Answer:
In a chemical equilibrium, the forward and reverse reactions occur at equal rates, and the concentrations of products and reactants remain constant. A catalyst speeds up the rate of a chemical reaction, but has no effect upon the equilibrium position for that reaction.
Explanation:
C₀=8.10M
c₁=5.28M
v₀=1.58 L
v₁-?
n=c₀v₀=c₁v₁
v₁=c₀v₀/c₁
v₁=8.10*1.58/5.28=2.42 L
Answer: a: reactants Na-2 Cl-2
Products: Na-2 Cl-2;
b: reactants P-1 Cl-13 H-6 Products P-1 H-6 Cl-13
c: reactants P-4 H-12 O-16
Products H-12 P-4 O-16
Explanation: since these equations are balanced the atoms on of element on the reactants side will be same as the atoms of the same element of the product side
Answer:
# 5
Explanation:
The question describes silver being "poured" into a mold and cools to become a solid bar. This is the phase of liquid to solid. When a element cools down below it's freezing points to become a solid.
<u>Liquid to Solid Definition:</u>
Freezing, or solidification, is a phase transition in which a liquid turns into a solid when its temperature is lowered to or below its freezing point. All known liquids, except helium, freeze when the temperature is low enough.
Answer:
The answers are in the explanation
Explanation:
A buffer is the mixture of a weak acid with its conjugate base or vice versa. Thus:
<em>1)</em> Mixing 100.0 mL of 0.1 M HF with 100.0 mL of 0.05 M mol KF. <em>Will </em>result in a buffer because HF is a weak acid and KF is its conjugate base.
<em>2)</em> Mixing 100.0 mL of 0.1 M NH₃ with 100.0 mL of 0.1 M NH₄Br. <em>Will not </em>result in a buffer because NH₃ is a strong base.
<em>3) </em>Mixing 100.0 mL of 0.1 M HCN with 100.0 mL of 0.05 M KOH. <em>Will </em>result in a buffer because HCN is a weak acid and its reaction with KOH will produce CN⁻ that is its conjugate base.
<em>4)</em> Mixing 100.0 mL of 0.1 M HCl with 100.0 mL of 0.1 M KCl <em>Will not </em>result in a buffer because HCl is a strong acid.
<em>5)</em> Mixing 100.0 mL of 0.1 M HCN with 100.0 mL of 0.1 M KOH <em>Will not </em>result in a buffer because each HCN will react with KOH producing CN⁻, that means that you will have just CN⁻ (Conjugate base) without HCN (Weak acid).
I hope it helps!
