The physical explanation is that increasing temperature increases the kinetic energy of the gas molecules. Hence, their random motion breaks more intermolecular bonds and the gas is less dissolved in the solvent. In contrast, solid solutes in water have increased solubility with increased temperatures.
<u>Answer:</u>
2400 mL
<u>Explanation:</u>
According to this equation, the stoichiometric ratio between 
and 
for the complete reaction is 1:2.
We know that the number of moles of can be calculated using the mole formula. (<em>number of moles = mass / molar mass</em>)
Moles of Calcium = 
= 1.5 mol
So the moles of = 
= 3.0 mol
<em>Volume of HCl solution = Moles of HCl/ concentration of HCl</em>
Volume of HCl solution = 
= 2400 mL
Answer:
Statement 1: All living matter at the smallest level is made of cells
Explamation:
All living things are made of cells; the cell itself is the smallest fundamental unit of structure and function in living organisms.
Hope this helps
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
Molar mass (NH₄)₂CO₃ = <span>96.09 g/mol
1 mole ---------> 96.09 g
0.500 moles ----> ?
0.500 * 96.09 = 48.045 g of </span><span>(NH₄)₂CO₃
hope this helps!
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