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tamaranim1 [39]
3 years ago
15

How much heat (kJ) is absorbed by 229.1 g of water in order for the temperature to increase from 25.00∘C to 32.50∘C?

Chemistry
1 answer:
hammer [34]3 years ago
4 0

Answer:

(Q1) 9.42 kJ.

(Q2) 1.999 kJ

Explanation:

Heat: This is a form of Energy that brings about the sensation of warmth.

The S.I unit of Heat is Joules (J).

The heat of a body depend on the mass of the body, specific heat capacity, and temperature difference. as shown below

Q = cm(t₂-t₁) ........................ Equation 1

(Q1)

Q = cm(t₂-t₁)

Where Q = amount of heat absorbed, c = specific heat capacity of water, m = mass of water, t₁ = initial temperature, t₂ = final temperature.

Given: m = 229.1 g = 0.2991 kg, t₁ = 25.0 °C, 32.50 °C

Constant: c = 4200 J/kg.°C

Substituting into equation 1

Q = 0.2991×4200(32.5-25)

Q = 1256.22(7.5)

Q = 9421.65 J

Q = 9.42 kJ.

Hence the heat absorbed = 9.42 kJ

(Q2)

Q = cm(t₂-t₁)

Where Q = amount of heat required, c = specific heat capacity of water, m = mass of water, t₁ = initial temperature, t₂ = final temperature.

Given: m = 34 g = 0.034 kg, t₁ = 9 °C, t₂ = 23 °C

Constant: c = 4200 J/kg.°C

Q = 0.034×4200(23-9)

Q = 142.8(14)

Q = 1999.2 J

Q = 1.999 kJ.

Thus the Heat required = 1.999 kJ

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LenKa [72]

We need to measure 20.0 grams of CaCl₂ to prepare 500 mL of 0.360 M solution.

First, we need to determine the required moles of CaCl₂. We have 500 mL (0.500 L) of a 0.360 M solution (0.360 moles of CaCl₂ per liter of solution).

0.500 L \times \frac{0.360mol}{L} = 0.180 mol

Then, we will convert 0.180 moles to grams using the molar mass of CaCl₂ (110.98 g/mol).

0.180 mol \times \frac{110.98g}{mol} = 20.0 g

To prepare the solution, we weigh 20.0 g of CaCl₂ and add it to a beaker with enough distilled water to dissolve it. We stir it, heat it if necessary, and when we have a solution, we transfer it to a 500 mL flask and complete it to the mark with distilled water.

We need to measure 20.0 grams of CaCl₂ to prepare 500 mL of 0.360 M solution.

You can learn more about solutions here: brainly.com/question/2412491

4 0
2 years ago
Consider a 100-gram sample of radioactive cobalt-60.
svetlana [45]

It will take 5.2 years to decay.

The half life of cobalt-60 is 5.2 years. The half life is the time taken for the mass of the substance to decrease by a half.

here, the amount of remaining substance is 50%,

so, (\frac{1}{2} )^{n} = 0.5

n. log (0.5) = log (0.5)

n = 1

So it would take 1 half lives to decay this much, which is 1 x 5.2 which is 5.2 years.

what do you mean by radioactive decay ?

The process through which an unstable atomic nucleus loses energy via radiation is known as radioactive decay, also known as nuclear decay, radioactivity, radioactive disintegration, or nuclear disintegration. A material that has unstable nuclei is considered as radioactive.

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7 0
2 years ago
For 2,663 kg of a compound with the formula Al(SO), determine the following quantities (4 pts each); a) The number of moles of t
Nastasia [14]

Answer:

a) 35.485 moles of Al(SO)

b) 35.485 moles of S atoms

c) 2.136197(10^{25}) Al atoms

d) 567.723 g of O

Explanation:

Let's define the following terms :

1 mol = 6.02.(10^{23}) elemental units

For example :

1 mol of oxygen atoms = 6.02.(10^{23}) oxygen atoms

Now, our compound has the following formula

Al(SO)

Where Al is aluminium

S is sulfur

And O is oxygen

All the subscripts are 1 so we can say the following :

1 molecule of Al(SO) has 1 atom of Al , 1 atom of S and 1 atom of O

In terms of moles :

1 mol of Al(SO) has 1 mol of Al , 1 mol of S and 1 mol of O

The molar masses of Al, S and O are

molarmass_{(Al)}=26.982\frac{g}{mol}

molarmass_{(S)}=32.065 \frac{g}{mol}

molarmass_{(O)}=15.999\frac{g}{mol}

If we sum all the molar masses =(26.982+32.065+15.999)\frac{g}{mol}=75.046\frac{g}{mol}

Finally, 75.046 g of Al(S0) is 1 mol of Al(SO) which contains 26.982 g of Al, 32.065 g of S and 15.999 g of O.

1 mol of Al(SO) contains 1 mol of Al, 1 mol of S and 1 mol of O.

Now we can calculate a),b),c) and d)

For a)

2.663 kg=2663g

75.046 g of Al(SO) = 1 mol of Al(SO)

2663 g of Al(SO) = x

x=\frac{2663}{75.046}mol=35.485 mol

2.663 kg of Al(SO) contains 35.485 moles of Al(SO)

b) and c) 1 mol of Al(SO) molecules contains 1 mol of S atoms and 1 mol of Al atoms

We have 35.485 moles of Al(SO) molecules so

We have 35.485 moles of S atoms

And 35.485 moles of Al atoms

If 1 mol = 6.02(10^{23})

35.485 moles of Al have (35.485)(6.02)(10^{23})=2.136197(10^{25}) Al atoms

d) 75.046 g of Al(SO) contains 15.999 g of O

2663 g of Al(SO) contains x g of O

x=\frac{(2663).(15.999)}{75.046} g

x = 567.723 g of O

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