Answer: a = 2 ; f = 5 ; b = 2 ; g = 2 ; c = 2 ; h = 2 ; d = 4 ; i = 5 ; e = 3 ; j = 7
Explanation: Some rules to follow while calculating sig figs is
1. If a number like 4500 is present, only two sig figs are counted, but none of the zeros are, but if 4500. has a decimal point present, then you should count all the numbers available.
2. If a number like .0005 is present, only count 5 as a sig fig, however if the number is .00050, count the 0 after the 5 in this example (this would then have two sig figs.
Answer:
a) 1.61 mol
b) Al is limiting reactant
c) HBr is in excess
Explanation:
Given data:
Moles of Al = 3.22 mol
Moles of HBr = 4.96 mol
Moles of H₂ formed = ?
What is limiting reactant =
What is excess reactant = ?
Solution:
Chemical equation:
2Al + 2HBr → 2AlBr + H₂
Now we will compare the moles:
Al : H₂
2 : 1
3.22 : 1/2×3.22 = 1.61 mol
HBr : H₂
2 : 1
4.96 : 1/2×4.96 = 2.48 mol
The number of moles of H₂ produced by Al are less it will be limiting reactant while HBr is present in excess.
Moles of H₂ :
Number of moles of H₂ = 1.61 mol
Answer:
B. wave 1 has a larger wavelength than wave 2
Could you attach a picture because I can tell you didn't post the entire question.