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vovikov84 [41]
3 years ago
14

For the following reaction,

Chemistry
1 answer:
jeka943 years ago
5 0

Answer:

The answer to your question is:

1.- CO

2.- 0.414 moles of CO2

Explanation:

Data

                               2CO   + O2      ⇒    2CO2

CO = 0.414 moles

O2 = 0.418

 

Process

theoretical ratio   CO/O2 = 2/1 = 1

experimental ratio  CO/O2 = 0.414/0.418 = 0.99

Then the limiting reactant is CO

2.-

                    2 moles of CO ---------------  2 moles of CO2

                    0.414 moles of CO ---------  x

                   x = (0.414 x 2) / 2

                   x = 0.414 moles of CO2

                   

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Brass is a substitutional alloy consisting of a solution of copper and zinc. A particular sample of red brass consisting of 79.0
Bas_tet [7]

<u>Answer:</u> The molality and molarity of zinc in the solution is 4.07 m and 28.11 M respectively.

<u>Explanation:</u>

Solute is the substance which is present in smaller proportion in a mixture and solvent is the substance which is present in larger proportion in a mixture.

We are given:

(m/m) % of Cu = 79 %

This means that 79 g of copper is present in 100 grams of alloy.

(m/m) % of Zn = 21 %

This means that 21 g of zinc is present in 100 grams of alloy.

As, zinc is present in smaller proportion. So, it is solute and copper is the solvent.

  • <u>Calculating molality of zinc:</u>

To calculate molality of the zinc, we use the equation:

Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}

where,

m_{solute} = Given mass of solute (Zinc ) = 21 g

M_{solute} = Molar mass of solute (Zinc) = 65.3 g/mol

W_{solvent} = Mass of solvent (copper) = 79 g

Putting values in above equation, we get:

\text{Molality of Zinc}=\frac{21\times 1000}{65.3\times 79}\\\\\text{Molality of zinc}=4.07m

  • <u>Calculating molarity of zinc:</u>

To calculate volume of a substance, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of solution = 8740kg/m^3=\frac{8740kg\frac{1000g}{1kg}}{1m^3\times \frac{10^6cm^3}{1m^3}}=\frac{8740g}{1000cm^3}=8.740g/cm^3

(Conversion factors used are:  1 kg = 1000 g &  1m^3=10^6cm^3  )

Mass of solution = 100 g

Putting values in above equation, we get:

8.740g/cm^3=\frac{100.0g}{\text{Volume of zinc}}\\\\\text{Volume of zinc}=11.44cm^3

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

We are given:

Molar mass of zinc = 65.3 g/mol

Volume of solution = 11.44cm^3=11.44mL     (Conversion factor:  1cm^3=1mL  )

Mass of zinc = 21.0 g

Putting values in above equation, we get:

\text{Molarity of zinc}=\frac{21\times 1000}{65.3\times 11.44}=28.11M

Hence, the molality and molarity of zinc in the solution is 4.07 m and 28.11 M respectively.

4 0
3 years ago
Imagine you had HCl with a concentration of exactly 0.10 mol/dm3. If 0.023 dm3 of a sodium hydroxide solution, NaOH (aq), could
polet [3.4K]

Answer:

Explanation:

Step 1: Calculate the amount of sodium hydroxide in moles

Volume of sodium hydroxide solution = 25.0 ÷ 1,000 = 0.0250 dm3

Rearrange:

Concentration in mol/dm3 =

Amount of solutein mol = concentration in mol/dm3 × volume in dm3

Amount of sodium hydroxide = 0.100 × 0.0250

= 0.00250 mol

Step 2: Find the amount of hydrochloric acid in moles

The balanced equation is: NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)

