The powder is made through a chemical reaction (saponification) between an alkaline base with oils is to be expected that there will always be a residual alkali. PH around <span>11.5
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Answer: Option A) 83.9g
Explanation:
KCl is the chemical formula of potassium chloride.
Given that,
Amount of moles of KCl (n) = ?
Volume of KCl solution (v) = 0.75L
Concentration of KCl solution (c) = 1.5M
Since concentration (c) is obtained by dividing the amount of solute dissolved by the volume of solvent, hence
c = n / v
make n the subject formula
n = c x v
n = 1.5M x 0.75L
n = 1.125 mole
Now given that,
Amount of moles of KCl (n) = 1.125
Mass of KCl in grams = ?
For molar mass of KCl, use the molar masses of:
Potassium, K = 39g;
Chlorine,Cl = 35.5g
KCl = (39g + 35.5g)
= 74.5g/mol
Since, amount of moles = mass in grams / molar mass
1.125 mole = m / 74.5g/mol
m = 1.125 mole x 74.5g/mol
m = 83.81g
Thus, 83.9 grams of KCl are needed to prepare 0.750 L of a 1.50 M solution in water
Answer:
Explanation:
There is a formula for this:
M = DRT/P where M = molar mass. This just derived from PV = nRT where you say n = grams/molar mass. However, just with this formula, we can get D which is density at STP (1 atm and 273K). We find that D = 6.52g/L.
Molar mass NaHCO₃ = 23 + 1 + 12 + 16 x 3 = 84 g/mol
1 mole ---------- 84 g
? mole ---------- 110 g
moles NaHCO₃ = 110 . 1 / 84
moles NaHCO₃ = 110 / 84
= 1.309 moles
hope this helps!
Answer:
Option (1) Br– is the catalyst, and the reaction follows a faster pathway with Br– than without
Explanation:
Let us consider the equation below:
Step 1:
H2O2(aq) + Br–(aq) → H2O(l) + BrO–(aq)
Step 2:
BrO–(aq) + H2O2(aq) → H2O(l) + O2(g) + Br–(aq)
From the above equation, we can see that Br– is unchanged.
This implies that Br– is the catalyst as catalyst does not take part in a chemical reaction but they create an alternate pathway to lower the activation energy in order for the reaction to proceed at a much faster rate to arrive at the products.
