Answer:
30 mL VOLUME OF 3.0 M HCl SHOULD BE USED BY THE STUDENT TO MAKE A 1.80 M IN 50 mL OF HCl.
Explanation:
M1 = 3.00 M
M2 = 1.80 M
V2 = 50 .0 mL = 50 /1000 L = 0.05 L
V1 = unknown
In solving this question, we know that number of moles of a solution is equal to the molar concentration multiplied by the volume. To compare two samples, we equate both number of moles and substitute for the required component.
So we use the equation:
M1 V1 = M2 V2
V1 = M2 V2 / M1
V2 = 1.80 * 0.05 / 3.0
V2 = 0.09 /3.0
V2 = 0.03 L or 30 mL
To prepare the sample of 1.80 M HCl in 50.0 mL from a 3.0 M HCl, 30 mL volume should be used.
Answer:
the stoichiometric coefficient for cobalt is 3
Explanation:
the unbalanced reaction would be
Co(NO₃)₂+ Al → Al(NO₃)₃ + Co
One way to solve is to build a system of linear equations for each element (or group as NO₃) , knowing that the number of atoms of each element is conserved.
For smaller reactions a quick way to solve it can be:
- First the Co as product and as reactant needs to have the same stoichiometric coefficient
- Then the Al as product and as reactant needs to have the same stoichiometric coefficient
- After that we look at the nitrates . There are 2 as reactants and 3 as products . Since the common multiple is 6 then multiply the reactant by 3 and the product by 2.
Finally the balanced equation will be
3 Co(NO₃)₂+ 2 Al → 2 Al(NO₃)₃ + 3 Co
then the stoichiometric coefficient for cobalt is 3
Molar mass
C₂H₄O₂ = 60.0 g/mol
n = mass / molar mass
3.00 = mass / 60.0
m = 3.00 * 60.0
m = 180 g of <span>C₂H₄O₂
hope this helps!</span>
