Answer:- 171 g
Solution:- It asks to calculate the grams of sucrose required to make 1 L of 0.5 Molar solution of it.
We know that molarity is moles of solute per liter of solution.
If molarity and volume is given then, moles of solute is molarity times volume in liters.
moles of solute = molarity* liters of solution
moles of solute = 0.5*1 = 0.5 moles
To convert the moles to grams we multiply the moles by molar mass.
Molar mass of sucrose = 12(12) + 22(1) + 11(16)
= 144 + 22 + 176
= 342 grams per mol
grams of sucrose required = moles * molar mass
grams of sucrose required = 0.5*342 = 171 g
So, 171 g of sucrose are required to make 1 L of 0.5 molar solution.
Answer: The name given to
is Gallium (III) sulfate.
Explanation: This is an ionic compound because in aqueous solution it dissociates into its respective ions.
Naming of Ionic compounds.
- Name the cation first and then write its oxidation number in roman numerical.
- Then name the anion or polyatomic ions without writing any prefix of the number of atoms present in it.
- The name of the anion should have a suffix '-ide' like for chlorine, the name will be chloride etc..
- For polyatomic ions, the suffix used will be '-ate' like for
the name will be sulfate etc..
Name of the given ionic compound is Gallium (III) Sulfate.
The molecular formula : C₁₈H₁₈N₈
<h3>Further explanation</h3>
Given
62.41% C, 5.24% H, and 32.36% N
Required
The molecular formula
Solution
mol ratio
C : 62.41/12.0096 = 5.1967
H : 5.24/1.00784 = 5.1992
N : 32.36/14.0067 = 2.310
Divide by 2.310(smallest)
C : 5.1967/2.31=2.25
H : 5.1992/2.31 = 2.25
N : 2.31/2.31 = 1
Multiplied by 4
C : H : N = 9 : 9 : 4
The empirical formula : C₉H₉N₄
(C₉H₉N₄)n=346.40 g/mol
(12.0096 x 9 + 1.00784 x 9 + 14.0067 x 4)n=346.4
(108.0864+9.07056+56.0268)n=346.4
(173.184)n=346.4
n=2
<em>The molecular formula : C₁₈H₁₈N₈</em>
Answer:
0.720 M/s
Explanation:
Let's consider the following balanced equation.
5Br⁻(aq) + BrO⁻₃(aq) + 6H⁺(aq) → 3 Br₂(aq) + 3H₂O(l)
The molar ratio of Br⁻ to Br₂ is 5:3, that is, when 5 moles of Br⁻ are consumed, 3 moles of Br₂ are produced. If the average rate of consumption of Br⁻ is 1.20 × 10⁻⁴ M/s, the average rate of formation of Br₂ is:
