one mole of P weights about 31 grams
in one mole there are 6.022*10^23 atoms
we use the rule of threes
6.022*10^23atoms......weight..........31 grams
3.45*10^23 atoms.........weight...........x grams
x=(3.45*10^23*31)/6.022*10^23
x=106.95/6.022=<u><em>17.76 grams</em></u>
<h3><u>Answer;</u></h3>
Directly proportional
<h3><u>Explanation;</u></h3>
- <em><u>Concentration is one of the factors that determine the rate of a reaction. Reaction rates increases with increase in the concentration of the reactants, which means they are directly proportional.</u></em>
- An increase in the concentration of reactants produces more collisions and thus increasing the rate at which the reaction is taking place. Therefore, <u>Increasing the concentration of a reactant increases the frequency of collisions between reactants and will cause an increase in the rate of reaction.</u>
Answer:
The rate of disappearance of C₂H₆O = 2.46 mol/min
Explanation:
The equation of the reaction is given below:
2 K₂Cr₂O₇ + 8 H₂SO₄ + 3 C₂H₆O → 2 Cr₂(SO₄)₃ + 2 K₂SO₄ + 11 H₂O
From the equation of the reaction, 3 moles of C₂H₆O is used when 2 moles of Cr₂(SO₄)₃ are produced, therefore, the mole ratio of C₂H₆O to Cr₂(SO₄)₃ is 3:2.
The rate of appearance of Cr₂(SO₄)₃ in that particular moment is given 1.64 mol/min. This would than means that C₂H₆O must be used up at a rate which is approximately equal to their mole ratios. Thus, the rate of of the disappearance of C₂H₆O can be calculated from the mole ratio of Cr₂(SO₄)₃ and C₂H₆O.
Rate of disappearance of C₂H₆O = 1.64 mol/min of Cr₂(SO₄)₃ * 3 moles of C₂H₆O / 2 moles of Cr₂(SO₄)₃
Rate of disappearance of C₂H₆O = 2.46 mol/min of C₂H₆O
Therefore, the rate of disappearance of C₂H₆O = 2.46 mol/min
2-butane is the correct answer
(C.) Each carbon atom in ethane forms 4
