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BabaBlast [244]
3 years ago
10

Find the mass in grams of 2.65 x 10^24 molecules of cl2

Chemistry
1 answer:
Alina [70]3 years ago
8 0
Moles of Cl2 = 2.65 x 10^24 / 6.02 x 10^23 = 4.4 moles.

Mass of Cl2 in grams = molar mass of Cl2 x moles of Cl2
                                     = 70.906 x 4.4
                                     = 311.98 grams.

Hope this helps!
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One mole of an ideal gas with a volume of 1.0 L and a pressure of 5.0 atm is allowed to expand isothermally into an evacuated bu
Deffense [45]

Answer:

w= - 1.7173 kJ, q= 1.7173 kJ, q(rev) = 1717.3 J = 1.7173 kJ.

Explanation:

Okay, from the question we are given the information below;

Number of moles, n= 1 mole; initial volume, v(1) = 1.0 litres (L); pressure (p) = 5atm, final volume(v2) = 2.0 Litres(L) ; the workdone, w= not given; the heat, q and q(rev)= not given and the gas was said to expand isothermally.

So, this question is a question from the part of chemistry known as thermodynamics. Therefore, grip yourself we are delving into thermodynamics 'waters' now.

For expansion isothermally; the workdone, w= -nRT ln v2/v1.

Where T= temperature= 25° C = 298 k and R= gas constant.

Therefore; workdone, w = - 1 × 8.314 × 298 × ln(2/1).

Workdone,w= - 1717.32204643. =

- 1717.3 Joules (J).

==> Workdone,w= - 1.7173 kJ.

Then, we are to find q. q can be solved by using the first law of thermodynamics, which by mathematical representation is:

∆U= q + w. Where ∆U= change in internal enegy. Since the question is dealing with isothermal expansion, there is this rule that says for an isothermal expansion ∆U = 0.

Hence, 0 =q + [- 1717.3 Joules (J)].

q=1717.3 J = 1.7173 kJ.

Finally, the q(rev) which is= nRT ln (v2/V1).

q(rev) = 1 × 8.314 × 298 ln (2/1).

q(rev) = 1717.3 J = 1.7173 kJ.

PS: please note the negative signs in the workdone and the positive sign in the q(rev).

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why do liquids flow? the particles are tightly packed together the particles do not move the particles vibrate the particles are
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The particles are quite tightly packed together but still have enough room to be able to move and flow, their bonds aren't as strong as a solids are
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Alisiya [41]
The homologous structures and the analogous structures of different species.
6 0
2 years ago
A chemist fills a reaction vessel with mercurous chloride solid, mercury (I) aqueous solution, and chloride aqueous solution at
makvit [3.9K]

Answer:

ΔG° = -533.64 kJ

Explanation:

Let's consider the following reaction.

Hg₂Cl₂(s) ⇄ Hg₂²⁺(aq) + 2 Cl⁻(aq)

The standard Gibbs free energy (ΔG°) can be calculated using the following expression:

ΔG° = ∑np × ΔG°f(products) - ∑nr × ΔG°f(reactants)

where,

ni are the moles of reactants and products

ΔG°f(i) are the standard Gibbs free energies of formation of reactants and products

ΔG° = 1 mol × ΔG°f(Hg₂²⁺) + 2 mol × ΔG°f(Cl⁻) - 1 mol × ΔG°f(Hg₂Cl₂)

ΔG° = 1 mol × 148.85 kJ/mol + 2 mol × (-182.43 kJ/mol) - 1 mol × (-317.63 kJ/mol)

ΔG° = -533.64 kJ

3 0
3 years ago
When 10.0 grams of sulfur reacts with fluorine gas at a pressure of 2.69 atmosphere in a 5.00 L container at 0.00 degrees Celsiu
Gwar [14]

Answer:

74.1%

Explanation:

Based on the reaction:

S₈ + 16F₂ → 8SF₄

<em>1 mole of sulfur reacts with 16 moles of F₂ to produce 8 moles of SF₄</em>

<em />

To solve this question we must find the moles of each reactant in order to find the moles of SF₄. Thus, we can find the theoretical mass produced. Percent yield is:

Percent yield = Actual yield (25.0g) / Theoretical yield * 100

<em>Moles S₈: 256.52g/mol</em>

10.0g * (1mol / 256.52g) = 0.0390 moles

<em>Moles F₂:</em>

<em>PV = nRT</em>

PV/RT = n

<em>Where P is pressure in atm, V is volume in liters, R is gas constant and T is absolute temperature (0°C = 273.15K)</em>

2.69atm*5.00L / 0.082atmL/molK*273.15K = n

0.600 moles = n

For a complete reaction of 0.600 moles F₂ are required:

0.600mol F₂ * (1mol S₈ / 8 mol F₂) = 0.075 moles S₈

As there are just 0.0390 moles, S₈ is limiting reactant.

The theoretical moles and mass of SF₄ -Molar mass: 108.07g/mol- is:

0.0390 moles S₈ * (8mol SF₄ / 1mol S₈) = 0.312 moles SF₄ * (108.07g) =

33.7g

Percent yield = 25.0g / 33.7g * 100

= 74.1%

6 0
2 years ago
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