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tekilochka [14]
3 years ago
10

When 47.2 J of heat is added to 13.4 g of a liquid, its

Chemistry
1 answer:
leva [86]3 years ago
8 0

Answer: 2.02 J/g°C

Explanation:

To find the heat capacity, we have to manipulate the equation for heat.

q=mCΔT becomes C=q/(mΔT) to find heat capacity. Since we are given our values, we can plug in to find C.

C=\frac{47.2J}{13.4g*1.74°C}

*Please ignore the capital A in front of the °C. In order to have ° in the equaiton, the A pops up.

C=2.02J/g°C

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To find the formula of a compound composed of iron and carbon monoxide, Fex(CO)y, the compound is burned in pure oxygen, an reac
Anarel [89]

Answer:

The empirical formula is: Fe(CO)₅

Explanation:

According the global reaction:

Feₓ(CO)y + O₂ → Fe₂O₃ + CO₂

You should calculate Fe₂O₃ and CO₂ moles, thus:

0,799 Fe₂O₃ grams  × \frac{1 mole}{159.69 Fe2O3 g} = 5,00×10⁻³ Fe₂O₃ moles

2,200 CO₂ grams  × \frac{1 mole}{44,01 CO2 g} = 5,00×10⁻²CO₂ moles

The ratio between Fe₂O₃ moles and CO₂ moles is 1:10. Thus ratio between x and y must be 1:5 because Fe₂O₃ has 2 irons but CO₂ has just one carbon.

Assuming the formula is Fe₁(CO)₅ the molecular weight is 195,9 g/mol. Thus:

1,959 Fe(CO)₅ grams  × \frac{1 mole}{195,9 Fe(CO)5 g} = 1,00×10⁻² Fe(CO)₅ moles

Thus, assuming 1,00×10⁻² moles as basis for calculation, the global reaction is:

1 Fe(CO)₅ + ¹³/₂O₂ → ¹/₂ Fe₂O₃ + 5 CO₂

With this balanced equation the moles produced have sense, thus, the empirical formula is: Fe(CO)₅

I hope it helps!

8 0
3 years ago
What is the major source of energy for the production of electricity in this facility ​
MariettaO [177]
Fossil fuels are mostly the major source of energy
4 0
2 years ago
Write the formula for calcium chloride​
Sauron [17]

CaCI2 is  for calcium chloride

7 0
3 years ago
A 23.5g aluminum block is warmed to 65.9°C and plunged into an insulated beaker containing 55.0g water initially at 22.3°C. The
atroni [7]

Answer:

25.97oC

Explanation:

Heat lost by aluminum = heat gained by water

M(Al) x C(Al) x [ Temp(Al) – Temp(Al+H2O) ] = M(H2O) x C(H2O) x [ Temp(Al+H2O) – Temp(H2O) ]

Where M(Al) = 23.5g, C(Al) = specific heat capacity of aluminum = 0.900J/goC, Temp(Al) = 65.9oC, Temp(Al+H2O)= temperature of water and aluminum at equilibrium = ?, M(H2O) = 55.0g, C(H2O)= specific heat capacity of liquid water = 4.186J/goC

Let Temp(Al+H2O) = X

23.5 x 0.900 x (65.9-X) = 55.0 x 4.186 x (X-22.3)

21.15(65.9-X) = 230.23(X-22.3)

1393.785 - 21.15X = 230.23X – 5134.129

230.23X + 21.15X = 1393.785 + 5134.129

251.38X = 6527.909

X = 6527.909/251.38

X = 25.97oC

So, the final temperature of the water and aluminum is = 25.97oC

4 0
3 years ago
A sample of helium gas is placed in a rigid cylinder that has a movable piston. The volume of the gas is varied by moving the pi
Nataliya [291]
Helium is a light gas.  Movement can be predicted by which chamber has the lighter gas(es) and which chamber has the heavier.  constant changes can also be a variable.  If the gas is constant, then changes are more easily predicted.
5 0
3 years ago
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