Answer:
It's true :) Hope that helps
Answer:
Chloroform= limiting reactant
0.209mol of CCl4 is formed
And 32.186g of CCl4 is formed
Explanation:
The equation of reaction
CHCl3 + Cl2= CCl4 + HCl
From the equation 1 mol of
CHCl3 reacts with 1mol Cl2 to yield 1mol of CCl4
From the question
25g of CHCl3 really with Cl2
Molar mass of CHCl3= 119.5
Molar mass of Cl2 = 71
Hence moles of CHCl3= 25/119.5 = 0.209mol
Moles of Cl2 = 25/71 = 0.352mol
Hence CHCl3 is the limiting reactant
Since 1 mole of CHCl3 gave 1mol of CCl4
It implies that 0.209moles of CHCl3 will also give 0.209mol of CCl4
Mass of CCl4 formed = moles× molar mass= 0.209×154= 32.186g
pH=2.7
<h3>Further explanation</h3>
Acetic acid = weak acid
![\tt [H^+]=\sqrt{Ka.M}](https://tex.z-dn.net/?f=%5Ctt%20%5BH%5E%2B%5D%3D%5Csqrt%7BKa.M%7D)
Ka = acid ionization constant
M = molarity
Ka for Acetic acid(CH₃COOH) : 1.8 x 10⁻⁵
![\tt [H^+]=\sqrt{1.8\times 10^{-5}\times 0.222}\\\\=0.001998=1.998\times 10^{-3}](https://tex.z-dn.net/?f=%5Ctt%20%5BH%5E%2B%5D%3D%5Csqrt%7B1.8%5Ctimes%2010%5E%7B-5%7D%5Ctimes%200.222%7D%5C%5C%5C%5C%3D0.001998%3D1.998%5Ctimes%2010%5E%7B-3%7D)

Answer:
14.7°C
Explanation:
Q = m·ΔT·c
ΔT = 
ΔT =
= 1320 J / ((230 g) * (.39 J/g°C)
ΔT = 14.7 °C
0.1 moles of chloride ions were involved in the reaction if 25.0 mL of a 2.00 M CaCl2 solution is used for the reaction.
Explanation:
Data given:
volume of Ca
= 25 ml 0r 0.025
molarity of the calcium chloride solution = 2M
number of chloride ions =?
Balance chemical reaction:
Ca + 2
⇒
number of moles in 25 ml is calculated as:
molarity = 
number of moles of calcium chloride = molarity x volume
putting the values in the equation:
number of moles = 2 x 0.025
= 0.05 moles of calcium chloride
1 mole of Ca
decomposes as 1 calcium ion and 2 chloride ions
so 0.05 moles will have x moles of chloride ion
= 
x= 0.1
0.1 moles of chloride ions will be involved in the reaction.