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3 years ago

How many moles of ethane gas, C2H6(g), will occupy 15.0 L at 30 degrees Celsius and 5.0 atm?

Chemistry

1 answer:

Alenkinab [10]3 years ago

8 0

Answer:

3 mol

Explanation:

Given data:

Volume of ethane = 15.0 L

Temperature = 30°C

Pressure = 5.0 atm

Number of moles of ethane = ?

Solution:

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K

T = temperature in kelvin

Now we will convert the temperature.

30+273 = 303 K

5.0 atm × 15.0 L = n×0.0821 atm.L/ mol.K × 303 K

75 atm.L = n× 24.87 atm.L/ mol

n = 75 atm.L / 24.87 atm.L/ mol

n = 3 mol

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Examine the incomplete statement.

hjlf

Answer: Option (c) is the correct answer.

Explanation:

In liquids, molecules are held by slightly less strong intermolecular forces of attraction as compared to solids.

Hence, molecules of a liquid are able to slide past each other as they have more kinetic energy than the molecules of a solid.

As a result, liquids are able to occupy the shape of container in which they are placed. Also, liquids have fixed volume but no fixed shape.

Thus, we can conclude that liquids have a variable shape and a fixed volume.

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3 years ago

Read 2 more answers

Metallic elements can be recovered from ores that are oxides, carbonates, halides, or sulfides. Give an example of each type.

vredina [299]

Metallic elements which can be recovered from ores that are oxides, carbonates, halides, or sulfides are iron, zinc, silver and lead respectively.

Ores which is deposited in earth's crust which contain minerals and metals. Metals can be obtained economically and sold commercially.

<h3>How metals obtained from sulphide or carbonate ore? </h3>

As we get to know that it is easy to obtain metals from their oxides. So, firstly ores which is found in the form of carbonates and sulphide are converted into their oxides by using the process of calcination and roasting.

The metals which is in the middle of the activity series are moderately reactive. These metals are found in the crust of the earth is mainly found as oxides, sulphides, or carbonates.

A metal which can occur in the form of sulphide ore is lead.

A metal which can occur in the form of oxide ore is iron.

A metal which can occur in the form of carbonate ore is zinc.

A metal which can occur in the form of halides ore is silver.

Thus, we concluded that the metallic elements which can be recovered from ores that are oxides, carbonates, halides, or sulfides are iron, zinc, silver and lead respectively.

learn more about ore:

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1 year ago

The Average Speed of the orbiting space shuttle is

maxonik [38]

Explanation:

It is given that, the Average Speed of the orbiting space shuttle is 17500 miles/hour.

We need to convert the speed in kilometers/ second

We know that,

1 mile = 1.609 km

1 km = 0.621 miles

1 hour = 3600 seconds

$17500\ \dfrac{\text{miles}}{\text{hour}}=17500\ \dfrac{\text{miles}}{\text{h}}\times \dfrac{1\ h}{3600\ s}\\\\=17500\times \dfrac{\text{miles}}{3600\ s}$

Now cancel the miles in numerator.

$17500\times \dfrac{\text{miles}}{3600\ s}=17500\times \dfrac{\text{miles}}{3600\ s}\times \dfrac{1.609\ km}{1\ \text{miles}}\\\\=17500\times \dfrac{1.609}{3600}\ km/s\\\\=7.82\ km/s$

So, 17500 miles/hour is equal to 7.82 km/s.

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3 years ago

What unusual thing can the bull shark do?

kotegsom [21]

Bull sharks have the unique ability of keeping salt in their bodies even freshwater

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3 years ago

What volume of 0.500 M HNO3(aq) must completely react to neutralize 100.0 milliliters of 0.100 M KOH(aq)?

asambeis [7]

KOH+ HNO3--> KNO3+ H2O<span>
From this balanced equation, we know that 1 mol HNO3= 1 mol KOH (keep in mind this because it will be used later).

We also know that 0.100 M KOH aqueous solution (soln)= 0.100 mol KOH/ 1 L of KOH soln (this one is based on the definition of molarity).

First, we should find the mole of KOH:
100.0 mL KOH soln* (1 L KOH soln/ 1,000 mL KOH soln)* (0.100 mol KOH/ 1L KOH soln)= 1.00*10^(-2) mol KOH.

Now, let's find the volume of HNO3 soln:
1.00*10^(-2) mol KOH* (1 mol HNO3/ 1 mol KOH)* (1 L HNO3 soln/ 0.500 mol HNO3)* (1,000 mL HNO3 soln/ 1 L HNO3 soln)= 20.0 mL HNO3 soln.

The final answer is </span>(2) 20.0 mL.<span>

Also, this problem can also be done by using dimensional analysis.

Hope this would help~ </span>

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3 years ago