Answer:
The amount of dissolved gases in the body. Have a good day! =)
Explanation:
Answer:
44.6millilitres
Explanation:
Using the general gas law equation as follows:
P1V1/T1 = P2V2/T2
Where;
P1 = initial pressure (atm)
V1 = initial volume (L)
T1 = initial temperature (K)
P2 = final pressure (atm)
V2 = final volume (L)
T2 = final temperature (K)
According to this question;
V1 = 30mL
T1 = 273K (STP)
P1 = 1 atm (STP)
V2 = ?
T2 = 300K
P2 = 75.0 kPa = 75 × 0.00987 = 0.74atm
Using P1V1/T1 = P2V2/T2
1 × 30/273 = 0.74×V2/300
30/273 = 0.74V2/300
Cross multiply
300 × 30 = 273 × 0.74V2
9000 = 202.02V2
V2 = 9000/202.02
V2 = 44.55
V2 = 44.6millilitres.
Answer: hre
Explanation:
N2(g) + 3H2-> 2NH3(g) This is the balanced equation
Note the mole ratio between N2, H2 and NH3. It is 1 : 3 : 2 This will be important.
moles N2 present = 28.0 g N2 x 1 mole N2/28 g = 1 mole N2 present
moles H2 present = 25.0 g H2 x 1 mole H2/2 g = 12.5 moles H2 present
Based on mole ratio, N2 is limiting in this situation because there is more than enough H2 but not enough N2.
moles NH3 that can be produced = 1 mole N2 x 2 moles NH3/mole N2 = 2 moles NH3 can be produced
grams of NH3 that can be produced = 2 moles NH3 x 17 g/mole = 34 grams of NH3 can be produced
NOTE: The key to this problem is recognizing that N2 is limiting, and therefore limits how much NH3 can be produced.
Answer:
(a) rate = -(1/3) Δ[O₂]/Δt = +(1/2) Δ[O₃]/Δt
(b) Δ[O₃]/Δt = 1.07x10⁻⁵ mol/Ls
Explanation:
By definition, t<u>he reaction rate for a chemical reaction can be expressed by the decrease in the concentration of reactants or the increase in the concentration of products:</u>
aX → bY (1)
<em>where, a and b are the coefficients of de reactant X and product Y, respectively. </em>
(a) Based on the definition above, we can express the rate of reaction (2) as follows:
3O₂(g) → 2O₃(g) (2)
(3)
(b) From the rate of disappearance of O₂ in equation (3), we can find the rate of appearance of O₃:
So the rate of appearance of O₃ is 1.07x10⁻⁵ mol/Ls.
Have a nice day!