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Natasha2012 [34]
2 years ago
8

Rutherford’s gold foil experiment demonstrated that

Chemistry
1 answer:
iVinArrow [24]2 years ago
4 0

Answer:

Explanation:

The space between the large nucleus and the electrons is huge. We know this because the alpha particles shot at the gold foil most went right on through. That means that the space between hold atoms is very large.

B

Not only that but the deflection that takes place is not frequent further telling us that the the nucleus must be positively charged. That observation comes from the deflection itself. The charge on the nucleus must be the same as the alpha particle. If they were different, and the nucleus was negative, those particles that were deflected would now be absorbed.

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Please only answer if you are 100% sure it's a major grade.<br> .<br> 35 POINTS
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The answer is B. Changes in air pressure and temperature. The changes in air temperature are often influenced by the increase in altitude, where the increase in altitude leads to a decrease in temperature, the same is true for the opposite!
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Where is an earthquake most likely to occur?
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The body's nervous system is made up of the brain,the spinal cord, the retina ,and many other nerves. The circulatory system is
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What is observed when a solution of potassium iodide is added to silver nitrate solution ? a) No reaction takes place (b) White
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Answer:

c) yellow precipitate of Agl is formed

Explanation:

KI + AgNO3 = AgI (s)+ KNO3

c) yellow precipitate of Agl is formed

6 0
3 years ago
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
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