All categories

2 years ago

The volume of hcl gas required to react with excess ca to produce 11.4 l of hydrogen gas at 1.62 atm and 62.0 °c is ________ l.

Chemistry

1 answer:

seropon [69]2 years ago

3 0

Answer:

22.8 L

Step-by-step explanation:

We can use <em>Gay-Lussac's Law of Combining Volumes</em> to solve this problem:

Gases <em>at the same temperature and pressure</em> react in simple whole-number ratios.

1. Write the chemical equation.

Ratio: 2 L 1 L

Ca(s) + 2HCl(g) ⟶ CaCl₂(s) + H₂(g)

V/L: 11.4

2. Calculate the volume of HCl.

According to the law, 2 L of HCl form 1 L of H₂.

Then, the conversion factor is (2 L HCl/1 L H₂).

Volume of HCl = 11.4 L H₂ × (2 L HCl/1 L H₂)

= 22.8 L HCl

You might be interested in

Octane is a liquid component of gasoline. Given the following vapor pressures of octane at various temperatures, estimate the bo

Hitman42 [59]

Answer:

110.8 ºC

Explanation:

To solve this problem we will make use of the Clausius-Clayperon equation:

lnP = - ΔHºvap/RT + C

where P is the pressure, ΔHºvap is the enthalpy of vaporization, R is the gas constant, T is the temperature, and C is a constant of integration.

Now this equation has a form y = mx + b where

y = lnP

x = 1/T

m = -ΔHºvap/R

Now we have to assume that ΔHºvap remains constant which is a good asumption given the narrow range of temperatures in the data ( 104-125) ºC

Thus what we have to do is find the equation of the best fit for this data using a software as excel or your calculator.

T ( K) 1/T ln P

377 0.002653 5.9915

384 0.002604 6.2115

390 0.002564 6.3969

395 0.002532 6.5511

398 0.002513 6.6333

The best line has a fit:

y = -4609.5 x + 18.218

with R² = 0.9998

Now that we have the equation of the line, we simply will substitute for a pressure of 496 mm in Leadville.

ln(496) = -4609.5(1/Tb) + 18.218

6.2066 = -4609.5(1/Tb) +18.218

⇒ 1/Tb = (18.218 - 6.2066)/4609.5 = 0.00261

Tb = 383.76 K = (383.76 -273)K = 110.8 ºC

Notice we have touse up to 4 decimal places since rounding could lead to an erroneous answer ( i.e boiling temperature greater than 111, an impossibility given the data in the question). This is as a result of the value 496 mmHg so close to 500 mm Hg.

Perhaps that is the reason the question was flagged.

7 0

2 years ago

Miriam notices when she goes to the beach that sometimes the water rises as high as the pier. At other times of the day, the wat

kondor19780726 [428]

High tides and low tides are caused by the moon. The moon's gravitational pull generates something called the tidal force. The tidal force causes Earth—and its water—to bulge out on the side closest to the moon and the side farthest from the moon. These bulges of water are high tides.

3 0

2 years ago

Stabiliţi numerele de oxidare ale tuturor elementelor prezente în următoarele substanţe chimice, ţinând cont de principalele reg

natita [175]

Answer:

N = +2

O = -2

Na = +1

O = -2

N = +5

O = -2

Al = +3

P = +5

O = -2

Ca = +2

C = +4

O = -2

Mn = +7

K = +1

O = -2

K = +1

Cr = +6

O = -2

Cl = +7

Cr = +1

O = -2

H = +1

Cr = +5

O = -2

H = +1

Cr = +3

O = -2

H = +1

Cr = +1

Explanation:

