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3 years ago

6

The atomic mass of sulfur is 32.1 amu, and the atomic mass of oxygen is 16.0 amu. To the nearest tenth of a percent, what is the

percent by mass of sulfur in sulfur trioxide (SO3)?

1 answer:

Zielflug [23.3K]3 years ago

6 0

Answer:

$\%\ Composition\ of\ sulfur=40.1\ \%$

Explanation:

Percent composition is percentage by the mass of element present in the compound.

Given , Mass of sulfur= 32.1 amu

Mass of oxygen = 16.0 amu

Mass of sulfur trioxide $SO_3$ = 32.1 amu + 3*16.0 amu = 80.1 amu

$\%\ Composition\ of\ sulfur=\frac{Mass_{sulfur}}{Mass_{SO_3}}\times 100$

$\%\ Composition\ of\ sulfur=\frac{32.1\ amu}{80.1\ amu}\times 100=40.1\ \%$

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If 150 g dimethylhydrazine reacts with excess dinitrogen tetroxide and the product gases are collected at 127oC in an evacuated

zhuklara [117]

Answer : The the partial pressure of the nitrogen gas is 0.981 atm.

The total pressure in the tank is 2.94 atm.

Explanation :

The balanced chemical reaction will be:

$(CH_3)_2N_2H_2(l)+2N_4O_4(l)\rightarrow 3N_2(g)+4H_2O(g)+2CO_2(g)$

First we have to calculate the moles of dimethylhydrazine.

Mass of dimethylhydrazine = 150 g

Molar mass of dimethylhydrazine =60.104 g/mole

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

$\text{Moles of dimethylhydrazine}=\frac{150g}{60.104g/mole}=2.49mole$

Now we have to calculate the moles of $N_2$ gas.

From the balanced chemical reaction we conclude that,

As, 1 mole of $(CH_3)_2N_2H_2$ react to give 3 moles of $N_2$ gas

So, 2.49 mole of $(CH_3)_2N_2H_2$ react to give $2.49\times 3=7.47$ moles of $N_2$ gas

Now we have to calculate the partial pressure of nitrogen gas.

Using ideal gas equation :

$PV=nRT\\\\P_{N_2}=\frac{nRT}{V}$

where,

P = Pressure of $N_2$ gas = ?

V = Volume of $N_2$ gas = 250 L

n = number of moles $N_2$ gas = 7.47 mole

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

T = Temperature of $N_2$ gas = $127^oC=273+127=400K$

Putting values in above equation, we get:

$P_{N_2}=\frac{(7.47mole)\times (0.0821L.atm/mol.K)\times 400K}{250L}=0.981atm$

Thus, the partial pressure of the nitrogen gas is 0.981 atm.

Now we have to calculate the total pressure in the tank.

Formula used :

$P_{N_2}=X_{N_2}\times P_T$

$P_T=\frac{1}{X_{N_2}}\times P_{N_2}$

$P_T=\frac{1}{(\frac{n_{N_2}}{n_T})}\times P_{N_2}$

$P_T=\frac{n_{T}}{n_{N_2}}\times P_{N_2}$

where,

$P_T$ = total pressure = ?

$P_{N_2}$ = partial pressure of nitrogen gas = 0.981 atm

$n_{N_2}$ = moles of nitrogen gas = 3 mole (from the reaction)

$n_{T}$ = total moles of gas = (3+4+2) = 9 mole (from the reaction)

Now put all the given values in the above formula, we get:

$P_T=\frac{9mole}{3mole}\times 0.981atm=2.94atm$

Thus, the total pressure in the tank is 2.94 atm.

8 0

3 years ago

What is the best reason for the difference in properties of LiCI and C6H14O

vitfil [10]

One is covalent, the other is ionic

7 0

3 years ago

A qué periodo y grupo pertenecen los átomos de un elemento cuyos átomos presentan 12 electrones caracterizados con ml = 0. Dar c

Klio2033 [76]

Answer:

Periodo: 3.

Grupo: IIA.

Explanation:

Hola.

