If 0.400 moles CO and 0.400 moles O2 completely react, 17.604 grams of CO2 would be produced.
First, let us look at the balanced equation of reaction:
According to the equation, the mole ratio of CO and O2 is 2:1. But in reality, the mole ratio supplied is 1:1. Thus, CO is the limiting reactant while O2 is in excess.
Also from the equation, the ratio of CO consumed to that of CO2 produced is 1:1. Thus, 0.400 moles of CO2 would also be produced from 0.400 moles of CO.
Recall that: mole = mass/molar mass
Therefore, the mass in grams of CO2 that would be produced from 0.400 moles can be calculated as:
Mass = mole x molar mass
= 0.400 x 44.01
= 17.604 grams
More on calculating mass from number of moles can be found here: brainly.com/question/12513822
<span>M(NO3)2 ==> [M2+] + 2 [NO3-]
0.202 M ==> 0.202 M
M(OH)2 ==> [M2+] + 2[OH-]
5.05*10^-18 ===> s + [2s]^2
5.05*10^-18 ===> 0.202 + [2s]^2
5.05*10^-18 = 0.202 * 4s^2
4s^2 = 25*10^-18
s^2 = 6.25*10^-18
s = 2.5*10^-9
So, the solubility is 2.5*10^-9</span>
The molarity of a solution equals to the mole number of the solute/the volume of the solution. For NH4Br, we know that the mole mass is 98. So the molarity is (14/98) mol /0.15 L=0.95 mol/L.
Answer:
According to Coulomb’s law, the Ca and Se ions have 4 times the attractive force (2+ × 2-) than that of the K and Br ions (1+ × 1-).
Explanation:
From Coulomb's law, the attractive force between calcium and selenium ions is four times the attractive force between potassium and bromide ions.
This has something to do with size and magnitude of charge. Calcium ions and selenide ions are smaller and both carry greater charge magnitude than potassium and bromide ions. This paves way for greater electrostatic attraction between them when the distance of the charges apart is minimal. Hence a greater lattice energy.
Answer:
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