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2 years ago

If you have 0.56 moles of an ideal gas at 87 ° C and a pressure of 569 torr, what volume will the gas take up?

2 answers:

4 0

The ideal gas law: PV = nRT

Conversions:
87°C = 360K
P = 0.75 atm
R = 0.0821
n = 0.56

0.75V = (0.56)(0.0821)(360)

V = 22 L

NikAS [45]2 years ago

4 0

Explanation:

The given data is as follows.

n = 0.56 moles, T = $87^{o}C$ = (87 + 273) K = 360 K

P = 569 torr = 0.748 atm (as 1 torr = 0.0013 atm)

According to ideal gas equation, PV = nRT

Therefore, putting given values into the above formula as follows.

PV = nRT

$0.748 \times V = 0.56 \times 0.0821 L atm/mol K \times 360 K$

V = 22.12 L

Thus, we can conclude that the given gas will take up 22.12 L of volume.

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How many moles of gas Does it take to occupy 520 mL at a pressure of 400 torr and a temperature of 340 k

Ann [662]

Answer would be B. I provided work on an image attached. Message me if u have any other questions on how to do it

6 0

2 years ago

Assuming equal concentrations of conjugate base and acid, which one of the following mixtures is suitable for making a buffer so

BartSMP [9]

Answer:

NH₃/NH₄Cl

Explanation:

We can calculate the pH of a buffer using the Henderson-Hasselbalch's equation.

$pH=pKa+log\frac{[base]}{[acid]}$

If the concentration of the acid is equal to that of the base, the pH will be equal to the pKa of the buffer. The optimum range of work of pH is pKa ± 1.

Let's consider the following buffers and their pKa.

CH₃COONa/CH3COOH (pKa = 4.74)

NH₃/NH₄Cl (pKa = 9.25)

NaOCl/HOCl (pKa = 7.49)

NaNO₂/HNO₂ (pKa = 3.35)

NaCl/HCl Not a buffer

The optimum buffer is NH₃/NH₄Cl.

4 0

2 years ago

Determine the PH at the point in the titration of 40.0ml of 0.200M HC4H7o2 with 0.100 M Sr(OH)2 after 100ml of the strong base h

Dmitriy789 [7]

Answer:

Check the explanation

Explanation:

Mols HC4H7O2 = (volume in L)*(molarity) = (40.0 mL)*(0.200 M)

= (40.0 mL)*(1 L)/(1000 mL)*(0.200 M)

= 8.00*10-3 mol.

Mols Sr(OH)2 corresponding to 10.0 mL of 0.100 M solution =

(volume in L)*(molarity)

= (10.0 mL)*(0.100 M)

= (10.0 mL)*(1 L)/(1000 mL)*(0.100 M)

= 1.00*10-3 mol.

Consider the ionization of Sr(OH)2 as below.

Sr(OH)2 (aq) ----------> Sr2+ (aq) + 2 OH- (aq)

As per the stoichiometric equation,

1 mol Sr(OH)2 = 2 mols OH-.

Therefore,

0.0010 mol Sr(OH)2 = [0.0010 mol Sr(OH)2]*(2 mols OH-)/[1 mole Sr(OH)2]

= 0.0020 mol

= 2.00*10-3 mol

Set up the ICE charts as below.

HC4H7O2 (aq) + OH- (aq) ------------> H2O (l) + C4H7O2- (aq)

Before (mol) 8.00*10-3 2.00*10-3 - -

Change (mol) -2.00*10-3 -2.00*10-3 - +2.00*10-3

After (mol) 6.00*10-3 0 - 2.00*10-3

The change in a pure substance, e.g., H2O is not considered in an acid-base reaction.

Volume of the solution = (40.0 + 10.0) mL = 50.0 mL = (50.0 mL)*(1 L)/(1000 mL) = 0.05 L.

The initial concentrations are obtained by dividing the numbers of moles by the volume, 0.05 L.

Set up the ICE charts as below.

HC4H7O2 (aq) + OH- (aq) ------------> H2O (l) + C4H7O2- (aq)

Initial (M) 0.160 0.0400 - -

Change (M) -0.0400 -0.0400 - +0.0400

Equilibrium (M) 0.120 0 - 0.0400

The acid-ionization constant is written as

Ka = [H3O+][C4H7O2-]/[HC4H7O2] = 1.5*10-5

Plug in the known values and get

Ka = [H3O+]*(0.0400)/(0.120) = 1.5*10-5

======> [H3O+] = (1.5*10-5)*(0.120)/(0.0400) (ignore units)

======> [H3O+] = 4.5*10-5

The proton concentration of the solution is 4.5*10-5 M.

pH = -log (4.5*10-5 M)

= 4.346

≈ 4.35 (ans).

8 0

2 years ago

A mixture of gases containing 0.20 mol of SO2 and 0.20 mol of O2 in a 4.0 L flask reacts to form SO3. If the temperature is 25ºC

diamong [38]

Answer : The pressure in the flask after reaction complete is, 2.4 atm

Explanation :

To calculate the pressure in the flask after reaction is complete we are using ideal gas equation.

$PV=n_TRT\\\\P=(n_1+n_2)\times \frac{RT}{V}$

where,

P = final pressure in the flask = ?

R = gas constant = 0.0821 L.atm/mol.K

T = temperature = $25^oC=273+25=298K$

V = volume = 4.0 L

$n_1$ = moles of $SO_2$ = 0.20 mol

$n_2$ = moles of $O_2$ = 0.20 mol

Now put all the given values in the above expression, we get:

$P=(0.20+0.20)mol\times \frac{(0.0821L.atm/mol.K)\times (298K)}{4.0L}$

$P=2.4atm$

Thus, the pressure in the flask after reaction complete is, 2.4 atm

5 0

2 years ago

Adjacent water molecules interact through the ________. a. sharing of electrons between the hydrogen of one water molecule and t

Yakvenalex [24]

Answer: Option (c) is the correct answer.

Explanation:

A water molecule is made up of two hydrogen atoms and one oxygen atom. Due to the difference in electronegativity of hydrogen and oxygen, the electrons are pulled more towards oxygen atom.

As a result, a partial positive charge will develop on hydrogen atom and a partial negative charge will develop on oxygen atom.

Thus, we can conclude that adjacent water molecules interact through the electrical attraction between the hydrogen of one water molecule and the oxygen of another water molecule.

7 0

2 years ago