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3 years ago

How many more oxygen atoms are in Na2HPO4H(disodium phosphate) then in 2H3PO(Phosphoric Acid)?

Chemistry

2 answers:

Serggg [28]3 years ago

3 0

Answer:

in disodium phosphate = 4 oxygen atoms

in phosphoric acid = 1 oxygen atom

Ann [662]3 years ago

3 0

Answer: There are 3 more oxygen atoms in Na2HPO4H than 2H3PO

Explanation:

There are 4 oxygen atoms in Na2HPO4H and 1 in 2H3PO. You subtract 4 - 1 =3

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Y_Kistochka [10]

Answer:

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3 years ago

15g of FeCI3 is dissolved in 450 mL of solution. What is the concentration of [CI-]?

storchak [24]

The concentration of [CI-] : 0.617 M

<h3>Further explanation</h3>

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Reaction

FeCl₃⇒Fe³⁺+3Cl⁻

mol FeCl₃(MW=162,2 g/mol)

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3 years ago

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3 years ago

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3 years ago

A mixture of 15.0 g of the anesthetic halothane (C2HBrClF3 197.4 g/mol) and 22.6 g of oxygen gas has a total pressure of 862 tor

AlexFokin [52]

Answer : The partial pressure of $C_2HBrClF_3$ and $O_2$ are, 84 torr and 778 torr respectively.

Explanation : Given,

Mass of $C_2HBrClF_3$ = 15.0 g

Mass of $O_2$ = 22.6 g

Molar mass of $C_2HBrClF_3$ = 197.4 g/mole

Molar mass of $O_2$ = 32 g/mole

First we have to calculate the moles of $C_2HBrClF_3$ and $O_2$ .

$\text{Moles of }C_2HBrClF_3=\frac{\text{Mass of }C_2HBrClF_3}{\text{Molar mass of }C_2HBrClF_3}=\frac{15.0g}{197.4g/mole}=0.0759mole$

and,

$\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{22.6g}{32g/mole}=0.706mole$

Now we have to calculate the mole fraction of $C_2HBrClF_3$ and $O_2$ .

$\text{Mole fraction of }C_2HBrClF_3=\frac{\text{Moles of }C_2HBrClF_3}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.0759}{0.0759+0.706}=0.0971$