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Savatey [412]
3 years ago
14

11) What is the volume of 4.92 grams of hydrogen gas at STP? (Show the work using dimensional analysis)

Chemistry
1 answer:
lorasvet [3.4K]3 years ago
4 0

Answer:

The volume of 4.92 grams of hydrogen gas at STP is 9.105 litter.

Explanation:

We know that, the molecular weight of hydrogen gas (H_{2} ) = 1\times 2 = 2.

Moles of hydrogen gas = \frac{weight}{molecular weight}  = \frac{4.92}{2} = 2.46 moles.

At STP, 1 mole of hydrogen gas = occupies 22.4 litter

    So, 2.46 mole of hydrogen gas = occupies \frac{22.4}{2.46} = 9.105 litter.

Hence, the volume of 2.46 grams of hydrogen gas at STP is 9.105 litter.

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In an experiment to study the photoelectric effect, a scientist measures the kinetic energy of ejected electrons as afunction of
crimeas [40]

Answer:

a) v₀ = 4.41 × 10¹⁴ s⁻¹

b) W₀ = 176 KJ/mol of ejected electrons

c) From the graph, light of frequency less than v₀ will not cause electrons to break free from the surface of the metal. Electron kinetic energy remains at zero as long as the frequency of incident light is less than v₀.

d) When frequency of the light exceeds v₀, there is an increase of electron kinetic energy from zero steadily upwards with a constant slope. This is because, once light frequency exceeds, v₀, its energy too exceeds the work function of the metal and the electrons instantaneously gain the energy of incident light and convert this energy to kinetic energy by breaking free and going into motion. The energy keeps increasing as the energy and frequency of incident light increases and electrons gain more speed.

e) The slope of the line segment gives the Planck's constant. Explanation is in the section below.

Explanation:

The plot for this question which is attached to this solution has Electron kinetic energy on the y-axis and frequency of incident light on the x-axis.

a) Wavelength, λ = 680 nm = 680 × 10⁻⁹ m

Speed of light = 3 × 10⁸ m/s

The frequency of the light, v₀ = ?

Frequency = speed of light/wavelength

v₀ = (3 × 10⁸)/(680 × 10⁻⁹) = 4.41 × 10¹⁴ s⁻¹

b) Work function, W₀ = energy of the light photons with the wavelength of v₀ = E = hv₀

h = Planck's constant = 6.63 × 10⁻³⁴ J.s

E = 6.63 × 10⁻³⁴ × 4.41 × 10¹⁴ = 2.92 × 10⁻¹⁹J

E in J/mol of ejected electrons

Ecalculated × Avogadros constant

= 2.92 × 10⁻¹⁹ × 6.023 × 10²³

= 1.76 × 10⁵ J/mol of ejected electrons = 176 KJ/mol of ejected electrons

c) Light of frequency less than v₀ does not possess enough energy to cause electrons to break free from the metal surface. The energy of light with frequency less than v₀ is less than the work function of the metal (which is the minimum amount of energy of light required to excite electrons on metal surface enough to break free).

As evident from the graph, electron kinetic energy remains at zero as long as the frequency of incident light is less than v₀.

d) When frequency of the light exceeds v₀, there is an increase of electron kinetic energy from zero steadily upwards with a constant slope. This is because, once light frequency exceeds, v₀, its energy too exceeds the work function of the metal and the electrons instantaneously gain the energy of incident light and convert this energy to kinetic energy by breaking free and going into motion. The energy keeps increasing as the energy and frequency of incident light increases and electrons gain more speed.

e) The slope of the line segment gives the Planck's constant. From the mathematical relationship, E = hv₀,

And the slope of the line segment is Energy of ejected electrons/frequency of incident light, E/v₀, which adequately matches the Planck's constant, h = 6.63 × 10⁻³⁴ J.s

Hope this Helps!!!

