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3 years ago

How many grams of propane (c3h8) are required to produce 39.4g of water

Chemistry

1 answer:

Ivahew [28]3 years ago

4 0

Answer:

96.12 grams

Explanation:

C3H8 + H2O

Mass 96.12 39.4

RFM 44 18

Moles 2.19 2.19

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What is the ph of a solution that is 0.50 m in propanoic acid and 0.40 m in sodium propanoate. (ka for propanoic acid = 1.3 x 10

max2010maxim [7]

The pH is given by the Henderson - Hasselbalch equation:
 pH = pKa + log([A-]/[HA])
 pH = -log(1.3 x 10^-5) + log(0.50/0.40)
 pH = 4.98
The answer to this question is 4.98.

3 0

3 years ago

Which concept can explain the acidic character of Carbon dioxide?

Mamont248 [21]

Answer:

The concept of lewis acids and bases can explain the acidic nature of CO2. A lewis acid is a compound which can accept an electron pair whereas a lewis base is a compound which can donate an electron pair. CO2 acts as a lewis acid.

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2 years ago

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dexar [7]

http://www.atnf.csiro.au/outreach/education/senior/cosmicengine/stars_hrdiagram.html. This will give u the answer, just go to link

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3 years ago

Modify the given fatty acid so that it represents the 18‑carbon fatty acid designated 18:2(Δ9,12). Draw any double bonds in the

Mekhanik [1.2K]

Answer:

See explanation

Explanation:

In this case, we have to remember the meaning of the nomenclature "18:2Δ9,12". Where 18 is the number of carbon atoms, 2 is the number of double bonds, and the numbers successive to Δ "delta" the position of the double bonds starting to count from the carboxylic -COOH end of the molecule.

In other words, the main functional group is a carboxylic acid. We have a total of 18 carbons. Additionally, we have 2 double bonds. On carbons 9 and 12.

Lets see figure 1

I hope it helps!

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3 years ago

The activation energy for proline isomerization of a peptide depends on the identity of the preceding residue and obeys Arrheniu

MAVERICK [17]

Answer:

Activation energy of phenylalanine-proline peptide is 66 kJ/mol.

Explanation:

According to Arrhenius equation- $k=Ae^{\frac{-E_{a}}{RT}}$ , where k is rate constant, A is pre-exponential factor, $E_{a}$ is activation energy, R is gas constant and T is temperature in kelvin scale.

As A is identical for both peptide therefore-

$\frac{k_{ala-pro}}{k_{phe-pro}}=e^\frac{[E_{a}^{phe-pro}-E_{a}^{ala-pro}]}{RT}$

Here $\frac{k_{ala-pro}}{k_{phe-pro}}=\frac{0.05}{0.005}$ , T = 298 K , R = 8.314 J/(mol.K) and $E_{a}^{ala-pro}=60kJ/mol$

So, $\frac{0.05}{0.005}=e^{\frac{[E_{a}^{phe-pro}-(60000J/mol)]}{8.314J.mol^{-1}.K^{-1}\times 298K}}$

$\Rightarrow$ $E_{a}^{phe-pro}=65705J/mol=66kJ/mol$ (rounded off to two significant digit)

So, activation energy of phenylalanine-proline peptide is 66 kJ/mol

7 0

3 years ago