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2 years ago

A copper sulfate solution contained 0.100 moles of copper sulfate dissolved

Chemistry

1 answer:

ludmilkaskok [199]2 years ago

7 0

The mass copper sulphate CuSO₄ in 30ml solution is 957.6g.

<h3>What is the solution?</h3>

A solution is a homogeneous mixture of one or more solutes dissolved in a solvent.

To determine the mass of copper sulphate in the given solution, compare both the molarity given:

M₁V₁=M₂V₂

$\frac{n_1}{V_1} =\frac{n_2}{V_2}$

0.1/0.5=n₂/30 cm³=ml

n₂=6 moles

6×159.5=957.6g

Multiply the calculated moles with the relative mass/molecular mass to get the mass of CuSO₄.

Hence, the mass of CuSO₄ is 957.6g

Learn more about the solution equation.

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how much heat, in terms in q, would it take to produce the change in temperature indicated in the picture? what is your reasonin

STALIN [3.7K]

Answer:

1. q.

2. 2q.

3. 3q.

4. 6q.

Explanation:

We'll begin by calculating the specific heat capacity of the liquid. This can be obtained as follow:

Mass (m) = 25 g

Change in temperature (ΔT) = 20 °C

Heat (Q) = q

Specific heat capacity (C) =?

Q = MCΔT

q = 25 × C × 20

q = 500C

Divide both side by 500

C = q/500

C = 2×10¯³ qg°C

Therefore, the specific heat capacity of liquid is 2×10¯³ qg°C

Now, we shall determine the heat required to produce the various change in temperature as follow:

2. Mass (m) = 50 g

Change in temperature (ΔT) = 20 °C

Specific heat capacity (C) = 2×10¯³ qg°C

Heat (Q) =?

Q = MCΔT

Q = 50 × 2×10¯³ × 20

Q = 2q.

Therefore, the heat required is 2q.

3. Mass (m) = 25 g

Change in temperature (ΔT) = 60 °C

Specific heat capacity (C) = 2×10¯³ qg°C

Heat (Q) =?

Q = MCΔT

Q = 25 × 2×10¯³ × 60

Q = 3q.

Therefore, the heat required is 3q.

4. Mass (m) = 50 g

Change in temperature (ΔT) = 60 °C

Specific heat capacity (C) = 2×10¯³ qg°C

Heat (Q) =?

Q = MCΔT

Q = 50 × 2×10¯³ × 60

Q = 6q.

Therefore, the heat required is 6q.

4 0

2 years ago

A 225-g sample of aluminum was heated to 125.5 oc, then placed into 500.0 g water at 22.5 oc. (the specific heat of aluminum is

Sidana [21]

when heat gained = heat lost

when AL is lost heat and water gain heat

∴ (M*C*ΔT)AL = (M*C*ΔT) water

when M(Al) is the mass of Al= 225g

C(Al) is the specific heat of Al = 0.9

ΔT(Al) = (125.5 - Tf)

and Mw is mass of water = 500g

Cw is the specific heat of water = 4.81

ΔT = (Tf - 22.5)

so by substitution:

∴225* 0.9 * ( 125.5 - Tf) = 500 * 4.81 * (Tf-22.5)

∴Tf = 30.5 °C

7 0

3 years ago

Read 2 more answers

Please help asap loll

BlackZzzverrR [31]

Explanation:

Metals -

Left side of periodic table
Malleable
Ductile
Good conductor
Mostly solid
Shinly lusture

Non-metals -

Poor conductor
Dull
Brittle
Right side of periodic table

Metalloids -

Properties of metals and non-metals.
Found on the 'staircase' of the periodic table.

5 0

3 years ago

0.446 g of hydrogen gas fills a 5.0 L bag determine the density of hydrogen

Oduvanchick [21]

The density of hydrogen : ρ = 0.0892 g/L

<h3>Further explanation</h3>

Given

mass of Hydrogen : 0.446 g

Volume = 5 L

Required

The density

Solution

Density is a quantity derived from the mass and volume

Density is the ratio of mass per unit volume

The unit of density can be expressed in g/cm³, kg/m³, or g/L

Density formula:

$\large {\boxed {\bold {\rho ~ = ~ \frac {m} {V}}}}$

Input the value :

ρ = m : V

ρ = 0.446 g : 5 L

ρ = 0.0892 g/L

5 0

3 years ago

Without constants you would not know which variable affected the

bonufazy [111]

Two independent variables could change at the same time, and you would not know which variable affected the dependent variable

4 0

3 years ago