hints

aleksklad [387]

The answer:

the overall balanced equation for the precipitation reaction occurring between silver nitrate and calcium bromide:
CaBr2 + AgNO3-----------> AgBr2 + CaNO3, so among the given choices, the
true answe is
agno3(aq)+cabr2(aq) (as reactants)

4 0

2 years ago

Consider the following system at equilibrium where H° = -87.9 kJ, and Kc = 83.3, at 500 K. PCl3(g) + Cl2(g) PCl5(g) If the VOLUM

7nadin3 [17]

Answer:

The value of Kc C. remains the same.

The value of Qc C. is less than Kc.

The reaction must: A. run in the forward direction to reestablish equilibrium

The number of moles of Cl2 will B. decrease.

Explanation:

Le Chatelier's Principle states that if a system in equilibrium undergoes a change in conditions, it will move to a new position in order to counteract the effect that disturbed it and recover the state of equilibrium.

A decrease in volume causes the system to evolve in the direction in which there is less volume, that is, where the number of gaseous moles is less.

But temperature is the only variable that, in addition to influencing equilibrium, modifies the value of the constant Kc. So if the volume of the equilibrium system is suddenly decreased at constant temperature: The value of Kc remains the same.

As mentioned, if the volume of an equilibrium gas system decreases, the system moves to where there are fewer moles. In this case, being:

PCl₃(g) + Cl₂(g) ⇔ PCl₅(g)

The equilibrium in this case then shifts to the right because there is 1 mole in the term on the right, compared to the two moles on the left. So, The reaction must: A. run in the forward direction to reestablish equilibrium.

By decreasing the volume, and so that Kc remains constant, being:

$Kc=\frac{[PCl_{5} ]}{[PCl_{3}]*[Cl_{2} ]}=\frac{\frac{nPCl_{5} }{Volume} }{\frac{nPCl_{3}}{Volume}*\frac{nCl_{2} }{Volume} } =\frac{nPCl_{5}}{nPCl_{3}*nCl_{2}} *Volume$

where nPCl₅, nPCl₃ and nCl₂ are the moles in equilibrium of PCl₅, PCl₃ and Cl₂

so, the number of moles of Cl₂ should decrease.The number of moles of Cl2 will B. decrease.

If the reaction quotient is less than the equilibrium constant, Qc <Kc, the system will evolve to the right, the direct reaction prevailing, to increase the concentration of products. So in this case, if the reaction moves to the right, the value of Qc C. is less than Kc.

3 0

2 years ago