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4 years ago

How many phosphorus atoms are present ?

Chemistry

1 answer:

Law Incorporation [45]4 years ago

3 0

Answer:

The answer is option 2.

Explanation:

(PO4)2 = P2O8

Therefore, there are 2 Phosphorus atoms present .

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The first coin has a density of approximately 19. What kind of metal is it

umka21 [38]

Answer: Its a A Cent.

4 0

3 years ago

A mixture of A and B is capable of being ignited only if the mole percent of A is 6 %. A mixture containing 9.0 mole% A in B flo

atroni [7]

Explanation:

As it is given that mixture (contains 9 mol % A and 91% B) and it is flowing at a rate of 800 kg/h.

Hence, calculate the molecular weight of the mixture as follows.

Weight = $0.09 \times 16.04 + 0.91 \times 29$

= 27.8336 g/mol

And, molar flow rate of air and mixture is calculated as follows.

$\frac{800}{27.8336}$

= 28.74 kmol/hr

Now, applying component balance as follows.

$0.09 \times 28.74 + 0 \times F_{B} = 0.06F_{p}$

$F_{p}$ = 43.11 kmol/hr

$F_{A} + F_{B} = F_{p}$

$F_{B}$ = 43.11 - 28.74

= 14.37 kmol/hr

So, mass flow rate of pure (B), is $F_{B}$ = $14.37 \times 29$

= 416.73 kg/hr

According to the product stream, 6 mol% A and 94 mol% B is there.

Molecular weight of product stream = $Mol. weight \times 43.11 kmol/hr$

= $0.06 \times 16.04 + 0.94 \times 29$

= 28.22 g/mol

Mass of product stream = 1216.67 kg/hr

Hence, mole of $O_{2}$ into the product stream is as follows.

$0.21 \times 0.94 \times 43.11$

= $8.5099 kmol/hr \times 329 g/mol$

= 272.31 kg/hr

Therefore, calculate the mass % of $O_{2}$ into the stream as follows.

$\frac{272.31}{1216.67} \times 100$

= 22.38%

Thus, we can conclude that the required flow rate of B is 272.31 kg/hr and the percent by mass of $O_{2}$ in the product gas is 22.38%.

4 0

3 years ago

What formula results when Ca+2 and Br -1 ions bond?

olga nikolaevna [1]

CaBr2
Because of the criss cross method of switching the charges

3 0

3 years ago

The reaction 2HgO (s)→2Hg (I)+O2 (g) has a percent yield of 50%. You want to produce 100 g of Hg.

nevsk [136]

The mass of HgO needed for the reaction is 216 g

The correct answer to the question is Option C. 216 g

We'll begin by calculating the theoretical yield of Hg.

Actual yield of Hg = 100 g

Percentage yield = 50%

Theoretical yield of Hg =?

Theoretical yield = Actual yield / percentage yield

Theoretical yield = 100 / 50%

Theoretical yield of Hg = 200 g

Finally, we shall determine the mass of HgO needed for the reaction.

2HgO → 2Hg + O₂

Molar mass of HgO = 201 + 16 = 217 g/mol

Mass of HgO from the balanced equation = 2 × 217 = 434 g

Molar mass of Hg = 201 g/mol

Mass of Hg from the balanced equation = 2 × 201 = 402 g

From the balanced equation above,

402 g of Hg were produced from 434 g of HgO.

Therefore

200 g of Hg will be produce by = (200 × 434) / 402 = 216 g of HgO.

Thus, 216 g of HgO is needed for the reaction.

Learn more about stoichiometry:

brainly.com/question/24426334

4 0

2 years ago

Read 2 more answers

2C4H10 + 13O2 Right arrow. 8CO2 + 10H2O<br><br> What is the mole ratio of butane to carbon dioxide?

Zielflug [23.3K]

2C4H10 + 13O2 Right arrow. 8CO2 + 10H2O

What is the mole ratio of butane to carbon dioxide?

ANSWER:

Butane is an organic compound with the formula C4H10 and carbon dioxide is compound (gas) with the formula CO2. So, the balanced coefficient of butane is 2, and that of carbon dioxide is 8. According to thid, the mole ratio of butane to carbon dioxide is 2:8 or 1:4.

7 0

3 years ago