5 0
3 years ago
If 1. 3618 moles of AsF3 are allowed to react with 1. 0000 mole of C2Cl6, what would be the theoretical yield of AsCl3, in moles
VARVARA [1.3K]

Answer:

AsF3:C2CI6

4:3

1.3618 moles: 1.02135 moles(1.3618÷4×3)

C2CI6 is the limting reagent

So the number of moles for AsCI3 is 0.817 moles( number of moles of the limting reagant) ÷3 ×4 (according to ratio by balancing chemical equation)=1.09 moles(3 s.f.)

or

Balanced equation

4AsF3 + 3C2Cl6 → 4AsCl3 + 3C2Cl2F4

Use stoichiometry to calculate the moles of AsCl3 that can be produced by each reactant.

Multiply the moles of each reactant by the mole ratio between it and AsCl3 in the balanced equation, so that the moles of the reactant cancel, leaving moles of AsCl3.

Explanation:

5 0
2 years ago
What is true about the structure or function of the plasma membrane?
den301095 [7]

Answer: hi theree, the answer is c. the processes of endocytosis and exocytosis occur here

hope this helps, have a good day :)

Explanation:

5 0
3 years ago
We will abbreviate malonic acid CH2(CO2H)2, a diprotic acid, as H2A (pK1 = 2.847 and pK2 = 5.696). Find the pH in (a) 0.200 M H2
larisa [96]

Answer:

a. pH  → 1.77

b. pH → 4.27

Explanation:

Malonic acid is a dyprotic acid. It releases two protons:

H₂A  +  H₂O →  H₃O⁺   +  HA⁻     Ka1

HA⁻   +  H₂O →  H₃O⁺   +  A⁻²          Ka2

Let's find the first pH:

We expose the mass balance:

Ca = [HA]  +  [HA⁻] + [A⁻²] = 0.2 M

We can not consider the [A⁻²] so → Ca =  [HA]  +  [HA⁻] = 0.2M

As the acid is so concentrated, we can not consider the HA- so:

Ca = [HA] = 0.2 M

Charge balance →  [H⁺]  =  [HA⁻]  + [OH⁻]

H₂A  +  H₂O →  H₃O⁺   +  HA⁻     Ka1

Ka = H₃O⁺ . HA⁻ / H₂A

We need the HA⁻ value to put on the charge balance. We re order the Ka expression:

HA⁻ = Ka . H₂A / H₃O⁺           (notice that H₃O⁺ = H⁺)

We replace:  H⁺¨ = Ka . H₂A / H⁺

(H⁺)² = Ka . Ca

H⁺ = √(Ka . Ca)         We determine Ka from pKa → 10²'⁸⁴⁷ = 1.42×10⁻³

H⁺ = √(1.42×10⁻³ . 0.2)  =  0.016866

- log [H⁺] = pH  → - log 0.016866 = 1.77

b. NaHA is the salt from the weak acid, where the HA⁻ works as an amphoterous, this means that can be an acid or a base:

HA⁻  +  H₂O  ⇄  A⁻²  +  H₃O⁺     Ka₂

HA⁻  +  H₂O ⇄  H₂A  +  OH⁻      Kb2

There is a formula than can predict the pH, so now we need to compare the Ka₂ and Kb₂ data.

Ka₂ = 10⁻⁵'⁶⁹⁶ = 2.01×10⁻⁶

Kb₂ = 1×10⁻¹⁴ / 1.42×10⁻³  = 7.04×10⁻¹²

So Ka₂ > Kb₂. In conclussion the pH will be acid.

[H⁺] = √(Ka1 .  Ka2)  →  [H⁺] = √ 1.42×10⁻³ . 2.01×10⁻⁶ = 5.34×10⁻⁵

- log 5.34×10⁻⁵  = pH → 4.27

3 0
3 years ago
The old car left on the side of the road showed signs of corrosion. We call this common chemical reaction rusting. The equation
Alex73 [517]
The answer is C) iron and oxygen. The reactants are on the left side of the equation
6 0
3 years ago
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