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user100 [1]
2 years ago
12

Using the reaction below: 2 CO2(g) + 2 H2O(l) → C2H4(g) + 3 O2(g) ΔHrxn= +1411.1 kJ What would be the heat of reaction for this

reaction? 0.5 C2H4(g) + 1.5 O2(g) → CO2(g) + H2O(l) ΔHrxn= ??? KJ Question 6 options: a) Not enough information is given b) -2822.2 kJ c) +1411.1 kJ d) -705.55 kJ e) -1411.1 kJ
Chemistry
1 answer:
maw [93]2 years ago
7 0

Answer:  d) -705.55 kJ

Explanation:

Heat of reaction is the change of enthalpy during a chemical reaction with all substances in their standard states.

2CO_2(g)+2H_2O(l)\rightarrow C_2H_4(g)+3O_2(g) \Delta H=+1411.1kJ

Reversing the reaction, changes the sign of \Delta H

C_2H_4(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(l)

\Delta H=-1411.1kJ

On multiplying the reaction by \frac{1}{2} , enthalpy gets half:

0.5C_2H_4(g)+1.5O_2(g)\rightarrow CO_2(g)+H_2O(l)\Delta H=\frac{1}{2}\times -1411.1kJ=-705.55kJ/mol

Thus the enthalpy change for the given reaction is -705.55kJ

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Why is a ylid a stable nucleophile? Select all that apply. Group of answer choices The proximity of the positively charged phosp
pogonyaev

Answer:

1)The proximity of the positively charged phosphorous and negatively charged carbon stabilizes the charges.

2) Inductive effects and resonance stabilize the negative charge

Explanation:

both atoms have full octets of electrons( I.e Carbon and say phosphorus). The result can be viewed as a structure in which two adjacent atoms are connected by both a covalent and an ionic bond; normally written X+–Y−. Ylides are thus 1,2-dipolar compounds, and a subclass of zwitterions

6 0
2 years ago
Two species, A and B, are separated on a 2.00 m long column which has 5.000 x 103 plates when the flow rate is 15.0 mL/min. A sp
sergiy2304 [10]

Explanation:

The given data is as follows.

         t_{m} = 30.0 sec,     t_{r1} = 5 min = 5 \times 60 sec = 300 sec

         t_{r2} = 12.0 min = 12 \times 60 sec = 720 sec

Formula for adjusted retention time is as follows.

      t'_{r} = t_{r} - t_{m}

                 = 300 sec - 30.0 sec

                 = 270 sec

   t'_{r2} = 720 sec - 30 sec

                            = 690 sec

Formula for relative retention (\alpha) is as follows.

          \alpha = \frac{t'_{r2}}{t'_{r1}}

                     = \frac{690 sec}{270 sec}

                     = 2.56

Thus, we can conclude that the relative retention is 2.56.

4 0
2 years ago
Calculate the enthalpy of combustion, δh∘comb, for c6h14. you'll first need to determine the balanced chemical equation for the
Vedmedyk [2.9K]
<span>Answer: Enthalpy Change = (6 x -393.5) + (7 x -285.8) - (-204.6) + (19/2) 0.....??? like.. (6 x Enth CO2) + ( 7 x Enth H2O) - (Enth C6H14) + (19/2) Enth O2</span>
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2 years ago
A flask contains 1/3 mole of h2 and 2/3 mol of he. Compare the force on the wall per impact of h2 relative to that for he.
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The force on the wall is actually the pressure exerted by gas molecules

Higher the pressure more the force exerted on the walls of container

The pressure depends upon the number of molecules of a gas

In a mixture of gas the pressure depends upon the mole fraction of the gas

As given the mole fraction of He is more than that of H2 therefore He will exert more pressure on the wall

The ratio of impact will be

H2 / He = 2/3 / 1/3 = 2: 1

8 0
2 years ago
Read 2 more answers
Suppose that 0.323 g of an unknown sulfate salt is dissolved in 50 mL of water. The solution is acidified with 6M HCl, heated, a
geniusboy [140]

Answer:

1) 41.16 % = 0.182 grams

2) The alkali cation is K+ , to form the salt K2SO4

Explanation:

Step 1: Data given

Mass of unknown sulfate salt = 0.323 grams

Volume of water = 50 mL

Molarity of HCl = 6M

Step 2: The balanced equation

SO4^2- + BaCl2 → BaSO4 + 2Cl-

Step 3: Calculate amount of SO4^2- in BaSO4

The precipitate will be BaSO4

The amount of SO4^2- in BaSO4 = (Molar mass of SO4^2-/Molar mass BaSO4)*100 %

The amount of SO4^2- in BaSO4 = (96.06 /233.38) * 100

= 41.16%

So in 0.443g of BaSO4 there will be 0.443 * 41.16 % = <u>0.182 grams</u>

<u />

<u />

2. If it is assumed that the salt is an alkali sulfate determine the identity of the alkali cation.

The unknown sulphate salt has 0.182g of sulphate. This means the alkali cation has a weight of 0.323-0.182 = 0.141g grams

An alkali cation has a chargoe of +1; sulphate has a charge of -2

The formula will be X2SO4 (with X = the unknown alkali metal).

Calculate moles of sulphate

Moles sulphate = 0.182 grams (32.1 + 4*16)

Moles sulphate = 0.00189 moles

The moles of sulphate = 0.182/(32.1+16*4)

The moles of sulphate = 0.00189 moles

X2SO4 → 2X+ + SO4^2-

For 2 moles cation we have 1 mol anion

For 0.00189 moles anion, we have 2*0.00189 = 0.00378 moles cation

Calculate molar mass

Molar mass = mass / moles

Molar mass = 0.141 grams / 0.00378 grams

Molar mass = 37.3 g/mol

The closest alkali metal is potassium. (K2SO4 )

3 0
2 years ago
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