A. NU

Numărul de oxidare a azotului = +2

Numărul de oxidare a oxigenului = -2

b. NaNO₃

Numărul de oxidare de Na = +1

Numărul de oxidare de O = -2

Numărul de oxidare de N = +5

c. AlPO₄

Numărul de oxidare de O = -2

Numărul de oxidare al Al = +3

Numărul de oxidare al P = +5

d. carbonat de calciu

Numărul de oxidare de O = -2

Numărul de oxidare de Ca = +2

Numărul de oxidare de C = +4

e. KMnO₄

Numărul de oxidare de O = -2

Numărul de oxidare al Mn = +7

Numărul de oxidare al lui K = +1

f. K₂Cr₂O₇

Numărul de oxidare de O = -2

Numărul de oxidare al lui K = +1

Numărul de oxidare al Cr = +6

g. HClO₄

Numărul de oxidare de O = -2

Numărul de oxidare al Cl = +7

Numărul de oxidare al Cr = +1

h. HClO₃

Numărul de oxidare de O = -2

Numărul de oxidare al lui H = +1

Numărul de oxidare al Cr = +5

i. HClO₂

Numărul de oxidare de O = -2

Numărul de oxidare al lui H = +1

Numărul de oxidare al Cr = +3

k. HClO

Numărul de oxidare de O = -2

Numărul de oxidare al lui H = +1

Numărul de oxidare al Cr = +1.

6 0

1 year ago

6. From the values of ΔH and ΔS, predict which of the following reactions would be spontaneous at 25ºC: Reaction A: ΔH = 10.5 kJ

qwelly [4]

Answer:

Both reaction A and reaction B are non spontaneous.

Explanation:

For a spontaneous reaction, change in gibbs free energy ( $\Delta G$ ) should be negative.

We know, $\Delta G=\Delta H-T\Delta S$ , where T is temperature in Kelvin scale.

Reaction A: $\Delta G=(10.5\times 10^{3})-(298\times 30)J/mol=1560J/mol$

As $\Delta G$ is positive therefore the reaction is non-spontaneous.

If at a temperature T K , the reaction is spontaneous then-

$\Delta H-T\Delta S< 0$

or, $T> \frac{\Delta H}{\Delta S}$

or, $T> \frac{10.5\times 10^{3}}{30}$

or, $T> 350$

So at a temperature greater than 350 K, the reaction is spontaneous.

Reaction B: $\Delta G=(1.8\times 10^{3})-(-113\times 298)J/mol=35474J/mol$

As $\Delta G$ is positive therefore the reaction is non-spontaneous.

If at a temperature T K , the reaction is spontaneous then-

$\Delta H-T\Delta S< 0$

or, $T> \frac{\Delta H}{\Delta S}$

or, $T> \frac{1.8\times 10^{3}}{-113}$

or, $T> -16$

So at a temperature greater than -16 K, the reaction is spontaneous.

3 0

2 years ago

The pH of a 0.30 M solution of a weak acid is 2.67. What is the Ka for this acid?

vova2212 [387]

Answer:

Ka → 1.5×10⁻⁵

Option E. None of the above

Explanation:

We propose the reaction of equlibrium

Weak ac.H + H₂O ⇄ Weak ac⁻ + H₃O⁺

Initially we have 0.30 moles of acid in 1 L

In equilibrium we would have:

Weak ac.H + H₂O ⇄ Weak ac⁻ + H₃O⁺

0.30 - x x x

We have the pH, where we can obtanined the x, the [H₃O⁺] in the equilibrium.

pH = - log [H₃O⁺] → [H₃O⁺] = 10^⁻(pH)

[H₃O⁺] = 10⁻²'⁶⁷ = 2.14×10⁻³

So let's determine the concentration of the acid, in the equilibrium

0.30 - 2.14×10⁻³ = 0.29786 → [Weak ac.H]

2.14×10⁻³ → [H₃O⁺] = Conjugate base (Weak ac.⁻)

Let's make the expression for Ka

Ka = [Weak ac.⁻] . [H₃O⁺] / [Weak ac.H]

Ka = x² / 0.30 - x

Ka = (2.14×10⁻³)² / 0.29786 → 1.5×10⁻⁵

6 0

2 years ago