All categories

3 years ago

Calculate the number of Aluminium ions present in 1.5 gram of Aluminium oxide

Chemistry

1 answer:

alexgriva [62]3 years ago

8 0

Answer:

1.771x10^22

Explanation:

6.02x10^23:avogadro's number

2: number of Al ions in Al2O3

1.5/102=0.0147:number of moles in the compound

moles=mass/relative mass number

so 0.0147x(6.02x10^23)x2=1.771x10^22

You might be interested in

Please help in Chemistry!!!! 15 points! I added a picture below...

krek1111 [17]

The answer would be choice B because the energy decreased by 20 J

5 0

3 years ago

Read 2 more answers

In your OWN words explain how season are formed

astra-53 [7]

Answer: Weather currents change wind tempature and there for change the weather of the seasons.

3 0

2 years ago

Read 2 more answers

I need help asap thank u

UNO [17]

It is b, i learned this last year hope this helps

4 0

3 years ago

Please answer this question

vova2212 [387]

Answer:

Yes

Explanation:

Natural gas is colorless and odorless, and explosive, so a sulfur-smell (similar to rotten eggs) is usually added for early detection of leaks. ... Natural gas is a fossil fuel. Natural gas is a non-renewable hydrocarbon used as a source of energy for heating, cooking, and electricity generation.

5 0

2 years ago

The ideal gas heat capacity of nitrogen varies with temperature. It is given by:

hammer [34]

Answer:

A) 1059 J/mol

B) 17,920 J/mol

Explanation:

Given that:

Cp = 29.42 - (2.170*10^-3 ) T + (0.0582*10^-5 ) T2 + (1.305*10^-8 ) T3 – (0.823*10^-11) T4

R (constant) = 8.314

We know that:

$C_p=C_v+R$

We can determine $C_v$ from above if we make $C_v$ the subject of the formula as:

$C_v$ $=C_p-R$

$C_V = 29.42-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4-8.314$

$C_V = 21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4$

A).

The formula for calculating change in internal energy is given as:

$dU=C_vdT$

If we integrate above data into the equation; it implies that:

$U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4\,) du$

$U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T/1+(5.82*10^{-7})T2/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)$

$U2-U1= 1059J/mol$

Hence, the internal energy that must be added to nitrogen in order to increase its temperature from 450 to 500 K = 1059 J/mol.

B).

If we repeat part A for an initial temperature of 273 K and final temperature of 1073 K.

then T = 273 K & T2 = 1073 K

∴

$U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T/1+(5.82*10^{-7})T2/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)$

$U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})273/1+(5.82*10^{-7})1073/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)$

$U2-U1= 17,920 J/mol$

3 0

3 years ago

Calculate the number of Aluminium ions present in 1.5 gram of Aluminium oxide​

Calculate the number of Aluminium ions present in 1.5 gram of Aluminium oxide