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3 years ago

Identify the combustion reaction. c4h12 + 7o2 ⟶ 6h2o + 4co2 2h2 + o2 ⟶ 2h2o al2s3 ⟶ 2al + 3s cl2 + 2kbr ⟶ 2kcl + br2

1 answer:

3 0

From the given equations, the combustion reaction is;
C₄H₁₂ + 7O₂ --> 4CO₂ + 6H₂O
Combustion reactions are when organic compounds react with O₂ to produce water and CO₂. From the given reactions, C₄H₁₂ is an organic compound that reacts with O₂ to produce water and CO₂.
Therefore this is the only reaction that follows the general equation for combustion.

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Which of the following statements is true?

miss Akunina [59]

C plasmas have a net negative charge

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4 years ago

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You have a solution of 600 mg of caffeine dissolved in 100 mL of water. The partition coefficient for aqueous caffeine extracted

klio [65]

Answer:

159 mg caffeine is being extracted in 60 mL dichloromethane

Explanation:

Given that:

mass of caffeine in 100 mL of water = 600 mg

Volume of the water = 100 mL

Partition co-efficient (K) = 4.6

mass of caffeine extracted = ??? (unknown)

The portion of the DCM = 60 mL

Partial co-efficient (K) = $\frac{C_1}{C_2}$

where; $C_1=$ solubility of compound in the organic solvent and $C_2$ = solubility in aqueous water.

So; we can represent our data as:

$K=(\frac{A_{(g)}}{60mL} )$ ÷ $(\frac{B_{(mg)}}{100mL} )$

Since one part of the portion is A and the other part is B

A+B = 60 mL

A+B = 0.60

A= 0.60 - B

4.6= $(\frac{0.6-B(mg)}{60mL} )$ ÷ $(\frac{B_{(mg)}}{100mL})$

4.6 = $\frac{(\frac{0.6-B(mg)}{60mL} )}{(\frac{B_{(mg)}}{100mL})}$

4.6 × $(\frac{B_{(mg)}}{100mL})$ = $(\frac{0.6-B(mg)}{60mL} )$

4.6 B $*\frac{60}{100}$ = 0.6 - B

2.76 B = 0.6 - B

2.76 + B = 0.6

3.76 B = 0.6

B = $\frac{0.6}{3.76}$

B = 0.159 g

B = 159 mg

∴ 159 mg caffeine is being extracted from the 100 mL of water containing 600 mg of caffeine with one portion of in 60 mL dichloromethane.

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3 years ago

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A patient administered 20 mg of iodine-131. How many mg of this radioactive isotope will remain the body after 24 days? NEEDS AN

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3 years ago

Which one of the following will change the value of anequilibrium constant?

IRINA_888 [86]

Answer:

d. changing temperature

Explanation:

The thermodynamic equilibrium constant K is defined as a quantity characterizing the equilibrium of a chemical reaction. For a reaction where concentrations are in equilibrium:

aA + bB ⇄ cC + dD

The equilibrium constant is:

$k = \frac{[C]^c[D]^d}{[A]^a[B]^b}$

Thus, the equilibrium constant will change if:

a. Varying the initial concentration of reactants . FALSE. The k constant doesn't depend of initial concentrations but concentration in equilibrium does.

b. Adding other substances that do not react with any of thespecies involved in the equilibrium . FALSE. The equilibrium constant just depends of substances that are involved in the equilibrium

c. Varying the initial concentration of products . FALSE. Again, equilibrium constant doesn't depend of initial concentrations.

d. Changing temperature . <em>TRUE. </em>As a thermodynamic constant, k depends of temperature thus:

$K = e^(-dG/RT)$

e. Changing the volume of the reaction vessel. FALSE. The changing in the volume of the reaction vessel will change just the initial concentrations of the reactants.

I hope it helps!

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3 years ago

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For this case, we must take into account the following conversion factors:

$1 mL = 1 cm ^ 3$