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Studentka2010 [4]
3 years ago
11

A compound contains 16.7 g of iridium and 10.3 g of selenium, what is its empirical formula?

Chemistry
1 answer:
Westkost [7]3 years ago
5 0

Answer:

THE EMPIRICAL FORMULA OF THE COMPOUND IS Ir Se2

Explanation:

Iridium = 16.7 g

Selenium = 10.3 g

Total mass = 16.7 + 10.3 = 27 g

To calculate the empirical formula, we first calculate the percentage com[position of the individual elements.

Percentage composition of iridium = 16.7 / 27 * 100 = 61.85 %

Percentage composition of selenium = 10.3 / 27 * 100 = 38.15 %

Next is to divide the percentage composition of each element by their respective molecular masses

Molecular mass of iridium = 192 g/mol

Molecular mass of selenium = 79g/mol

Iridium = 61.85 / 192 =  0.3221

Selenium = 38.15 / 79 = 0.4829

Next is to divide the values by the smaller of the two values

Iridium = 0.3221 / 0.3221 = 1

Selenium = 0.4829 / 0.3221 = 1.5

Next is to round up the values to a whole number

Iridium = 1

Selenium = 2

The empirical formula therefore is Ir Se2

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Anybody good at chemistry?
Luba_88 [7]

Answer:

Explanation:

1)

Given data:

Mass of lead = 25 g

Initial temperature = 40°C

Final temperature = 95°C

Cp = 0.0308 j/g.°C

Heat required = ?

Solution:

Specific heat capacity: Cp

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = Final temperature = initial temperature

ΔT = 95°C -  40°C

ΔT = 55°C

Q = 25 g × 0.0308 j/g.°C  × 55°C

Q = 42.35 j

2)

Given data:

Mass  = 3.1 g

Initial temperature = 20°C

Final temperature = 100°C

Cp = 0.385 j/g.°C

Heat required = ?

Solution:

Specific heat capacity: Cp

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = Final temperature = initial temperature

ΔT = 100°C -  20°C

ΔT = 80°C

Q = 3.1 g × 0.385 j/g.°C  × 80°C

Q = 95.48 j

3)

Given data:

Mass of Al = ?

Initial temperature = 60°C

Final temperature = 30°C

Cp = 0.897 j/g.°C

Heat released = 120 j

Solution:

Specific heat capacity: Cp

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = Final temperature = initial temperature

ΔT = 30°C -  60°C

ΔT = -30°C

120 j = m × 0.897 j/g.°C  × -30°C

120 j = m × -26.91  j/g

m = 120 j / -26.91  j/g

m =  4.46 g

negative sign show heat is released.

4)

Given data:

Mass of ice = 1.5 g

Change in temperature  = ?

Cp = 0.502 j/g.°C

Heat added= 30.0 j

Solution:

Specific heat capacity: Cp

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = Final temperature = initial temperature

30.0 j = 1.5 g × 0.502 j/g.°C  × ΔT

30.0 j = 0.753 j/°C  × ΔT

30.0 j /0.753 j/°C  = ΔT

39.84 °C  =  ΔT

3 0
3 years ago
A particular balloon can be stretched to a maximum surface area of 1257 cm2. The balloon is filled with 3.1 L of helium gas at a
lianna [129]

Answer:

P = 0.6815 atm

Explanation:

Pressure = 754 torr

The conversion of P(torr) to P(atm) is shown below:

P(torr)=\frac {1}{760}\times P(atm)

So,  

Pressure = 754 / 760 atm = 0.9921 atm

Temperature = 294 K

Volume = 3.1 L

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

0.9921 atm × 3.1 L = n × 0.0821 L.atm/K.mol × 294 K  

⇒n of helium gas= 0.1274 moles

Surface are = 1257 cm²

For a sphere, Surface area = 4 × π × r² = 1257 cm²

r² = 1257 / 4 × π ≅ 100 cm²

r = 10 cm

The volume of the sphere is :

V=\frac {4}{3}\times \pi\times r^3

Where, V is the volume

r is the radius

V=\frac {4}{3}\times \frac {22}{7}\times {10}^3

V = 4190.4762 cm³

1 cm³ = 0.001 L

So, V (max) = 4.19 L

T = 273 K

n = 0.1274 moles

Using ideal gas equation as:

PV=nRT

Applying the equation as:

P × 4.19 L = 0.1274 × 0.0821 L.atm/K.mol × 273 K  

<u>P = 0.6815 atm</u>

<u></u>

4 0
3 years ago
A solution contains an unknown mass of dissolved barium ions. When sodium sulfate is added to the solution, a white precipitate
AlladinOne [14]

The given question is incomplete. The complete question is as follows.

A solution contains an unknown mass of dissolved barium ions. When sodium sulfate is added to the solution, a white precipitate forms. The precipitate is filtered and dried and then found to have a mass of 212 mg. What mass of barium was in the original solution? (Assume that all of the barium was precipitated out of solution by the reaction.)

Explanation:

When Ba^{2+} and Na_{2}SO_{4} are added  then white precipitate forms. And, reaction equation for this is as follows.

       Ba^{2+} + SO^{2-}_{4} \rightarrow BaSO_{4}

It is given that mass (m) is 212 mg or 0.212 g (as 1 g = 1000 mg). Molecular weight of BaSO_{4} is 233.43.

Now, we will calculate the number of moles as follows.

  No. of moles = mass × M.W

                        = \frac{0.212}{233.43}

                        = 0.00091 mol of BaSO_{4}

Hence, it means that 0.00091 mol of Ba^{2+}. Now, we will calculate the mass as follows.

       Mass = moles × MW

                 = 0.00091 \times 137.327

                 = 0.124 grams or 124 mg of barium

Thus, we can conclude that mass of barium into the original solution is 124 mg.

8 0
3 years ago
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