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3 years ago

Given the unbalanced equation:

Chemistry

1 answer:

sergij07 [2.7K]3 years ago

5 0

Answer:

3 is the coefficient in front of the $CaSO_{4}$

The smallest whole-number coefficient is 2.

Explanation:

The balanced chemical equation is as follows.

$Al_{2}(SO_{4})_{3}(aq)+3Ca(OH)_{2}(s) \rightarrow 2Al(OH)_{3}(s)+3CaSO_{4}(s)$

3 is the coefficient in front of the $CaSO_{4}$

The smallest whole-number coefficient is 2.

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Virus Multiplication-Viruses can make copies of themselves only inside a living____cell.

shutvik [7]

Answer:

Host

Explanation:

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2 years ago

7. NH2CO2NH4(s) when heated to 450 K undergoes the following reaction to produce a system which reaches equilibrium: NH2CO2NH4(s

Taya2010 [7]

Answer:

Value of equilibrium constant is 0.0888

Explanation:

Both $NH_{3}$ and $CO_{2}$ are gaseous. Hence equilibrium constant depends upon partial pressures of $NH_{3}$ and $CO_{2}$ .

Initially no $NH_{3}$ and $CO_{2}$ were present.

Hence mole fraction of $NH_{3}$ and $CO_{2}$ at equilibrium can be calculated from coefficient of $NH_{3}$ and $CO_{2}$ in balanced equation.

Mole fraction of $NH_{3}$ = (number of moles of $NH_{3}$ )/(total number of moles of $NH_{3}$ and $CO_{2}$ ) = $\frac{2moles}{(2+1)moles}=\frac{2}{3}$

Mole fraction of $CO_{2}$ = (number of moles of $CO_{2}$ )/(total number of moles of $NH_{3}$ and $CO_{2}$ ) = $\frac{1moles}{(2+1)moles}=\frac{1}{3}$

Let's assume both $CO_{2}$ and $NH_{3}$ behaves ideally.

Therefore partial pressure of $NH_{3}$ , $P_{NH_{3}}= x_{NH_{3}}.P_{total}$ and P_{CO_{2}}= x_{CO_{2}}.P_{total}

Where x represents mole fraction

So, $P_{NH_{3}}=\frac{2}{3}\times 0.843atm=0.562atm$

$P_{CO_{2}}=\frac{1}{3}\times 0.843atm=0.281atm$

So, $K_{p}=P_{NH_{3}}^{2}.P_{CO_{2}}=(0.562)^{2}\times 0.281=0.0888$

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3 years ago

(02.01 MC)

Rina8888 [55]

Answer:

C. Particle size

Explanation:

The sand, which has smaller particles, will go through the sieve, while the rice (with a larger particle size) will not

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2 years ago

What mass of oxygen contains the same number of molecules as 42g of nitrogen?

lakkis [162]

Answer:

Given: 42 g of N2

Solve for O2 mass that contains the same number of molecules to 42 g of N2.

Solve for the number of moles in 42 g of N2

1 mole of N2 = (14 * 2) g = 28 g so the number of moles in 42 g of N2 is equal to 42 g / 28 g per mole = 1.5 moles

Solve for mass of 1 mole of oxygen

1 mole of O2 = 16 g * 2 = 32 g per mole

Solve for the mass of 1.5 moles of oxygen

mass of 1.5 moles of O2 = 32 g per mole * 1.5 moles

mass of 1.5 moles of O2 = 48 g

So 48 g of O2 contains the same number of molecules as 42 g of N2

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3 years ago

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3 years ago