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Lostsunrise [7]
3 years ago
5

A compound is 2.00% H by mass, 32.7% S by mass, and 65.3% O by mass. What is its empirical formula? The second step is to calcul

ate the number of moles of each element present in the sample. To do this, you must divide the mass of each element by its molar mass. The masses of the elements present are: mass H = g mass S = g mass O = g
Chemistry
2 answers:
Free_Kalibri [48]3 years ago
8 0
Lets take 100 g of this compound,
so it is going to be 2.00 g H, 32.7 g S and 65.3 g O.

2.00 g H *1 mol H/1.01 g H ≈ 1.98 mol H
32.7 g S *1 mol S/ 32.1 g S ≈ 1.02 mol S
65.3 g O * 1 mol O/16.0 g O ≈ 4.08 mol O

1.98 mol H : 1.02 mol S : 4.08 mol O = 2 mol H : 1 mol S : 4 mol O

Empirical formula
H2SO4
lbvjy [14]3 years ago
3 0

the masses of the elements present are:

mass H = 2.00 g

mass S = 32.7 g

mass O = 65.3 g

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The instruction booklet for your pressure cooker indicates that its highest setting is 11.8 psipsi . You know that standard atmo
zavuch27 [327]

<u>Answer:</u> The temperature at which the food will cook is 219.14°C

<u>Explanation:</u>

To calculate the final temperature of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

\frac{P_1}{T_1}=\frac{P_2}{T_2}

where,

P_1\text{ and }T_1 are the initial pressure and temperature of the gas.

P_2\text{ and }T_2 are the final pressure and temperature of the gas.

We are given:

P_1=14.7psi\\T_1=273K\\P_2=(14.7+11.8)psi=26.5psi\\T_2=?

Putting values in above equation, we get:

\frac{14.7psi}{273K}=\frac{26.5psi}{T_2}\\\\T_2=\frac{26.5\times 273}{14.7}=492.14K

Converting the temperature from kelvins to degree Celsius, by using the conversion factor:

T(K)=T(^oC)+273

492.14=T(^oC)+273\\\\T(^oC)=219.14^oC

Hence, the temperature at which the food will cook is 219.14°C

7 0
3 years ago
What is the composition, in atom percent, of an alloy that consists of a) 5.5 wt% Pb and b) 94.5 wt% of Sn? Assume that the atom
Anastaziya [24]

Answer : The percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.

Explanation :

First we have to calculate the number of atoms in 5.5 wt% Pb and 94.5 wt% of Sn.

As, 207.2 g of lead contains 6.022\times 10^{23} atoms

So, 5.5 g of lead contains \frac{5.5}{207.2}\times 6.022\times 10^{23}=1.59\times 10^{22} atoms

and,

As, 118.71 g of lead contains 6.022\times 10^{23} atoms

So, 94.5 g of lead contains \frac{94.5}{118.71}\times 6.022\times 10^{23}=4.79\times 10^{23} atoms

Now we have to calculate the percent composition of Pb and Sn in atom.

\% \text{Composition of Pb}=\frac{\text{Atoms of Pb}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100

\% \text{Composition of Pb}=\frac{1.59\times 10^{22}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=3.21\%

and,

\% \text{Composition of Sn}=\frac{\text{Atoms of Sn}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100

\% \text{Composition of Sn}=\frac{4.79\times 10^{23}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=96.8\%

Thus, the percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.

6 0
3 years ago
A sample of helium has a volume of 50.00 L at STP. How many helium atoms are in the sample? Show work
Setler79 [48]

Answer:

Explanation:

molar volume at STP=22.4 L

given volume=50.0 L

number of moles=given volume/molar volume

number of moles=50.0/22.4

number of moles=2.2

1 mole of helium =6.023*10^23 atoms

2.2 moles of helium =6.023*10^23*2.2=1.3*10^24

therefore 50.0 L of helium contain 1.33*10^24 atoms

4 0
3 years ago
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Salsk061 [2.6K]

Answer:

a compound,typically a crystalline one,in which water molecules are chemically bound to another compound or an element

3 0
3 years ago
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The answer is C
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