Answer:
0.6258 g
Explanation:
To determine the number grams of aluminum in the above reaction;
- determine the number of moles of HCl
- determine the mole ratio,
- use the mole ratio to calculate the number of moles of aluminum.
- use RFM of Aluminum to determine the grams required.
<u>Moles </u><u>of </u><u>HCl</u>
35 mL of 2.0 M HCl
2 moles of HCl is contained in 1000 mL
x moles of HCl is contained in 35 mL
We have 0.07 moles of HCl.
<u>Mole </u><u>ratio</u>
6HCl(aq) + 2Al(s) --> 2AlCl3(aq) + 3H2(g)
Hence mole ratio = 6 : 2 (HCl : Al
- but moles of HCl is 0.07, therefore the moles of Al;
Therefore we have 0.0233333 moles of aluminum.
<u>Grams of </u><u>Aluminum</u>
We use the formula;
The RFM (Relative formula mass) of aluminum is 26.982g/mol.
Substitute values into the formula;
The number of grams of aluminum required to react with HCl is 0.6258 g.
Answer:
(FeSCN⁺²) = 0.11 mM
Explanation:
Fe ( NO3)3 (aq) [0.200M] + KSCN (aq) [ 0.002M] ⇒ FeSCN+2
M (Fe(NO₃)₃ = 0.200 M
V (Fe(NO₃)₃ = 10.63 mL
n (Fe(NO₃)₃ = 0.200*10.63 = 2.126 mmol
M (KSCN) = 0.00200 M
V (KSCN) = 1.42 mL
n (KSCN) = 0.00200 * 1.42 = 0.00284 mmol
Total volume = V (Fe(NO₃)₃ + V (KSCN)
= 10.63 + 1.42
= 12.05 mL
Limiting reactant = KSCN
So,
FeSCN⁺² = 0.00284 mmol
M (FeSCN⁺²) = 0.00284/12.05
= 0.000236 M
Excess reactant = (Fe(NO₃)₃
n(Fe(NO₃)₃ = 2.126 mmol - 0.00284 mmol
=2.123 mmol
For standard 2:
n (FeSCN⁺²) = 0.000236 * 4.63
=0.00109
V(standard 2) = 4.63 + 5.17
= 9.8 mL
M (FeSCN⁺²) = 0.00109/9.8
= 0.000111 M = 0.11 mM
Therefore, (FeSCN⁺²) = 0.11 mM
Some bad people will tend to forge dollar bills and make fake bills, giving them to cashiers and asking for it to be split into a different amount of bills so they are given real cash and wont be caught or they will buy things, return them and be given back real money and get away with it. This is obviously a crime because it's fake.
Answer:
because it is two substances reacting together to create a new substance
Explanation:
when iron rusts, iron molecules react with the oxygen molecules creating iron oxide aka rust.
Answer:
There are 3600 pages to 2 significant figures
Explanation:
Working only with the information given, we have that there are 23 books each with about 31 chapters.
This means that if 1 book has 31 chapter, the 23 science books will have
31 X 23 chapters =713 chapters in total.
If each chapter contains 5 pages, for the 713 chapters we will have a total of 713 X 5 pages = 3565 pages.
Rounding off the answer to 2 significant figures , we are looking to have only two non-zero figures in our answer. To do this, we count two numbers from the left. This will be numbers 3 and 5.
Next, we check the net number after 5 to see if it is greater than 5 if it is, we will have to add the number (1 ) to the 5. This shows that we are approximating upwards, to account for the fact that that number has passed the middle point.
After this, we change the remaining two numbers (6 and 5) to zeroes.
This will give us 3600 pages.
