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3 years ago

Explain the connections between reactants, products, and limiting reactant.

1 answer:

6 0

Reactants is a starting substance in a chemical reaction; (reactants appear on the left-hand side of a chemical equation).
Products is a substance produced in a chemical reaction; (products appear on the right-hand side of a chemical equation).
Limiting reactant is the reactant that has the smallest stoichiometric amount in a reactant mixture and consequently limits the amount of product in a chemical reaction.

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A piston–cylinder assembly contains 5.0 kg of air, initially at 2.0 bar, 30 oF. The air undergoes a process to a state where the

vladimir2022 [97]

Answer:

The work and heat transfer for this process is = 270.588 kJ

Explanation:

Take properties of air from an ideal gas table. R = 0.287 kJ/kg-k

The Pressure-Volume relation is <em>PV</em> = <em>C</em>

<em>T = C </em> for isothermal process

Calculating for the work done in isothermal process

<em>W</em> = <em>P</em>₁<em>V</em>₁ $ln[\frac{P_{1} }{P_{2} }]$

= <em>mRT</em>₁ $ln[\frac{P_{1} }{P_{2} }]$ [∵<em>pV</em> = <em>mRT</em>]

= (5) (0.287) (272.039) $ln[\frac{2.0}{1.0}]$

= 270.588 kJ

Since the process is isothermal, Internal energy change is zero

Δ<em>U</em> = $mc_{v}(T_{2} - T_{1} ) = 0$

From 1st law of thermodynamics

Q = Δ<em>U </em>+ <em>W</em>

= 0 + 270.588

= 270.588 kJ

4 0

3 years ago

4.0 kg objects mive a distance of 7.8 m under the action of a cinstant gorce of 5.6 N. how much work is dine on object?

alexdok [17]

Answer:

43.68 J

Explanation:

Distance moved= 7.8 m

Force = 5.6 N

Work Done= Distance moved * Force

= 7.8 *5.6

=43.68 Joules

6 0

3 years ago

One string of a certain musical instrument is 70.0 cm long and has a mass of 8.79 g . It is being played in a room where the spe

Svetach [21]

To solve this problem we will apply the concepts of linear mass density, and the expression of the wavelength with which we can find the frequency of the string. With these values it will be possible to find the voltage value. Later we will apply concepts related to harmonic waves in order to find the fundamental frequency.

The linear mass density is given as,

$\mu = \frac{m}{l}$

$\mu = \frac{8.79*10^{-3}}{70*10^{-2}}$

$\mu = 0.01255kg/m$

The expression for the wavelength of the standing wave for the second overtone is

$\lambda = \frac{2}{3} l$

Replacing we have

$\lambda = \frac{2}{3} (70*10^{-2})$

$\lambda = 0.466m$

The frequency of the sound wave is

$f_s = \frac{v}{\lambda_s}$

$f_s = \frac{344}{0.768}$

$f_s = 448Hz$

Now the velocity of the wave would be

$v = f_s \lambda$

$v = (448)(0.466)$

$v = 208.768m/s$

The expression that relates the velocity of the wave, tension on the string and linear mass density is

$v = \sqrt{\frac{T}{\mu}}$

$v^2 = \frac{T}{\mu}$

$T= \mu v^2$

$T = (0.01255kg/m)(208.768m/s)^2$

$T = 547N$

The tension in the string is 547N

PART B) The relation between the fundamental frequency and the $n^{th}$ harmonic frequency is

$f_n = nf_1$

Overtone is the resonant frequency above the fundamental frequency. The second overtone is the second resonant frequency after the fundamental frequency. Therefore

$n=3$