Answer:
2.645
Explanation:
Rate of diffusion formula:
Sqrt(mass2/mass1)
>>sqrt(14/2)
(Note:Hydrogen must exist in dwiatomic, [H2])
Answer:
Volume of stock solution needed = 6.0299 mL
Explanation:
<u>
</u>Dilution consists of lowering the amount of solute per unit volume of solution. It is achieved by adding more diluent to the same amount of solute.
This is deduced when thinking that both the dissolution at the beginning and at the end will have the same amount of moles.
<u>Data:</u>
M1 = 6.01 M stock solution concentration
M2 = 0.3624 M diluted solution concentration
V2 =100 mL diluted solution volume
V1 = ? stock solution volume
M1 * V1 = M2 * V2

6 miles of H2O is equal to 12 Hydrogen molecules and 6 oxygen molecules. Equaling 18 in total.
Answer:
Chemical properties, such as combustibility, are generally observed as the identity of a substance changes and one or more new substances form.
Explanation:
Chemical change involves formation of new substances. Therefore, the correct answer is "Chemical properties, such as combustibility, are generally observed as the identity of a substance changes and one or more new substances form."
Answer:
0.912 mL
Explanation:
3 Na2S(aq) + 2 FeCl3(aq) → Fe2S3(s) + 6 NaCl(aq)
FeCl3 is the limiting reactant.
Number of moles of iron III sulphide produced= 3.75g/87.92 g/mol = 0.043 moles
Hence actual yield of Iron III sulphide = 0.043 moles
Theoretical yield of Iron III sulphide = actual yield ×100%/ %yield
Theoretical yield of iron III sulphide= 0.043 ×100/75 = 0.057 moles of Iron III sulphide
From the reaction equation,
2moles of iron III chloride produced 1 mole of iron III sulphide
x moles of iron III chloride, will produce 0.057 of iron III sulphide
x= 2× 0.057= 0.114 moles of iron III chloride
But
Volume= number of moles/ concentration
Volume= 0.114/0.125
Volume= 0.912 mL