<span>Answer:
(16.2 g C2H6O2) / (62.0678 g C2H6O2/mol) / (0.0982 kg) = 3.9704 mol/kg = 3.9704 m
a.)
(3.9704 m) x (1.86 °C/m) = 7.38 °C change
0.00°C - 7.38 °C = - 7.38 °C
b.)
(3.9704 m) x (0.512 °C/m) = 2.03 °C change
100.00°C + 2.03 °C = 102.03 °C</span>
Just search up how much the weight is equal to another, then multiply it, thats what i do lol sorry if im no help
Answer:
Explanation:
Given parameters:
pH = 3.50
Unknown:
concentration of [H₃0⁺] = ?
concentration of [OH⁻] = ?
Solution:
In order to find the unknown, we use some simple expressions which best explains the pH scale and the equilibrium systems of aqueous solutions.
pH = -log₁₀[H₃O⁺]
[H₃O⁺] = inverse log₁₀ (-pH) =
= 
[H₃O⁺] = 3.2 x 10⁻⁴moldm⁻³
For the [OH⁻]:
we use : pOH = -log₁₀ [OH⁻]
Recall: pOH + pH = 14
pOH = 14 - pH = 14 - 3.5 = 10.5
Now we plug the value of pOH into pOH = -log₁₀ [OH⁻]
[OH⁻] = 
[OH⁻] =
= 3.2 x 10⁻¹¹moldm⁻³
The solution is acidic as the concentration of H₃0⁺ is more than that of the OH⁻ ions.
The balanced chemical reaction would be as follows:
<span>5P4O6 +8I2 ---> 4P2I4 +3P4O10
We are given the amount of reactants used for the reaction. We first need to determine the limiting reactant from the given amounts. We do as follows:
8.80 g P4O6 (1 mol / </span><span>219.88 g) = 0.04 mol P4O6
12.37 g I2 ( 1 mol / </span><span>253.809 g ) = 0.05 mol I2
Therefore, the limiting reactant is iodine since less it is being consumed completely in the reaction. We calculate the amount of P2I4 prepared as follows:
0.05 mol I2 ( 4 mol P2I4 / 8 mol I2 ) (</span><span>569.57 g / 1 mol) = 14.24 g P2I4</span>