Answer:
Mass = 29.23 g
Explanation:
Given data:
Volume of solution = 814.2 mL 814.2/1000 = 0.8142 L)
Molarity of solution = 0.227 M
Mass of solute in gram = ?
Solution:
Molarity = number of moles / volume in L
By putting values,
0.227 M = number of moles / 0.8142 L
Number of moles = 0.227 M × 0.8142 L
Number of moles = 0.184 mol
Mass in gram:
Mass = number of moles × molar mass
Molar mass of calcium acetate = 158.17 g/mol
Mass = 0.184 mol × 158.17 g/mol
Mass = 29.23 g
Answer:
22 mol
Explanation:
Given data:
Number of atoms of Cl = 2.65×10²⁵ atom
Number of moles of Cl = ?
Solution:
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
2.65×10²⁵ atom × 1 mol / 6.022 × 10²³ atoms
0.44×10² mol
22 mol
Answer:
Coefficient of 
Coefficient of
=8
Explanation:
We are given that a reaction in which
reacts with 
We have to find the coefficient of each reactants in balanced reaction

Coefficient is defined the constant value multiplied with a reactant in a reaction.
Coefficient of
=3
Coefficient of 
Coefficient of 
Coefficient of 
Coefficient of 
Coefficient of KOH=2
Hence, Coefficient of
and coefficient of 
Answer:
2Ag⁺ (aq) + 2OH⁻ (aq) → Ag₂O (s) + H₂O (l)
Explanation:
Step 1: RxN
2AgNO₃ + 2NaOH → Ag₂O + 2NaNO₃ + H₂O
Step 2: Define states of matter
2AgNO₃ (aq) + 2NaOH (aq) → Ag₂O (s) + 2NaNO₃ (aq) + H₂O (l)
Step 3: Total Ionic Equation
2Ag⁺ (aq) + 2NO₃⁻ (aq) + 2Na⁺ (aq) + 2OH⁻ (aq) → Ag₂O (s) + 2Na⁺ (aq) + 2NO₃⁻ (aq) + H₂O (l)
Step 4: Cancel out spectator ions
2Ag⁺ (aq) + 2OH⁻ (aq) → Ag₂O (s) + H₂O (l)
Answer:
<em>so mass in gram=560grams</em>
Explanation:
number of moles=10moles
molar mass=56grams/moles
mass in gram of Fe=?
as we know that

<em>evaluating the formula</em>
<em>number of moles×molar mass=mass in gram</em>
<em>mass in gram=10moles×56grams/moles</em>
<em>mass in gram=560grams</em>
<em>i hope this will help you :)</em>