Answer:
Mass of solute = 0.0036 g
Explanation:
Given data:
Concentration of Cl⁻ = 15.0 ppm
Volume of water = 240 mL
Mass of Cl⁻ present = ?
Solution:
1 mL = 1 g
240 mL = 240 g
Formula:
ppm = mass of solute / mass of sample ×1,000,000
by putting values,
15.0 ppm = (mass of solute / 240 g) ×1,000,000
Mass of solute = 15.0 ppm × 240 g / 1,000,000
Mass of solute = 0.0036 g
Answer:
58.9g of SO2 is produced
8g of oxygen remains unconsumed
Explanation:
The burning of Carbon disulfide (CS2) in oxygen. gives the reaction:
CS2 (g) + 3O2 (g) → CO2 (g) + 2SO2 (g)
Molar mass of CS2 = 76.139 g/mol
Molar mass of O2 = 15.99 g/mol
Molar mass of SO2 = 64.066 g/mol
Number of moles of CS2 = 35g/ 76.139 g/mol =0.46 moles
Number of moles of O2 = 30g/15.999 g/mol =1.88 moles
From the chemical reaction
1 mole of CS2 react with 3 moles of O2 to give 2 moles of SO2
Thus 0.46 moles of CS2 reacts to form 2× 0.46 = 0.92 moles of SO2
Mass of SO2 produced = 0.92×64.07 = 58.9g of SO2 is produced
thus 0.46 moles of CS2 reacts with 3 × 0.46 moles of O2 which is =1.38 moles of O2
Thus oxygen is the limiting reactant with 1.88 - 1.38 = 0.496~~0.5 mole remaining
Or 8g of oxygen
58.9g of SO2 is produced
oxygen is the limiting
Answer:
0.80 seconds (2 significant figures)
Explanation:
The equation of the reaction is given as;
CICH2CH2Cl (g) --> CH2CHCI (g) + HCl(g)
Rate constant (k) = 2.01 s^-1
From the units of the rate constant, this is a first order reaction.
Initial Concentration = 1.34 M
t = ?
Final concentration = 20% of 1.34 = 0.268 M
The integrated rate law for a first order reaction is given as;
ln[A] = ln[A]o - kt
ln(0.268) = ln(1.34) - 2.01(t)
-2.01(t) = - 1.6094
t = 0.8007 ≈ 0.80 seconds (2 significant figures)
NO is the limiting reagent and 4.34 g is the amount of the excess reagent that remains after the reaction is complete
<h3>What is a limiting reagent?</h3>
The reactant that is entirely used up in a reaction is called as limiting reagent.
The reaction:
→ 
Moles of nitrogen monoxide
Molecular weight:
=30g/mol



Moles of hydrogen
Molecular weight:
=30g/mol



Hydrogen gas is in excess.
NO is the limiting reagent.
The amount of the excess reagent remains after the reaction is complete.
(2.9 mol- 0.73 mol NO x
) x 
4.34 g
Learn more about limiting reagents here:
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