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3 years ago

What is the mass of 5.50 mol of pure liquid bromine ?

Chemistry

1 answer:

GREYUIT [131]3 years ago

5 0

Answer:

m Br = 439.472 g

Explanation:

mass Br = ?

∴ mol Br = 5.50 mol

∴ molar mass Br 79.904 g/mol

mass = (mol)*(g/mol)

⇒ m Br = (5.50 mol)*(79.904 g/mol)

⇒ m Br = 439.472 g

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How many grams of F2 gas are there in a 5.00-L cylinder at 4.00 × 10^3 mm Hg and 23°C?

GuDViN [60]

Answer:

41.17g

Explanation:

We are given the following parameters for Flourine gas(F2).

Volume = 5.00L

Pressure = 4.00× 10³mmHG

Temperature =23°c

The formula we would be applying is Ideal gas law

PV = nRT

Step 1

We find the number of moles of Flourine gas present.

T = 23°C

Converting to Kelvin

= °C + 273k

= 23°C + 273k

= 296k

V = Volume = 5.00L

R = 0.08206L.atm/mol.K

P = Pressure (in atm)

In the question, the pressure is given as 4.00 × 10³mmHg

Converting to atm(atmosphere)

1 mmHg = 0.00131579atm

4.00 × 10³ =

Cross Multiply

4.00 × 10³ × 0.00131579atm

= 5.263159 atm

The formula for number of moles =

n = PV/RT

n = 5.263159 atm × 5.00L/0.08206L.atm/mol.K × 296K

n = 1.0834112811moles

Step 2

We calculate the mass of Flourine gas

The molar mass of Flourine gas =

F2 = 19 × 2

= 38 g/mol

Mass of Flourine gas = Molar mass of Flourine gas × No of moles

Mass = 38g/mol × 1.0834112811moles

41.169628682grams

Approximately = 41.17 grams.

3 0

2 years ago

The equilibrium 2NO(g)+Cl2(g)⇌2NOCl(g) is established at 500 K. An equilibrium mixture of the three gases has partial pressures

QveST [7]

Answer:

For A: The $K_p$ for the given reaction is $4.0\times 10^1$

For B: The $K_c$ for the given reaction is 1642.

Explanation:

The given chemical reaction follows:

$2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)$

For A:

The expression of $K_p$ for the above reaction follows:

$K_p=\frac{(p_{NOCl})^2}{(p_{NO})^2\times p_{Cl_2}}$

We are given:

$p_{NOCl}=0.24 atm\\p_{NO}=9.10\times 10^{-2}atm=0.0910atm\\p_{Cl_2}=0.174atm$

Putting values in above equation, we get:

$K_p=\frac{(0.24)^2}{(0.0910)^2\times 0.174}\\\\K_p=4.0\times 10^1$

Hence, the $K_p$ for the given reaction is $4.0\times 10^1$

For B:

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = $4.0\times 10^1$

$K_c$ = equilibrium constant in terms of concentration = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = 500 K

$\Delta ng$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-3=-1$

Putting values in above equation, we get:

$4.0\times 10^1=K_c\times (0.0821\times 500)^{-1}\\\\K_c=\frac{4.0\times 10^1}{(0.0821\times 500)^{-1})}=1642$

Hence, the $K_c$ for the given reaction is 1642.

7 0

2 years ago

4. Synthesis problems. Propose a synthetic route from the starting material to the product. In your answer, show each individual

Troyanec [42]

The reduction of alkyne to an alkene in the first step allows the best reagent to be chosen for each subsequent step.

Describe reagents.

A reagent is merely an essential component of a chemical reaction, it should be mentioned. It is an ingredient that speeds up the reaction.

With H2 and Lindlar's catalyst, an alkyne is reduced to alkene as the initial step in this process. Alkene will then be brominated to produce allyl bromide as the next step.

In this instance, the required allyl alcohol will be produced via the reaction of allyl bromide with NaOH.

Learn more about reagent here brainly.com/question/23342137

#SPJ4.

5 0

1 year ago

Fill in the blank Cells break down sugar and release ___________ into the plant.

katrin2010 [14]

Answer:

oxygen

Explanation:

8 0

2 years ago

FIND THE RATIO BY MASS OF THE COMBINING ELEMENTS IN THE FOLLOWING COMPOUNDS

boyakko [2]

The given compound is Aluminum sulfate, Al2(SO4)3:

Molar masses:

Aluminum = 27 g/mol
Sulfur = 32 g/mol
Oxygen = 16 g/mol

The total molar mass is 342 g/mol
The ratio by mass of the elements:

Aluminum = 27*2/342
= 0.16
Sulfur = (32*3)/342
= 0.28
Oxygen = (16*12)/342
= 0.56

5 0

2 years ago