<span> The formula for </span>measuring<span> density is Density = </span>Mass/Volume<span>, or D=M/V. The / means “per” or “for each,” which in math is the same as “divided by.”</span>
When CaSO4 → Ca2+ + SO4
So when we have Ksp = [Ca2+][SO4]
when Ksp = 4.93 x 10^-5
and [SO4] = 0.02 M
so by substitution we can get [Ca2+]
4.93x10^-5 = [Ca2+] [0.02]
∴ [Ca2+] = 0.0025 mol/L
∴ the moles of calcium chloride = 0.0025 mol / L * 1.5 L
= 0.00167 mol
Answer:
-7.34 kilo Joules is the change in enthaply when 20.0 grams of nitrogen triiodide decomposes.
Explanation:
Mass of nitrogen triiodide = 20.0 g
Moles of nitrogen triiodide = 
According to reaction, 2 moles of nitrogen triiodide gives 290.0 kilo Joules of heat on decomposition ,then 0.05063 moles of nitrogen triiodide will give :
-7.34 kilo Joules is the change in enthaply when 20.0 grams of nitrogen triiodide decomposes.
Answer:
pH = 3.3
Explanation:
Buffer solutions minimize changes in pH when quantities of acid or base are added into the mix. The typical buffer composition is a weak electrolyte (wk acid or weak base) plus the salt of the weak electrolyte. On addition of acid or base to the buffer solution, the solution chemistry functions to remove the acid or base by reacting with the components of the buffer to shift the equilibrium of the weak electrolyte left or right to remove the excess hydronium ions or hydroxide ions is a way that results in very little change in pH of the system. One should note that buffer solutions do not prevent changes in pH but minimize changes in pH. If enough acid or base is added the buffer chemistry can be destroyed.
In this problem, the weak electrolyte is HNO₂(aq) and the salt is KNO₂(aq). In equation, the buffer solution is 0.55M HNO₂ ⇄ H⁺ + 0.75M KNO₂⁻ . The potassium ion is a spectator ion and does not enter into determination of the pH of the solution. The object is to determine the hydronium ion concentration (H⁺) and apply to the expression pH = -log[H⁺].
Solution using the I.C.E. table:
HNO₂ ⇄ H⁺ + KNO₂⁻
C(i) 0.55M 0M 0.75M
ΔC -x +x +x
C(eq) 0.55M - x x 0.75M + x b/c [HNO₂] / Ka > 100, the x can be
dropped giving ...
≅0.55M x ≅0.75M
Ka = [H⁺][NO₂⁻]/[HNO₂] => [H⁺] = Ka · [HNO₂]/[NO₂⁻]
=> [H⁺] = 6.80x010⁻⁴(0.55) / (0.75) = 4.99 x 10⁻⁴M
pH = -log[H⁺] = -log(4.99 x 10⁻⁴) -(-3.3) = 3.3
Solution using the Henderson-Hasselbalch Equation:
pH = pKa + log[Base]/[Acid] = -log(Ka) + log[Base]/[Acid]
= -log(6.8 x 10⁻⁴) + log[(0.75M)/(0.55M)]
= -(-3.17) + 0.14 = 3.17 + 0.14 = 3.31 ≅ 3.3
Answer:
The greatest acceleration when the unbalanced force is applied will be experienced in :
A) The box with a mass of 2 kg
Explanation:
According to second law of motion the external unbalanced force is directly proportional to rate of change of momentum.
F = (Final momentum - initial momentum)/time
or
Force is equal to the product of mass and acceleration
F = m x a
Here a= acceleration
m = mass of the object
If Force is constant then acceleration is inversely proportional to mass
A) The box with a mass of 2kg
F = 8 N
a = 4 m/s2
B) The box with the mass of 4kg
a = 2 m/s2
C) The box with a mass of 6kg
a = 1.33 m/s2
D) The box with a mass of 8kg
a = 1 m/s2
