Answer:
The answer to your question is 0.005
Explanation:
Data
Volume of NaOH = 25 ml
[NaOH] = 0.2 M
moles of NaOH = ?
To solve this problem is not necessary to have the chemical reaction. Just use the formula of Molarity and solve it for moles.
Formula
Molarity = moles / volume
-Solve for moles
moles = Molarity x volume
-Convert volume to liters
1000 ml ---------------- 1 l
25 ml ---------------- x
x = (25 x 1) / 1000
x = 0.025 l
-Substitution
moles = 0.2 x 0.025
-Result
moles = 0.005
Your closes answer would be a.10
Answer:
14.33 g
Explanation:
Solve this problem based on the stoichiometry of the reaction.
To do that we need the molecular weight of the masses involved and then calculate the number of moles, find the limiting reagent and finally calculate the mass of AgCl.
2 AgNO₃ + CaCl₂ ⇒ Ca(NO₃)₂ + 2 AgCl
mass, g 6.97 6.39 ?
MW ,g/mol 169.87 110.98 143.32
mol =m/MW 0.10 0.06 0.10
From the table above AgNO₃ is the limiting reagent and we will produce 0.10 mol AgCl which is a mass :
0.10 mol x 143.32 g/mol = 14.33 g
The empirical formula : C₁₂H₄F₇
The molecular formula : C₂₄H₈F₁₄
<h3>Further explanation</h3>
mol C (MW=12 g/mol)
mol H(MW=1 g/mol) :
mol F(MW=19 g/mol)
mol ratio of C : H : O =1.52 : 0.51 : 0.89=3 : 1 : 1.75=12 : 4 : 7
Empirical formula : C₁₂H₄F₇
(Empirical formula)n=molecular formula
( C₁₂H₄F₇)n=562 g/mol
(12.12+4.1+7.19)n=562
(281)n=562⇒ n =2
Molecular formula : C₂₄H₈F₁₄
1.205 × 10²³ atoms of oxygen will be present in 7.51 grams of glycine with formula C₂H5O2N. Details about number of atoms can be found below.
How to calculate number of atoms?
The number of atoms of a substance can be calculated by multiplying the number of moles of the substance by Avogadro's number.
However, the number of moles of oxygen in glycine can be calculated using the following expression:
Molar mass of C₂H5O2N = 75.07g/mol
Mass of oxygen in glycine = 32g/mol
Hence; 32/75.07 × 7.51 = 3.2grams of oxygen in glycine
Moles of oxygen = 3.2g ÷ 16g/mol = 0.2moles
Number of atoms of oxygen = 0.2 × 6.02 × 10²³ = 1.205 × 10²³ atoms
Therefore, 1.205 × 10²³ atoms of oxygen will be present in 7.51 grams of glycine with formula C₂H5O2N.
Learn more about number of atoms at: brainly.com/question/8834373
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