The volume of 0.160 m Li2S solution required to completely react with 130 ml of 0.160 CO(NO3)2 is calculated as below
write the reacting equation
Co(NO3)2 + Li2S = 2LiNO3 + COS
find the moles of CO(NO3)2 = molarity x volume
= 130 ml x 0.160=20.8 moles
since the reacting moles between CO(NO3)2 to LiS is 1:1 the moles of LiS is also 20.8 moles
volume of Lis is therefore = moles of Lis/ molarity of LiS
= 20.8/0.160 = 130 Ml
Answer:
159.3 grams Al
Explanation:
1 Mol of Al = 27 grams, so:
5.9 Mol Al (27g/1mol) = 159.3 grams Al
The Mol gets canceled out, leaving the unit of grams.
Answer:
591.25ml
Explanation:
4 cups=946ml
0.5cups=946÷8=118.25ml
2.5cups=118.25ml*5=591.25ml
Since you are provided with the pressure and initial volume and told a standard temperature, you should use the Combined gas law. This is (P1 V1)/ T1 = (P2 V2)/ T2
STP is 273 k for temperature, so plug in the values.
(3.5 Pa * 25 L)/ 273 K =1.3 Pa * X L)/ 273 K.
(this is the left side>)0.32051= 1.3 X/ 273. Multiply both sides by 273 to get rid of it on the right, which leaves you with
87.4999 = 1.3 X. Divide both sides by 1.3 and there you go,
67.308 L