Answer:
c) 2.5 mL
Explanation:
Solution
Doctors order = 0.125g
and
The liquid suspension concentration = 250 mg/5ml
= 0.250g/5ml
Or 0.05g/ml
Amount of ml of suspension required = 0.125g/(0.05g/ml) = 2.5ml
Mass number is the total number of protons and neutrons in the nucleus
Answer:
The mass of copper(II) sulfide formed is:
= 81.24 g
Explanation:
The Balanced chemical equation for this reaction is :

given mass= 54 g
Molar mass of Cu = 63.55 g/mol

Moles of Cu = 0.8497 mol
Given mass = 42 g
Molar mass of S = 32.06 g/mol

Moles of S = 1.31 mol
Limiting Reagent :<em> The reagent which is present in less amount and consumed in a reactio</em>n
<u><em>First find the limiting reagent :</em></u>

1 mol of Cu require = 1 mol of S
0.8497 mol of Cu should require = 1 x 0.8497 mol
= 0.8497 mol of S
S present in the reaction Medium = 1.31 mol
S Required = 0.8497 mol
S is present in excess and <u>Cu is limiting reagent</u>
<u>All Cu is consumed in the reaction</u>
Amount Cu will decide the amount of CuS formed

1 mole of Cu gives = 1 mole of Copper sulfide
0.8497 mol of Cu = 1 x 0.8497 mole of Copper sulfide
= 0.8497
Molar mass of CuS = 95.611 g/mol


Mass of CuS = 0.8497 x 95.611
= 81.24 g
Answer:
0.022 mol O
Explanation:
Mg3(Si2O5)2(OH)2
We can see that 1 mol of this substance has 3 mol of Mg.
Oxygen altogether is 5*2 (from (Si2O5)2) + 2(from(OH)2) = 10 +2 = 12
So, 1 mol of this substance has 12 mol oxygen.
So, 1 mol of this substance contains 3 mol Mg and 12 mol O, or
ratio Mg : O = 3 : 12 = 1 : 4
1 mol Mg ----- 4 mol O
0.055 mol Mg ---x mol O
x = 0.055*4/1 = 0.220 mol O