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vaieri [72.5K]
3 years ago
12

A car travelling

Chemistry
1 answer:
Jet001 [13]3 years ago
8 0

Answer:

816.33 kg

Explanation:

To solve this problem we'll need to keep in mind the <em>formula for the kinetic energy of an object</em>:

  • E = 0.5 * m * v²

Where E is the kinetic energy, m is the object's mass and v its velocity.

We<u> input the data given by the problem</u>:

  • 500000 J = 0.5 * m * (35 m/s)²

And <u>solve for m</u>:

  • m = 816.33 kg
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20 Ca :

1s², 2s², 2p⁶, 3s², 3p⁶, 4s²<span>
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Answer D

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3 years ago
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Question 3. A batch chemical reactor achieves a reduction in
kotykmax [81]

Answer:

Rate constant for zero-order kinetics: 1, 58 [mg/L.s]

Rate constant for first-order kinetics: 0,05 [1/s]

Explanation:

The reaction order is the relationship between the concentration of species and the rate of the reaction. The rate law is as follows:

r = k [A]^{x} [B]^{y}

where:

  • [A] is the concentration of species A,
  • x is the order with respect to species A.
  • [B] is the concentration of species B,
  • y is the order with respect to species B
  • k is the rate constant

The concentration time equation gives the concentration of reactants and products as a function of time. To obtain this equation we have to integrate de velocity law:

v(t) = -\frac{d[A]}{dt} = k [A]^{n}

For the kinetics of zero-order, the rate is apparently independent of the reactant concentration.

<em>Rate Law:                                    rate = k</em>

<em>Concentration-time Equation:   [A]=[A]o - kt</em>

where

  • k: rate constant [M/s]
  • [A]: concentration in the time <em>t</em> [M]
  • [A]o: initial concentration [M]
  • t: elapsed reaction time [s]

For first-order kinetics, we have:

<em>Rate Law:                                        rate= k[A]</em>

<em>Concentration -Time Equation:      ln[A]=ln[A]o - kt</em>

where:

  • K: rate constant [1/s]
  • ln[A]: natural logarithm of the concentration in the time <em>t </em>[M]
  • ln[A]o: natural logarithm of the initial concentration [M]
  • t: elapsed reaction time [s]

To solve the problem, wee have the following data:

[A]o = 100 mg/L

[A] = 5 mg/L

t = 1 hour = 60 s

As we don't know the molar mass of the compound A, we can't convert the used concentration unit (mg/L) to molar concentration (M). So we'll solve the problem using mg/L as the concentration unit.

Zero-order kinetics

we use:                        [A]=[A]o - Kt

we replace the data:   5 = 100 - K (60)

we clear K:                 K = [100 - 5 ] (mg/L) /60 (s)  = 1, 583 [mg/L.s]

First-order kinetics

we use:                                  ln[A]=ln[A]o - Kt

we replace the data:               ln(5)  = ln(100) - K (60)

we clear K:                                   K = [ln(100) - ln(5)] /60 (s)  = 0,05 [1/s]

4 0
3 years ago
Which process is an oxidation?
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<span>The problem has to do with oxidation states of the matter. The oxidation state of oxygen will always be -2 with the exception of peroxides which will have a state of -1. The overall balanced state of chemical compounds will be 0, so the oxidation state of Mn in MnO2 will be +4. The oxidation state of MnO4- will then be +7 to balance out to the negative one charge. The state change from +4 to +7 is 3, thus three electrons have to be lost in order for this to happen; a loss of a charge of -3 results in an increase of charge of 3. Oxidation is always the process of 'losing' electrons.

</span><span>E] MnO2(s) MnO4-(aq</span>
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Answer:

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Vegetable soup should be the correct answer. The reason for this is that it can easily be split, where as the rest would be very difficault if not imposible because of chemical change.

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