Answer:
10.7 g of KOH
Explanation:
First of all, we determine the reaction:
2K (s) + 2H₂O(l) → H₂(g) + 2KOH(aq)
We convert the mass of K, to moles → 7.5 g . mol/39.1 g = 0.192 moles
Ratio is 2:2, so the moles I have of K must produce the same moles of KOH. In this case, the produces moles of KOH are 0.192 moles.
We convert the moles to mass, to finish the answer:
0.192 mol . 56.1g /1mol = 10.7 g of KOH
your answer is B low-pressure systems that cause stormy weather
The reaction uses B) 9.0 g Br₂.
iron + bromine ⟶ product
2.0 g + <em>x</em> g ⟶ 11.0 g
According to the <em>Law of Conservation of Mass</em>, the total mass of the reactants must equal the total mass of the products.
∴2.0 + <em>x</em> = 11.0
<em>x</em> = 11.0 – 2.0 = 9.0
The reaction uses 9.0 g Br₂.
Answer:
15.4 g of sucrose
Explanation:
Formula to be applied for solving these question: colligative property of freezing point depression. → ΔT = Kf . m
ΔT = Freezing T° of pure solvent - Freezing T° of solution
Let's replace data given: 0°C - (-0.56°C) = 1.86 C/m . m
0.56°C / 1.86 m/°C = m → 0.301 mol/kg
m → molality (moles of solute in 1kg of solvent)
Our mass of solvent is not 1kg, it is 150 g. Let's convert it from g to kg, to determine the moles of solute: 150 g. 1kg/1000g = 0.150 kg
0.301 mol/kg . 0.150kg = 0.045 moles.
We determine the mass of sucrose, by the molar mass:
0.045 mol . 342 g/1mol = 15.4 g
Yes, that is true. in order for it to be a redox reaction, both oxidation and reduction must be occurring.
