Question

4. A typical silicon chip such as these in electronie calculators weighs 21 10 in such a chip? How many silicone What mass of calcium chloride must be dissolved in water to prepare 200 ml of a solution with concentration of 0.50 M 6. What is the coefficient for exygen when this equation is balanced! H.SC) _0x) SOC) - HOM (A) 2 (B) 3 (C)4 (D) 5 7. Ammonia is prepared by the Haber Process NT HE NH. If 2.0 mol of N, and 3.0 mol of its are combined other, what is the percent of yield if 15 mol of NH, is obtained 8. What would be the volume of oxygen gas produced at 25°C and 10 st from the electrolysis of 20 kg of 2H00) - H.() 0 ()

please do each problem except for #6 and explain how you got them. dumb it down please.

Answer 1

Answer 4-

Given,

Mass of Slicon chip = 2.3 * 104 g [It is not visible so, i am taking 104]

Molar Mass of Silicon = 28.0855 u

Number of Atoms in Silicon = ?

We know that,

Moles = mass / Molar Mass

So, moles of Silicon = 2.3 * 104 g / 28.0855 u

moles of Silicon = 818.93 mol

Now,

1 mol = 6.022 * 1023 atoms

So, atoms in silicon chip = 818.93 * 6.022 * 1023 atoms

atoms in silicon chip = 4.9 * 1026 atoms [ANSWER]

Answer 5 -

Given,

Volume of solution = 200 mL or 0.2 L

Molarity of solution = 0.50 M

Molar Mass of Calcium Chloride = 110.98 g/mol

Mass of Calcium Chloride needed = ?

We know that,

Molarity = Moles / Volume in L

So, moles = Molarity * Volume in L

moles = 0.50 M * 0.2 L

moles = 0.1 mol

Also,

Moles = mass / Molar mass

Mass = Moles * Molar Mass

Mass of Calcium Chloride = 0.1 mol * 110.98 g/mol

Mass of Calcium Chloride = 11.1 g [ANSWER]

Answer 7 -

Given,

Moles of N2 = 2.0 mol

Moles of H2 = 3.0 mol

Actual Yield = 1.5 mol

Percentage Yield = ?

N2 + 3 H2 $\rightleftharpoons$ 2 NH3

First, we have to find the Limiting reagent.

For this, divide the available moles by their stiochiometric coefficients, the one which is smaller is the Limiting Reagent.

For N2 = 2.0 mol/1 = 2.0 mol

For H2 = 3.0 mol /3 = 1.0 mol

So, H2, is the Limiting Reagent.

Now,

Using Stiochiometry, it can be analyzed that for 3 moles of H2, 2 moles of NH3 are produced.

i.e.

Moles of NH3 produced = 2/3 * moles of H2

Moles of NH3 produced = 2/3 * 3.0 mol

Moles of NH3 produced = 2 mol (THEORETICAL)

Now,

Percentage Yield = (Actual Yield/Theoretical Yiled)*100

Percentage Yield = (1.5 mol/2 mol)*100

Percentage Yield = 75 % [ANSWER]

Answer 8 -

Given,

Temperature = 25 C or 298.15 K

Pressure = 1.0 atm

Mass of water = 2 kg or 2000 g

Molar Mass of Water = 18.01 g/mol

2 H2O (l) $\rightarrow$ 2 H2 (g) + O2 (g)

We know that,

Moles = Mass / Molar Mass

So, Moles of water = 2000g/18.01 g/mol

Moles of water = 111.05 mol

Using Stiochiometry, it can be analyzed that, for 2 moles of water, 1 moles of Oxygen gas is produced.

i.e.

Mole os O2 produced = 1/2 * moles of H2O

So,

Mole os O2 produced = 55.52 mol

Also,

PV = nRT

where, P = Pressure

V = Volume

n = moles

R = Gas constant (0.08205 L⋅atm⋅K−1⋅mol−1)

T = Temperature

Put the values,

1.0atm * V = 55.52 mol * 0.08205 L⋅atm⋅K−1⋅mol−1 * 298.15 K

V = 1358.2 L [ANSWER]