Question #1
Potasium hydroxide (known)
volume used is 25 ml
Molarity (concentration) = 0.150 M
Moles of KOH used
0.150 × 25/1000 = 0.00375 moles
Sulfuric acid (H2SO4)
volume used = 15.0 ml
unknown concentration
The equation for the reaction is
2KOH (aq)+ H2SO4(aq) = K2SO4(aq) + 2H2O(l)
Thus, the Mole ratio of KOH to H2SO4 is 2:1
Therefore, moles of H2SO4 used will be;
0.00375 × 1/2 = 0.001875 moles
Acid (sulfuric acid) concentration
0.001875 moles × 1000/15
= 0.125 M
Question #2
Hydrogen bromide (acid)
Volume used = 30 ml
Concentration is 0.250 M
Moles of HBr used;
0.25 × 30/1000
= 0.0075 moles
Sodium Hydroxide (base)
Volume used 20 ml
Concentration (unknown)
The equation for the reaction is
NaOH + HBr = NaBr + H2O
The mole ratio of NaOH : HBr is 1 : 1
Therefore, moles of NaOH used;
= 0.0075 moles
NaOH concentration will be
= 0.0075 moles × 1000/20
= 0.375 M
The last answer is correct
We use the gas law named Charle's law for the calculation of the second temperature. The law states that,
V₁T₂ = V₂T₁
Substituting the known values,
(0.456 L)(65 + 273.15) = (3.4 L)(T₁)
T₁ = 45.33 K
I will, I am in a chemistry class right now. What’s up?
The rate of a reaction would be one-fourth.
<h3>Further explanation</h3>
Given
Rate law-r₁ = k [NO]²[H2]
Required
The rate of a reaction
Solution
The reaction rate (v) shows the change in the concentration of the substance (changes in addition to concentrations for reaction products or changes in concentration reduction for reactants) per unit time.
Can be formulated:
Reaction: aA ---> bB
or
The concentration of NO were halved, so the rate :
![\tt r_2=k[\dfrac{1}{2}No]^2[H_2]\\\\r_2=\dfrac{1}{4}k.[No]^2[H_2]\\\\r_2=\dfrac{1}{4}r_1](https://tex.z-dn.net/?f=%5Ctt%20r_2%3Dk%5B%5Cdfrac%7B1%7D%7B2%7DNo%5D%5E2%5BH_2%5D%5C%5C%5C%5Cr_2%3D%5Cdfrac%7B1%7D%7B4%7Dk.%5BNo%5D%5E2%5BH_2%5D%5C%5C%5C%5Cr_2%3D%5Cdfrac%7B1%7D%7B4%7Dr_1)
