How many moles are in 2.8 g of Mg?

a 125 g chunk of aluminum at 182 degrees Celsius was added to a bucket filled with 365 g of water at 22.0 degrees Celsius. Ignor

Diano4ka-milaya [45]

<h3>Answer:</h3>

32.98°C

<h3>Explanation:</h3>

We are given the following;

Mass of Aluminium as 125 g

Initial temperature of Aluminium as 182°C

Mass of water as 265 g

Initial temperature of water as 22°C

We are required to calculate the final temperature of the two compounds;

First, we need to know the specific heat capacity of each;

Specific heat capacity of Aluminium is 0.9 J/g°C

Specific heat capacity of water is 4.184 J/g°C

<h3>Step 1: Calculate the Quantity of heat gained by water.</h3>

Assuming the final temperature is X°C

we know, Q = mcΔT

Change in temperature, ΔT = (X-22)°C

therefore;

Q = 365 g × 4.184 J/g°C × (X-22)°C

= (1527.16X-33,597.52) Joules

<h3>Step 2: Calculate the quantity of heat released by Aluminium </h3>

Using the final temperature, X°C

Change in temperature, ΔT = -(X°- 182°)C (negative because heat was lost)

Therefore;

Q = 125 g × 0.90 J/g°C × (182°-X°)C

= (20,475- 112.5X) Joules

<h3>Step 3: Calculating the final temperature</h3>

We need to know that the heat released by aluminium is equal to heat absorbed by water.

Therefore;

(20,475- 112.5X) Joules = (1527.16X-33,597.52) Joules

Combining the like terms;

1639.66X = 54072.52

X = 32.978°C

= 32.98°C

Therefore, the final temperature of the two compounds will be 32.98°C

7 0

4 years ago

How many grams of methionine (MW = 149.21) are needed to make 20 mL of a 150 mM solution?

Leni [432]

Answer:

0.45 g

Explanation:

Step 1: Given data

Molar mass of methionine (M): 149.21 g/mol

Volume of the solution (V): 20 mL

Concentration of the solution (C): 150 mM

Step 2: Calculate the moles of methionine (n)

We will use the following expression.

n = C × V

n = 150 × 10⁻³ mol/L × 20 × 10⁻³ L

n = 3.0 × 10⁻³ mol

Step 3: Calculate the mass of methionine (m)

We will use the following expression.

m = n × M

m = 3.0 × 10⁻³ mol × 149.21 g/mol

m = 0.45 g

7 0

3 years ago

What is the product of radioactive decay 4/2he called?

riadik2000 [5.3K]

A large unstable nucleus becomes tinier and emits an alpha particle during Alpha decay. 4/2He or the alpha particle that consists of two protons and two neutrons can be discharged from the nucleus during the disintegration of radioactive decay. It has a charge of positive 2.

3 0

3 years ago

Read 2 more answers

A group of students removed the plastic insulation around a wire. They wrapped the uninsulated wire around a steel rivet 50 time

maksim [4K]

Answer:

d. add another d-cell to the circuit

Explanation:

To make the electromagnet pick something, the student should add another d-cell to the circuit.

Simply, the electromagnet set up is generating very weak magnetic fields.

To produce an electromagnet, there must be an interaction between electric fields an conductor.
As the wire turns around in the vicinity of the electric field, it induces magnetism.
The number of turns of wire here is pretty considerable.
But the electric field is very weak.
Additional d-cell in the circuit will circumvent this problem

8 0

3 years ago

What is the volume of 3.66 x 1032 molecules of fluorine gas at STP?

kipiarov [429]

Answer:

1.36x10^10L

Explanation:

Step 1:

Determination of the mole of fluorine that contains 3.66x10^32 molecules. This is shown below:

From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.02x10^23 molecules. This implies that 1 mole of fluorine also contains 6.02x10^23 molecules.

Now if 1 mole of fluorine contains 6.02x10^23 molecules,

Therefore, Xmol of fluorine will contain 3.66x10^32 molecules i.e

Xmol of fluorine = 3.66x10^32/6.02x10^23

Xmol of fluorine = 6.08x10^8 moles

Step 2:

Determination of the volume occupied by 6.08x10^8 moles of fluorine.

1 mole of any gas occupy 22.4L at stp. This means that 1 mole of fluorine also occupy 22.4L at stp.

Now if 1 mole of fluorine occupies 22.4L at stp,

Then 6.08x10^8 moles of fluorine will occupy = 6.08x10^8 x 22.4 = 1.36x10^10L

6 0

3 years ago