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3 years ago

How many moles are in 1.05 g of gold (Au)?

Chemistry

1 answer:

Wittaler [7]3 years ago

7 0

Answer:

0.005 mol

Explanation:

Moles is denoted by given mass divided by the molecular mass ,

Hence ,

n = w / m

n = moles ,

w = given mass ,

m = molecular mass .

From the question ,

w = given mass of Gold = 1.05 g ,

m = molecular mass of Gold = 197 g/mol

Hence , moles can be calculated as -

n = w / m = 1.05 g / 197 g/mol = 0.005 mol

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Explanation:

3CaBr2 + 2K3N → 6KBr + Ca3N2

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3 years ago

A gas sample occupies 350.0 mL at 546 mm Hg. What volume does the gas occupy at 652 mm Hg?

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Answer:

293.1 mL.

Explanation:

Boyle's law states that: at a constant temperature the pressure of a given mass of an ideal gas is inversely proportional to its volume.

It can be expressed as: P₁V₁ = P₂V₂,

P₁ = 546.0 mm Hg, V₁ = 350.0 mL.

P₂ = 652.0 mm Hg, V₂ = ??? mL.

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2 years ago

Suppose 0.795 g of sodium iodide is dissolved in 100. mL of a 39.0 m M aqueous solution of silver nitrate.

lubasha [3.4K]

Answer:

The final molarity of iodide anion is 0.053 M

Explanation:

Step 1: Data given

Mass of sodium iodide (NaI) = 0.795 grams

Volume of the solution = 100 mL = 0.1 L

Molarity of aqueous solution of silver nitrate (AgNO3) = 39 mM = 0.039M

The molecular mass of sodium iodide is 149.89 g/mol.

Step 2: The balanced equation

AgNO3(aq) + NaI(aq) → AgI(s) + NaNO3(aq)

Step 3: Calculate number of moles of sodium iodide

Moles NaI = mass NaI / Molar mass NaI

Moles NaI = 0.795 grams / 149.89 g/mol

Moles NaI = 0.0053 moles

For 1 mole AgNO3 consumed, we need 1 mole NaI to produce 1 mole AgI and 1 mole NaNO3

The sodium iodide will dissociate as followed:

NaI(aq) → Na+(aq) + I-(aq)

Step 4: Calculate iodide ions

For 1 mole NaI, we have 1 mole of I-

For 0.0053 moles of NaI we'll have 0.0053 moles I-

Step 5: Calculate molarity of iodide ion

Molarity = moles I- / volume

Molarity I- = 0.0053 moles / 0.1 L

Molarity I- = 0.053 M

The final molarity of iodide anion is 0.053 M

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3 years ago

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3 years ago

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