Answer is: 8568.71 of baking soda.
Balanced chemical reaction: H₂SO₄ + 2NaHCO₃ → Na₂SO₄ + 2CO₂ + 2H₂O.
V(H₂SO₄) = 17 L; volume of the sulfuric acid.
c(H₂SO₄) = 3.0 M, molarity of sulfuric acid.
n(H₂SO₄) = V(H₂SO₄) · c(H₂SO₄).
n(H₂SO₄) = 17 L · 3 mol/L.
n(H₂SO₄) = 51 mol; amount of sulfuric acid.
From balanced chemical reaction: n(H₂SO₄) : n(NaHCO₃) = 1 :2.
n(NaHCO₃) = 2 · 51 mol.
n(NaHCO₃) = 102 mol, amount of baking soda.
m(NaHCO₃) = n(NaHCO₃) · M(NaHCO₃).
m(NaHCO₃) = 102 mol · 84.007 g/mol.
m(NaHCO₃) = 8568.714 g; mass of baking soda.
The law of supply states, things remaining equal, an increase in price results in an increase in quantity supplied. The direct relation which is between the price and quantity.
Therefore, when the price of wood increases, then the producers increases the quantity which is produced as they are willing to sell more at a higher price. Each group might respond to the change in price in that, the law of demand states that all things are equal.
A decrease in a quantity leads to an increase in price which is demanded. The existence in an inverse relation which is between price and quantity.
When the demand for good increases the consumer demand decreases.
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
Answer:
12.7g of Cu
Explanation:
First let us generate a balanced equation for the reaction. This is illustrated below:
Zn + CuSO4 —> ZnSO4 + Cu
Molar Mass of Cu = 63.5g/mol
Molar Mass of CuSO4 = 63.5 + 32 + (16x4) = 63.5 + 32 + 64 = 159.5g/mol
From the equation,
159.5g of CuSO4 produced 63.5g of Cu.
Therefore, 31.9g of CuSO4 will produce = (31.9 x 63.5) / 159.5 = 12.7g of Cu
In NaMnO₄, Mn has the highest oxidation number.
The question is incomplete, the complete question is;
Which of the following species contains manganese with the highest oxidation number?
A) Mn
B) MnF₂
C) Mn₃(PO₄)₂
D) MnCl₄
E) NaMnO₄
In order to ascertain the specie that contains manganese with the highest oxidation number, we must calculate the oxidation number of manganese in each of the species one after the other.
1) For Mn, the oxidation number of Mn is zero because the atom is uncombined.
2) For MnF₂;
Mn has an oxidation number of +2
3) For Mn₃(PO₄)₂
Mn has an oxidation number of +2
4) For MnCl₄
Mn has an oxidation number of +4
5) For NaMnO₄
Mn has an oxidation number of +7
Hence in NaMnO₄, Mn has the highest oxidation number.
Learn more: brainly.com/question/10079361
