You would get 35.................
Answer:
Noble gases are a unique set of elements in the periodic table because they don't naturally bond with other elements.
Explanation:
HAVE A GOOD DAY!
Answer:
d and e
Explanation:
We have 5 solutions with different molar concentrations, that is, the quotient between the number of moles of solute and the liters of solution. This can be expressed as mol/L or M. The most dilute would be the one having the less number of moles of solute per liters of solution, that is, solution d or e, which have the same concentration. If we order them from the most diluted to the most concentrated, we get:
d = e < a < b < c
Answer: 1) Maximum mass of ammonia 198.57g
2) The element that would be completely consumed is the N2
3) Mass that would keep unremained, is the one of the excess Reactant, that means the H2 with 3,44g
Explanation:
- In order to calculate the Mass of ammonia , we first check the Equation is actually Balance:
N2(g) + 3H2(g) ⟶2NH3(g)
Both equal amount of atoms side to side.
- Now we verify which reagent is the limiting one by comparing the amount of product formed with each reactant, and the one with the lowest number is the limiting reactant. ( Keep in mind that we use the molecular weight of 28.01 g/mol N2; 2.02 g/mol H2; 17.03g/mol NH3)
Moles of ammonia produced with 163.3g N2(g) ⟶ 163.3g N2(g) x (1mol N2(g)/ 28.01 g N2(g) )x (2 mol NH3(g) /1 mol N2(g)) = 11.66 mol NH3
Moles of ammonia produced with 38.77 g H2⟶ 38.77 g H2 x ( 1mol H2/ 2.02 g H2 ) x (2 mol NH3 /3 mol H2 ) = 12.79 mol NH3
- As we can see the amount of NH3 formed with the N2 is the lowest one , therefore the limiting reactant is the N2 that means, N2 is the element that would be completey consumed, and the maximum mass of ammonia will be produced from it.
- We proceed calculating the maximum mass of NH3 from the 163.3g of N2.
11.66 mol NH3 x (17.03 g NH3 /1mol NH3) = 198.57 g NH3
- In order to estimate the mass of excess reagent, we start by calculating how much H2 reacts with the giving N2:
163.3g N2 x (1mol N2/28.01 g N2) x ( 3 mol H2 / 1 mol N2)x (2.02 g H2/ 1 mol H2) = 35.33 g H2
That means that only 35.33 g H2 will react with 163.3g N2 however we were giving 38.77g of H2, thus, 38.77g - 35.33 g = 3.44g H2 is left
Answer:
1.18×10²³ atoms.
Explanation:
From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.02×10²³ atoms.
From the above concept, 1 mole of sodium also contains 6.02×10²³ atoms.
1 mole of sodium = 23 g.
Thus,
23 g of sodium contains 6.02×10²³ atoms.
Therefore, 4.5 g of sodium will contain = (4.5 × 6.02×10²³)/23 = 1.18×10²³ atoms.
From the above calculation,
4.5 g of sodium contains 1.18×10²³ atoms.
