Answer:
The answers are explained below
Explanation:
a)
Given: concentration of salt/base = 0.031
concentration of acid = 0.050
we have
PH = PK a + log[salt]/[acid] = 1.8 + log(0.031/0.050) = 1.59
b)
we have HSO₃⁻ + OH⁻ ------> SO₃²⁻ + H₂O
Moles i............0.05...................0.01.................0.031.....................0
Moles r...........-0.01.................-0.01................0.01........................0.01
moles f...........0.04....................0....................0.041.....................0.01
c)
we will use the first equation but substituting concentration of base as 0.031 + 10ml = 0.031 + 0.010 = 0.041
Hence, we have
PH = PK a + log[salt]/[acid] = 1.8 + log(0.041/0.050) = 1.71
d)
pOH = -log (0.01/0.510) = 1.71
pH = 14 - 1.71 = 12.29
e)
Because the buffer solution (NaHSO3-Na2SO3) can regulate pH changes. when a buffer is added to water, the first change that occurs is that the water pH becomes constant. Thus, acids or bases (alkali = bases) Additional may not have any effect on the water, as this always will stabilize immediately.
Molarity is defined as number of moles of solute in 1 L of solution.
Here, 0.1025 g of Cu is reacted with 35 mL of HNO_{3} to produced Cu^{2+} ions.
The balanced reaction will be as follows:
Cu+3HNO_{3}\rightarrow Cu(NO_{3})_{2}+NO_{2}+H_{2}O
From the above reaction, 1 mole of Cu produces 1 mole of Cu^{2+}, convert the mass of Cu into number of moles as follows:
n=\frac{m}{M}
molar mass of Cu is 63.55 g/mol thus,
n=\frac{0.1025 g}{63.55 g/mol}=0.0016 mol
Now, total molarity of solution, after addition of water is 200 mL or 0.2 L can be calculated as follows:
M=\frac{n}{V}=\frac{0.0016 mol}{0.2 L}=0.008 mol/L=0.008 M
Thus, molarity of Cu^{2+} is 0.008 M.
Answer:
See explanation below for answers
Explanation:
We know that the balance is tared, so the innitial weight would be zero. Now, let's answer this by parts.
a) mass of displaced water.
In this case all we need to do is to substract the 0.70 with the 0.13 g. so:
mW = 0.70 - 0.13
mW = 0.57 g of water
b) Volume of water.
In this case, we have the density of water, so we use the formula for density and solve for volume:
d = m/V
V = m/d
Replacing:
Vw = 0.57/0.9982
Vw = 0.5710 mL of water
c) volume of the metal weight
In this case the volume would be the volume displaced of water, which would be 0.5710 mL
d) the mass of the metal weight.
In this case, it would be the mass when the metal weight hits the bottom which is 0.70 g
e) density.
using the above formula of density we calculate the density of the metal
d = 0.70 / 0.5710
d = 1.2259 g/mL
B. Which is the kinetic energy
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