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Cloud [144]
2 years ago
14

What is the volume of 0.200 mol of an ideal gas at 200. kPa and 400. K?

Chemistry
1 answer:
Alexxandr [17]2 years ago
3 0

3.3256 Liters

See the image I have shared to you above

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Given the following sets of values, calculate the unknown
weqwewe [10]

Answer:

3.91 L

Explanation:

Using the ideal gas law equation as follows:

PV = nRT

Where:

P = pressure (atm)

V = volume (L)

n = number of moles (mol)

R = gas law constant (0.0821 Latm/molK)

T = temperature (K)

Based on the information given in this question,

P = 5.23 atm

V= ?

n = 0.831 mol

T = 27°C = 27 + 273 = 300K

Using PV = nRT

V = nRT/P

V = (0.831 × 0.0821 × 300) ÷ 5.23

V = 20.47 ÷ 5.23

V = 3.91 L

8 0
3 years ago
Classify each of the following as energy primarily transferred as HEAT or energy primarily transferred as WORK.
IgorLugansk [536]
number 1 is work number 2 is heat
6 0
3 years ago
Valence Bond theory predicts that tin will use what hybrid orbitals in Snf5 -1​
LUCKY_DIMON [66]

Answer:

sp3d

Explanation:

The ground state electronic configuration of tin is written as; [Kr] 5s²4d¹⁰5p². Hybridization is a concept used to explain the combination of orbitals of appropriate energy to produce suitable orbitals that could be used for bonding.

In forming the compound Snf5^ -1, we have to hybridize the following orbitals on tin; 5p, 5d and 6s orbitals. This gives us a set of sp3d hybrid  hence the answer.

6 0
2 years ago
there is a total of seventeen constitutional isomers for the molecular formula c5h13n. draw the skeletal formula of all three co
MAXImum [283]
Amines are derivatives of Ammonia (NH₃) in which atleast one hydrogen atom is replaced by an alkyl group. Amines are further classifies as;

Primary Amines:
                          In primary amines the nitrogen atom is attached to two hydrogen atoms and one alkyl group.

Secondary Amines:
                             In secondary amines the nitrogen atom is attached to two alkyl groups and one hydrogen atom.

Tertiary Amines:
                         In tertiary amines the nitrogen atom is attached to three alkyl groups, hence it has no hydrogen atom.

Below are three isomers of tertiary amines with molecular formula C₅H₁₃N.

5 0
3 years ago
A 25.0-mL sample of 0.150 M hydrocyanic acid is titrated with a 0.150 M NaOH solution. The Ka of hydrocyanic acid is 4.9 × 10-10
lara [203]

Answer:

The pOH = 1.83

Explanation:

Step 1: Data given

volume of the sample = 25.0 mL

Molarity of hydrocyanic acid = 0.150 M

Molarity of NaOH = 0.150 M

Ka of hydrocyanic acid = 4.9 * 10^-10

Step 2: The balanced equation

HCN + NaOH → NaCN + H2O

Step 3: Calculate the number of moles hydrocyanic acid (HCN)

Moles HCN = molarity * volume

Moles HCN = 0.150 M * 0.0250 L

Moles HCN = 0.00375 moles

Step 3: Calculate moles NaOH

Moles NaOH = 0.150 M * 0.0305 L

Moles NaOH = 0.004575 moles

Step 4: Calculate the limiting reactant

0.00375 moles HCN will react with 0.004575 moles NaOH

HCN is the limiting reactant. It will completely be reacted. There will react 0.00375 moles NaOH. There will remain 0.004575 - 0.00375 = 0.000825 moles NaOH

Step 5: Calculate molarity of NaOH

Molarity NaOH = moles NaOH / volume

Molarity NaOH = 0.000825 moles / 0.0555 L

Molarity NaOH = 0.0149 M

Step 6: Calculate pOH

pOH = -log [OH-]

pOH = -log (0.0149)

pOH = 1.83

The pOH = 1.83

6 0
3 years ago
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