Answer:
8.88 x 10⁻² M/s
Explanation:
The rate of reaction for:
NO(g) + Cl₂ (g) ⇒ 2NOCl(g)
is rate = -ΔNO/Δt = -ΔCl2/Δt = 1/2 ΔNOCl/Δt
so ΔNOCl/Δt = 2 ΔCl2/Δt = 2 x 4.44 × 10⁻² M/s = 8.88 x 10⁻² M/s
In general given a reaction
aA + bB ⇒ cC + dD
rate = -1/a ΔA/Δt = -1/b ΔB/Δt = 1/c ΔC/Δt = 1/d ΔD/Δt
Answer:
O, N, C, H
Explanation:
Electronegativity of an element is the property that combines the ability of its atom to lose or gain electrons. It measures the relative tendency with which the atoms of the element attracts valence electrons in a chemical bond.
On the periodic table, Electronegativity increases across the period but decreases down a group.
To solve the given problem, let us use thE Pauling's table of electronegativities to compare the electronegativities of the elements.
On the table:
C = 2.5
H = 2.1
O = 3.5
N = 3.0
In terms of decreasing electronegativities, the atoms are arranged as:
O N C H
<span>The balanced chemical equation for this reaction is:
2NaOH (aq)+H2SO4 (aq) → Na2SO4 (aq)+2H2O (l)
According to question, 60 ml of NaOH solution was used for neutralizing 40 ml of 0.50M H2SO4.
The no. of moles of H2SO4 is calculated using the equation:
mol solute = (molarity) (L soln)
mol H2SO4 = 0.50 M x 0.040 L = 0.02 moles of H2SO4
As per the equation, the number of moles of NaOH used is:
0.02 moles of H2SO4 (2 mol NaOH) (1 mol H2SO4) = 0.04 moles of NaOH
Therefore, using the given volume of NaOH, the concentration or molarity of NaOH can be calculated using the formula :
Molarity = mol solute/L soln = 0.04 mol NaOH/0.06 L = 0.67 M
Therefore, the concentration of NaOH is 0.67 M.</span>
N₂ + 3H₂ ⇄ 2NH₃ + heat
In the given equilibrium, we notice that the heat is on the right. which means that if the heat requirements don't meet, the reactants on the right will no longer react due to the lack of heat
but because the reactants on the left don't have such weaknesses, they will keep reacting hence producing more and more ammonia until a new equilibrium is reached
where there will be more ammonia and less nitrogen and hydrogen as compared to the equilibrium we had initially
