Answer:
The answer to your question is the third option
Explanation:
Chemical reaction
4NH₃ + 5O₂ ⇒ NO + H₂O
Count the number of elements in the reactants and products
Reactants Elements Products
4 N 1
12 H 2
10 O 2
All the elements are unbalanced
Let's add the following coefficients
4NH₃ + 5O₂ ⇒ 4NO + 6H₂O
Reactants Elements Products
4 N 4
12 H 12
10 O 10
Now the reaction is balanced
Hello my friend,
since atomic no. = no. of electrons
there will be 40 electrons in that atom.
if and element has 111 protons, atomic no. will be 111.
no. of neutron = 161 - 55 = 106 neutrons.
Answer:
0.5 × 10²³ atoms of iodine
Explanation:
Given data:
Mass of calcium iodide = 12.75 g
Number of atoms of iodine = ?
Solution:
First of all we will calculate the number of moles of calcium iodide.
Number of moles = mass/ molar mass
Number of moles = 12.75 g/ 293.9 g/mol
Number of moles = 0.04 mol
In one mole of calcium iodide there are two moles of iodine.
Thus in 0.04 moles:
0.04 mol × 2 = 0.08 moles of iodine
Now we will use the 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 atoms
0.08 moles of iodine × 6.022 × 10²³ atoms / 1 mol
0.5 × 10²³ atoms of iodine.
Well, there are 1,000 milligrams in a gram so 0.625 grams is 625 milligrams
<span>this is a limiting reagent problem.
first, balance the equation
4Na+ O2 ---> 2Na2O
use both the mass of Na and mass of O2 to figure out how much possible Na2O you could make.
start with Na and go to grams of Na2O
55.3 gNa x (1molNa/23.0gNa) x (2 molNa2O/4 molNa) x (62.0gNa2O/1molNa2O) = 75.5 gNa2O
do the same with O2
64.3 gO2 x (1 molO2/32.0gO2) x (2 molNa2O/1 mol O2) x (62.0gNa2O/1molNa2O) = 249.2 g Na2O
now you must pick the least amount of Na2O for the one that you actually get in the reaction. This is because you have to have both reacts still present for a reaction to occur. So after the Na runs out when it makes 75.5 gNa2O with O2, the reaction stops.
So, the mass of sodium oxide is
75.5 g</span>