Ka and Kb values of weak acids and weak bases are small.
This is because weak acids and weak bases do not dissociate completely, favoring the reactants more than dissociating into the product of H+ or OH-.
Answer:
344.21 g/mol
General Formulas and Concepts:
<u>Chemistry - Atomic Structure</u>
Explanation:
<u>Step 1: Define</u>
Cr₂(SO₃)₃
<u>Step 2: Identify</u>
Molar Mass of Cr - 52.00 g/mol
Molar Mass of S - 32.07 g/mol
Molar Mass of O - 16.00 g/mol
<u>Step 3: Find</u>
Molar Mass of Cr₂(SO₃)₃ - 2(52.00) + 3(32.07) + 9(16.00) = 344.21 g/mol
The atmospheric pressure will be:
The pressure of the atmosphere resulting from the mercury column is 0.959 atm
What is atmospheric pressure?
The force that an object experiences from the weight of the air above it per unit area are known as atmospheric pressure.
Given: Height of mercury column = 729 mm Hg
To find: The pressure of the atmosphere
Calculation:
The atmospheric column resulting from the mercury column is calculated as follows:
1 atm =760 mm Hg
So, we can convert the 729 mm Hg to atm, and we get
Atmospheric pressure = 729 x 1 atm / 760 = 0.959 atm
Learn more about atmospheric pressure here,
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Answer:
1. 15.71 g CO2
2. 38.19 % of efficiency
Explanation:
According to the balanced reaction (2 CO(g) + O2(g) → 2 CO2(g)), it is clear that the CO is the limitant reagent, because for every 2 moles of CO we are using only 1 mole of O2, so even if we have the same quantity for both reagents, not all of the O2 will be consumed. This means that we can just use the stoichiometric ratios of the CO and the CO2 to solve this question, and for that we need to convert the gram units into moles:
For CO:
C = 12.01 g/mol
O = 16 g/mol
CO = 28.01 g/mol
(10.0g CO) x (1 mol CO/28.01 g) = 0.3570 mol CO
For CO2:
C = 12.01 g/mol
O = 16 x 2 = 32 g/mol
CO2 = 44.01 g/mol
We now that for every 2 moles of CO we are going to get 2 moles of CO2, so we resolve as follows:
(0.3570 mol CO) x (2 mol CO2/2 mol CO) = 0.3570 moles CO2
We are obtaining 0.3570 moles of CO2 with the 10g of CO, now lets convert the CO2 moles into grams:
(0.3570 moles CO2) x (44.01 g/1 mol CO2) = 15.71 g CO2
Now for the efficiency question:
From the previous result, we know that if we produce 15.71 CO2 with all the 10g of CO used, we would have an efficiency of 100%. So to know what would that efficiency be if we would only produce 6g of CO2, we resolve as follows,
(6g / 15.71g) x 100 = 38.19 % of efficiency
