The dilution factor : 100
Further explanationThe dilution factor(DF) : ratio of concentration of stock solution and diluted solution or ratio of the volume of final solution to the initial volume from stock solution
From equation of dilution :
M₁V₁=M₂V₂
[tex]\tt DF=\dfrac{M_1}{M_2}=\dfrac{V_2}{V_1}[/tex]
Volume of aliquot = 0.1 ml
Volume of diluent = 9.9 ml
So final volume :
[tex]\tt V_f(final~volume)=volume~aliquot+volume~diluent\\\\V_f=0.1+9.9=10~ml[/tex]
The dilution factor (DF) :
[tex]\tt D_f=\dfrac{V_f}{V_i}\\\\D_f=\dfrac{10}{0.1}=100[/tex]
Will give brainliest
Answer:
c. both gases have the same number of molecules
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How many grams are in 2.3 x 1024 formula units of KNO3?
390 g KNO₃
General Formulas and Concepts:ChemistryAtomic Structure
Reading a Periodic TableUsing Dimensional AnalysisAvogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.MathPre-Algebra
Order of Operations: BPEMDAS
Brackets Parenthesis Exponents Multiplication Division Addition Subtraction Left to Right Explanation:Step 1: Define
2.3 × 10²⁴ formula units KNO₃
Step 2: Identify Conversions
Avogadro's Number
Molar Mass of K - 39.10 g/mol
Molar Mass of N - 14.01 g/mol
Molar Mass of O - 16.00 g.mol
Molar Mass of KNO₃ - 39.10 + 14.01 + 3(16.00) = 101.11 g/mol
Step 3: Convert
[tex]2.3 \cdot 10^{24} \ formula \ units \ KNO_3(\frac{1 \ mol \ KNO_3}{6.022 \cdot 10^{23} \ formula \ units \ KNO_3} )(\frac{101.11 \ g \ KNO_3}{1 \ mol \ KNO_3} )[/tex] = 386.172 g KNO₃
Step 4: Check
We are given 2 sig figs. Follow sig fig rules and round.
386.172 g KNO₃ ≈ 390 g KNO₃
How many moles of copper are in 6,000,000 atoms of copper?
Answer: There will be 9.9632 × 10⁻¹⁸ moles of Copper in 6,000,000 atoms of Copper.
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