The mass of water that will be produced if 10.54 g of H₂ react with 95.10 g of O₂ is 94.86 g
2H₂ + O₂ —> 2H₂O
Molar mass of H₂ = 2 × 1 = 2 g/mol
Mass of H₂ from the balanced equation = 2 × 2 = 4 g
Molar mass of O₂ = 2 × 16 = 32 g/mol
Mass of O₂ from the balanced equation = 1 × 32 = 32 g
Molar mass of H₂O = (2×1) + 16 = 18 g/mol
Mass of H₂O from the balanced equation = 2 ×18 = 36 g
SUMMARY
From the balanced equation above,
4 g of H₂ reacted with 32 g of O₂ to produce 36 g of H₂O
How to determine the limiting reactant
From the balanced equation above,
4 g of H₂ reacted with 32 g of O₂
Therefore,
10.54 g of H₂ will react with = (10.54 × 32) / 4 = 84.32 g of O₂
From the calculation made above, we can see that only 84.32 g out of 95.10 g of O₂ given, is needed to react completely with 10.54 g of H₂.
Therefore, H₂ is the limiting reactant.
From the balanced equation above,
4 g of H₂ reacted to produce 36 g of H₂O
Therefore,
10.54 g of H₂ will react to produce = (10.54 × 36) / 4 = 94.86 g of H₂O
Thus, 94.86 g of H₂O were obtained from the reaction.
Learn more about stoichiometry:
Explanation:
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Answer:
pH= 0.369
Explanation:
The formic acid reacts with NaOH as
HCOOH + NaOH= HCOONa + H2O
Apply CaVa/ CbVb = Na/Nb
Ca×25/(29.8×0.3587) = 1/1
Ca= (29.8×0.3587)/25= 0.428M
pH = - log(H+)
Since only 0ne H+ is in the stoichiometric equation, it means H+ = 0.428M
pH = -log(0.428) =0.369
- When [S] << Km, the reaction is second order and V0 depends on [S] and [Et].
- Their kcat is a second order rate constant.
- The lower their Km, the better they recognize their substrate, but the lower their reaction rate.
- When [S] << Km, V0 depends on [S] and [Et].
Answer:
1. True. 2. True. 3. Not true. 4. True. 5. True
Explanation:
1. Yes, because if the amount of substrate i much greater than of competitive inhibitor then the probability of substrate to bind to ferment is much higher than of inhibitor (if we have noncompetitive inhibitor it damages the structure of active site and the substrate concentration does not have a role in reaction rate).
2. Yeah, because then the michaelis-menten equation will transform into [tex} V0=(kcat*[E]*[S])/Km [/tex] and it is a second order equation.
3. No, because it is measured in sec-1 and that means it is 1 rate constant.
4. True, if the lower Km the better is binding and due to that rate is slower because it's harder for substrate to unbind.
5. The same as question two.
Explanation:
1s 2s 2p
1s 2s 2p1s2 2s1
1s 2s 2p1s2 2s1 2 1 1
1s 2s 2p1s2 2s1 2 1 11s2 2s2
2) Columnar
3) Squamous
The type of epithelial cells that would be found in an area of the body where diffusion of nutrients needs to occur is squamous.
In conclusion, squamous can be found on the surface of the skin.
Read related link on:
Answer: C. Squamous
Explanation:
I got it right on my test...lol
Answer: The volume that gas occupy will be 820.04mL.
Explanation:
To calculate the volume of the gas at different temperature, we will use the equation given by Charles' Law.
This law states that volume is directly proportional to the temperature of the gas at constant pressure and number of moles.
or,
where,
are the initial volume and initial temperature of the gas.
are the final volume and final temperature of the gas.
We are given:
Putting values in above equation:
Hence, the volume that gas occupy will be 820.04mL.