First, we need to get the molar mass of:
KClO3 = 39.1 + 35.5 + 3*16 = 122.6 g/mol
KCl =39.1 + 35.5 = 74.6 g/mol
O2 = 16*2 = 32 g/mol
From the given equation we can see that:
every 2 moles of KClO3 gives 3 moles of O2
when mass = moles * molar mass
∴ the mass of KClO3 = (2mol of KClO3*122.6g/mol) = 245.2 g
and the mass of O2 then = 3 mol * 32g/mol = 96 g
so, 245.2 g of KClO3 gives 96 g of O2
A) 2.72 g of KClO3:
when 245.2 KClO3 gives → 96 g O2
2.72 g KClO3 gives → X
X = 2.72 g KClO3 * 96 g O2/245.2 KClO3
= 1.06 g of O2
B) 0.361 g KClO3:
when 245.2 g KClO3 gives → 96 g O2
0.361 g KClO3 gives → X
∴ X = 0.361g KClO3 * 96 g / 245.2 g
= 0.141 g of O2
C) 83.6 Kg KClO3:
when 245.2 g KClO3 gives → 96 g O2
83.6 Kg KClO3 gives → X
∴X = 83.6 Kg* 96 g O2 /245.2 g KClO3
= 32.7 Kg of O2
D) 22.4 mg of KClO3:
when 245.2 g KClO3 gives → 96 g O2
22.4 mg KClO3 gives → X
∴X = 22.4 mg * 96 g O2 / 245.2 g KClO3
= 8.8 mg of O2
Answer:
8.1433 g of XeF₆ are required.
Explanation:
Balanced chemical equation;
XeF₆ (s) + 3H₂ (g) → Xe (g) + 6HF (g)
Given data:
Volume of hydrogen = 0.579 L
Pressure = 4.46 atm
Temperature = 45 °C (45+273= 318 k)
Solution:
First of all we will calculate the moles of hydrogen
PV = nRT
n = PV/ RT
n = 4.46 atm × 0.579 L / 0.0821 atm. dm³. mol⁻¹. K⁻¹ × 318 K
n = 2.6 atm . L / 26.12 atm. dm³. mol⁻¹
n = 0.0995 mol
Mass of hydrogen:
Mass = moles × molar mass
Mass = 0.0995 mol × 2.016 g/mol
Mass = 0.2006 g
Now we will compare the moles of hydrogen with XeF₆ from balance chemical equation.
H₂ : XeF₆
3 : 1
0.0995 : 1/3× 0.0995 = 0.0332 mol
Now we will calculate the mass of XeF₆.
Mass = moles × molar mass
Mass = 0.0332 mol × 245.28 g/mol
Mass = 8.1433 g
A less intense wave will have fewer OSCILLATING AMPLITUDE than a more intense wave.
The intensity of a wave is the power transferred per unit area, where the area is measured on the plane perpendicular to the direction of propagation of the energy. Intense sounds are characterized by the particles of the medium vibrating back and forth with large amplitude.<span />
Answer:
Mass of Sodium = 574.75 g
Mass of Chlorine = 886.25 g
Explanation:
The balance chemical equation for the synthesis of NaCl is,
2 Na + Cl₂ → 2 NaCl
Step 1: <u>Find out moles of each reactant required,</u>
According to balance chemical equation,
2 moles of NaCl is produced by = 2 moles of Na
So,
25 moles of NaCl will be produced by = X moles of Na
Solving for X,
X = 25 mol × 2 mol / 2 mol
X = 25 moles of Na
Similarly for Cl₂,
According to balance chemical equation,
2 moles of NaCl is produced by = 1 mole of Cl₂
So,
25 moles of NaCl will be produced by = X moles of Cl₂
Solving for X,
X = 25 mol × 1 mol / 2 mol
X = 12.5 moles of Cl₂
Step 2: <u>Convert each moles to mass as;</u>
Mass = Moles × Atomic Mass
For Na,
Mass = 25 mol × 22.99 g/mol
Mass = 574.75 g
For Cl₂,
Mass = 12.5 mol × 70.90 g/mol
Mass = 886.25 g
