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joja [24]
3 years ago
5

Which equation is derived from the combined gas law? OV, T = P₂72 O Pq V, T, - P ₂ V2 T 2

Chemistry
1 answer:
Olin [163]3 years ago
8 0

Answer:

\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2}

Explanation:

The equation of state (combined gas law) for an ideal gas states that

pV=nRT

where

p is the gas pressure

V is the volume of the gas

n is the number of moles of the gas

R is the gas constant

T is the absolute temperature of the gas

The equation can be rewritten as

\frac{pV}{T}=nR

For an ideal gas that undergoes a transformation, if the amount of gas remains the same, the term on the right of the equation remains constant. This means that the ratio

\frac{pV}{T}

for the gas remains constant, and therefore we can rewrite the equation as

\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2}

where 1 indicates the initial conditions of the gas while 2 indicates the conditions of the gas after the transformation.

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poizon [28]

Answer:

<h2>Actin and myosin.</h2>

Explanation:

The cells that allow your bones to move, the movement of thick (myosin) and thin (actin) filaments during contraction .

During a contraction thick and thin filaments do not shorten but increase their overlap  of each other.

Thin filaments slide past thick filaments extending more deeply into the A band.

The I bands and H bands decrease in lenght as Z discs are come closer together .

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8 0
3 years ago
"How much NH4Cl, when present in 2.00 liters of 0.200 M ammonia, will give a solution with pH = 8.20? For NH3, Kb = 1.8 x 10-5"
Andru [333]

Answer:

245.66g of NH₄Cl is the mass we need to add to obtain the desire pH

Explanation:

The mixture of NH3/NH4Cl produce a buffer. We can find the pH of a buffer using H-H equation:

pH = pKa + log [A⁻] / [HA]

<em>Where [A⁻] is the molar concentration of the base, NH₃, and [HA] molar concentration of the acid, NH₄⁺. This molar concentration can be taken as the moles of each chemical</em>

<em />

First, we need to find pKa of NH₃ using Kb. Then, the moles of NH₃ and finally replace these values in H-H equation to solve moles of NH₄Cl we need to obtain the desire pH.

  • <em>pKa NH₃/NH₄⁺</em>

pKb = - log Kb

pKb = -log 1.8x10⁻⁵ = <em>4.74</em>

pKa = 14 - pKb

pKa = 14 - 4.74

pKa = 9.26

  • <em>Moles NH₃</em>

<em>2.00L ₓ (0.200mol NH₃ / L) = 0.400 moles NH₃</em>

  • <em>H-H equation:</em>

pH = pKa + log [NH₃] / [NH₄Cl]

8.20 = 9.26 + log [0.400 moles] / [NH₄Cl]

-1.06 =  log [0.400 moles] / [NH₄Cl]

0.0087 =  [0.400 moles] / [NH₄Cl]

[NH₄Cl] = 0.400 moles / 0.0087

[NH₄Cl] = 4.59 moles of NH₄Cl we need to add to original solution to obtain a pH of 8.20. In grams (Using molar mass NH₄Cl=53.491g/mol):

4.59 moles NH₄Cl ₓ (53.491g / mol) =

<h3>245.66g of NH₄Cl is the mass we need to add to obtain the desire pH</h3>

<em />

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enot [183]

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<span>The addition of a catalyst to a chemical reaction provides an alternate pathway that c</span>atalysts lowers the activation energy.
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