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Describe the properties of noble gases. Based on their electronic arrangement, explain whether they can exist alone in nature.

Marizza181 [45]

The properties of noble gases include:

They are nonreactive
They have hight ionisation energy
They have low boiling point
They conduct electricity at low pressure.

The noble gases are those elements that makes up the group 18 of the periodic table.

They are non as the inert gases because they are relatively nonreactive.

According to the electronic arrangement, which is according to how the electrons are arranged on their shells, they can exist alone.

The noble gases are nonreactive, have high ionisation energy and low boiling point because in its electronic arrangement, the outer shell is completely full.

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3 0

3 years ago

How many liters of C3H6O are present in a sample weighing 25.6 grams?

lawyer [7]

To Find :

Number of moles of C₃H₆O present in a sample weighing 25.6 grams.

Solution :

Molecular mass of C₃H₆O is :

M = (6×12) + (6×1) + (16×1) grams

M = 94 grams/mol

We know, number of moles of 25.6 grams of C₃H₆O is :

$n = \dfrac{Given \ Mass \ Of \ C_3H_6O }{Molar\ Mass \ Of \ C_3H_6O }\\\\n = \dfrac{25.6}{94}\ mole\\\\n = 0.27 \ mole$

Hence, this is the required solution.

4 0

3 years ago

CH4 with pressure 1 atm and volume 10 liter at 27°C is passed into a reactor with 20% excess oxygen, how many moles of oxygen is

BaLLatris [955]

Answer : The moles of $O_2$ left in the products are 0.16 moles.

Explanation :

First we have to calculate the moles of $CH_4$ .

Using ideal gas equation:

$PV=nRT$

where,

P = pressure of gas = 1 atm

V = volume of gas = 10 L

T = temperature of gas = $27^oC=273+27=300K$

n = number of moles of gas = ?

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

Now put all the given values in the ideal gas equation, we get:

$(1atm)\times (10L)=n\times (0.0821L.atm/mol.K)\times (300K)$

$n=0.406mole$

Now we have to calculate the moles of $O_2$ .

The balanced chemical reaction will be:

$CH_4+2O_2\rightarrow CO_2+2H_2O$

From the balanced reaction we conclude that,

As, 1 mole of $CH_4$ react with 2 moles of $O_2$

So, 0.406 mole of $CH_4$ react with $2\times 0.406=0.812$ moles of $O_2$

Now we have to calculate the excess moles of $O_2$ .

$O_2$ is 20 % excess. That means,

Excess moles of $O_2$ = $\frac{(100 + 20)}{100}$ × Required moles of $O_2$

Excess moles of $O_2$ = 1.2 × Required moles of $O_2$

Excess moles of $O_2$ = 1.2 × 0.812 = 0.97 mole

Now we have to calculate the moles of $O_2$ left in the products.

Moles of $O_2$ left in the products = Excess moles of $O_2$ - Required moles of $O_2$

Moles of $O_2$ left in the products = 0.97 - 0.812 = 0.16 mole

Therefore, the moles of $O_2$ left in the products are 0.16 moles.

7 0

3 years ago

A patient receives 3.3 L of glucose solution intravenously (IV). If 100. mL of the solution

artcher [175]

Answer:

660kcal

Explanation:

The question is missing the concentration of the glucose solution. Standard glucose concentration for IV solution is 5% or 5g of glucose every 100mL of solution.

We need to determine how many grams of glucose are there inside the solution. The number of glucose in 3.3L solution will be:

3.3L * (1000mL / L) * (5g/100mL)= 165 g.

If glucose will give 4kcal/ g, then the total calories 165g glucose give will be: 165g * 4kcal/ g= 660kcal.

6 0

4 years ago

How many moles of sulfur trioxide will be produced when the complete combustion of 100.0 g of sulfur dioxide takes place

allsm [11]

Answer:

1.563 moles of SO3.

Explanation:

We begin by calculating the number of mole present in 100g of sulphur dioxide, SO2. This can be obtained as follow:

Molar mass of SO2 = 32 + (16x2) = 64g/mol

Mass of SO2 = 100g

Mole of SO2 =..?

Mole = mass/Molar mass

Mole of SO2 = 100/64

Mole of SO2 = 1.563 mole

Now, we can obtain the number of mole of sulphur trioxide, SO3 produce from the reaction as follow:

2SO2 + O2 —> 2SO3

From the balanced equation above,

2 moles of SO2 reacted to produce 2 moles of SO3.

Therefore, 1.563 moles of SO2 will also react to produce 1.563 moles of SO3.

Therefore, 1.563 moles of SO3 is obtained from the reaction.

8 0

3 years ago