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2 years ago

Calculate the molarity of a solution of Nach if it contains 7.2.g Nach in 100.0 mL of solution. andver: m Nach . .

Chemistry

1 answer:

Norma-Jean [14]2 years ago

5 0

Answer:

1.23 M

Explanation:

Molarity of a substance , is the number of moles present in a liter of solution .

M = n / V

M = molarity

V = volume of solution in liter ,

n = moles of solute ,

Moles is denoted by given mass divided by the molecular mass ,

Hence ,

n = w / m

n = moles ,

w = given mass ,

m = molecular mass .

From the question ,

w = given mass of NaCl = 7.2 g

As we know , the molecular mass of NaCl = 58.5 g/mol

Moles is calculated as -

n = w / m = 7.2 g / 58.5 g/mol = 0.123 mol

Molarity is calculated as -

V = 100ml = 0.1 L (since , 1 ml = 1/1000L )

M = n / V = 0.123 mol / 0.1 L = 1.23 M

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2 years ago

What is the density of lead (in g/cm^3 3 ) if a rectangular bar measuring 0.500 cm in height, 1.55 cm in width, and 25.00 cm in

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Answer:

Density = 11.4 g/cm³

Explanation:

Given data:

Density of lead = ?

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Length of lead bar = 25.00 cm

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2 years ago

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2 years ago

Read 2 more answers

Suppose the gas resulting from the sublimation of 1.00 g carbon dioxide is collected over water at 25.0◦c into a 1.00 l containe

jeyben [28]

Answer:

0.55 atm

Explanation:

First of all, we need to calculate the number of moles corresponding to 1.00 g of carbon dioxide. This is given by

$n=\frac{m}{M_m}$

where

m = 1.00 g is the mass of the gas

Mm = 44.0 g/mol is the molar mass of the gas

Substituting,

$n=\frac{1.00 g}{44.0 g/mol}=0.0227 mol$

Now we can find the pressure of the gas by using the ideal gas law:

$pV=nRT$

where

p is the gas pressure

V = 1.00 L is the volume

n = 0.0227 mol is the number of moles

R = 0.082 L/(atm K mol) is the gas constant

T = 25.0 C + 273 = 298 K is the temperature of the gas

Solving the formula for p, we find

$p=\frac{nRT}{V}=\frac{(0.0227 mol)(0.082 L/(atm K mol))(298 K)}{1.00 L}=0.55 atm$

8 0

3 years ago

How many moles are in 3.113 g of Au?Molar mass of Au=197 g/mol

SVEN [57.7K]

<h3>Answer:</h3>

0.0157 g Au

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

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table
Using Dimensional Analysis

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<h3>Explanation:</h3>

<u>Step 1: Define</u>

3.113 g Au

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

Molar Mass of Au - 197.87 g/mol

<u>Step 3: Convert</u>

<u /> $3.113 \ g \ Au(\frac{1 \ mol \ Au}{197.87 \ g \ Au} )$ = 0.015733 g Au

<u>Step 4: Check</u>

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

0.015733 g Au ≈ 0.0157 g Au

3 0

2 years ago