Answer:
A
Explanation:
because CO2 is carbon dioxide and CO is carbon monoxide. mono meaning one which in this case is monoxide.
Answer:
Explanation:
Ratio of mass of C , N , H and O
= .8007 :0.9333:0.2016:2.133
Ratio of moles of C , N , H and O
= .8007/12 : .9333 / 14 : 0.2016 / 1 : 2.133/16
= .0667 : .0667: .2016 : .1333
= .0667 / .0667 : .0667 / .0667 : .2016 /.0667 : .1333 / .0667
= 1 : 1 : 3: 2
Hence empirical formula = CNH₃O₂
7 .
Weight of titanium Ti = 1.916 g
Weight of oxygen = 3.196 - 1.916 = 1.28 g
Ratio of weight of Ti and O
= 1.916 : 1.28
Ratio of moles of Ti and O
1.916/48 : 1.28/16 [ Molecular weight of Titanium is 48 ]
= .04 : .08
= .04/.04 : .08/.04
= 1 :2 .
Empirical formula
TiO₂
Answer:
V2~0.4839M
Explanation:
We're going to use Boyles law to answer the question.
Boyle's law:
P1V1=P2V2
P1=151mmHg
P2=166mmHg
V1=0.532L
V2=?
V2=(P1 x V1)/P2
V2=(151 x 0.532)/166
V2~0.4839M
Hope it helps:)
Answer:
Final concentration of C at the end of the interval of 3s if its initial concentration was 3.0 M, is 3.06 M and if the initial concentration was 3.960 M, the concentration at the end of the interval is 4.02 M
Explanation:
4A + 3B ------> C + 2D
In the 3s interval, the rate of change of the reactant A is given as -0.08 M/s
The amount of A that has reacted at the end of 3 seconds will be
0.08 × 3 = 0.24 M
Assuming the volume of reacting vessel is constant, we can use number of moles and concentration in mol/L interchangeably in the stoichiometric balance.
From the chemical reaction,
4 moles of A gives 1 mole of C
0.24 M of reacted A will form (0.24 × 1)/4 M of C
Amount of C formed at the end of the 3s interval = 0.06 M
If the initial concentration of C was 3 M, the new concentration of C would be (3 + 0.06) = 3.06 M.
If the initial concentration of C was 3.96 M, the new concentration of C would be (3.96 + 0.06) = 4.02 M
No, because there’s 6 Oxygens going in and only 3 going out. To balance:
2S+3O2 -> 2SO3
