Identify each process numbered 1-6

A solution is prepared by dissolving 15.0 g of NH3 in 250.0 g of water. The density of the resulting solution is 0.974 g/mL. The

iVinArrow [24]

Answer:

[NH₃] → 3.24 M

Explanation:

Our solute: Ammonia

Our solvent: Water

Solution's mass = Mass of solute + Mass of solvent

Solution's mass = 15 g + 250 g = 265g

We use density to determine, the volume.

D = mass /volume → Volume = m / D → 265 g /0.974 g/mL = 272.07 mL.

We convert the mL to L → 272.07 mL . 1L /1000mL = 0.27207 L

To determine molarity we need the moles of solute in 1 L of solution.

Moles of solute are: 15g / 17g/mol = 0.882 moles

[NH₃] = 0.882mol /0.27207 L → 3.24 M

3 0

3 years ago

What mass of oxygen is needed to burn 4g of hydrogen, given the equation below

aleksandrvk [35]

32g of oxygen is required to burn 4g of hydrogen.

Define molecular mass.

A specific molecule's mass is expressed in daltons and is known as the molecular mass (m) (Da or u). Due to the varying isotopes of an element that they contain, multiple molecules of the same substance can have distinct molecular weights.

The total atomic mass of every atom in a molecule, calculated using a scale with hydrogen, carbon, nitrogen, and oxygen having atomic masses of 1, 12, 14, and 16, respectively. For instance, water has a molecular mass of 18 (2 + 16), which consists of two hydrogen atoms and one oxygen atom. known also as molecular weight.

In ,2H2+O2-----> 2H2O

H 2 molecules have a mass of 2 g/mol.

The molecular weight of oxygen is 32 g/mol.

When the chemical equation is balanced,

To totally react, 32 g of oxygen are needed for every 22=4 g of hydrogen.

To know more about molecular mass use link below:

brainly.com/question/21334167

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7 0

1 year ago

stomach acid contains dilute hydrochloric acid write a chemical equation to show the neutralization of stomach acid by milk of m

Eddi Din [679]

Answer:

Mg(OH)2(s) + 2HCl(aq) → 2H2O(l) + MgCl2(aq)

Explanation:

4 0

3 years ago

For the reaction: CH₃OH(g) + heat ↔ CO(g) + 2H₂(g) How will the equilibrium shift if the overall pressure is decreased?

MakcuM [25]

Answer:

To the right

Explanation:

CH₃OH(g) + heat <=> CO(g) + 2H₂(g)

According to Le Chatelier's principle, a decrease in pressure will shift the equilibrium position to the side where there is a higher volume.

From the balanced equation above,

Volume of reactant = 1

Volume of product = 1 + 2 = 3

From the above, we can see that the volume of the gasous product is higher than the volume of the gasous reactant.

Therefore, a decrease in the pressure of the system will shift the equilibrium position to the right.

7 0

3 years ago

What is the concentration of NaCl in a solution if titration of 15.00 mL of the solution with 0.2503 M AgNO3 requires 20.22 mL o

Nina [5.8K]

Answer:

The concentration of NaCl = 0.3374 M

Explanation:

Given :

Molarity of AgNO₃ = 0.2503 M

Volume of AgNO₃ = 20.22 mL

The conversion of mL into L is shown below:

$1 mL= 10^{-3} L$

Thus, volume of the solution = 20.22×10⁻³ L

Molarity of a solution is the number of moles of solute present in 1 L of the solution.

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

The formula can be written for the calculation of moles as:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Thus,

$Moles\ of\ AgNO_3 =Molarity \times {Volume\ of\ the\ solution}$

$Moles\ of\ AgNO_3 =0.2503 \times {20.22\times 10^{-3}}\ moles$

$Moles\ of\ AgNO_3 = 5.0611 \times 10^{-3} moles$

The chemical reaction taking place:

$AgNO_3_(aq) + NaCl_(aq) \rightarrow AgCl_(s) + NaNO_3_(aq)$

According to reaction stoichiometry:

1 mole of AgNO₃ reacts with 1 mole of NaCl

Thus,

5.0611×10⁻³ moles of AgNO₃ reacts with 5.0611×10⁻³ moles of NaCl

Thus, moles of NaCl required = 5.0611×10⁻³ moles

Volume of NaCl required = 15.00 mL

The conversion of mL into L is shown below:

$1 mL= 10^{-3} L$

Thus, volume of the solution = 15.00×10⁻³ L

Applying in the formula of molarity as:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Molarity\ of\ NaCl=\frac{5.0611\times 10^{-3}}{15.00\times 10^{-3}}$

$Molarity\ of\ NaCl= 0.3374 M$

Thus, the concentration of NaCl = 0.3374 M

6 0

3 years ago