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sweet [91]
3 years ago
13

Which of the following elements can't have an expanded octet? answers A. oxygen B. phosphorous C. chlorine d. sulfer

Chemistry
2 answers:
dybincka [34]3 years ago
6 0

answer is oxygen .

oxygen is an exception in octet rule

kupik [55]3 years ago
4 0

Answer:

Oxygen

Explanation:

Sulfer, Phosphorus, and chlorine can have an expanded octet.

Oxygen can't have an expanded octet.

So, the answer is oxygen.

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If all of the energy from burning 281.0 g of propane (ΔHcomb,C3H8 = –2220 kJ/mol) is used to heat water, how many liters of wate
lapo4ka [179]

This problem is providing us with the mass of propane, its enthalpy of combustion, and the initial and final temperature of water that can be heated from the burning of this fuel. At the end, the result turns out to be 42.27 L.

<h3>Combustion:</h3>

In chemistry, combustion reactions are based on the burning of fuels by using oxygen and producing both carbon dioxide and water. For propane, we will have:

C_3H_8+5O_2\rightarrow 3CO_2+4H_2O

Hence, we can calculate the heat released from this reaction by using the mass, which has to be converted to moles, and the given enthalpy of combustion:

Q=281.0g*\frac{1mol}{44.09g}*-2220\frac{kJ}{mol}*\frac{1000J}{1kJ}\\ \\ Q=-1.415x10^7 J

<h3>Calorimetry:</h3>

In chemistry, we can analyze the mass-specific heat-temperature-heat relationship via the most general heat equation:

Q=mC\Delta T

Thus, since Q was obtained from the previous problem, but the sign change because the released heat is now absorbed by the water, one can calculate the mass of water that rises from 20.0°C to 100.0°C with this heat:

m=\frac{Q}{C\Delta T} =\frac{1.415x10^7J}{4.184\frac{J}{g\°C}(100.0\°C-20.0\°C)}\\ \\m=4.227x10^4g

Finally, we convert it to liters as required:

V=4.227 x10^4g*\frac{1mL}{1.00g}*\frac{1L}{1000mL}  \\\\V=42.27L

Learn more about calorimetry: brainly.com/question/1407669

4 0
2 years ago
How many moles of Cl– ions are in 2.20 L of 1.50 M of sodium-chloride aqueous solution?
xxMikexx [17]

Answer:

How many moles of Cl– ions are in 2.20 L of 1.50

7 0
3 years ago
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The answer is letter A i think
3 0
3 years ago
Tin(IV) sulfide, SnS2, a yellow pigment, can be produced using the following reaction.
Maksim231197 [3]

The theoretical yield of SnS_2 will be 4.20 grams while the percent yield will be 7.93%

<h3>How is yield calculated?</h3>

From the equation of the reaction, the mole ratio of SnBr_4 to Na_2S is 1:2.

Mole of 48.1 mL, 0.478 M  SnBr_4 = 0.478 x 48.1/100 = 0.023 mols

Mole of 48.8 mL, 0.160 M   Na_2S = 0.160 x 48.8/1000 = 0.0078 moles

SnBr_4Na_2S is the limiting reactant.

Mole ratio of  SnBr_4  and SnS_2 = 1:1

Equivalent mole of  SnS_2 = 0.023 moles

Mass of 0.023 noles SnS_2= 0.023 x 182.81 = 4.20 grams

With 0.0333 g of SnS_2 recovered, percent yield = 0.333/4.2 x 100 = 7.93%

More on yields of reactions can be found here: brainly.com/question/17042787

#SPJ1

Tin(IV) sulfide, SnS2, a yellow pigment, can be produced using the following reaction.

SnBr4(aq)+2Na2S(aq)⟶4NaBr(aq)+SnS2(s)

Suppose a student adds 48.1 mL of a 0.478 M solution of SnBr4 to 48.8 mL of a 0.160 M solution of Na2S.

1) Calculate the theoretical yield of SnS2. ;

2) The student recovers 0.333 g of SnS2. Calculate the percent yield of SnS2 that the student obtained.

7 0
2 years ago
1) 2C2H6 + 7O2 --&gt; 4CO2 + 6H2O
Lena [83]

Answer:

a).

2 \: moles \: of \: ethane \: react \: with \: 7 \: moles \: of \: oxygen \\ 24 \: moles \: react \: with \: ( \frac{24 \times 7}{2} ) \: moles \\  = 84 \: moles

b).

2 \: moles \: react \: with \: 7 \: moles \\ ( \frac{12 \times 2}{7} ) \: moles \: react \: with \: 12 \: moles \\  = 3 .43 \: moles \\ 1 \: mole \: weighs \: 30 \: g \\ 3.43 \: moles \: weigh \: (3.43 \times 30) \\  = 102.9 \: g

c).

746.7 g

7 0
3 years ago
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