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Montano1993 [528]
3 years ago
15

ΔH for the reaction below is -826.0 kJ/mol. Calculate the heat change when a 69.03-g sample of iron is reacted.4Fe(s) + 3 O2(g)

--> Fe2O3(s)a. - 255.2 kJb. -510.5 kJc. -1020.9 kJd. -2042 kJe. -2.851 x 10^4 kJ
Chemistry
1 answer:
Anuta_ua [19.1K]3 years ago
4 0

Answer:

c. -1020.9 kJ

Explanation:

4Fe (s) + 3 O₂ (g) --> 2 Fe₂O₃(s)         ΔH  =  -826.0 kJ/mol.

atomic weight of iron = 56

69.03 g = 69.03 / 56

= 1.23268 moles

Heat released by 1.23268 moles

= 1.23268 x 826.0

= -1020.9 kJ .

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Given the balanced equation below, calculate the moles of aluminum that are needed to react completely with 28.7 moles of FeO. Y
Alexxx [7]

Answer:

43.05 moles of Al needed to react with 28.7 moles of FeO.

Explanation:

Given data:

Moles of FeO = 28.7 mol

Moles of Al needed to react with FeO = ?

Solution:

Chemical equation:

2Al + 3FeO → 3Fe + Al₂O₃

Now we will compare the moles of Al with FeO.

                            FeO        :           Al

                             2            :            3

                          28.7          :         3/2×28.7 = 43.05 mol

Thus 43.05 moles of Al needed to react with 28.7 moles of FeO.

8 0
3 years ago
Incomplete dash of fuels produces many poisonous dash​
tangare [24]

Answer:

Incomplete combustion of fuels produces a very poisonous gas called carbon monoxide:Excessive inhaling of carbon dioxide gas can kill a person.

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6 0
3 years ago
Identify the type of molecule shown in the picture
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3 years ago
At 2°C, the vapor pressure of pure water is 23.76 mmHg and that of a certain seawater sample is 23.09 mmHg. Assuming that seawat
wariber [46]

Answer:

0.808  M

Explanation:

Using Raoult's Law

\frac{P_s}{Pi}= x_i

where:

P_s = vapor pressure of sea water( solution) = 23.09 mmHg

P_i = vapor pressure of pure water (solute) = 23.76 mmHg

x_i = mole fraction of water

∴

\frac{23.09}{23.76}= x_i

x_i = 0.9718

x_i+x_2=1

x_2 = 1- x_i

x_2 = 1- 0.9718

x_2 = 0.0282

x_i = \frac{n_i}{n_i+n_2}  ------ equation (1)

x_2 = \frac{n_2}{n_i+n_2}  ------ equation (2)

where; (n_2) = number of moles of sea water

(n_i) = number of moles of pure water

equating above equation 1 and 2; we have :

\frac{n_2}{n_i}= \frac{0.0282}{0.9178}

= 0.02901

NOW, Molarity =  \frac{moles of sea water}{mass of pure water }*1000

= \frac{0.02901}{18}*1000

= 0.001616*1000

= 1.616 M

As we assume that the sea water contains only NaCl, if NaCl dissociates to Na⁺ and Cl⁻; we have \frac{1.616}{2} =0.808 M

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3 years ago
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