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sattari [20]
3 years ago
6

A molecule has two hydrogen atoms, one carbon atom and three oxygen atoms.

Chemistry
2 answers:
IceJOKER [234]3 years ago
7 0

Answer:

C.

Explanation:

Marianna [84]3 years ago
7 0

Answer:

actually its d

Explanation:

J

J h2co3

two hydrogen(H2)

1 carbon (C)

3 oxygen (o3)

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What can be said about a reaction with H = 62.4 kJ/mol and S = 0.145 kJ/(mol·K)?
Zigmanuir [339]
Answer:

At 430.34 K the reaction will be at equilibrium, at  T > 430.34 the reaction will be spontaneous, and at T < 430.4K the reaction will not occur spontaneously.

Explanation:

1) Variables:

G = Gibbs energy
H = enthalpy
S = entropy

2) Formula (definition)

G = H + TS

=> ΔG = ΔH - TΔS

3) conditions

ΔG < 0 => spontaneous reaction
ΔG = 0 => equilibrium
ΔG > 0 non espontaneous reaction

4) Assuming the data given correspond to ΔH and ΔS

ΔG = ΔH - T ΔS = 62.4 kJ/mol + T 0.145 kJ / mol * K

=>  T = [ΔH - ΔG] / ΔS

ΔG = 0 =>  T = [ 62.4 kJ/mol - 0 ] / 0.145 kJ/mol*K = 430.34K

This is, at 430.34 K the reaction will be at equilibrium, at  T > 430.34 the reaction will be spontaneous, and at T < 430.4K the reaction will not occur spontaneously.

3 0
3 years ago
Read 2 more answers
Many scientist compare the parts of a cell to the parts of a factory. Do you think this comparison is fair and useful?
Fittoniya [83]
Yes. Parts of a cell work together just like stations in a factory.
4 0
3 years ago
HELP When 400 J of heat is added to 5.6 g of olive oil at 23*C, the temperature increases to 87*C. What is the specific heat of
babunello [35]

Q=mc*temperature change

400=5.6/1000*c*(87-23)

c=1160.07J/kg/k

4 0
3 years ago
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.

4 0
3 years ago
Read 2 more answers
What is the number of Cl atoms in a 784 gram pure sample of NC13?
LuckyWell [14K]

Answer:

idk

Explanation:

idk cool pee bee mee nee hee gee fee kee

5 0
3 years ago
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