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2 years ago

If a 67.3G rock is dissolved in 2.00L of acid, what is the molar concentration of gold in the acid solution

Chemistry

1 answer:

brilliants [131]2 years ago

6 0

Answer:

[Au] = 0.171 M

Explanation:

For this question, we assume the rock is 100 % gold.

First of all, we determine the moles of gold

67.3 g . 1mol/ 196.97g = 0.342 moles

Molar concentration is defined as the moles of solute, contained in 1L of solution.

Our solution volume is 2L.

M = 0.342 mol / 2L = 0.171

Molar concentration, also called molarity of solution is the most typical unit of concentration.

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What cause the ratio of PV/RT to drop below 1 when 1 mole of

pentagon [3]

Answer:

pressure and temperature are directly proportional.

Explanation:

At constant temperature and pressure the volume of a gas is directly proportional to the number of moles of gas. Also given a constant temperature and volume the pressure of a gas is directly proportional to the number of moles of gas.

This means that if V is constant then

P = n (RT/V) then n= PV/RT. As P is increased, T is also increased thereby decreasing the value of n since pressure and temperature are directly proportional.

6 0

3 years ago

The net ionic equation for the reaction between aqueous nitric acid and aqueous sodium hydroxide is ________. h+ (aq) + na+ (aq)

Nana76 [90]

HNO3+NaOH ----> H2O
H⁺ +NO3⁻+Na⁺+OH⁻ ---> Na⁺ +NO3⁻ +H2O
H⁺ (aq)+OH⁻(aq)----> H2O(l)

7 0

3 years ago

Determine the freezing point and boiling point of a solution that has 68.4 g of sucrose

Ymorist [56]

Answer:

Freezing T° of solution = - 3.72°C

Boiling T° of solution = 101.02°C

Explanation:

To solve this we apply colligative properties. Firstly, freezing point depression:

ΔT = Kf . m . i

ΔT = Freezing T° of pure solvent - Freezing T° of solution

Kf = Cryoscopic constant, for water is 1.86 °C/m

m = molality (moles of solute in 1kg of solvent)

i = Ions dissolved in solution

Our solute is sucrose, an organic compound so no ions are defined. i = 1.

Let's determine the moles: 68.4 g . 1mol/ 342g = 0.2 moles

molality = 0.2 mol / 0.1kg of water = 2 m

We replace data: ΔT = 1.86°C/m . 2m . 1

Freezing T° of solution = - 3.72°C

Now, we apply elevation of boiling point: ΔT = Kb . m . i

ΔT = Boiling T° of solution - Boiling T° of pure solvent

Kf = Ebulloscopic constant, for water is 0.512 °C/m

We replace:

Boiling T° of solution - Boiling T° of pure solvent = 0.512 °C/m . 2 . 1

Boiling T° of solution = 0.512 °C/m . 2 . 1 + 100°C → 101.02°C

6 0

2 years ago

What is the volume, in liters, of 0.500 mol of c3h8 gas at stp? (hint..use avogadro’s principle to solve this)?

wolverine [178]

When we are at STP conditions, we can use this conversion: 1 mol= 22.4 L

0.500 mol C₃H₈ (22.4 L/ 1 mol)= 11.2 L

3 0

3 years ago

Read 2 more answers

Agustina was standing on the edge of a cliff of unknown height holding a bowling ball. She wanted to find the height of the clif

nlexa [21]

Answer:

433 m

Explanation:

Since the fall represents motion under gravity, we use the equation

s = ut - 1/2gt² where s = height of cliff or distance bowling ball falls through, u = initial velocity of bowling ball = 0 m/s(since it starts from rest), t = time = 9.4 s and g = acceleration due to gravity = -9.8 m/s².

So, substituting the values of the variables into the equation, we have

s = 0 m/s × 9.4 s - 1/2 × 9.8 m/s² × (9.4 s)²

s = 0 m - 1/2 × -9.8 m/s² × 88.36 s²

s = 1/2(865.928 m)

s = 432.964

s ≅ 433 m

5 0

3 years ago