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3 years ago

What is the molarity of a solution that contains 2.38 g of h2c2o4 * 2h2o in exactly 300 ml of solution?

1 answer:

6 0

Data Given:
Mass of H₂C₂O₂.2H₂<span>O = 2.38 g

M.mass of </span>H₂C₂O₂.2H₂O = 126 g/mol

Volume = 300 ml = 0.3 dm³

Solution:
Molarity is given as,

Molarity = Moles / Vol. of Sol.
Or,
Molarity = (Mass / M.mass) × (1 / Vol)
Putting Values,
Molarity = ( 2.38 g ÷ 126 g/mol ) × ( 1 ÷ 0.3 dm³ )

Molarity = 0.063 mol/dm³

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(01.07 но) An experiment was conducted to determine the density of a liquid. The table shows a partial record of the experiment

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Answer:

How do you find the density of a liquid experiment?

To measure the density of a liquid you do the same thing you would for a solid. Mass the fluid, find its volume, and divide mass by volume. To mass the fluid, weigh it in a container, pour it out, weigh the empty container, and subtract the mass of the empty container from the full container.

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2 years ago

Why is Cellular Respiration so important?

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3 years ago

A sample of metal has a mass of 5. 2 g and absorbs 20. 0 j of energy as it is heated from 30. 0°c to 40. 0°c. What is the identi

kiruha [24]

The identity of the metal is copper.

<h3>What is specific heat?</h3>

The amount of energy needed to raise the temperature of one gram of a substance by one degree Celsius.

Using the formula of specific heat

H = mcdT

Where, H = Heat absorbed

m = mass of the metal

c = specific heat capacity of the metal

dT = temperature change

Putting the values in the equation

20 J = 0.0052 Kg × c × ( 40.0°C - 30.0°C)

c = 20 J/0.0052 Kg × ( 40.0°C - 30.0°C)

c = 385 JKg-1°C-1

Thus, the metal is copper.

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2 years ago

A student calculated the density of a sample of graphite to be 2.3 g/cm3. Show a numerical setup for calculating the student’s p

LUCKY_DIMON [66]

Answer:

Percent error = 1.5%

Explanation:

Given data:

Measured value of density of graphite = 2.3 g/cm³

Percent error = ?

Solution:

Formula:

Percent error = [Measured value - Actual value / actual value] × 100

Actual/accepted value of density of graphite = 2.266 g/cm³

Now we will put the values:

Percent error = [2.3 g/cm³ - 2.266 g/cm³ / 2.266 g/cm³] × 100

Percent error = [0.034 g/cm³ / 2.266 g/cm³] × 100

Percent error = 0.015 × 100

Percent error = 1.5%

8 0

3 years ago

Read 2 more answers

1‑Propanol ( P⁰ 1 = 20.9 Torr at 25 ⁰C ) and 2‑propanol ( P⁰ 2 = 45.2 Torr at 25 ⁰C ) form ideal solutions in all proportions. L

anastassius [24]

Answer : The mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.

Explanation : Given,

Vapor presume of 1‑Propanol $(P^o_1)$ = 20.9 torr

Vapor presume of 2‑Propanol $(P^o_2)$ = 45.2 torr

Mole fraction of 1‑Propanol $(x_1)$ = 0.540

Mole fraction of 2‑Propanol $(x_2)$ = 1-0.540 = 0.46

First we have to calculate the partial pressure of 1‑Propanol and 2‑Propanol.

$p_1=x_1\times p^o_1$

where,

$p_1$ = partial vapor pressure of 1‑Propanol

$p^o_1$ = vapor pressure of pure substance 1‑Propanol

$x_1$ = mole fraction of 1‑Propanol

$p_1=(0.540)\times (20.9torr)=11.3torr$

and,

$p_2=x_2\times p^o_2$

where,

$p_2$ = partial vapor pressure of 2‑Propanol

$p^o_2$ = vapor pressure of pure substance 2‑Propanol

$x_2$ = mole fraction of 2‑Propanol

$p_2=(0.46)\times (45.2torr)=20.8torr$

Thus, total pressure = 11.3 + 20.8 = 32.1 torr

Now we have to calculate the mole fraction of vapor phase 1‑Propanol and 2‑Propanol.

$\text{Mole fraction of 1-Propanol}=\frac{\text{Partial pressure of 1-Propanol}}{\text{Total pressure}}=\frac{11.3}{32.1}=0.352$

and,

$\text{Mole fraction of 2-Propanol}=\frac{\text{Partial pressure of 2-Propanol}}{\text{Total pressure}}=\frac{20.8}{32.1}=0.648$

Thus, the mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.

3 0

3 years ago