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3 years ago

How many grams of phosphorus are in a sample of Ca3(PO4)2 that contains 50.0 g of calcium? Answer: ________ grams of phosphorus.

Chemistry

1 answer:

wel3 years ago

6 0

Answer: The mass of phosphorus that is present for given amount of calcium is 28.53 g.

Explanation:

We are given:

Mass of calcium = 50 grams

The chemical formula of calcium phosphate is $Ca_3(PO_4)_2$

Molar mass of calcium = 40 g/mol

Molar mass of phosphorus = 31 g/mol

In 1 mole of calcium phosphate, 120 grams of calcium is combining with 62 grams of phosphorus.

So, 50 grams of calcium will combine with = $\frac{62}{120}\times 50=28.53g$ of phosphorus.

Hence, the mass of phosphorus that is present for given amount of calcium is 28.53 g.

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What name is given to the amount of substance in an object?

STatiana [176]

Answer:

mole

Explanation:

is a unit of amount of substance in the International System of Units, defined (since 2019) by fixing the Avogadro constant at the given value. Sometimes, the amount of substance is referred to as the chemical amount

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2 years ago

NO LINKS RIGHT ANSWERS ONLY!!!!

Mumz [18]

I think the answer is D

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3 years ago

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The reaction 2NO(g)+O2(g)−→−2NO2(g) is second order in NO and first order in O2. When [NO]=0.040M, and [O2]=0.035M, the observed

Oksanka [162]

Answer:

(a) The rate of disappearance of $O_{2}$ is: $4.65*10^{-5}$ M/s

(b) The value of rate constant is: 0.83036 $M^{-2}s^{-1}$

(d) The rate will increase by a factor of 3.24

Explanation:

The rate of a reaction can be expressed in terms of the concentrations of the reactants and products in accordance with the balanced equation.

For the given reaction:

$2NO(g)+O_{2}->2NO_{2}$

rate = $-\frac{1}{2} \frac{d}{dt}[NO]$ = $-\frac{d}{dt}[O_{2}]$ = $\frac{1}{2}\frac{d}{dt}[NO_{2}]$ -----(1)

According to the question, the reaction is second order in NO and first order in $O_{2}$ .

Then we can say that, rate = k $[NO]^{2}[O_{2}]$ -----(2)

where k is the rate constant.

The rate of disappearance of NO is given:

$-\frac{d}{dt}[NO]$ = $9.3*10^{-5}$ M/s.

(a) From (1), we can get the rate of disappearance of $O_{2}$ .

Rate of disappearance of $O_{2}$ = $-\frac{d}{dt}[O_{2}]$ = (0.5)*( $9.3*10^{-5}$ ) M/s = $4.65*10^{-5}$ M/s.

(b) The rate of the reaction can be obtained from (1).

rate = $-\frac{1}{2} \frac{d}{dt}[NO]$ = (0.5)*( $9.3*10^{-5}$ )

rate = $4.65*10^{-5}$ M/s

The value of rate constant can be obtained by using (2).

rate constant = k = $\frac{rate}{[NO]^{2}[O_{2}]}$

k = $\frac{4.65*10^{-5}}{(0.040)^{2}(0.035)}$ = 0.83036 $M^{-2}s^{-1}$

k = $\frac{rate}{[NO]^{2}[O_{2}]}$

Substituting the units of rate as M/s and concentrations as M, we get:

$\frac{Ms^{-1} }{M^{3}}$ = $M^{-2}s^{-1}$

(d) The reaction is second order in NO. Rate is proportional to square of the concentration of NO.

$rate\alpha [NO]^{2}$

If the concentration of NO increases by a factor of 1.8, the rate will increase by a factor of $(1.8)^{2}$ = 3.24

5 0

3 years ago

A certain gas is present in a 13.0 l cylinder at 4.0 atm pressure. if the pressure is increased to 8.0 atm , the volume of the g

ella [17]

Boyle's Law is k = PV so Initial k = 13.0 L x 4.0 atm = 52 L atm Final kf = 6.5 L x 8 atm = 52 L atm The gas obeys Boyle's Law The answer with two significant figures separated by a comma is k = 52, kf = 52.

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3 years ago

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Which two statements describe what happens to the nuclei of atoms during a fusion reaction? O A. Large nuclei break apart into t

pashok25 [27]

Answer:

D. Nuclei with small masses combine to form nuclei with larger masses.

B. A small amount of mass in the nuclei that combine is converted to energy

Explanation:

A nuclear fusion, in contrary to fission, is the process by which the nuclei of two atoms combine to form a much larger atom with a large nuclei. Likewise, during a fusion reaction, a large amount of energy is released from the small amount of mass in the nuclei (two) that combines.

According to this question, the following are true of a fusion reaction:

- Nuclei with small masses combine to form nuclei with larger masses.

- A small amount of mass in the nuclei that combine is converted to enormous energy.

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2 years ago

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