Chemistry College

How many grams of Mg(NO3), are necessary to
make 1500 mL of a 0.50 M solution?

How Many Grams Of Mg(NO3), Are Necessary Tomake 1500 ML Of A 0.50 M Solution?

Answers

Answer 1

Answer:

1.1 × 10² g

Explanation:

Step 1: Given data

Concentration of the solution (C): 0.50 M (0.50 mol/L)Volume of solution (V): 1500 mL (1.500 L)

Step 2: Calculate the moles of Mg(NO₃)₂ (solute)

Molarity is equal to the moles of solute (n) divided by the liters of solution.

C = n/V

n = C × V

n = 0.50 mol/L × 1.500 L = 0.75 mol

Step 3: Calculate the mass corresponding to 0.75 moles of Mg(NO₃)₂

The molar mass of Mg(NO₃)₂ is 148.3 g/mol.

0.75 mol × 148.3 g/mol = 1.1 × 10² g

Related Questions

How does the entropy change in the reaction 2C3H6(g) + 9O2(g) → 6C02(g) + 6H2O (g)?

I will mark brainliest!! Thank you so much!!

Answers

Answer:

The entropy increases!!!

Explanation:

a pex

The entropy increases in the reaction.

What is entropy?Entropy is defined as the measure of the disorder of a system.Entropy is an extensive property of a thermodynamic system, to put it in simple words, its value changes depending on the amount of matter that is present.Entropy is denoted by the letter S and has units of joules per kelvin    (JK−1)

The entropy increases in the reaction if the total number of product molecules are greater than the total number of reactant molecules.

2C3H6(g) + 9O2(g) → 6C02(g) + 6H2O (g)

In the above reaction, the product molecules are greater than the reactant molecules. Hence, entropy increases.

Hence, we can conclude that option A is the answer.

To learn more about entropy here

https://brainly.com/question/22861773

#SPJ2

Please give me the answer please

Answers

Answer:

A. 30cm³

Explanation:

Based on the chemical reaction:

CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

1 mol of calcium carbonate reacts with 2 moles of HCl to produce 1 mol of CO₂

To solve this question we must convert the mass of each reactant to moles. With the moles we can find limiting reactant and the moles of CO₂ produced. Using PV = nRT we can find the volume of the gas:

Moles CaCO₃ -Molar mass: 100.09g/mol-

1.00g * (1mol / 100.09g) = 9.991x10⁻³ moles

Moles HCl:

50cm³ = 0.0500dm³ * (0.05 mol / dm³) = 2.5x10⁻³ moles

For a complete reaction of 2.5x10⁻³ moles HCl there are necessaries:

2.5x10⁻³ moles HCl * (1mol CaCO₃ / 2mol HCl) = 1.25x10⁻³ moles CaCO₃. As there are 9.991x10⁻³ moles, HCl is limiting reactant.

The moles produced of CO₂ are:

2.5x10⁻³ moles HCl * (1mol CO₂ / 2mol HCl) = 1.25x10⁻³ moles CO₂

Using PV = nRT

Where P is pressure = 1atm assuming STP

V volume in L

n moles = 1.25x10⁻³ moles CO₂

R gas constant = 0.082atmL/molK

T = 273.15K at STP

V = nRT / P

1.25x10⁻³ moles * 0.082atmL/molK*273.15K / 1atm = V

0.028L = V

28cm³ = V

As 28cm³ ≈ 30cm³

Right option is:

A. 30cm³

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