How many grams of sodium nitrate are needed to make 3L of 9.05 Solution?

Four gases were combined in a gas cylinder with these partial pressures: 3.5 atm N2, 2.8 atm O2, 0.25 atm Ar, and 0.15 atm He.

Goryan [66]

Answer:

This can be solved using Dalton's Law of Partial pressures. This law states that the total pressure exerted by a gas mixture is equal to the sum of the partial pressure of each gas in the mixture as if it exist alone in a container. In order to solve, we need the partial pressures of the gases given. Calculations are as follows:

Explanation:

P = 3.00 atm + 2.80 atm + 0.25 atm + 0.15 atm

P = 6.8 atm

3.5 atm = x (6.8 atm)

x = 0.51

8 0

3 years ago

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What is the correct balanced equation for the reaction that occurs when aqueous ammonium chloride reacts with aqueous barium nit

Setler [38]

Answer:

$2NH_{4}Cl + Ba(NO_{3} )_{2}$ ⇒ $BaCl_{2} + 2NH_{4}NO_{3}$

Explanation:

In balancing a chemical equation we make sure the number of atoms of each element on the reactant side of the equation equals that on the product side

$NH_{4}Cl + Ba(NO_{3} )_{2}$ ⇒ $BaCl_{2} + NH_{4}NO_{3}$

The equation above represents an unbalanced equation of the reaction of aqueous ammonium chloride with aqueous barium nitrate .we can balance this equation by adding the right coefficients to reactants and product.

$2NH_{4}Cl + Ba(NO_{3} )_{2}$ ⇒ $BaCl_{2} + 2NH_{4}NO_{3}$

5 0

3 years ago

6. How will you obtain ? (a) Magnesium oxide from magnesium. (b) Silver chloride from silver nitrate. (c) Nitrogen dioxide from

Reil [10]

Answer:

a) reaction with oxygen

2mg +o2---------2mgo

b) Agno3+NaCl ----------AgCl+NaNo3

8 0

2 years ago

A 1.00 g sample of a metal X (that is known to form X ions in solution) was added to 127.9 mL of 0.5000 M sulfuric acid. After a

Semenov [28]

Answer: The metal having molar mass equal to 26.95 g/mol is Aluminium

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

Molarity of NaOH solution = 0.5000 M

Volume of solution = 0.03340 L

Putting values in equation 1, we get:

$0.5000M=\frac{\text{Moles of NaOH}}{0.03340L}\\\\\text{Moles of NaOH}=(0.5000mol/L\times 0.03340L)=0.01670mol$

The chemical equation for the reaction of NaOH and sulfuric acid follows:

$2NaOH+H_2SO_4\rightarrow Na_2SO_4+H_2O$

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of sulfuric acid

So, 0.01670 moles of NaOH will react with = $\frac{1}{2}\times 0.01670=0.00835mol$ of sulfuric acid

Excess moles of sulfuric acid = 0.00835 moles

Calculating the moles of sulfuric acid by using equation 1, we get:

Molarity of sulfuric acid solution = 0.5000 M

Volume of solution = 127.9 mL = 0.1279 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$0.5000M=\frac{\text{Moles of }H_2SO_4}{0.1279L}\\\\\text{Moles of }H_2SO_4=(0.5000mol/L\times 0.1279L)=0.06395mol$

Number of moles of sulfuric acid reacted = 0.06395 - 0.00835 = 0.0556 moles

The chemical equation for the reaction of metal (forming $M^{3+}$ ion) and sulfuric acid follows:

$2X+3H_2SO_4\rightarrow X_2(SO_4)_3+3H_2$

By Stoichiometry of the reaction:

3 moles of sulfuric acid reacts with 2 moles of metal

So, 0.0556 moles of sulfuric acid will react with = $\frac{2}{3}\times 0.0556=0.0371mol$ of metal

To calculate the molar mass of metal for given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Mass of metal = 1.00 g

Moles of metal = 0.0371 moles

Putting values in above equation, we get:

$0.0371mol=\frac{1.00g}{\text{Molar mass of metal}}\\\\\text{Molar mass of metal}=\frac{1.00g}{0.0371mol}=26.95g/mol$

Hence, the metal having molar mass equal to 26.95 g/mol is Aluminium

6 0

3 years ago

If Salt is NaCl, what atoms does the compound contain?

allochka39001 [22]

Na is the elemental abbreviation for sodium, and Cl is the abbreviation for chlorine.
When the two are combined, you get sodium-chloride, or table salt.
Hope that helped =)

8 0

3 years ago

How many grams of sodium nitrateare needed to make 3L of 9.05Solution?​

How many grams of sodium nitrateare needed to make 3L of 9.05Solution?