The answer is definitely 4. It has fewer electrons. This is because, whenever the electrons are few, they are positive. Positive electrons are anions, while the negative ones are cations. Hope i helped. Have a nice day.
The resistance of the appliance is 64.1 Ω.
<u>Explanation:</u>
As per the Ohm's law, which states that the electric current is in direct proportion to the voltage and is in inverse proportion to the resistance. It is given by the expression as,
V = IR
Where V is the voltage (V) = 150.0 V
I is the current (amps) = 2.34 amps
R is the resistance (ohm) or Ω = ?
Now we have to rearrange the equation to get the resistance as,
Now we have to plug in the values as,
= 64.1 Ω
So the resistance of the appliance is 64.1 Ω.
Missing question: what is the density of 53.4 wt% aqueous NaOH if 16.7 mL of the solution diluted to 2.00L gives 0.169 M NaOH?
Answer is: density is 1.52 g/mL.
c₁(NaOH) = ?; molarity of concentrated sodium hydroxide.
V₁(NaOH) = 16.7 mL; volume of concentrated sodium hydroxide.
c₂(NaOH) = 0.169 M; molarity of diluted sodium hydroxide.
V₂(NaOH) = 2.00 L · 1000 mL/L = 2000 mL; volume of diluted sodium hydroxide.
Use equation: c₁V₁ = c₂V₂.
c₁ = c₂V₂ / V₁.
c₁ = 0.169 M · 2000 mL / 16.7 mL.
c₁(NaOH) = 20.23 M.
m(NaOH) = 20.23 mol · 40 g/ml.
m(NaOH) = 809.53 g.
The mass fraction is the ratio of one substance (in this example sodium hydroxide) with mass to the mass of the total mixture (solution).
Make proportion: m(NaOH) : m(solution) = 53.4 g : 100 g.
m(solution) = 1516 g in one liter of solution.
d(solution) = 1516 g/L = 1.52 g/mL.
Answer:
The correct answer is 574.59 grams.
Explanation:
Based on the given information, the number of moles of NH₃ will be,
= 2.50 L × 0.800 mol/L
= 2 mol
The given pH of a buffer is 8.53
pH + pOH = 14.00
pOH = 14.00 - pH
pOH = 14.00 - 8.53
pOH = 5.47
The Kb of ammonia given is 1.8 * 10^-5. Now pKb = -logKb,
= -log (1.8 ×10⁻⁵)
= 5.00 - log 1.8
= 5.00 - 0.26
= 4.74
Based on Henderson equation:
pOH = pKb + log ([salt]/[base])
pOH = pKb + [NH₄⁺]/[NH₃]
5.47 = 4.74 + log ([NH₄⁺]/[NH₃])
log([NH₄⁺]/[NH₃]) = 5.47-4.74 = 0.73
[NH₄⁺]/[NH₃] = 10^0.73= 5.37
[NH₄⁺ = 5.37 × 2 mol = 10.74 mol
Now the mass of dry ammonium chloride required is,
mass of NH₄Cl = 10.74 mol × 53.5 g/mol
= 574.59 grams.
