Answer:
Explanation:
General reaction of acid in water is as follows:
HCl + H2O = H3O+ + Cl-
Thus Acids increase the concentration of hydronium ions in solution by donating hydrogen ions to water molecules is true
Answer is "B - 700,000".<span>
<span>Kinetic energy of a single particle (atom or molecule)<span> is directly proportional to its
temperature according to the following equation.</span></span>
KE = (3kT)/2
<span>Where </span>KE<span> is the
kinetic energy of a single atom/molecule (</span>J<span>), </span>k<span> is the Boltzmann
constant (</span>1.381 × 10</span>⁻²³ J/K<span>) and </span>T<span> is the temperature (</span>K<span>) </span><span>
When temperature increases, then the kinetic
energy increases.
<span>If kinetic
energy of atoms increases, then there will be more motions which create many
collisions.</span></span>
The reaction is given as:
Here, two moles of copper nitrate reacts with four moles of potassium iodide to give two moles of copper iodide, one mole of iodine and four moles of potassium nitrate.
First, calculate the number of moles of copper nitrate.
Number of moles is equal to the product of molarity and volume of solution in litre.
Number of moles = 
(1 L =1000 mL)
= 
Copper nitrate requires = 
mole of potassium iodide
= 
of potassium iodide
Volume of solution in litre = 
Thus, volume of potassium iodide is =
= 
1 L =1000 mL
Volume of potassium iodide in mL =
Hence, 0.2089 M potassium iodide consist of sufficient potassium iodide to react with copper nitrate in 3.88 mL of a 0.3842 M solution of copper nitrate .
Given reaction represents dissociation of bromine gas to form bromine atoms
Br2(g) ↔ 2Br(g)
The enthalpy of the above reaction is given as:
ΔH = ∑n(products)Δ
- ∑n(reactants)Δ
where n = number of moles
Δ
= enthalpy of formation
ΔH = [2*ΔH(Br(g)) - ΔH(Br2(g))] = 2*111.9 - 30.9 = 192.9 kJ/mol
Thus, enthalpy of dissociation is the bond energy of Br-Br = 192.9 kJ/mol
2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O
c₁=2.00 mol/L
v₁=0.25 L
v₂=2.00 L
c₂-?
n(NaOH)=c₂v₂
n(H₂SO₄)=c₁v₁
n(NaOH)=2n(H₂SO₄)
c₂v₂=2c₁v₁
c₂=2c₁v₁/v₂
c₂=2*2.00*0.25/2.00=0.5 mol/L
0.5 M NaOH
