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2 years ago

If I have 340 mL of a 1.5 M NaBr solution, what will the concentration be if I add 560 mL more water to it?

Chemistry

1 answer:

ipn [44]2 years ago

5 0

Answer:

0.5667 M ≅ 0.57 M.

Explanation:

It is known that the no. of millimoles of a solution before dilution is equal to the no. of millimoles of the solution after the dilution.

It can be expressed as:

(MV) before dilution = (MV) after dilution.

M before dilution = 1.5 M, V before dilution = 340 mL.

M after dilution = ??? M, V after dilution = 340 mL + 560 mL = 900 mL.

∴ M after dilution = (MV) before dilution/(V) after dilution = (1.5 M)(340 mL)/(900 mL) = 0.5667 M ≅ 0.57 M.

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LuckyWell [14K]

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8 0

3 years ago

10. The density of aluminum is 2.7 g/cm². A 27g piece of aluminum will have what volume?

kotegsom [21]

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3 0

3 years ago

Part 1. A chemist reacted 18.0 liters of F2 gas with NaCl in the laboratory to form Cl2 gas and NaF. Use the ideal gas law equat

Alika [10]

Answer:

Part 1

The mass of the NaCl that reacted with F₂ at 290.K and 1.5 atm is approximately 132.6 gams

Part 2

The mass of NaCl that can react with the same volume of gas at STP is approximately 93.77 grams

Explanation:

Part 1

The volume of F₂ gas in the reaction, V = 18.0 liters

The ideal gas equation is P·V = n·R·T

∴ n = P·V/(R·T)

The pressure, P = 1.5 atm

The temperature, T = 290 K

The universal gas constant, R = 0.0820573 L·atm/(mol·K)

∴ n = 1.5×18/(0.0820573 × 290) ≈ 1.134615

The number of moles of F₂ in the reaction n ≈ 1.134615 moles

The chemical reaction is given as follows;

F₂ + 2NaCl → Cl₂ + 2NaF

1 mole of F₂ reacts with 2 moles of NaCl

Therefore;

1.134615 moles of F₂ reacted with 2 × 1.134615 moles ≈ 2.26923 moles of NaCl

1 mole of NaCl = The molar mass of NaCl, MM = 58.44 g/mol

The mass, of 2.26923 moles of NaCl, m = Number of moles × MM

∴ m ≈ 2.26923 moles × 58.44 g/mol ≈ 132.6 grams

The mass of the NaCl ≈ 132.6 gams

Part 2

The volume occupied by 1 mole of all gases at STP = 22.4 l/mole

Therefore, the number of moles of F₂ in 18.0 L of F₂ = 18.0 L/(22.4 L/mole) ≈ 0.804 moles

Therefore;

The number of moles of NaCl, in the reaction n = 2 × The number of moles of F₂ ≈ 2×0.804 moles = 1.608 moles

The number of moles of NaCl, in the reaction n ≈ 1.608 moles

The mass of NaCl in the reaction, m = n × MM

∴ m ≈ 1.608 moles × 58.44 g/mol ≈ 93.97 grams

The mass of NaCl that can react with the same volume of gas at STP ≈ 93.77 grams

8 0

3 years ago

You mix 285.0 mL of 1.20 M lead(II) nitrate with 300.0 mL of 1.60 M potassium iodide. The lead(II) iodide is insoluble. Which of

SIZIF [17.4K]

Answer:

D. The final concentration of NO3– is 0.821 M.

Explanation:

Considering:

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

Or,

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

Given :

For potassium iodide :

Molarity = 1.60 M

Volume = 300.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 300.0×10⁻³ L

Thus, moles of potassium iodide :

$Moles=1.60 \times {300.0\times 10^{-3}}\ moles$

Moles of potassium iodide = 0.48 moles

For lead(II) nitrate :

Molarity = 1.20 M

Volume = 285 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 285×10⁻³ L

Thus, moles of lead(II) nitrate :

$Moles=1.20\times {285\times 10^{-3}}\ moles$

Moles of lead(II) nitrate = 0.342 moles

According to the given reaction:

$2KI_{(aq)}+Pb(NO_3)_2_{(aq)}\rightarrow PbI_2_{(s)}+2KNO_3_{(aq)}$

2 moles of potassium iodide react with 1 mole of lead(II) nitrate

1 mole of potassium iodide react with 1/2 mole of lead(II) nitrate

0.48 moles potassium iodide react with 0.48/2 mole of lead(II) nitrate

Moles of lead(II) nitrate = 0.24 moles

Available moles of lead(II) nitrate = 0.342 moles

Limiting reagent is the one which is present in small amount. Thus, potassium iodide is limiting reagent.

Also, consumed lead(II) nitrate = 0.24 moles (lead ions precipitate with iodide ions)

Left over moles = 0.342 - 0.24 moles = 0.102 moles

Total volume = 300 + 285 mL = 585 mL = 0.585 L

So, Concentration = 0.102/0.585 M = 1.174 M

Statement A is correct.

The formation of the product is governed by the limiting reagent. So,

2 moles of potassium iodide gives 1 mole of lead(II) iodide

1 mole of potassium iodide gives 1/2 mole of lead(II) iodide

0.48 mole of potassium iodide gives 0.48/2 mole of lead(II) iodide

Mole of lead(II) iodide = 0.24 moles

Molar mass of lead(II) iodide = 461.01 g/mol

Mass of lead(II) chloride = Moles × Molar mass = 0.24 × 461.01 g = 111 g

Statement B is correct.

Potassium iodide is the limiting reagent. So all the potassium ion is with potassium nitrate . Thus,

2 moles of Potassium iodide on reaction forms 2 moles of potassium ion

0.48 moles of Potassium iodide on reaction forms 0.48 moles of potassium ion

Total volume = 300 + 285 mL = 585 mL = 0.585 L

So, Concentration = 0.48/0.585 M = 0.821 M

Statement C is correct.

Nitrate ions are furnished by lead(II) nitrate . So,

1 mole of lead(II) nitrate produces 2 moles of nitrate ions

0.342 mole of lead(II) nitrate produces 2*0.342 moles of nitrate ions

Moles of nitrate ions = 0.684 moles

So, Concentration = 0.684/0.585 M = 1.169 M

Statement D is incorrect.

4 0

3 years ago

If 185 g of KBr are dissolved in 1.2 kg of water, what would be the expected change in boiling point? The boiling point constant

Fudgin [204]

Answer:

ΔTb = 0.66 C

Explanation:

Given

Mass of KBr = 185 g

Mass of water = 1.2 kg

Kb = 0.51 C/m

Explanation:

The change in boiling point (ΔTb) is given by the product of molality (m) of the solution and the boiling point constant (Kb)

$\Delta T_{b}= K_{b}* m$

$Molality = \frac{moles\ KBr}{Kg\ water} \\\\moles KBr = \frac{mass\ KBr}{Mol.wt\ KBr} = \frac{185}{119} = 1.555\\\\Molality (m) = \frac{1.555 }{1.2} =1.296 m\\$

[tex]\Delta T_{b}= 0.51 C/m * 1.296 m = 0.66 C[\tex]

6 0

3 years ago

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