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3 years ago

Please I need help on these 3 questions. Thank You.

Chemistry

1 answer:

Liula [17]3 years ago

6 0

V = 200 mL (volume)

c = 3 M = 3 mol/L (concentration)

First we convert mL to L:

200 mL = 0.2 L

Then we calculate the moles using the formula: n = V × c = 0.2 L × 3 mol = 0.6 mol

Finally, we just use the molar mass of CaF2 to calculate the actual mass:

molar mass = 78 g/mol

The formula is: m = n × mm (mass = moles × molar mass)

m = 0.6 mol × 78 g/mol = 46.8 g

For this question the steps are exactly like the first question.

V = 50mL = 0.05 L

c = 12 M = 12 mol/L

n = V × c = 0.05 L × 12 mol/L = 0.6 mol

molar mass (HCl) = 36.5 g/mol

m = n × mm = 0.6 mol × 36.5 g/mol = 21.9 g.

The steps for this question are the opposite way.

m(K2CO3) = 250 g

molar mass = 138 g/mol

n = m ÷ mm = 1.81 mol

c = 2 mol/L

V = n ÷ c = 1.81 mol ÷ 2 mol/L = 0.905 L = 905 mL

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Can you look at the picture Look at the picture ASAP and help please?

Murrr4er [49]

Answer:

Volume of the reaction vessel is increased - shift to the left

The reaction is cooled down - shift to the right

H2 is added to the system - shift to the right

The pressure of the system is decreased - shift to the left

A catalyst is added to the system - no change

Water is removed from the system - shift to the right

Explanation:

When a constraint such as a change in temperature, pressure or volume is imposed on a reaction system in equilibrium, the equilibrium position will shift in such a way as to annul the constraint.

When the volume of a reaction system is increased, the equilibrium position shifts in the direction in which there is the highest total volume. This is the left hand side.

Since the reaction is exothermic (heat is given out) when the reaction is cooled down, the forward reaction is favoured.

Adding of reactants shifts the equilibrium position to the right hand side hence when H2 is added, the equilibrium position shifts to the right.

Decreasing the pressure shifts the equilibrium position to the direction of higher total volume hence the equilibrium shifts to the left when pressure is decreased.

A catalyst has no effect on the equilibrium position. It increases the rate of forward and reverse reaction to the same extent hence the equilibrium position is unaffected.

Removal of water from the system increases the rate of forward reaction since a product is being removed from the reaction system.

7 0

3 years ago

A power plant is driven by the combustion of a complex fossil fuel having the formula C11H7S. Assume the air supply is composed

AlekseyPX

(a) 4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2 + 20.68N_2;

(b) 4C_11 H_7S + 66O_2 → 44CO_2 + 14H_2O + 4SO_2 + 248.2N_2 + 11O_2;

(a) Balanced equation including N_2 from air

The balanced equation ignoring N_2 from air is

4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2

Moles of N_2 =55 mol O_2 × (3.76 mol N_2/1 mol O_2) = 206.8 mol N_2

Including N_2 from air, the balanced equation is

4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2 + 206.8N_2

(b) Balanced equation for 120 % stoichiometric combustion

Moles of O_2 = 55 mol O_2 × 1.20 = 66.00 mol O_2

Excess moles O_2 = (66.00 – 55) mol O_2 = 11.00 mol O_2

Moles of N_2 = 66.00 mol O_2 × (3.76 mol N_2/1 mol O_2) = 248.2 mol N_2

The balanced equation is

4C_11 H_7S + 66O_2 → 44CO_2 + 14H_2O + 4SO_2 + 248.2N_2 + 11O_2

Moles of O_2 required = 1700 kg C_11 H_7S

× (1 kmol C_11 H_7S/185.24 kg C_11 H_7S) × (55 kmol O_2/4 kmol C_11 H_7S)

= 126.2 kmol O_2

Mass of O_2 = 126.2 kmol O_2 × (32.00 kg O_2/1 kmol O_2) = 4038 kg O_2

Mass of N_2 required = 126.2 kmol O_2 × (3.76 kmol N_2/1 kmol O_2)

× (28.01 kg N_2/1 kmol N_2) = 13 285 kg N_2

Mass of air = Mass of N_2 + mass of O_2 = (4038 + 13 285) kg = 17 300 kg air

(d) Air:fuel mass ratio for 100 % combustion

Air:fuel = 17 300 kg/1700 kg = 10.2 :1

(e) Air:fuel mass ratio for 120 % combustion

Mass of air = 17 300 kg × 1.20 = 20 760 kg air

Air:fuel = 20 760 kg/1700 kg = 12.2 :1

6 0

2 years ago

In an experiment, 3.25 g of NH3 are allowed to react with 3.50 g of O2. 4NH3 + 5O2 > 4NO + 6H2O a. Which reactant is the limi

cluponka [151]

Answer:

2.61 g of NO will be formed

The limiting reagent is the O₂

Explanation:

The reaction is:

4NH₃ + 5O₂ → 4NO + 6H₂O

We convert the mass of the reactants to moles:

3.25g / 17 g/mol = 0.191 moles of NH₃

3.50g / 32 g/mol =0.109 moles of O₂

Let's determine the limiting reactant by stoichiometry:

4 moles of ammonia react with 5 moles of oxygen

Then, 0.191 moles of ammonia will react with (0.191 . 5) / 4 = 0.238 moles of oxygen. We only have 0.109 moles of O₂ and we need 0.238, so as the oxygen is not enough, this is the limiting reagent

Ratio with NO is 5:4

5 moles of oxygen produce 4 moles of NO

0.109 moles will produce (0.109 . 4)/ 5 = 0.0872 moles of NO

We convert the moles to mass, to get the answer

0.0872 mol . 30g / 1 mol = 2.61 g

7 0

2 years ago

. Write the following isotope in nuclide notation (e.g., “ ”): copper-70

vekshin1

Answer:

${\boxed{\text{$_{29}^{70}${Cu}}}$

Explanation:

The atomic number (Z) of copper is 29 and this isotope has an atomic mass (A) of 70.

The general symbol for an isotope E is $_{Z}^{A}\text{E}$ .

The atomic number is a left subscript, and the atomic mass is a left superscript.

$\rm {\text{The nuclide notation for copper-70 is }}{\boxed{\textbf{$_{29}^{70}${Cu}}}$

4 0

3 years ago

How many moles of Hions are present in the following aqueous solution?

Fittoniya [83]

Answer:

Nitric acid is a strong acid meaning it completely dissociates in water. Therefore, you can say that the concentration of H+ ions in solution is the same as the concentration of the acid (molar concentration). So once you have that, you can use the pH formula to find the pH of the acid.

pH = -log[H+]

So if you have .15 M HNO3 the pH is 0.82.

But, if you have a different concentration, the pH will change.

Explanation:

6 0

3 years ago

Please I need help on these 3 questions. Thank You.​

Please I need help on these 3 questions. Thank You.