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3 years ago

The materials or compounds that are changed into a new compounds in a chemical reaction are know as

Chemistry

1 answer:

OLga [1]3 years ago

4 0

Answer:

Reactants

Explanation:

They become known as reactants

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Using the data below, calculate the enthalpy for the combustion of C to CO

Leona [35]

Answer:

ΔH3 = -110.5 kJ.

Explanation:

Hello!

In this case, by using the Hess Law, we can manipulate the given equation to obtain the combustion of C to CO as shown below:

C(s) + 1/2O2(g) --> CO(g)

Thus, by letting the first reaction to be unchanged:

C(s) + O2(g)--> CO2 (g) ; ΔH1 = -393.5 kJ

And the second one inverted:

CO2(g) --> CO(g) + 1/2O2(g) ; ΔH2= 283.0kJ

If we add them, we obtain:

C(s) + O2(g) + CO2(g) --> CO(g) + CO2 (g) + 1/2O2(g)

Whereas CO2 can be cancelled out and O2 subtracted:

C(s) + 1/2O2(g) --> CO(g)

Therefore, the required enthalpy of reaction is:

ΔH3 = -393.5 kJ + 283.0kJ

ΔH3 = -110.5 kJ

Best regards!

6 0

3 years ago

Consider the fermentation reaction of glucose: C6H12O6 → 2C2H5OH + 2CO2 A 1.00-mol sample of C6H12O6 was placed in a vat with 10

inessss [21]

Answer:

% yield = 73.48 %

Explanation:

The fermentation reaction is:

C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂

The percent yield of C₂H₅OH is given by:

$\% yield = \frac{m_{E}}{m_{T}} * 100$

where $m_{E}$ : is the obtained mass of C₂H₅OH = 67.7g and $m_{T}$ : is the theoretical mass of C₂H₅OH.

The theoretical mass of C₂H₅OH is calculated knowing that 1 mol of C₆H₁₂O₆ produces 2 moles of C₂H₅OH:

$m_{T} = mol * M$

where M: is the molar mass of C₂H₅OH = 46.068 g/mol

$m_{T} = 2 moles * 46.068 g/mol = 92.136 g$

Hence, the percent yield of C₂H₅OH is:

$\% yield = \frac{67.7 g}{92.136 g}*100 = 73.48 \%$

I hope it helps you!

3 0

3 years ago

If an atom loses two electrons what charge will it have ? +2 -1 +1 -2

Citrus2011 [14]

Answer:

Explanation:

Because an electron has a negative charge.

6 0

4 years ago

Propanoic acid, ch3ch2cooh, has a pka =4.9. draw the structure of the conjugate base of propanoic acid and give the ph above whi

xenn [34]

Conjugate base of Propanoic acid ( $CH_{3}CH_{2}COOH$ is propanoate where -COOH group gets converted to -CO $O^{1-}$ . The structure of conjugate base of Propanoic acid is shown in the diagram.

The $p^{H}$ above which 90% of the compound will be in this conjugate base form can be determined using Henderson's equation as propanoic acid is weak acid and it can form buffer solution on reaction with strong base.