So the mole ratio NaOH:HCl is 1:1

Therefore 0.00250 mol of NaOH reacts with 0.00250 mol of HCl

Step 3: Calculate the concentration of hydrochloric acid in mol/dm3

Volume of hydrochloric acid = 20.00 ÷ 1000 = 0.0200 dm3

Concentration in mol/dm3 =

Concentration in mol/dm3 =

= 0.125 mol/dm3

Step 4: Calculate the concentration of hydrochloric acid in g/dm3

Relative formula mass of HCl = 1 + 35.5 = 36.5

Mass = relative formula mass × amount

Mass of HCl = 36.5 × 0.125

= 4.56 g

So concentration = 4.56 g/dm3

8 0
2 years ago
(1) The solubility of Salt AB2(S) IS 5mol/dm^3.
Svetlanka [38]

Answer:

a. Ksp = 4s³

b. 5.53 × 10⁴ mol³/dm⁹

Explanation:

a. Obtain an expression for the solubility product of AB2(S),in terms of s.

AB₂ dissociates to give

AB₂ ⇄ A²⁺ + 2B⁻

Since 1 mole of AB₂ gives 1 mole of A and 2 moles of B, we have the mole ratio as

AB₂ ⇄ A²⁺ + 2B⁻

1 : 1 : 2

Since the solubility of AB₂ is s, then the solubility of A is s and that of B is 2s

So, we have

AB₂ ⇄ A²⁺ + 2B⁻

[s]        [s]    [2s]

So, the solubility product Ksp = [A²⁺][B⁻]²

= (s)(2s)²

= s(4s²)

= 4s³

b. Calculate the Ksp of AB₂, given that solubility is 2.4 × 10³ mol/dm³

Given that the solubility of AB is 2.4 × 10³ mol/dm³ and the solubility product Ksp = [A²⁺][B⁻]² = 4s³ where s = solubility of AB = 2.4 × 10³ mol/dm³

Substituting the value of s into the equation, we have

Ksp = 4s³

= 4(2.4 × 10³ mol/dm³)³

= 4(13.824 × 10³ mol³/dm⁹)

= 55.296 × 10³ mol³/dm⁹

= 5.5296 × 10⁴ mol³/dm⁹

≅ 5.53 × 10⁴ mol³/dm⁹

Ksp = 5.53 × 10⁴ mol³/dm⁹

6 0
3 years ago
Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid. HCl(aq), as described by
melisa1 [442]

Answer:

0.88 g

Explanation:

Using ideal gas equation to calculate the moles of chlorine gas produced as:-

PV=nRT

where,

P = pressure of the gas = 805 Torr

V = Volume of the gas = 235 mL = 0.235 L

T = Temperature of the gas = 25^oC=[25+273]K=298K

R = Gas constant = 62.3637\text{torr}mol^{-1}K^{-1}

n = number of moles of chlorine gas = ?

Putting values in above equation, we get:

805torr\times 0.235L=n\times 62.3637\text{ torrHg }mol^{-1}K^{-1}\times 298K\\\\n=\frac{805\times 0.235}{62.3637\times 298}=0.01017\ mol

According to the reaction:-

MnO_2+4HCl\rightarrow MnCl_2+2H_2O+Cl_2

1 mole of chlorine gas is produced when 1 mole of manganese dioxide undergoes reaction.

So,

0.01017 mole of chlorine gas is produced when 0.01017 mole of manganese dioxide undergoes reaction.

Moles of MnO_2 = 0.01017 moles

Molar mass of MnO_2 = 86.93685 g/mol

So,

Mass=Moles\times Molar\ mass

Applying values, we get that:-

Mass=0.01017moles \times 86.93685\ g/mol=0.88\ g

<u>0.88 g of MnO_2(s) should be added to excess HCl (aq) to obtain 235 mL of Cl_2(g) at 25 degrees C and 805 Torr.</u>

6 0
3 years ago
Which process can be used to power a home? solar thermal energy neither solar thermal energy nor solar electric energy both sola
faltersainse [42]

Answer:

solar electric energy

Explanation:

5 0
2 years ago
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