Answer:
10 L of CO₂.
Explanation:
The balanced equation for the reaction is given below:
2CO + O₂ —> 2CO₂
From the balanced equation above,
2 L of CO reacted to produce 2 L of CO₂.
Finally, we shall determine the volume of CO₂ produced by the reaction of 10 L CO. This can be obtained as follow:
From the balanced equation above,
2 L of CO reacted to produce 2 L of CO₂.
Therefore, 10 L of CO will also react to produce 10 L of CO₂.
Thus, 10 L of CO₂ were obtained from the reaction.
Answer: 122 moles
Procedure:
1) Convert all the units to the same unit
2) mass of a penny = 2.50 g
3) mass of the Moon = 7.35 * 10^22 kg (I had to arrage your numbers because it was wrong).
=> 7.35 * 10^22 kg * 1000 g / kg = 7.35 * 10^ 25 g.
4) find how many times the mass of a penny is contained in the mass of the Moon.
You have to divide the mass of the Moon by the mass of a penny
7.35 * 10^ 25 g / 2.50 g = 2.94 * 10^25 pennies
That means that 2.94 * 10^ 25 pennies have the mass of the Moon, which you can check by mulitiplying the mass of one penny times the number ob pennies: 2.50 g * 2.94 * 10^25 = 7.35 * 10^25.
5) Convert the number of pennies into mole unit. That is using Avogadros's number: 6.022 * 10^ 23
7.35 * 10^ 25 penny * 1 mol / (6.022 * 10^ 23 penny) = 1.22* 10^ 2 mole = 122 mol.
Answer: 122 mol
Mass to moles
5.2 mol/Ca(no3)2 to mol
5.2 mol/Ca(no3)2 / molar mass
5.2 mol/Ca(no3)2 / 164.1= 0.032 g/Ca(no3)2
Answer:
126.73 mL
Explanation:
The total pressure of the gas mixture is the sum of the vapor pressure of its constituents. So, the vapor pressure of N₂O(p) can be calculated:
750 = 18.85 + p
p = 750 - 18.85
p = 731.15 torr
It means that for 731.15 torr, N₂O occupied 130 mL. For the general gas equation, we know that
Where <em>p</em> is the pressure, <em>V</em> is the volume, <em>T</em> is the temperature, 1 is the initial state, and 2 the final state. For the same temperatue (21ºC), the equation results on Boyle's law:
p1V1 = p2V2, so:
731.15x130 = 750xV2
750V2 = 95049.5
V2 = 126.73 mL
Explanation:
Transcribed image text: H26.25 - Level 2 Homework. Unanswered Match reagents as starting materials for the synthesis of ethyl 3-phenyl-3-oxopropanate. You can draw out the structures on your own to help you answer this question. Premise Response Drag and drop to match 1 Methyl benzoate = A Acetophenone 2 Diethyl carbonate = B Ethyl acetate
