When a solid mixture of mgco3 and caco3 is heated strongly, carbon dioxide gas is given off and a solid mixture of mgo and cao i

Answer:

Mass released = 8.6 g

Explanation:

Provided information:

Starting moles of nitrogen= 0.950 mol

Starting volume = 25.5 L

Final nitrogen mass released = ?

Final volume = 17.3 L

Calculation:

Equation:

V₁/n₁ = V₂/n₂

25.5 L / 0.950 mol = 17.3 L/n₂

n₂ = 17.3 L× 0.950 mol/25.5 L

n₂ = 16.435 L.mol /25.5 L

n₂ = 0.644 mol

Starting mass of nitrogen:

Mass = moles × molar mass

Mass = 0.950 mol × 28 g/mol

Mass = 26.6 g

Ending mass of nitrogen:

Mass = moles × molar mass

Mass = 0.644 mol × 28 g/mol

Mass = 18.0 g

Mass released = initial mass - ending mass

Mass released = 26.6 g - 18.0 g

Mass released = 8.6 g

An element of evidence indicating that the color of the flame is attributed to the metal ion rather than the chemical is that none of the flames produced by different metals shared the same color (each metal produced its unique flame color). Although most tested metals had chloride, the flame colors were all distinct. The two flames that contained copper (one from copper (II) chloride and the other from copper (II) sulfate) showed similar colors; one was green-blue and the other was bright green. This suggests a close resemblance, and any slight variation could be attributed to error.

Answer: 0.0164 molar concentration of hydrochloric acid in the resulting solution.

Explanation:

1) Molarity of 0.250 L HCl solution: 0.0328 M

The amount of HCl in the 0.250 L solution = 0.0082 moles

2) Molarity of 0.100 L NaOH solution: 0.0245 M

The amount of NaOH in the 0.100 L solution = 0.00245 moles

3) Determining the concentration of hydrochloric acid in the final solution.

0.00245 moles of NaOH will neutralize 0.00245 moles of HCl from the original 0.0082 moles of HCl.

The total volume of the mixture becomes 0.100 L + 0.250 L = 0.350 L

Remaining moles of unreacted HCl = 0.0082 moles - 0.00245 moles = 0.00575 moles

Concentration of the remaining HCl:

0.0164 molar concentration of hydrochloric acid in the resulting solution.

Answer:

Refer to the explanation.

Explanation:

Formation reactions involve the creation of one mole of a compound from its elements in their standard states.

NaBr (s)

The equation for the standard formation is

Na (s) + (1/2)Br₂ (g) → NaBr (s)

As per appendix C, the standard heat of formation for NaBr(s) is

ΔH∘f = -359.8 kJ/mol.

SO₃ (g)

The equation for the standard formation is

S (s) + (3/2) O₂ (g) → SO₃ (g)

ΔH∘f = -395.2 kJ/mol.

Pb(NO₃)₂ (s)

The equation for the standard formation is

Pb (s) + N₂ (g) + 3O₂ (g) → Pb(NO₃)₂ (s)

According to appendix C, the standard heat of formation for Pb(NO₃)₂(s) is

ΔH∘f = -451.9 kJ/mol.

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