Answer:
Chemical energies
Explanation:
We get chemical energy from foods, which we use to run about, and move and talk (kinetic and sound energy). Chemical energies are stored in fuels which we burn to release thermal energy - this is one way of making electricity, see Electricity for more information.
The molarity of the acid sample H₂SO₄ is 0.052M .
<h3>What is Molarity ?</h3>
Molarity (M) is the amount of a substance in a certain volume of solution.
Molarity is defined as the moles of a solute per liters of a solution.
Molarity is also known as the molar concentration of a solution
Now to determine the molarity of the acid sample
V( H₂SO₄) = 24.0 mL in liters = 24.0 / 1000 = 0.024 L
M(H₂SO₄) = ?
V(NaOH) = 20.0 mL = 20.0 / 1000 = 0.02 L
M(NaOH) = 0.125 M
Number of moles NaOH :
n = M x V
n = 0.125 x 0.02
n = 0.0025 moles of NaOH
H₂SO₄(aq) + 2 NaOH(aq) = Na₂SO₄(aq) + 2 H₂O(l)
1 mole H₂SO₄ ---------- 2 mole NaOH
? mole H₂SO₄ ---------- 0.0025 moles NaOH
moles = 0.0025 * 1 / 2
= 0.00125 moles of H₂SO₄
M(H₂SO₄) = n / V
M = 0.00125 / 0.024
= 0.052 M
Therefore the molarity of the acid sample H₂SO₄ is 0.052M .
To know more about molarity
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Answer:
Ka = 0.1815
Explanation:
Chromic acid
pH = ?
Concentration = 0.078 M
Ka = ?
HCl
conc. = 0.059M
pH = -log(H+)
pH = -log(0.059) = 1.23
pH of chromic acid = 1.23
Step 1 - Set up Initial, Change, Equilibrium table;
H2CrO4 ⇄ H+ + HCrO4−
Initial - 0.078M 0 0
Change : -x +x +x
Equilibrium : 0.078-x x x
Step 2- Write Ka as Ratio of Conjugate Base to Acid
The dissociation constant Ka is [H+] [HCrO4−] / [H2CrO4].
Step 3 - Plug in Values from the Table
Ka = x * x / 0.078-x
Step 4 - Note that x is Related to pH and Calculate Ka
[H+] = 10^-pH.
Since x = [H+] and you know the pH of the solution,
you can write x = 10^-1.23.
It is now possible to find a numerical value for Ka.
Ka = (10^-1.23))^2 / (0.078 - 10^-1.23) = 0.00347 / 0.0191156
Ka = 0.1815
Answer:
1. CaO + H₂O ----> Ca(OH)₂
Compound ----- Compound
2. 2 Na + Cl₂ ----> 2 NaCl
Element ----- Element
3. 2 SO₂ + O₂ ----> 2 SO₃
Element ----- Compound
