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1 year ago

Calculate the mass (in grams) of 250mL of ether at 25 oC. The density of

Chemistry

1 answer:

leonid [27]1 year ago

6 0

Volume=250mL
Density=0.71g/ml

$\boxed{\sf Density=\dfrac{Mass}{Volume}}$

$\\ \sf{:}\implies Mass=Density(Volume)$

$\\ \sf{:}\implies Mass=0.71(250)$

$\\ \sf{:}\implies Mass=177.5g$

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Answer:

Atomic number = 10

Mass number = 20

Explanation:

Mass number = neutrons+protons

Mass number = 10+10 = 20

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2 years ago

Read 2 more answers

Calculate the molarity of each solution.

weeeeeb [17]

Q1)
molarity is defined as the number of moles of solute in 1 L solution
the number of moles of LiNO₃ - 0.38 mol
volume of solution - 6.14 L
since molarity is number of moles in 1 L
number of moles in 6.14 L - 0.38 mol
therefore number of moles in 1 L - 0.38 mol / 6.14 L = 0.0619 mol/L
molarity of solution is 0.0619 M

Q2)
the mass of C₂H₆O in the solution is 72.8 g
molar mass of C₂H₆O is 46 g/mol
number of moles = mass present / molar mass of compound
the number of moles of C₂H₆O - 72.8 g / 46 g/mol
number of C₂H₆O moles - 1.58 mol
volume of solution - 2.34 L
number of moles in 2.34 L - 1.58 mol
therefore number of moles in 1 L - 1.58 mol / 2.34 L = 0.675 M
molarity of C₂H₆O is 0.675 M

Q3)

Mass of KI in solution - 12.87 x 10⁻³ g
molar mass - 166 g/mol
number of mole of KI = mass present / molar mass of KI
number of KI moles = 12.87 x 10⁻³ g / 166 g/mol = 0.0775 x 10⁻³ mol
volume of solution - 112.4 mL
number of moles of KI in 112.4 mL - 0.0775 x 10⁻³ mol
therefore number of moles in 1000 mL- 0.0775 x 10⁻³ mol / 112.4 mL x 1000 mL
molarity of KI - 6.90 x 10⁻⁴ M

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1 year ago

Why is it important to have regular supervision of the weight and measurements in the market

Sergio [31]

Answer:

Supervision of weights and measures promotes accurate measurements of goods and services to ensure that everybody gets a fair trade in the marketplace. Not so coincidentally it also is a deterrent to ensure that traders are being honest in their trade practises.

Explanation:

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1 year ago

A 5.00 liter gas sample is collected at a temperature and pressure of 27.0 degree C and 1.20 atm. It is desired to transfer the

kipiarov [429]

Answer:

The temperature of the gas in the 3.00 liter container, must be 150K

Explanation:

Let's apply the Ideal Gas Law, to find out the moles

P . V = n . R . T

1,20 atm . 5L = n . 0,082 L.atm/mol.K . 300 K

(1,20 atm . 5L) / (0,082 mol.K/L.atm . 300 K) = n

6/24,6 mol = n = 0,244 moles

We have the moles now, so let's find the temperature in our new conditions.

P . V = n . R . T

1 atm . 3L = 0,244 moles . 0,082 L.atm/mol.K . T° in K

(1 atm . 3L / 0,244 moles . 0,082 mol.K/L.atm) = T° in K

3/20,008 K = T° in K = 150K

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1 year ago

What volume in<br><br> L<br><br> of a 0.724 M Nal solution contains 0.405 mol of Nal ?

Amanda [17]

Answer:

$0.5594\ \text{L}$

Explanation:

Mol of NaI = 0.405 mol

Molarity of solution = 0.724 M

Molarity is given by

$M=\dfrac{\text{mol}}{\text{Volume of solution in }NaI}\\\Rightarrow \text{Volume of solution in }NaI=\dfrac{0.405}{0.724}\\\Rightarrow \text{Volume of solution in }NaI=0.5594\ \text{L}$

The required volume is $0.5594\ \text{L}$ .

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1 year ago