Answer: Use Avogadro's number, NA = 6.02 × 1023 atoms/mol.
Explanation:
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
A beta particle is an electron and it has a -1 charge and zero mass.
Beta decay by
emitting an electron is called as β⁻
decay. When this happens, a neutron of the element converts into a proton by
emitting an electron. Hence, the mass of daughter nucleus is same as parent
atom but atomic number/number of protons is higher by 1 than atomic number of
parent atom.
In a β⁻ decay, the symbol is used as ₋₁⁰β or ₋₁⁰e.
-1 is for charge
<span> 0 is for the mass of the particle
</span>
Answer:
25.89 × 10²³ molecules
Explanation:
Given data:
Mass of CoCl₂ = 560 g
Number of molecules present = ?
Solution:
Number of moles of CoCl₂:
Number of moles = mass/molar mass
Number of moles = 560 g/ 129.84 g/mol
Number of moles = 4.3 mol
Avogadro number:
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance. The number 6.022 × 10²³ is called Avogadro number.
1 mole = 6.022 × 10²³ molecules
4.3 mol × 6.022 × 10²³ molecules /1 mol
25.89 × 10²³ molecules
Answer: iron (oxides of iron)
Explanation: manganese nodules usually contain layers of iron and manganese oxides in a concentric arrangement and are called polymetallic nodules.
