Answer:
c. 80 g
Explanation:
metal alloy:
wt% = (mass compound/mass alloy)×100
for antimony:
⇒ 16% = (g antimony / g alloy)×100
⇒ 0.16 = g antimony / g alloy
g alloy = 500 g
⇒ (0.16)×(500 g) = g antimony
⇒ g antimony = 80 g
Answer:
I think it's D one. Because all others are wrong
Potassium hydroxide is a strong base and hydrobromic acid is a strong acid. This implies that the pH of the end-point [neutralization] of their titration will be around pH 7. A good indicator for this kind of pH is bromthymol blue. This is because this indicator changes its colour at pH 7.
Answer:
42.9%
Explanation:
Step 1: Write the balanced decomposition reaction
PbO₂ ⇒ Pb + O₂
Step 2: Calculate the theoretical yield of O₂ from 5.77 g of PbO₂
According to the balanced equation, the mass ratio of PbO₂ to O₂ is 239.2:32.00.
5.77 g PbO₂ × 32.00 g O₂/239.2 g PbO₂ = 0.772 g O₂
Step 3: Calculate the percent yield of O₂
The real yield of O₂ is 0.331 g. The percent yield of O₂ is:
%yield = real yield / theoretical yield × 100%
%yield = 0.331 g / 0.772 g × 100% = 42.9%
Because the easier a metal donates an electron (Low Ionization Energy), the more readily a non-metal with high electromagnetically will accept it.
Thus they match well, - one readily donates, the other readily accepts. This is how ionic bonding occurs.
Hope this helps.