The initial temperature of the metal = 35 °C
<h3>Further explanation</h3>Heat can be formulated :
Q = m . c . ΔT
Q = heat, J
c = specific heat, J/g C
ΔT = temperature, °C
m = 20 g
c = 5 J/(g°C)
Q = 500 J
T₁ = 40 C
the initial temperature :
g Enter your answer in the provided box. If 30.8 mL of lead(II) nitrate solution reacts completely with excess sodium iodide sol
1 answer:
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The complete table is inserted.
A table is given,
Formulas used:
pH= -log(H⁺)
pOH= -log(OH⁻)
pH+ pOH=14
Calculations:
For A: (H⁺)=2×10⁻⁸M
Using the pH formula:
pH= -log(H⁺)=-log(2×10⁻⁸)=7.69
pOH=14 - 7.69=6.3
Calculating OH concentration,
pOH= -log(OH⁻)
6.3= -log(OH⁻)
(OH⁻)=5.011×10⁻⁷M
Hence, the nature of A is basic.
Similarily,
For B,
(OH⁻)=1×10⁻⁷
Using the pH formula:
pOH= -log(OH⁻)= -log(1×10⁻⁷)=7
pH=14-7=7
Calculating H concentration,
pH= -log(H⁺)
7= -log(H⁺)
(H⁺)=1×10⁻⁷M
Hence, the nature of B is neutral.
Similarily,
For C,
pH=12.3
Using the pH formula:
pOH=14-12.3=1.7
Calculating H concentration,
pH= -log(H⁺)
12.3= -log(H⁺)
(H⁺)=5.011×10⁻¹³M
Calculating OH concentration,
pOH= -log(OH⁻)
1.7= -log(OH⁻)
(OH⁻)=1.99×10⁻²M
Hence, the nature of C is Basic.
Similarily,
For D,
pOH=6.8
Using the pH formula:
pH=14-6.8=7.2
Calculating H concentration,
pH= -log(H⁺)
7.2= -log(H⁺)
(H⁺)=6.309×10⁻⁸M
Calculating OH concentration,
pOH= -log(OH⁻)
6.8= -log(OH⁻)
(OH⁻)=1.58×10⁻⁷M
Hence, the nature of D is basic.
Learn more about the acid and bases here:
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