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2 years ago

How many moles of pbcl2 are produced when 16 alcl3 are consumed

1 answer:

4 0

Using dimensional analysis and the mole ratios from the balanced equation, we can calculate moles of PbCl2 formed from 14 moles of AlCl3.

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It needs to be random so that there isn't bias that would skew the consistency

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How many milliliters of water at 25.0°C with a density of 0.997 g/mL must be mixed with 163 mL of coffee at 97.9°C so that the r

Stolb23 [73]

Answer:

248 mL

Explanation:

According to the law of conservation of energy, the sum of the heat absorbed by water (Qw) and the heat released by the coffee (Qc) is zero.

Qw + Qc = 0

Qw = -Qc [1]

We can calculate each heat using the following expression.

Q = c × m × ΔT

where,

c: specific heat

m: mass

ΔT: change in the temperature

163 mL of coffee with a density of 0.997 g/mL have a mass of:

163 mL × 0.997 g/mL = 163 g

From [1]

Qw = -Qc

cw × mw × ΔTw = -cc × mc × ΔTc

mw × ΔTw = -mc × ΔTc

mw × (54.0°C-25.0°C) = -163 g × (54.0°C-97.9°C)

mw × 29.0°C = 163 g × 43.9°C

mw = 247 g

The volume corresponding to 247 g of water is:

247 g × (1 mL/0.997 g) = 248 mL

8 0

3 years ago

A solution of barium nitrate has 61.2g of barium nitrate in 1 liter of solution. How many mg of barium are there in 7.5 quarts

Nuetrik [128]

Answer:

220.44g Ba²⁺ ions in solution

Explanation:

Given parameters:

Mass of barium nitrate = 61.2g

Volume of solution = 1 liter

Unkown:

Mass of barium in 7.5quarts of solution?

Solution

We must first convert quarts to its liter equivalence:

1 quarts = 0.95 liter

7.5 quarts = 0.95 x 7.5; 7.125liter

Now, let us find the mass of barium nitrate in a solution of 7.125liter:

Given:

1 liter of solution contains 61.2g of barium nitrate:

7.125 liter will contain 7.125 x 61.2 = 436.05g of barium nitrate.

The formula of the compound is Ba(NO₃)₂:

In solution we have Ba²⁺ + NO₃⁻

Ba(NO₃)₂ → Ba²⁺ + 2NO₃⁻

Number of moles of Ba(NO₃)₂ = $\frac{mass of Ba(NO₃)₂}{Molar mass of Ba(NO₃)₂}$

Molar mass of Ba(NO₃)₂ = 137 + 2[14 + 3(16)] = 271g/mol

Number of moles of Ba(NO₃)₂ = $\frac{436.05}{271}$ = 1.609mole

1 mole of Ba(NO₃)₂ will produce 1 mole of Ba²⁺ ions in solution

therefore, 1.609mole of Ba(NO₃)₂ will also yield 1.609mole of Ba²⁺ ions in solution

Mass of Ba²⁺ ions = Number of moles of Ba²⁺ ions x molar mass of Ba²⁺ ions

Mass of Ba²⁺ ions = 1.609 x 137 = 220.44g

3 0

3 years ago