Answer:
84.8%
Explanation:
Step 1: Given data
Bob measured out 1.60 g of Na. He forms NaCl according to the following equation.
Na + 1/2 Cl₂ ⇒ NaCl
According to this equation, he calculates that 1.60 g of sodium should produce 4.07 g of NaCl, which is the theoretical yield. However, he carries out the experiment and only makes 3.45 g of NaCl, which is the real yield.
Step 2: Calculate the percent yield.
We will use the following expression.
%yield = real yield / theoretical yield × 100%
%yield = 3.45 g / 4.07 g × 100% = 84.8%
Answer:
Chromium is electroplated on other metals. 2. Gold is electroplated on other cheap metals.
Answer:
1. d[H₂O₂]/dt = -6.6 × 10⁻³ mol·L⁻¹s⁻¹; d[H₂O]/dt = 6.6 × 10⁻³ mol·L⁻¹s⁻¹
2. 0.58 mol
Explanation:
1.Given ΔO₂/Δt…
2H₂O₂ ⟶ 2H₂O + O₂
-½d[H₂O₂]/dt = +½d[H₂O]/dt = d[O₂]/dt
d[H₂O₂]/dt = -2d[O₂]/dt = -2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = -6.6 × 10⁻³mol·L⁻¹s⁻¹
d[H₂O]/dt = 2d[O₂]/dt = 2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = 6.6 × 10⁻³mol·L⁻¹s⁻¹
2. Moles of O₂
(a) Initial moles of H₂O₂

(b) Final moles of H₂O₂
The concentration of H₂O₂ has dropped to 0.22 mol·L⁻¹.

(c) Moles of H₂O₂ reacted
Moles reacted = 1.5 mol - 0.33 mol = 1.17 mol
(d) Moles of O₂ formed

i believe
2 g/mL
good luck!