En este caso, es posible determinar el periodo y grupo por medio de la configuración electrónica del elemento que tiene 12 electrones:

$1s^2,2s^2,2p^6,3s^2$

De este modo, vemos que el termino final es $3s^2$ , por lo tanto, decimos que el periodo es 3, ya que este coincide con el nivel de energía. Adicionalmente, para $s^2$ tenemos que el átomo se encuentra en el grupo IIA ya que tiene dos electrones en su capa o nivel (3) más externa. Esto coincide con el número cuántico dado (magnético, ml=0) ya que cuando el término en la configuración electrónica tiene el subnivel $s$ , este tiene un valor de cero.

Así, el elemento en cuestión sería Magnesio.

¡Saludos!

3 0

4 years ago

consider one glucose unit in glycogen. what is the overall or net reaction for the conversion of this unit into 2 pyruvate, star

liubo4ka [24]

This cycle is known as Glycolysis

Glycolysis is the 10 step process, which occurs in cytoplasm of cell and is conversion of glucose to pyruvate.

There are several steps and enzymes that is required in glycolysis pathway.

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This is irreversible reaction.

Here glucose is phosphorylated to glucose 6 phosphate with the help of enzyme hexokinase and 1 ATP is utilized.

STEP 2 : ISOMERISATION

The isomerization of Glucose-6-phosphate to Fructose-6-phosphate, done with the help of enzyme phosphoglucoisomerase.

STEP 3 : SECOND PHOSPHORYLATION

Fructose-1,6-bisphosphate is phosphorylated to Fructose 1,6 bisphosphate which is catalyzed by phosphofructokinase and cost another ATP.

STEP 4: BREAKDOWN

The fructose-1,6 bisphosphate is breaken down too produce two 3carbon molecules - Glyceraldehyde-3-phosphate, or GADP, and a molecule of Dihydroxyacetone phosphate or DHAP.

The reaction is catalyzed by aldolase.

STEP 5 : CONVERSION OF DHAP INTO GADP

DHAP is oxidized to form GADP.

The reaction is catalyzed by triose phosphate isomerase enzyme.

STEP 6: OXIDATION

Here 2 mol. of GADP are oxidized.The GDAP is converted to 1,3 bisphosphoglycerate with the help of glyceraldehyde phosphate dehydrogenase. This requires NAD+ and a free phosphate.

STEP 7: DEPHOSPHORYLATION

First substrate level phosphorylation ( addition of phosphate to ADP to give ATP )

1,3 bisphosphoglycerate with the help of Phosphoglycerate kinase become 3-phosphoglycerate and will produce 1 ATP.

STEP 8: PHOSPHATE TRANSFER

The phosphate ester linkage in 3 phosphoglycerate is moved from 3 C to 2 , because of low free energy to form 2 phosphoglycerate with the help of phosphoglycerate mutase .

STEP 9: DEHYDRATION

2 phopshoglycerate is dehydrated by enolase to form Phosphoenolpyruvate ( PEP)

This is reversible reaction.

STEP 10: SECOND DEPHOSPHORYLATION

2 substrate level phosphorylation which gives out ATP.

Non - oxidative phosphorylation.

Here Phosphoenolpyruvate ( PEP) is converted to last product of glycolysis pyruvate releasing ATP by pyruvate kinase.

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To know more about glycolysis -

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5 0

1 year ago

Perform the following operation

Ivenika [448]

Answer:

$7.00\times 10^8-5.00\times 10^6$ = $6.95\times 10^8$

Explanation:

The given expression is :

$7.00\times 10^8-5.00\times 10^6$

We need to solve the above expression using operations and then express it in scientific notation.

First we will make the power of 10 same in both terms

$7.00\times 10^8-(5.00\times 10^6\times \dfrac{100}{100})\\\\=7.00\times 10^8-(0.05\times 10^8)\\\\=(7-0.05)\times 10^8\\\\=6.95\times 10^8$

Hence, the answer is $6.95\times 10^8$ .

7 0

3 years ago