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4 years ago

2. A student wishes to find the density of an irregularly shaped rock. He finds the mass to be 12.34

Chemistry

1 answer:

lukranit [14]4 years ago

7 0

Answer:

The answer is

Explanation:

The Density of a substance can be found by using the formula

$density = \frac{mass}{volume}$

From the question

mass of rock = 12.34 g

volume = final volume of water - initial volume of water

volume = 54.5 mL - 50 mL = 4.5 mL

So the density is

$density = \frac{12.34}{4.5} \\ = 2.744444444...$

We have the final answer as

Hope this helps you

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Suppose 0.245 g of sodium chloride is dissolved in 50. mL of a 18.0 m M aqueous solution of silver nitrate.

Bezzdna [24]

Answer:

$\large \boxed{\text{ 0.066 mol/L}}$

Explanation:

We are given the amounts of two reactants, so this is a limiting reactant problem.

1. Assemble all the data in one place, with molar masses above the formulas and other information below them.

Mᵣ: 58.44

NaCl + AgNO₃ ⟶ NaNO₃ + AgCl

m/g: 0.245

V/mL: 50.

c/mmol·mL⁻¹: 0.0180

2. Calculate the moles of each reactant

$\text{Moles of NaCl} = \text{245 mg NaCl} \times \dfrac{\text{1 mmol NaCl}}{\text{58.44 mg NaCl}} = \text{4.192 mmol NaCl}\\\\\text{ Moles of AgNO}_{3}= \text{50. mL AgNO}_{3} \times \dfrac{\text{0.0180 mmol AgNO}_{3}}{\text{1 mL AgNO}_{3}} = \text{0.900 mmol AgNO}_{3}$

3. Identify the limiting reactant

Calculate the moles of AgCl we can obtain from each reactant.

From NaCl:

The molar ratio of NaCl to AgCl is 1:1.

$\text{Moles of AgCl} = \text{4.192 mmol NaCl} \times \dfrac{\text{1 mmol AgCl}}{\text{1 mmol NaCl}} = \text{4.192 mmol AgCl}$

From AgNO₃:

The molar ratio of AgNO₃ to AgCl is 1:1.

$\text{Moles of AgCl} = \text{0.900 mmol AgNO}_{3} \times \dfrac{\text{1 mmol AgCl}}{\text{1 mmol AgNO}_{3}} = \text{0.900 mmol AgCl}$

AgNO₃ is the limiting reactant because it gives the smaller amount of AgCl.

4. Calculate the moles of excess reactant

Ag⁺(aq) + Cl⁻(aq) ⟶ AgCl(s)

I/mmol: 0.900 4.192 0

C/mmol: -0.900 -0.900 +0.900

E/mmol: 0 3.292 0.900

So, we end up with 50. mL of a solution containing 3.292 mmol of Cl⁻.

5. Calculate the concentration of Cl⁻

$\text{[Cl$^{-}$] } = \dfrac{\text{3.292 mmol}}{\text{50. mL}} = \textbf{0.066 mol/L}\\\text{The concentration of chloride ion is $\large \boxed{\textbf{0.066 mol/L}}$}$

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3 years ago

Which substance is a heterogeneous mixture? Suspension Solution Compound Colloid

aleksklad [387]

Answer:

Suspension

Explanation:

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4 years ago

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Answer:

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3 years ago

CH3OH can be synthesized by the reaction:

Leviafan [203]

Answer:

Explanation:

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3 years ago

Chlorine gas is held in a container with volume of 1.15L. The container initially held 3.53 x 10^−2 mol of chlorine gas, but aft

Paha777 [63]

Answer:

The grams that were removed are 0.787

Explanation:

We can raise two situations, with Ideal Gases Law.

First (S1): P. 1.15L =0.0353 . R .T

Second (S2): P. 0.790L = n . R .T

If we compare them, we must make a division, so we can know the moles of gas, that are still remained in the container.

S1/S2

P . 1.15L / P. 0.790L = 0.0353 . R.T / n . R. T

Pressure, temperature and R, are constant, so they can be cancelled.

So finally we have.

1.15L / 0.790L = 0.0353 moles / n

n = 0.0353 moles (0.790L /1.15L)

n = 0.0242 moles

Initially we have 0.0353 moles; afterwards we have 0.0242 moles. In the reaction we lost (0.0353 - 0.0242) = 0.0111 moles

The grams that were removed are:

Moles . molar mass = mass

0.0111 m . 70.9 g/m = 0.787 g

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3 years ago