Please help with this

what is the molar mass of a gaseous flouride of sulfur containing 70.4% F and having a density of approximately 4.5g/L at 20 deg

zlopas [31]

Answer:

The molar mass is 180.2 g/mol

Explanation:

Step 1: Data given

% of F = 70.4 %

Density = 4.5 g/L

Temperature = 20 °C

Pressure = 1 atm

Step 2: Calculate the number of moles

PV = nRT

⇒ with P = the pressure = 1.00 atm

⇒ with V = the volume = Assume this is 1L

⇒ with n = the number of moles = TO BE DETERMINED

⇒ R = the gas constant = 0.08206 L*atm/K*mol

⇒ T = the temperature : 20°C = 293 Kelvin

1 atm*1L= n(0.08206 L-atm/mol-K)*(293 K)

n = 0.04159 moles

Step 3: Calculate molar mass

Molar mass = Mass / moles

4.5 grams / 0.04159 moles = 108.2 g/mol

Step 4: Calculate moles of F

Moles = Mass / molar mass

Moles F = 70.4 g / 19 g/mol

Moles F = 3.70 moles

Moles S = 29.6g / 32.07 g/mol

Moles S = 0.923 moles S

Step 5: Divide by the smallest amount of moles

F = 3.70 / 0.923 = 4

S = 0.923 / 0.923 = 1

The empirical formula is SF4

The molar mass of SF4 = 32.07 + 4*19 = 108.07 g/mol

This means the empirical formula is the same as the molecular formula SF4

The molar mass is 180.2 g/mol

4 0

3 years ago

From his experiment j.j. Thomson concluded that

iragen [17]

Particles as small as atoms exist.

3 0

3 years ago

What is the quantity of heat (in kJ) associated with cooling 185.5 g of water from 25.60°C to ice at -10.70°C?Heat Capacity of S

Cerrena [4.2K]

Taking into account the definition of calorimetry, sensible heat and latent heat, the amount of heat required is 37.88 kJ.

<h3>Calorimetry</h3>

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

<h3>Sensible heat</h3>

Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).

<h3>Latent heat</h3>

Latent heat is defined as the energy required by a quantity of substance to change state.

When this change consists of changing from a solid to a liquid phase, it is called heat of fusion and when the change occurs from a liquid to a gaseous state, it is called heat of vaporization.

25.60 °C to 0 °C

First of all, you should know that the freezing point of water is 0°C. That is, at 0°C, water freezes and turns into ice.

So, you must lower the temperature from 25.60°C (in liquid state) to 0°C, in order to supply heat without changing state (sensible heat).

The amount of heat a body receives or transmits is determined by:

Q = c× m× ΔT

where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.

In this case, you know:

c= Heat Capacity of Liquid= 4.184 $\frac{J}{gC}$
m= 185.5 g
ΔT= Tfinal - Tinitial= 0 °C - 25.60 °C= - 25.6 °C

Replacing:

Q1= 4.184 $\frac{J}{gC}$ × 185.5 g× (- 25.6 °C)

Solving:

Q1= -19,868.98 J

Change of state

The heat Q that is necessary to provide for a mass m of a certain substance to change phase is equal to

Q = m×L

where L is called the latent heat of the substance and depends on the type of phase change.

In this case, you know:

n= 185.5 grams× $\frac{1mol}{18 grams}$ = 10.30 moles, where 18 $\frac{g}{mol}$ is the molar mass of water, that is, the amount of mass that a substance contains in one mole.

ΔHfus= 6.01 $\frac{kJ}{mol}$

Replacing:

Q2= 10.30 moles×6.01 $\frac{kJ}{mol}$

Solving:

Q2=61.903 kJ= 61,903 J

0 °C to -10.70 °C

Similar to sensible heat previously calculated, you know:

c = Heat Capacity of Solid = 2.092 $\frac{J}{gC}$
m= 185.5 g
ΔT= Tfinal - Tinitial= -10.70 °C - 0 °C= -10.70 °C

Replacing:

Q3= 2.092 $\frac{J}{gC}$ × 185.5 g× (-10.70) °C

Solving:

Q3= -4,152.3062 J

<h3>Total heat required</h3>

The total heat required is calculated as:

Total heat required= Q1 + Q2 +Q3

Total heat required=-19,868.98 J + 61,903 J -4,152.3062 J

Total heat required= 37,881.7138 J= 37.8817138 kJ= 37.88 kJ

In summary, the amount of heat required is 37.88 kJ.

Learn more about calorimetry:

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7 0

2 years ago

A sample of gas is observed to effuse through a pourous barrier in 4.98 minutes. Under the same conditions, the same number of m

kogti [31]

Answer:

The molar mass of the unknown gas is $\mathbf{ 51.865 \ g/mol}$

Explanation:

Let assume that the gas is O2 gas

O2 gas is to effuse through a porous barrier in time t₁ = 4.98 minutes.

Under the same conditions;

the same number of moles of an unknown gas requires time t₂ = 6.34 minutes to effuse through the same barrier.

From Graham's Law of Diffusion;

Graham's Law of Diffusion states that, at a constant temperature and pressure; the rate of diffusion of a gas is inversely proportional to the square root of its density.

i.e

$R \ \alpha \ \dfrac{1}{\sqrt{d}}$

$R = \dfrac{k}{d}$ where K = constant

If we compare the rate o diffusion of two gases;

$\dfrac{R_1}{R_2}= {\sqrt{\dfrac{d_2}{d_1}}$

Since the density of a gas d is proportional to its relative molecular mass M. Then;

$\dfrac{R_1}{R_2}= {\sqrt{\dfrac{M_2}{M_1}}$

Rate is the reciprocal of time ; i.e

$R = \dfrac{1}{t}$

Thus; replacing the value of R into the above previous equation;we have:

$\dfrac{R_1}{R_2}={\dfrac{t_2}{t_1}}$

We can equally say:

${\dfrac{t_2}{t_1}}= {\sqrt{\dfrac{M_2}{M_1}}$

${\dfrac{6.34}{4.98}}= {\sqrt{\dfrac{M_2}{32}}$

$M_2 = 32 \times ( \dfrac{6.34}{4.98})^2$

$M_2 = 32 \times ( 1.273092369)^2$

$M_2 = 32 \times 1.62076418$

$\mathbf{M_2 = 51.865 \ g/mol}$

7 0

3 years ago

Which element has a greater electronegativity? fluorine (9) or radium (88)

emmainna [20.7K]

Answer:

Fluorine

General Formulas and Concepts:

Chemistry

Reading a Periodic Table
Periodic Trends
Electronegativity - the tendency for an element to attract an electron to itself
Z-effective and Coulomb's Law, Forces of Attraction

Explanation:

The Periodic Trend for Electronegativity is up and to the right of the Periodic Table.

Fluorine is Element 9 and has 9 protons. Radium is Element 88 and has 88 protons. Therefore, Radium has a bigger Zeff than Flourine.

However, since Radium is in Period 7 while Fluorine is in Period 2, Radium has more core e⁻ than Fluorine does. This will create a much larger shielding effect, causing Radium's outermost e⁻ to have less FOA between them. Fluorine, since it has less core e⁻, the FOA between the nucleus and outershell e⁻ will be much stronger.

Therefore, Fluorine would attract an electron more than Radium, thus bringing us to the conclusion that Fluorine has a higher electronegativity.

7 0

3 years ago

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