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RideAnS [48]
4 years ago
10

Of the following gases which will have the lowest rate of effusion at a given temperature?A. SO3B. CH4C. NH3D. HBr E. HCl

Chemistry
1 answer:
egoroff_w [7]4 years ago
6 0

Answer: HBr has the lowest rate of effusion at a given temperature.

Explanation: The effusion rate usually increases with increase in temperature because the kinetic energy of the gaseous molecules increases. But it was not true for gases having heavier mass. This was explained by Graham's Law.

Graham's Law states that the rate of effusion of a gas is inversely proportional to the square root of its molecular weight.

(\text{Rate of effusion})_A\propto \frac{1}{\sqrt{\text{Mol. mass}_A}}

We are given different gases with different Molecular masses. The gas having larger Molecular mass will have the lowest rate of effusion.

Mol. Mass of SO_3 = 80 g/mol

Mol. Mass of CH_4 = 16 g/mol

Mol. Mass of NH_3 = 17 g/mol

Mol. Mass of HBr = 81 g/mol

Mol. Mass of HCl = 36 g/mol

As, Mol. mass of HBr is the highest, so its rate of effusion will be the lowest.

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The pressure of a gas contained in a cylinder with a movable
insens350 [35]

Answer:

B) 244.5

Explanation:

2.62 torr =349.3046 pascals.

349.3046*0.7 sq meter = 244.5 N

4 0
3 years ago
Which of the following reactions is an example of a single replacement reaction?
chubhunter [2.5K]
The answer is C. Since aluminum reacts with chloride displacing only Copper.
5 0
3 years ago
Determine the number of moles of C5H12 that are contained in 357.4 g of the compound.
AlladinOne [14]

Answer: 4.96 moles

Explanation:

C5H12 is the chemical formula for pentane, the fifth member of the alkane family.

Given that,

number of moles of C5H12 = ?

Mass in grams = 357.4 g

Molar mass of C5H12 = ?

To get the molar mass of C5H12, use the atomic mass of carbon = 12g; and Hydrogen = 1g

i.e C5H12 = (12 x 5) + (1 x 12)

= 60g + 12g

= 72g/mol

Now, apply the formula

Number of moles = Mass / molar mass

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= 4.96 moles

Thus, 4.96 moles of C5H12 that are contained in 357.4 g of the compound.

4 0
3 years ago
You are given a sulfuric acid solution of unknown concentration. You dispense 10.00 mL of the unknown solution into an Erlenmeye
dmitriy555 [2]

Answer:

"0.053457 M" of sulfuric acid.

Explanation:

The given values are:

V = 10 mL solution

V_{added} = 12.20 mL

V_{total} = 22.20 mL

then,

M 0.103 M of NaOH,

V_{rinsed} = experiment  will not be affected

V_{total \ base} = 10.38 mL

Now,

⇒  mol of NAOH = MV

                            = 0.103\times 10.38

                            =  1.06914  \ m

Whether Sulfuric acid, then

⇒  H_{2}SO_{4} + 2NaOH = Na_{2}SO_{4} + 2H_{2}O

⇒  mol \ of \ acid =\frac{1}{2}\times \ mol \   of  \ base

⇒  1.06914 \ m \ mol \ of \ base = \frac{1}{2}\times 1.06914 = 0.53457 \ m \ mol \ of \ acid

Before any dilution:

V_{sample} = 10  \ mL

⇒  M \ acid = \frac{m \ mol}{V}

                 =\frac{ 0.53457 }{10}

                 =0.053457 \ M (Sulfuric acid)

6 0
3 years ago
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