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m_a_m_a [10]
3 years ago
7

The speed of sound in a container of gas at 20.0°C is v. If the temperature of the gas is increased to 80.0°C, the new speed wil

l be?
Chemistry
1 answer:
Rufina [12.5K]3 years ago
5 0

Answer : The new speed will be, 1.098v

Explanation :

The formula used for speed of sound in a container of gas is:

c=\sqrt{\frac{\gamma RT}{M}}

where,

c = speed of sound

R = gas constant

\gamma = adiabatic constant

T = absolute temperature

M = molecular mass of gas

As, c\propto \sqrt{T}

or, we can say that

\frac{c_1}{c_2}=\sqrt{\frac{T_1}{T_2}}            .............(1)

Given:

T_1 = 20.0^oC=273+20.0=293K

T_1 = 80.0^oC=273+80.0=353K

c_1 = v

Now put all the given values in the above formula 1, we get:

\frac{v}{c_2}=\sqrt{\frac{293}{353}}

c_2=1.098v

Thus, the new speed will be, 1.098v

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A 250 ml sample of saturated a g o h solution was titrated with h c l , and the endpoint was reached after 2. 60 ml of 0. 0136 m h c l was dispensed. Based on this titration, what is the k s p of a g o h <u>. Ksp=1.9×10⁻⁸</u>

<h3>What is titration?</h3>

Titration is a typical laboratory technique for quantitative chemical analysis used to calculate the concentration of a specified analyte. It is also referred to as titrimetry and volumetric analysis (a substance to be analyzed). A standard solution with a known concentration and volume is prepared as the reagent, also known as the titrant or titrator. To ascertain the concentration of the analyte, the titrant reacts with an analyte solution (also known as the titrand). The titration volume is the amount of titrant that interacted with the analyte.

A typical titration starts with a beaker or Erlenmeyer flask being placed below a calibrated burette or chemical pipetting syringe that contains the titrant and a little amount of the indicator (such as phenolphthalein).

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1 year ago
Consider a voltaic cell where the anode half-reaction is Zn(s) → Zn2+(aq) + 2 e− and the cathode half-reaction is Sn2+(aq) + 2 e
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<u>Answer:</u> The concentration of Sn^{2+} in the cell is 9.0\times 10^{-3}M

<u>Explanation:</u>

We are given:

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<u>Reduction half reaction:</u>  Sn^{2+}(aq.)+2e^-\rightarrow Sn(s)   E^o_{Sn^{2+}/Sn}=-0.136V

The substance having highest positive E^o potential will always get reduced and will undergo reduction reaction. Here, fluorine will undergo reduction reaction will get reduced.

Here, tin will undergo reduction reaction and will get reduced.

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

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Putting values in above equation, we get:

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0.660=0.624-\frac{0.059}{2}\times \log(\frac{2.5\times 10^{-3}}{[Sn^{2+}})

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