When the Heat gain or lose = the mass * specific heat * ΔT
and when we have the mass of gold coin= 40 g
and the specific Heat of gold= 0.13 J/g°
and ΔT = (Tf- Ti) = 10°C - 50°C = -40 °C
so by substitution:
∴Heat H = 40 g * 0.13 J/g° * -40
= - 208 J
Answer:
5.01%
Explanation:
Density of vinegar = mass/volume
Mass of 10.00 mL = density x volume
= 1.006 x 10 = 10.06 g
From the equation of reaction:
1 mole pf CH3COOH requires 1 mole of NaOH for neutralization.
mole of NaOH = molarity x volume
= 0.5062 x 0.01658
= 0.008392796 mole
0.008392796 mole of NaOH will therefore require 0.008392796 mole of CH3COOH.
mass of CH3COOH = mole x molar mass
= 0.008392796 x 60.052
= 0.504 g
Percentage by mass of acetic acid in the vinegar = 0.504/10.06 x 100%
= 5.01%
The percent by mass of acetic acid in the vinegar is 5.01%
Answer:
The percentage of N in the compound is 0.5088
Explanation:
Mass of compound = 8.75 mg = 8.75×1000 = 8750 g
Mass of N2 = number of moles of N2 × MW of N2 = 1.59 × 28 = 44.52 g
% of N in the compound = (mass of N2/mass of compound) × 100 = (44.52/8750) × 100 = 5.088×10^-3 × 100 = 0.5088
He used that name because if you lit hydrogen on fire it would catch. therefore, it was dangerous to be around. think of it as a kind of warning. :) hope that helps.
Answer: option C) II < III < I
i.e [OH−] < [H3O+] < I
Explanation:
First, obtain the pH value of I and II, then compare both with III.
For I
Recall that pH = -log (H+)
So pH3O = -log (H3O+)
= - log (1x10−5)
= 4
For II
pOH = - log(OH-)
= - log(1x10−10)
= 9
For III
pH = 6
Since, pH range from 1 to 14, with values below 7 to be acidic, 7 to be neutral, above 7 to be alkaline: then, 9 < 6 < 4
Thus, the following solutions from least acidic to most acidic is II < III < I
