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4 years ago

If the pressure exerted by a gas at 0.00 °C in a volume of 0.0010 L of 5.00 atm, how many moles of gas are present?

Chemistry

1 answer:

bazaltina [42]4 years ago

7 0

Answer:

2.23 x 10^-4 mol

Explanation:

0°C =273 K

PV= nRT

n= PV/RT

n= (5.00 atm * 0.0010 L)/ (0.0821 * 273 K)

n= 2.23 x 10^-4

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sam records the mass of his evaporating dish as 8.365 g. he records the mass of the evaporating dish and the sample of hydrate a

guapka [62]

0.2701 moles of water were evaporated from the sample by the heating process

Mass of evaporating dish = 8.365 g

mass of dish + hydrate = 18.597 g

mass of hydrate = (mass of dish + hydrate) - mass of dish

hydrate = 18.597 - 8.365 = 10.232 g

Mass of dehydrated material = (mass of dish + dehydrate ) - mass of dish

dehydrated material = 18.735 - 8.365 = 5.37 g

Mass of water evaporated = hydrate - dehydrate

mass of water = 10.232 - 5.37 = 4.862 g

Molar mass of water = 18 gm/mol

no. of moles of water evaporated = mass / molecular mass

moles = 4.862/18 = 0.2701 moles

To learn more about mole click here:

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1 year 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

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3 years ago

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Answer:

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