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Rufina [12.5K]
2 years ago
8

What is the molar mass of H3PO4?

Chemistry
2 answers:
Tasya [4]2 years ago
6 0

Answer:

The above answer is correct

Explanation:

I took the test

SCORPION-xisa [38]2 years ago
4 0

Answer:

Molar mass of H3PO4  = 97.99 g/mol (Approx.)

Explanation:

Find:

Molar mass of H3PO4

Given;

Molar mass of H = 1.0079 g/mol

Molar mass of P = 30.974 g/mol

Molar mass of O = 15.999 g/mol

Computation:

Molar mass of H3PO4  = (1.0079)(3) + 30.974 + 15.999(4)

Molar mass of H3PO4  = 3.0237 + 30.974 + 63.996

Molar mass of H3PO4  = 97.9937

Molar mass of H3PO4  = 97.99 g/mol (Approx.)

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Given, 0.29 g of hydrocarbon produces 448ml of CO2 at STP. then, C2H5 is the emperical formula of hydrocarbon . n = 2 , hence, molecular formula will be C4H10

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Select the words that correctly fill in the blanks for this statement:
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The p sublevel has __three_orbitals that are _dunmbbell_-shaped.

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Between HClO3 and HIO3, which is stronger and why? Question 16 options: 1) HClO3 is stronger because chlorine is in a higher oxi
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2) HClO3 is stronger because chlorine is more electronegative than iodine.

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What is the defintion of speed of light
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What is the density of air at 75 F and latm pressure? Express in lbm/ft3 and kg/m3
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Answer : The density of air in lbm/ft^3 and kg/m^3 is, 0.0743lbm/ft^3 and 1.19kg/m^3 respectively.

Explanation :

PV=nRT\\\\PV=\frac{m}{M}RT\\\\P=\frac{m}{V}\frac{RT}{M}\\\\P=\rho \frac{RT}{M}\\\\\rho=\frac{PM}{RT}

where,

P = pressure of air = 1 atm

V = volume of air

T = temperature of air = 297 K

The conversion used for the temperature from Fahrenheit to degree Celsius is:

^oC=(^oF-32)\times \frac{5}{9}

^oC=(75-32)\times \frac{5}{9}=24^oC

The conversion used for the temperature from degree Celsius to Kelvin is:

K=273+^oC

K=273+24=297K

n = number of moles

m = mass of air

M = average molar mass of air = 28.97 g/mole

\rho = density of air = ?

R = gas constant = 0.0821 L.atm/mol.K

Now put all the given values in the above formula, we get:

\rho=\frac{PM}{RT}

\rho=\frac{(1atm)\times (28.97g/mole)}{(0.0821L.atm/mol.K)\times (297K)}

\rho=1.19g/L

Now we have to calculate density in lbm/ft^3.

Conversion used :

1g/L=0.0624lbm/ft^3

So,

1.19g/L=\frac{1.19g/L}{1g/L}\times 0.0624lbm/ft^3=0.0743lbm/ft^3

The density of air in lbm/ft^3 is, 0.0743lbm/ft^3

Now we have to calculate density in kg/m^3.

Conversion used :

1g/L=1kg/m^3

So,

1.19g/L=1.19kg/m^3

The density of air in kg/m^3 is, 1.19kg/m^3

8 0
3 years ago
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