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VLD [36.1K]
3 years ago
12

Carbon Dioxide in a 34 L container at a pressure of 387 mm Hg is transferred to a container with a pressure of 274 mm Hg. What i

s the new volume if the temperature changes from 211 K to 350 K?
Chemistry
1 answer:
AVprozaik [17]3 years ago
7 0

Answer:

V2= 0.0796m3

Explanation:

We have that for a real gas the following relation is fulfilled:

PV / T = constant

To use this formula I need to pass the pressure of mmHg to KPa and the volume of L to m3, knowing that

 1Kpa = 7.50062mmHg

1L = 0.001m3

So i can say

P1V1/T1 =P2V2/T2

So

V2= P1V1T2/P2T1

V2= 51.60KPa*0.034m3*350k/36.53KPa*211k

finally

V2= 0.0796m3

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Use the periodic table to identify each of the following bonds as polar or nonpolar.
Molodets [167]

Answer:

Polar  

Step-by-step explanation:

Electronegativity increases from <em>left to right</em> in the Periodic Table.

Cl is further right than C (both tinted pink) in the portion of the Periodic Table below.

Cl is <em>more electronegative</em> than C, so the Cl has a partial negative charge and the C has a partial positive charge.

The C-Cl bond is polar.

5 0
3 years ago
What is the energy of a photon of infrared radiation with a frequency of 2.53 × 1012 Hz? Planck’s constant is
Novosadov [1.4K]
It is letter b or c nit d
6 0
1 year ago
Describes the process by which gas
gavmur [86]

Answer:

condensation

Explanation:

hope it is helpful

4 0
2 years ago
Which of the following atoms contains the same number of neutrons as an atom of Calcium-40? (2 points) Select one: a. Potassium-
SVEN [57.7K]
First, let's find the number of neutrons in Calcium-40. to find the number of neutrons, you need to take the atomic mass minus atomic number

neutrons in Calcium---> 40 - 20 (atomic number) = 20 neutrons

potassium-40---> 40- 19= 21
scandium-44----> 44- 21= 23
argon-38---------> 38- 18= 20
chlorine-36------> 36- 17= 19

the answer is argon-38
5 0
3 years ago
Read 2 more answers
Methanol, ethanol, and n-propanol are three common alcohols. When 1.00 g of n-propanol (C3H7OH)is burned in air, -33.4 kJ of hea
Inga [223]

Answer : The heat of combustion of n-propanol is 0.554 kJ/mol

Explanation :

First we have to calculate the moles of n-propanol.

\text{Moles of n-propanol}=\frac{\text{Mass of n-propanol}}{\text{Molar mass of n-propanol}}

Molar mass of n-propanol = 60.09 g/mole

\text{Moles of n-propanol}=\frac{1.00g}{60.09g/mole}=0.0166mole

Now we have to calculate the heat of combustion of n-propanol.

As, 0.0166 mole of n-propanol liberated heat of combustion = -33.4 kJ

So, 1 mole of n-propanol liberated heat of combustion = 0.0166 × (-33.4 kJ)

                                                                                           = 0.554 kJ/mol

Therefore, the heat of combustion of n-propanol is 0.554 kJ/mol

5 0
2 years ago
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