<span>(a) Assuming the amount of O2(g) is not limiting the reaction, a mass of 4.23g of Ca(s) will produce an equal mass of CaO(s), hence it will produce 4.23g of CaO(s).
According to their respective molar masses, we have the following CaO molar mass :
Molar mass of Ca + Molar mass of O = 40.1 + 16 = 56.1 g/mol
4.23g of Ca will then produce : 4.23 / 56.1 = 0,07540107 mol of CaO.
(b) With the same reasonment as above, and assuming the amount of Ca is not limiting, we have :
2.87g of O2 will produce : 2.87 / 56.1 = 0,051158645 mol of CaO.
(c) From (a) and (b) answers, we can conclude that the reactant that produces less mol of CaO is limiting the reaction. Hence following the given masses, O2 is the limiting reactant.
(d) Knowing the molar mass of CaO is 56.1 g/mol and knowing that O2 is the limiting reactant, we also know the reaction can produce a maximum of 0,051158645 mol of CaO can be produced.
So we can conclude we will produce :
56.1 * 0,051158645 = 2.87g of CaO.</span>
um where the work nothing is popping up for me to answer
Answer:
Explanation:
You can infere the units of the rate constant for an <em>overall order 2 reaction </em>departing from the simplified case of a two identical reactants combined in a single elementary step:
- A + A → Products (single elementary step equation)
For that equation the differenttial form of the rate of reaction is:
- r = - d [A] / dt = K [A]²
The units of [A] are M¹ (i.e. M), and the units of - d [A] / dt are M¹ s⁻¹ . . . ( M stands for molar concentration)
Substituting the unis on both sides of the rate equation you get:
Solving for K:
- K = M¹ s⁻¹ M⁻² = M⁻¹ s⁻¹, which is the final result.
Answer:
(a) 0.0345 g/cm^3
Explanation:
<u>Given:</u>
Edge length of cube (a) = 0.750 m
Mass = 14.56 kg
<u>To determine:</u>
The density of the cube
<u>Calculation:</u>
Unit conversions:
1000 g = 1 kg
Therefore, the mass of cube in g = 14.56*10³ g
10⁶ cm³ = 1 m³
Therefore the volume of the cube in cm³=0.422*10⁶cm³
Based on equation (1)
The answer is B.
ATP + Oxygen ---> ADP + P