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Shalnov [3]
3 years ago
5

A sample of neon occupies a volume of 752 ml at 25 0

Chemistry
1 answer:
lyudmila [28]3 years ago
5 0
V_{initial} = 752\:mL
T_{initial} = 25.0^0C
converting to Kelvin
TK = TC + 273
TK = 25.0 + 273 → TK = 298.0 → T_{initial} = 298.0\:K
V_{final} = ? (in\:milliliters)
T_{final} = 50.0^0C
TK = TC + 273
TK = 50.0 + 273 → TK = 323.0 → T_{final} = 323.0\:K

By the first Law of Charles and Gay-Lussac, we have: 
\frac{ V_{i} }{ T_{i} } = \frac{ V_{f} }{ T_{f} }

Solving:
\frac{ V_{i} }{ T_{i} } = \frac{ V_{f} }{ T_{f} }
\frac{ 752 }{ 298.0 } = \frac{ V_{f} }{ 323.0 }
Product of extremes equals product of means:
298.0* V_{f} = 752*323.0
298.0 V_{f} = 242896
V_{f} = \frac{242896}{298.0}
\boxed{\boxed{V_{f} \approx 815.08\:mL}}\end{array}}\qquad\quad\checkmark
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A sample of gas occupies 10.0 L at 240°C under a pressure of
NISA [10]

Answer: 1090°C

Explanation: According to combined gas laws

(P1 × V1) ÷ T1 = (P2 × V2) ÷ T2

where P1 = initial pressure of gas = 80.0 kPa

V1 = initial volume of gas = 10.0 L

T1 = initial temperature of gas = 240 °C = (240 + 273) K = 513 K

P2 = final pressure of gas = 107 kPa

V2 = final volume of gas = 20.0 L

T2 = final temperature of gas

Substituting the values,

(80.0 kPa × 10.0 L) ÷ (513 K) = (107 kPa × 20.0 L) ÷ T2

T2 = 513 K × (107 kPa ÷80.0 kPa) × (20.0 L ÷ 10.0 L)

T2 = 513 K × (1.3375) × (2)

T2 = 1372.275 K

T2 = (1372.275 - 273) °C

T2 = 1099 °C

8 0
3 years ago
Read 2 more answers
Calculate the percent ionization of formic acid (hco2h) in a solution that is 0.311 m in formic acid and 0.189 m in sodium forma
ohaa [14]
<span>Answer: 0.094%


</span><span>Explanation:
</span>
<span></span><span /><span>
1) Equilibrium chemical equation:
</span><span />

<span>Only the ionization of the formic acid is the important part.
</span><span />

<span>HCOOH(aq) ⇄ HCOO⁻(aq) + H⁺(aq).
</span><span />

<span>2) Mass balance:
</span><span />

<span>                   HCOOH(aq)     HCOO⁻(aq)     H⁺(aq).

Start             0.311                 0.189

Reaction       - x                      +x                   +x

Final             0.311 - x          0.189 + x            x


3) Acid constant equation:
</span><span />

<span>Ka = [HCOO-] [N+] / [HCOOH] = (0.189 + x) x / (0.311 -x)
</span><span />

<span>= (0.189 + x )x / (0.311 - x) = 0.000177


4) Solve the equation:


You can solve it exactly (it will lead to a quadratic equation so you can use the quadratiic formula). I suggest to use the fact that x is much much smaller than 0.189 and 0.311.
</span><span />

<span>With that approximation the equation to solve becomes:


</span><span>0.1890x / 0.311 = 0.000177, which leads to:</span>
<span /><span>
x = 0.000177 x 0.311 / 0.189 = 2.91 x 10⁻⁴ M


5) With that number, the percent of ionization (alfa) is:
</span><span />

<span>percent of ionization = (moles ionized / initial moles) x 100 =
</span><span>
</span><span>
</span><span>percent ionization = (concentration of ions / initial concentration) x 100 =
</span><span>
</span><span>
</span><span>percent ionization = (0.000291 / 0.311)x 100 = 0.0936% = 0.094%
</span>
<span></span><span />
8 0
3 years ago
Heavy elements, like uranium, were/are created _______________. a. in supernovae and star collisions b. all of the above c. in t
Ivenika [448]

Answer:

a. in supernovae and star collisions

Explanation:

The periodical table contains some heavier elements, which are formed as neutron stars pairs hit eachother and erupt cataclysmically.

The star emitts very large quantities of energy and neutrons during supernova, which allow for the production of heavier elements than iron, such as uranium and gold. All these elements are ejected into space during the supernova explosion.

3 0
2 years ago
What is the maximum volume of 0.25 M sodium hypochlorite solution (NaOCl, laundry bleach) that can be prepared by dilution of 1.
ehidna [41]

Answer:

0.313L

Explanation:

0.25Mx1.00L=0.80Mx V2

so, 0.25x1.00/0.80=0.313L

8 0
3 years ago
Read 2 more answers
A 3.45 microgram sample of Uranium has a mass of how many grams?
Ivan
There are 3.45e-6 grams.
4 0
2 years ago
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