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3 years ago

What is the concentration, in m/v percent, of a solution prepared from 50 g NaCl and 2.5 L of solution?

1 answer:

4 0

When we convert the given mass in grams and volume in liters to m/v percent, we recall that m/v percent is expressed as grams/100 milliliters. In this case the expression becomes (50 grams/ 2500 L)*(0.1L/100ml), that is equal to 0.002 grams/ 100 mL. Hence the the concentration is equal to 0.2 m/v percent.

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A gas occupies 14.3 liters at a pressure of 45.0 mm Hg. What is the volume when the pressure is increased to

Simora [160]

Answer:

P1V1= P2V2

Explanation:

Inverse relationship

V2 = V1 X P1/P2

V2= 14.3 L x 45.0 mm Hg/63.0 mmHg= 8.99

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3 years ago

What are the formulas for the ionic compounds named sodium iodide and magnesium iodide

bixtya [17]

Sodium Iodide: NaI

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3 years ago

What is the mass in grams of 1.00 x 10 24 atoms of Mn?

Alex17521 [72]

<h3>Answer:</h3>

91.2 g Mn

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] 1.00 × 10²⁴ atoms Mn

<u>Step 2: Identify Conversions</u>

Avogadro's Numer

[PT] Molar Mass of Mn - 54.94 g/mol

<u>Step 3: Convert</u>

[DA] Set up: $\displaystyle 1.00 \cdot 10^{24} \ atoms \ Mn(\frac{1 \ mol \ Mn}{6.022 \cdot 10^{23} \ atoms \ Mn})(\frac{54.94 \ g \ Mn}{1 \ mol \ Mn})$
[DA] Multiply/Divide [Cancel out units]: $\displaystyle 91.2321 \ g \ Mn$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

91.2321 g Mn ≈ 91.2 g Mn

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3 years ago

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Klio2033 [76]

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3 years ago

PLEASE!!! I NEED HELP ASAP!!

Lana71 [14]

W=m₁/m₀=2^(-t/T)

t=4.6·10⁹ years
T=5·10¹⁰ years

w=2^(-4.6·10⁹/5·10¹⁰)

w=0.9382
w=93.82%

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3 years ago

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