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3 years ago

How did Niels Bohr describe electrons in his atomic model

2 answers:

4 0

Answer: Bohr proposed his quantized shell model of the atom to explain how electrons can have stable orbits around the nucleus.

so C would seems to be right.

Explanation: The energy of an electron depends on the size of the orbit and is lower for smaller orbits. Radiation can occur only when the electron jumps from one orbit to another. The atom will be completely stable in the state with the smallest orbit, since there is no orbit of lower energy into which the electron can jump.

oksano4ka [1.4K]3 years ago

4 0

Answer:

Explanation:

Just did it on edge

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Which formula shows a correct representation of the combined gas law?(1 point)

weqwewe [10]

Answer:

Explanation:

D. V1P1 / T1=V2P2 / T2 is correct

8 0

2 years ago

For the reaction A+B+C→D+E, the initial reaction rate was measured for various initial concentrations of reactants. The followin

erastovalidia [21]

Answer : The initial rate for a reaction will be $3.4\times 10^{-3}Ms^{-1}$

Explanation :

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

$A+B+C\rightarrow D+E$

Rate law expression for the reaction:

$\text{Rate}=k[A]^a[B]^b[C]^c$

where,

a = order with respect to A

b = order with respect to B

c = order with respect to C

Expression for rate law for first observation:

$6.0\times 10^{-5}=k(0.20)^a(0.20)^b(0.20)^c$ ....(1)

Expression for rate law for second observation:

$1.8\times 10^{-4}=k(0.20)^a(0.20)^b(0.60)^c$ ....(2)

Expression for rate law for third observation:

$2.4\times 10^{-4}=k(0.40)^a(0.20)^b(0.20)^c$ ....(3)

Expression for rate law for fourth observation:

$2.4\times 10^{-4}=k(0.40)^a(0.40)^b(0.20)^c$ ....(4)

Dividing 1 from 2, we get:

$\frac{1.8\times 10^{-4}}{6.0\times 10^{-5}}=\frac{k(0.20)^a(0.20)^b(0.60)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\3=3^c\\c=1$

Dividing 1 from 3, we get:

$\frac{2.4\times 10^{-4}}{6.0\times 10^{-5}}=\frac{k(0.40)^a(0.20)^b(0.20)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\4=2^a\\a=2$

Dividing 3 from 4, we get:

$\frac{2.4\times 10^{-4}}{2.4\times 10^{-4}}=\frac{k(0.40)^a(0.40)^b(0.20)^c}{k(0.40)^a(0.20)^b(0.20)^c}\\\\1=2^b\\b=0$

Thus, the rate law becomes:

$\text{Rate}=k[A]^2[B]^0[C]^1$

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

$6.0\times 10^{-5}=k(0.20)^2(0.20)^0(0.20)^1$

$k=7.5\times 10^{-3}M^{-2}s^{-1}$

Now we have to calculate the initial rate for a reaction that starts with 0.75 M of reagent A and 0.90 M of reagents B and C.

$\text{Rate}=k[A]^2[B]^0[C]^1$

$\text{Rate}=(7.5\times 10^{-3})\times (0.75)^2(0.90)^0(0.90)^1$

$\text{Rate}=3.4\times 10^{-3}Ms^{-1}$

Therefore, the initial rate for a reaction will be $3.4\times 10^{-3}Ms^{-1}$

8 0

3 years ago

Pricing the product low in order to stimulate demand and increase the installed base, then trying to make high profits on the sa

Amanda [17]

Answer:

razor and blade strategy

Explanation:

Razor and blade strategy -

It refers to the method of pricing , where the price of one of the item is reduced in order to increase the sale of another item , is referred to as razor and blade strategy .

It is a type of pricing tactics used by the company to indirectly earn profit for their goods and services .

Sometimes , fews goods are given free along with certain products , in order to earn profit .

Hence , from the given information of the question ,

The correct answer is razor and blade strategy .

5 0

3 years ago

for the reaction shown compute the theoretical yield of product in moles each of the initial quantities of reactants. 2 Mn(s)+3

Y_Kistochka [10]

Answer:

2 mole MnO₂

Explanation:

2Mn(s) + 2O₂(g) => 2MnO₂(s)

4 0

4 years ago

Read 2 more answers

What types of radiation cause the parent isotope to change into a different element?

kvv77 [185]

Answer:

Beta decay is most common in elements with a high neutron to proton ratio. Gamma decay follows the form: In gamma emission, neither the atomic number or the mass number is changed. A high energy gamma ray is given off when the parent isotope falls into a lower energy state.

Explanation:

pls mark me as brainliest

3 0

3 years ago