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3 years ago

The density of a gaseous chlorofluorocarbon (CFC) at 23.8 °C and 432 mmHg is 3.23 g/L. What is its molar mass?

Chemistry

1 answer:

Wewaii [24]3 years ago

6 0

Answer:

138.57 g/mol.

Explanation:

The following data were obtained from the question:

Temperature (T) = 23.8 °C

Pressure = 432 mmHg

Density (D) = 3.23 g/L

Next, we shall obtain an expression for the density in relation to molar mass, pressure and temperature.

This can be obtained by using the ideal gas equation as shown below:

PV = nRT.... (1)

Recall:

Mole (n) = maas(m) /Molar mass (M)

n = m/M

Substituting the value of n into equation 1

PV = nRT

PV = mRT/M

Divide both side by P

V = mRT/MP

Divide both side by m

V/m = RT/MP

Invert the above equation

m/V = MP /RT..... (2)

Recall:

Density (D) = mass(m) /volume (V)

D = m/V

Replace m/V with D in equation 2

m/V = MP /RT

D = MP /RT

Thus, with the above formula we can obtain the molar mass of chlorofluorocarbon (CFC) as shown below:

Temperature (T) = 23.8 °C = 23.8 °C + 273 = 296.8 K

Pressure = 432 mmHg = 432/760 = 0.568 atm

Density (D) = 3.23 g/L

Gas constant (R) = 0.0821 atm.L/Kmol

Molar mass (M) =..?

D = MP /RT

3.23 = M x 0.568 / 0.0821 x 296.8

Cross multiply

M x 0.568 = 3.23 x 0.0821 x 296.8

Divide both side by 0.568

M = (3.23 x 0.0821 x 296.8) / 0.568

M = 138.57 g/mol

Therefore the molar mass of CFC is 138.57 g/mol.

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How many joules of heat are removed from a 21.0 g sample of water if it is cooled from 34.0°C

yaroslaw [1]

Answer:

527.184 J of heat is removed from a 21 g water sample if it is cooled from 34.0 ° C to 28.0 ° C.

Explanation:

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

When the heat added or removed from a substance causes a change in temperature in it, this heat is called sensible heat.

In other words, the sensible heat of a body is the amount of heat received or transferred by a body when it undergoes a change in temperature without there being a change in physical state (solid, liquid or gaseous). The equation that allows to calculate this heat exchange is:

Q = c * m * ΔT

Where Q is the heat exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT=Tfinal-Tinitial is the change in temperature.

In this case:

c= 4.184 $\frac{J}{g*C}$
m=21 g
ΔT=Tfinal-Tinitial=28 °C - 34 °C=-6 °C

Replacing:

Q= 4.184 $\frac{J}{g*C}$ * 21 g* (-6 C)

Q= - 527.184 J

To lower the temperature, heat has to be given, for that the final temperature must be lower than the initial temperature; and it receives the name of transferred heat and has a negative value, as in this case.

527.184 J of heat is removed from a 21 g water sample if it is cooled from 34.0 ° C to 28.0 ° C.

4 0

3 years ago

Click the DeltaH is an Extensive Property button within the activity, and analyze the relationship between the two reactions tha

DaniilM [7]

Answer:

∴ΔH₂ = - 12,258 KJ

Explanation:

Enthalpy:

Enthalpy is a property of a thermodynamic system. Enthalpy of a system is equal to the sum of internal energy of the system and presser times volume of the system.

The heat absorbes or releases in a closed system is the change of enthalpy of the system.

Given reactions are:

Reaction 1: C₃H₈(g)+5O₂(g)→ 3CO₂(g)+4H₂O, ΔH₁= - 2043 KJ

Reaction 2: 6C₃H₈(g)+30 O₂(g)→ 18 CO₂(g)+24 H₂O, ΔH₂=?

Take a look at reaction 1 and reaction 2, the only difference is that 1 molecule of C₃H₈ is combusted in reaction 1 and 6 molecules of C₃H₈ is combusted in reaction 2.

We can think the reaction 2 as occurring 6 different container and each containers contains 1 molecule of C₃H₈. The enthalpy is an extensive property. Total enthapy of the 6 containers is = 6×(-2043 KJ)

= - 12,258 KJ

∴ΔH₂ = - 12,258 KJ

6 0

3 years ago

if 1.386 g of mg ribbon combusts to form 2.309 g of oxide product, calculate the experimental mass percent of oxygen from this d

Pavlova-9 [17]

1.386 g of Mg ribbon combusts to form 2.309 g of oxide product. The mass percent of oxygen in the oxide is 40.0 %.

Let's consider the reaction for the combustion of Mg.

Mg + 1/2 O₂ ⇒ MgO

1.386 g of Mg combusts to form 2.309 g of MgO. We want to determine the mass of oxygen in MgO. According to Lavoisier's law of conservation of mass, matter is not created nor destroyed over the course of a chemical reaction. Then, the mass of Mg in the reactants is equal to the mass of Mg in MgO. The mass of the magnesium oxide is the sum of the masses of magnesium and oxygen. The mass of oxygen in the oxide is:

$mMgO = mMg + mO\\mO = mMgO - mMg = 2.309 g - 1.386 g = 0.923 g$

We can calculate the mass percent of O in MgO using the following expression.

$\% O = \frac{mO}{mMgO} \times 100\% = \frac{0.923 g}{2.309g} \times 100\% = 40.0 \%$

You can learn more about mass percent here: brainly.com/question/14990953

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2 years ago

g Calculate the theoretical yield (in grams) of your product if you start with 0.50 grams of E-stilbene. The molecular weight of

sattari [20]

Answer:

0.9433g

Explanation:

Theoretical yield is defined as the mass produced assuming all reactant reacts producing the product.

Assuming the reaction is 1:1, we need to find the moles of E-stilbene (Reactant). If all reactant reacts, the moles of E-stilbene = Moles of product.

Using the molar mass of the product we can find the theoretical yield as follows:

Moles E-stilbene:

0.50g * (1mol/180.25g) = 0.00277 moles = Moles Product

Mass Product = Theoretical yield:

0.00277 moles * (340.058g/mol) = 0.9433g

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3 years ago

How many grams of N2 are in 44.8L at STP?

lutik1710 [3]

Answer: 10 i think

Explanation:

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3 years ago