how many moles of phosphoric acid are there in 658 grams of H3PO4

109.3 × 10⁵ mol

Explanation:

Given data:

Number of moles of phosphoric acid = ?

Number of formula units = 658 × 10²⁸

Solution:

The given problem will solve by using Avogadro number.

The number 6.022 × 10²³ is called Avogadro number.

1 mole contain 6.022 × 10²³ formula units,

658 × 10²⁸ formula units × 1 mol / 6.022 × 10²³ formula units

109.3 × 10⁵ mol

In one liter of solution, there are 4.194 moles of H₃PO₄

Explanation:

A molal solution contains 1 mole of solute in 1 Kg of solvent.

A  aqueous 5.257 molal solution of H₃PO₄ will contain 5.257 moles of  H₃PO₄ in 1 Kg or 1000 g of water .

Molar mass of H₃PO₄ = 98 g/mol

Therefore mass of H₃PO₄ = number of moles * molar mass

mass of H₃PO₄ = 5.257 mol * 98 g/mol = 515.2 g

mass of solution = mass of solute + mass of solvent

mass of solution = 515.2 + 1000 = 1515.2 g

Volume of solution = mass of solution / density of solution

Volume of solution = 1515.2 / 1.2089 = 1253.4 mL

Molarity of solution = number of moles / volume (L)

Molarity of solution = 5.257 / 1.2534 L = 4.194 mol/L

Therefore, in one liter of solution, there are 4.194 moles of H₃PO₄

In one liter of solution, there are 4.194 moles of H₃PO₄

Explanation:

A molal solution contains 1 mole of solute in 1 Kg of solvent.

A  aqueous 5.257 molal solution of H₃PO₄ will contain 5.257 moles of  H₃PO₄ in 1 Kg or 1000 g of water .

Molar mass of H₃PO₄ = 98 g/mol

Therefore mass of H₃PO₄ = number of moles * molar mass

mass of H₃PO₄ = 5.257 mol * 98 g/mol = 515.2 g

mass of solution = mass of solute + mass of solvent

mass of solution = 515.2 + 1000 = 1515.2 g

Volume of solution = mass of solution / density of solution

Volume of solution = 1515.2 / 1.2089 = 1253.4 mL

Molarity of solution = number of moles / volume (L)

Molarity of solution = 5.257 / 1.2534 L = 4.194 mol/L

Therefore, in one liter of solution, there are 4.194 moles of H₃PO₄

0.020 mol

Explanation:

Let's consider the following relations:

With these relations, we can calculate the moles of phosphoric acid in 27 drops of solution.

0.020 mol

Explanation:

Let's consider the following relations:

With these relations, we can calculate the moles of phosphoric acid in 27 drops of solution.

Answer:

Step-by-step explanation:

It is clear by the equation 2(27+3×35.5)= 267 gm of AlCl3 reacts with 6× 80 = 480 gm of Br2 . So 29.2 gm reacts = 480× 29.2/267= 52.6 gm

Answer:

Taking into accoun the ideal gas law, The volume of a container that contains 24.0 grams of N2 gas at 328K and 0.884 atm is 26.07 L.

An ideal gas is a theoretical gas that is considered to be composed of point particles that move randomly and do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:  

P×V = n×R×T

where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas. The universal constant of ideal gases R has the same value for all gaseous substances.

Explanation:

In this case, you know:

P= 0.884 atm

V= ?

n= 0.857 moles (where 28 g/mole is the molar mass of N₂, that is, the amount of mass that the substance contains in one mole.)

R=0.082

T= 328 K

Replacing in the ideal gas law:

0.884 atm×V= 0.857 moles× 0.082 ×328 K

Solving:

V= 26.07 L

The volume of a container that contains 24.0 grams of N2 gas at 328K and 0.884 atm is 26.07 L.