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Lelechka [254]
2 years ago
7

A particular sample of pure iron requires 0.612 kJ of energy to raise its temperature from 30.°C to 51°C. What must be the mass

of the sample? (See the table below.)
Chemistry
1 answer:
mixas84 [53]2 years ago
3 0

Answer:

m = 65.637 g

Explanation:

Heat = 0.612 kJ = 612 J ( Converting to J by multiplying by 1000)

Initial Temperature =  30.°C

Final Temperature = 51°C

Temperature change = Final Temperature - Initial Temperature = 51 - 30 = 21°C

Mass = ?

The relationship between these quantities is given by the equation;

H = mCΔT

where c = 0.444 J/g°C

Inserting the values in the equation;

612 = m *  0.444 * 21

m = 612 / (0.444 * 21)

m = 65.637 g

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If the distance covered by a jogger is 2,541meters through the park and the time it took to cover that distance was 43.6 minutes
Tatiana [17]
43.6x60=y amount of seconds.

speed = distance divided by time


Speed = 2541 divided by y amount of seconds
4 0
3 years ago
Use Le Chatelier’s principle to predict how the equilibrium concentration of the FeCl2+ ion will change when an aqueous solution
GarryVolchara [31]

Answer: towards the reactant

Explanation:

5 0
3 years ago
What is the percent composition of a 16.75 g sample of a compound containing 14.02 g oxygen and 2.73 g hydrogen? Select the corr
TEA [102]

Answer:

83.7% oxygen, 16.3% hydrogen

Explanation:

Mass of compound = 16.75 g

Mass of Oxygen = 14.02 g

Mass of Hydrogen =  2.73 g

Percentage composition = Mass of element /  Mass of compound    100%

Percentage composition of Oxygen = 14.02 g / 16.75 g    * 100%

Percentage composition of Oxygen = 0.837 * 100 = 83.7%

Percentage composition of Hydrogen = 2.73 g / 16.75 g    * 100%

Percentage composition of Hydrogen = 0.163 * 100 = 16.3%

The correct option is;

83.7% oxygen, 16.3% hydrogen

7 0
2 years ago
Read 2 more answers
How much carbon dioxide in kilograms is produced upon the complete combustion of 29.4 L of propane (approximate contents of one
Novosadov [1.4K]

Answer:

61kg

Explanation:

Like all other hydrocarbons, the combustion of propane will yield water and carbon iv oxide.

Let’s write a complete and balanced chemical equation for this.

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)

Now, we first convert the information into a mass.

We know that mass = density * volume

Let’s convert the volume here into ml since density is in mL. It must be remembered that 1000ml = 1L. Hence, 29.4L = 29,400ml

The mass is thus = 0.6921 * 29,400 = 20,347.74g

Now from the balanced equation, we can see that one mole of propane produced 3 moles of carbon iv oxide. This is the theoretical relation.

Let’s now calculate the actual number of moles. The number of moles of propane produced is the mass of propane divided by the molar mass of propane. The molar mass of propane is 3(12) + 8(1) = 36 + 8 = 44g/mol

The number of moles = 20,347.74/44 = 462.44

We simply multiply this number by 3 to get the actual number of moles of carbon dioxide produced. This is equals 1387.35 moles

Since we know this, we now calculate the mass of carbon iv oxide produced. The mass is equals the number of moles multiplied by the molar mass. The molar mass of carbon iv oxide is 44g/mol.

The mass produced is thus 1387.35 * 44 = 61,043g

To get the kilogram equivalent, we simply divide by 1000 = 61,043/1000 = 61kg

6 0
3 years ago
Three moles of neon expand isothermally from 0.153 m3 to 0.319 m3 while 4.86 × 103 J of heat flows into the gas. Assuming that n
goldfiish [28.3K]

Answer:

T= 265 degree

Explanation:

Given. n = 3

Vf = 0.319m3.

Vi = 0.153m3.

Q = 4.86*10^3 J

For external process, change in internal energy is zero

W= Q= 4.86*10^3

Also work done for isothermal process is given by

W= nRT in(vf/vi)

T= w/nR in(vf/vi)

T = 4.86*10^3/3*8.314in(0.319/0.153)

T = 265degree

4 0
3 years ago
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