Find the molality of a solution of 58.5 g NaCl dissolved in 0.556 kg of water

If you react 2.00 g of hydrogen completely using 15.87 g of oxygen to produce water, how much water (in grams) will you have?

leonid [27]

Answer:

The amount (mass) of water we will have is 17.869 grams

Explanation:

The molar mass of hydrogen gas H₂ = 2.016 grams/mole

The molar mass of oxygen gas = 31.999 g/mol

Therefore, 2.00 g of hydrogen will give;

2.00/2.016 = 0.9921 moles of H₂ gas and

15.87 g of O₂ will give;

15.87/31.999 = 0.49595 moles

The reaction is as follows;

2H₂ (g) + O₂ (g) → 2H₂O (l)

Two moles of H₂ react with one mole of O₂ to produce two moles of H₂O

Therefore 0.9921 moles of H₂ will react with 0.9921/2 or 0.49595 moles of O₂ to produce 0.9921 moles of H₂O

From the above we note that all the H₂ and O₂ are completely consumed to form 0.9921 moles of H₂O

Molar mass of H₂O = 18.01528 g/mol

Number of moles = Mass/(Molar mass)

∴ Mass of H₂O = (Molar mass) × (Number of moles)

= 18.01528 g/mol × 0.9921 moles = 17.869 grams

Therefore the amount (mass) of water we will have = 17.869 grams.

8 0

3 years ago

What is the systematic name of the following compound?

Ede4ka [16]

Answer: D. Potassium bromide

6 0

3 years ago

How is burning magnesium different than burning methane

olya-2409 [2.1K]

You can stop the burning of methane with water or carbon dioxide extinguishers but problems arise when you try to use this to stop the burning of the magnesium.

Explanation:

To burn magnesium (Mg) and methane (CH₄) you need to react them with oxygen:

2 Mg (s) + O₂ (g) → 2 MgO + heat

CH₄ (g) + 2 O₂ (g) → CO₂ (g) + 2 H₂O (g) + heat

However at that temperatures magnesium (Mg) is able to react with water (H₂O) and carbon dioxide (CO₂).

Mg (s) + 2 H₂O (l) → Mg(OH)₂ (s) + H₂ (g)

2 Mg (s) + CO₂ (g) → 2 MgO (s) + C (s)

So the safe option to stop the burning of the magnesium is to limit the oxygen in the air.

we have used the following notations:

(s) - solid

(g) - gas

(l) - liquid

Learn more about:

combustion reactions

brainly.com/question/13824679

#learnwithBrainly

6 0

3 years ago

1. The solubility of lead(II) chloride at some high temperature is 3.1 x 10-2 M. Find the Ksp of PbCl2 at this temperature.

solniwko [45]

Answer:

1) The solubility product of the lead(II) chloride is $1.2\times 10^{-4}$ .

2) The solubility of the aluminium hydroxide is $1.6\times 10^{-10} M$ .

3)The given statement is false.

Explanation:

1)

Solubility of lead chloride = $S=3.1\times 10^-2M$

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

S 2S

The solubility product of the lead(II) chloride = $K_{sp}$

$K_{sp}=[Pb^{2+}][Cl^-]^2$

$K_{sp}=S\times (2S)^2=4S^3=4\times (3.1\times 10^{-2})^3=1.2\times 10^{-4}$

The solubility product of the lead(II) chloride is $1.2\times 10^{-4}$ .

2)

Concentration of aluminium nitrate = 0.000010 M

Concentration of aluminum ion = $1\timed 0.000010 M=0.000010 M$

Solubility of aluminium hydroxide in aluminum nitrate solution = $S$

$Al(OH)_3(aq)\rightleftharpoons Al^{3+}(aq)+3OH^-(aq)$

S 3S

The solubility product of the aluminium nitrate = $K_{sp}=1.0\times 10^{-33}$

$K_{sp}=[Al^{3+}][OH^-]^3$

$1.0\times 10^{-33}=(0.000010+S)\times (3S)^3$

$S=1.6\times 10^{-10} M$

The solubility of the aluminium hydroxide is $1.6\times 10^{-10} M$ .

3.

$Molarity=\frac{Moles}{Volume (L)}$

Mass of NaCl= 3.5 mg = 0.0035 g

1 mg = 0.001 g

Moles of NaCl = $\frac{0.0035 g}{58.5 g/mol}=6.0\times 10^{-5} mol$

Volume of the solution = 0.250 L

$[NaCl]=\frac{6.0\times 10^{-5} mol}{0.250 L}=0.00024 M$

1 mole of NaCl gives 1 mole of sodium ion and 1 mole of chloride ions.

$[Cl^-]=[NaCl]=0.00024 M$

Moles of lead (II) nitrate = n

Volume of the solution = 0.250 L

Molarity lead(II) nitrate = 0.12 M

$n=0.12 M]\times 0.250 L=0.030 mol$

1 mole of lead nitrate gives 1 mole of lead (II) ion and 2 moles of nitrate ions.

$[Pb^{2+}]=[Pb(NO_2)_3]=0.030 M$

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

Solubility of lead(II) chloride = $K_{sp}=1.2\times 10^{-4}$

Ionic product of the lead chloride in solution :

$Q_i=[Pb^{2+}][Cl^-]^2=0.030 M\times (0.00024 M)^2=1.7\times 10^{-9}$

$Q_i$ ( no precipitation)

The given statement is false.

3 0

3 years ago

Which of these elements has the smallest atomic radius?

nadya68 [22]

Answer:

3. be

Explanation:

it shows it in the word search

7 0

3 years ago