Answer:
There is no air, wind, liquid water or life on the moons surface.
Explanation:
Explanation:
The given data is as follows.
Now, according to Michaelis-Menten kinetics,
![V_{o} = V_{max} \times [\frac{S}{(S + Km)}]](https://tex.z-dn.net/?f=V_%7Bo%7D%20%3D%20V_%7Bmax%7D%20%5Ctimes%20%5B%5Cfrac%7BS%7D%7B%28S%20%2B%20Km%29%7D%5D)
where, S = substrate concentration = 
M
Now, putting the given values into the above formula as follows.
![V_{o} = 6.8 \times 10^{-10} \mu mol/min \times [\frac{10.4 \times 10^{-6} M}{(10.4 \times 10^{-6}M + 5.2 \times 10^{-6} M)}]](https://tex.z-dn.net/?f=V_%7Bo%7D%20%3D%206.8%20%5Ctimes%2010%5E%7B-10%7D%20%5Cmu%20mol%2Fmin%20%5Ctimes%20%5B%5Cfrac%7B10.4%20%5Ctimes%2010%5E%7B-6%7D%20M%7D%7B%2810.4%20%5Ctimes%2010%5E%7B-6%7DM%20%2B%205.2%20%5Ctimes%2010%5E%7B-6%7D%20M%29%7D%5D)
= 
This means that 
would approache 
.
Answer:
CO32-(aq) + 2H+(aq) → CO2(g) + H2O(l)
Explanation:
According to this question, sodium carbonate reacts with sulfuric acid to form aqueous sodium sulfate, carbon dioxide and water. The balanced chemical equation is as follows:
Na2CO3(aq) + H2SO4(aq) → Na2SO4(aq) + CO2(g) + H2O(l)
- Next, split compounds that are aqueous into ions.
2Na+(aq) + CO32-(aq) + 2H+(aq) + SO42-(aq) → 2Na+(aq) + SO42-(aq) + CO2(g) + H2O(l)
- Next, we cancel out the spectator ions, which are ions that remain the same in the reactants and products side of a chemical reaction. The spectator ions in this equation are 2Na+(aq) and SO42-(aq).
CO32-(aq) + 2H+(aq) → CO2(g) + H2O(l)
- Hence, the balanced ionic equation is as follows:
CO32-(aq) + 2H+(aq) → CO2(g) + H2O(l)
C3H8+3O2--->3CO2+8H
Therefore for every 1:3 there are 3 Carbon dioxides that form. That means find the limiting reactant from the two reactants.
5.5g(1mole C3H8/44.03g of C3H8)=0.1249 moled of C3H8 and if for every one C3H8 we can form three CO2. We can assume 0.3747 miles of CO2 will be produced.
15g of O2(1 mole O2/32g of O2)=0.4685moles O2 and if for every three O2 we can produce three CO2 we may assume a 1:1 ratio.
This means C3H8 will be your limiting reactant. Therefore 0.3747 moles of CO2 will be produced.
0.3747 moles of CO2(48.01 g of CO2/1 mole of CO2)= 17.99 grams of CO2
