All categories

3 years ago

What is the mass, in grams, of 6.11 mol of sulfur trioxide?

Chemistry

2 answers:

Travka [436]3 years ago

8 0

Answer : The mass of sulfur trioxide is, 488.8 grams.

Explanation : Given,

Moles of sulfur trioxide = 6.11 mole

Molar mass of sulfur trioxide = 80 g/mole

Formula used :

$\text{Mass of }SO_3=\text{Moles of }SO_3\times \text{Molar mass of }SO_3$

Now put all the given values in this formula, we get the mass of sulfur trioxide.

$\text{Mass of }SO_3=6.11mole\times 80g/mole=488.8g$

Therefore, the mass of sulfur trioxide is, 488.8 grams.

Kazeer [188]3 years ago

7 0

It would be approximately 489 g, rounded for significant figures. The calculator given answer would be 489.20326. To get that, all you have to do is take your 6.11 moles and multiply it by 80.066 (the molar mass) divided by 1 mol to cancel out the unit.

You might be interested in

Consider the half reactions below for a chemical reaction.

ladessa [460]

Answer:

Option A:

Zn(s) + Cu^(2+) (aq) → Cu(s) + Zn^(2+)(aq)

Explanation:

The half reactions given are:

Zn(s) → Zn^(2+)(aq) + 2e^(-)

Cu^(2+) (aq) + 2e^(-) → Cu(s)

From the given half reactions, we can see that in the first one, Zn undergoes oxidation to produce Zn^(2+).

While in the second half reaction, Cu^(2+) is reduced to Cu.

Thus, for the overall reaction, we will add both half reactions to get;

Zn(s) + Cu^(2+) (aq) + 2e^(-) → Cu(s) + Zn^(2+)(aq) + 2e^(-)

2e^(-) will cancel out to give us;

Zn(s) + Cu^(2+) (aq) → Cu(s) + Zn^(2+)(aq)

7 0

2 years ago

A bird sits on her eggs to keep them warm

Troyanec [42]

Answer:

This would be Conduction

Explanation:

Conduction is the transfer of heat between two surfaces in contact with each other like the bird and the eggs.

5 0

2 years ago

Which of the following is true of science? (2 points)

Tju [1.3M]

Answer:

It is based on testable and replicable evidence.

7 0

2 years ago

63Ni decays by a first-order process via the emission of a beta particle. The 63Ni isotope has a half-life of 100. years. How lo

rewona [7]

Answer:

151.4863 years

Explanation:

Half life, t1/2 = 100 years

Initial concentration,[A]o = 100%

Final concentration, [A] = 35% (after 65% have been decayed)

Time = ?

Half life for a first Order reaction is given as;

t1/2 = ln (2) / k

k = ln(2) / 100

k = 0.00693y-1

The integral rate law for first order reactions is given as;

ln[A] = ln[A]o − kt

kt = ln[A]o - ln[A]

t = ( ln[A]o - ln[A]) / k

t = [ln(100) - ln(35)] /0.00693

t = 1.0498 / 0.00693

t = 151.4863 years

8 0

3 years ago

Automobile airbags contain solid sodium azide, NaN3, that reacts to produce nitrogen gas when heated, thus inflating the bag. 2N

Vitek1552 [10]

Answer : The value of work done for the system is 1144.69 J

Explanation :

First we have to calculate the moles of $NaN_3$

$\text{Moles of }NaN_3=\frac{\text{Mass of }NaN_3}{\text{Molar mass of }NaN_3}$

Molar mass of $NaN_3$ = 65.01 g/mole

$\text{Moles of }NaN_3=\frac{20.2g}{65.01g/mole}=0.311mole$

Now we have to calculate the moles of nitrogen gas.

The balanced chemical reaction is,

$2NaN_3(s)\rightarrow 2Na(s)+3N_2(g)$

From the balanced reaction we conclude that

As, 2 mole of $NaN_3$ react to give 3 mole of $N_2$

So, 0.311 moles of $NaN_3$ react to give $\frac{0.311}{2}\times 3=0.466$ moles of $N_2$

Now we have to calculate the volume of nitrogen gas.

Using ideal gas equation:

$PV=nRT$

where,

P = Pressure of $N_2$ gas = 1.00 atm

V = Volume of $N_2$ gas = ?

n = number of moles $N_2$ = 0.466 mole

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of $N_2$ gas = $22^oC=273+22=295K$

Putting values in above equation, we get:

$1.00atm\times V=0.466mole\times (0.0821L.atm/mol.K)\times 295K$

$V=11.3L$

As initially no nitrogen was present. So,

Volume expanded = Volume of nitrogen evolved

Thus,

Expansion work = Pressure × Volume

Expansion work = 1.00 atm × 11.3 L

Expansion work = 11.3 L.atm

Conversion used : (1 L.atm = 101.3 J)

Expansion work = 11.3 × 101.3 = 1144.69 J

Therefore, the value of work done for the system is 1144.69 J

5 0

3 years ago