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2 years ago

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The volume of a gas is decreased from 100 liters at 173.0°C to 50 liters at a constant pressure. After the decrease in volume, w

hat is the new temperature of the gas?

1 answer:

Minchanka [31]2 years ago

5 0

Answer:

223.08 K

Explanation:

First we <u>convert 173.0 °C to K</u>:

173.0 °C + 273.16 = 446.16 K

With the absolute temperature we can use <em>Charles' law</em> to solve this problem:

T₁V₂=T₂V₁

Where in this case:

T₁ = 446.16 K

V₂ = 50 L

T₂ = ?

V₁ = 100 L

We <u>input the data</u>:

446.16 K * 50 L = T₂ * 100 L

And <u>solve for T₂</u>:

T₂ = 223.08 K

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Some hydrogen gas is enclosed within a chamber being held at 200∘C∘C with a volume of 0.0250 m3m3. The chamber is fitted with a

Semenov [28]

Answer:

The final volume is 39.5 L = 0.0395 m³

Explanation:

Step 1: Data given

Initial temperature = 200 °C = 473 K

Volume = 0.0250 m³ = 25 L

Pressure = 1.50 *10^6 Pa

The pressure reduce to 0.950 *10^6 Pa

The temperature stays constant at 200 °C

Step 2: Calculate the volume

P1*V1 = P2*V2

⇒with P1 = the initial pressure = 1.50 * 10^6 Pa

⇒with V1 = the initial volume = 25 L

⇒with P2 = the final pressure = 0.950 * 10^6 Pa

⇒with V2 = the final volume = TO BE DETERMINED

1.50 *10^6 Pa * 25 L = 0.950 *10^6 Pa * V2

V2 = (1.50*10^6 Pa * 25 L) / 0.950 *10^6 Pa)

V2 = 39.5 L = 0.0395 m³

The final volume is 39.5 L = 0.0395 m³

3 0

2 years ago

Two weak acids, A and B, have pKa values of 4 and 6, respectively. Which statement is true?A) Acid A dissociates to a greater ex

zalisa [80]

Answer:

A) Acid A dissociates to a greater extent in water than acid B

Explanation:

A) Acid A dissociates to a greater extent in water than acid B

We are given the pKa for both acids, and we know

pKa = - log Ka

Taking antilog to both sides of the equation we can solve for Ka

⇒ -pKa = log Ka

-antilog pKa = Ka

10 ^-pka = Ka

So ka for acid A = 10⁻⁴

and

ka for acid B = 10⁻⁶

True the equilibrium constant for acid A is greater, so it dissociates more.

B) For solutions of equal concentration, acid B will have a lower pH.

We know the stronger acid is A, and it dissociates more. Since pH is the negative log of H₃O⁺ concentration, it follows that at equal concentrations the acid A will have at equilibrium a greater [H₃O⁺] and hence a lower pH

C) B is the conjugate base of A

False:

If B were the conjugate base of A, its Kb would have been given by:

Ka x Kb = Kw

Kb = 10⁻¹⁴/10⁻⁶ = 10⁻⁸ for the conjugate base of acid B

Kb = 10⁻¹⁴/10⁻⁴ = 10⁻¹⁰ for the conjugate base of acid A

which are not equal.

D) Acid A is more likely to be a polyprotic acid than acid B.

False

Just having the pkas for both acids one cannot know if any of the acids is polyprotic. We will need the formula for the acids.

E) The equivalence point of acid A is higher than that of acid B

False

The equivalence point depends on the the concentration of the acids and their volumes.

The equivalence point is reached in the titration when the number of equivalents of base equals the number of equivalents of acid:

# equivalents acid = # equivalents of base @ end point

and

# equivalents acid = Molarity of acid x Volume of acid

4 0

2 years ago

A 85.2 g copper bar was heated to 221.32 degrees Celsius and placed in a coffee cup calorimeter containing 4250 mL of water at 2

VLD [36.1K]

Answer:- 64015 J

Solution: There is 4250 mL of water in the calorimeter at 22.55 degree C.

density of water is 1 g per mL.

So, the mass of water = $4250mL(\frac{1g}{1mL})$ = 4250 g

Final temperature of water after adding the hot copper bar to it is 26.15 degree C.

So, $\Delta T$ for water = 26.15 - 22.55 = 3.60 degree C

Specific heat for water is 4.184 $\frac{J}{g.^0C}$

The heat gained by water is calculated by using the formula:

$q=mc\Delta T$

where, q is the heat energy, m is mass and c is specific heat.

Let's plug in the values in the formula and do the calculations:

$q=4250g*\frac{4.184J}{g.^0C}*3.60^0C$

q = 64015 J

So, 64015 J of heat is gained by the water.

5 0

2 years ago

trans-2-Butene does not exhibit a signal in the double-bond region of the spectrum (1600–1850 cm-1); however, IR spectroscopy is

spayn [35]

Answer:

The other signal that would indicate the presence of a C= C bond appears close to 3100 $cm^{-1}$ .

Explanation:

Bands that appear above 3000 $cm^{-1}$ are often unsaturation diagnoses suggest. The band at 3000- 3100 $cm^{-1}$ is characteristics for C-H stretching frequencies and normally is overlaps with the ones for alkanes because it is a band of weak intensity.

4 0

2 years ago

A 60.0 mL solution of 0.112 M sulfurous acid (H2SO3) is titrated with 0.112 M NaOH. The pKa values of sulfurous acid are 1.857 (

djverab [1.8K]

Answer:

a)4.51

b) 9.96

Explanation:

Given:

NaOH = 0.112M

H2S03 = 0.112 M

V = 60 ml

H2S03 pKa1= 1.857

pKa2 = 7.172

a) to calculate pH at first equivalence point, we calculate the pH between pKa1 and pKa2 as it is in between.

Therefore, the half points will also be the middle point.

Solving, we have:

pH = (½)* pKa1 + pKa2

pH = (½) * (1.857 + 7.172)

= 4.51

Thus, pH at first equivalence point is 4.51

b) pH at second equivalence point:

We already know there is a presence of SO3-2, and it ionizes to form

SO3-2 + H2O <>HSO3- + OH-

$Kb = \frac{[ HSO3-][0H-]}{SO3-2}$

$Kb = \frac{10^-^1^4}{10^-^7^.^1^7^2} = 1.49*10^-^7$

[HSO3-] = x = [OH-]

mmol of SO3-2 = MV

= 0.112 * 60 = 6.72

We need to find the V of NaOh,

V of NaOh = (2 * mmol)/M

= (2 * 6.72)/0.122

= 120ml

For total V in equivalence point, we have:

60ml + 120ml = 180ml

[S03-2] = 6.72/120

= 0.056 M

Substituting for values gotten in the equation $Kb=\frac{[HSO3-][OH-]}{[SO3-2]}$

We noe have:

$1.485*10^-^7=\frac{x*x}{(0.056-x)}$

$x = [OH-] = 9.11*10^-^5$

$pOH = -log(OH) = -log(9.11*10^-^5)$

=4.04

pH = 14- pOH

= 14 - 4.04

= 9.96

The pH at second equivalence point is 9.96

4 0

2 years ago