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2 years ago

Question 14 of 25

2 answers:

5 0

A. Areas within an energy level that are occupied by electrons

There are different energy levels where electrons (probably) are, they are hanging out in the orbitals, or energy levels around the nucleus.

We know the other options are wrong because protons are in the nucleus of an atom, neutrons have a neutral charge, and although B sounds right, A is the more accurate choice

algol132 years ago

3 0

Answer:

describes properties characteristic of no more than two electrons in the vicinity of an atomic nucleus or of a system of nuclei as in a molecule

You might be interested in

Look at the above table.

MariettaO [177]

Answer:

1. A. 0

2. B. 7

3. C. 4

Explanation:

1. charge is equal to the number of protons minus the number of electrons!

2. neutrons is equal to mass number minus atomic number!

3. valence electrons equal 4!

Hope this helped you! :)

7 0

3 years ago

Which of the following can an isosceles triangle not be? an acute triangle a scalene triangle an equilateral triangle a right tr

zimovet [89]

Answer;

A scalene

Explanation;

A triangle is a geometric figure with three sides and three angles that always add up to 180 degrees.

Based on properties triangles can be classified as equilateral, scalene, isosceles and right triangles.

Equilateral triangles has all the three sides and the angles equal, an isosceles triangle has two of its sides equal, and two angles that are known as the base angles equal. A scalene triangle has all the three angles and sides different from each other. A right triangle is a triangle in which one of the angle is right angle or is 90 degrees.

An isosceles triangle may be a right triangle, equilateral triangle, and also may be an acute triangle, however an isosceles triangle can never be a scalene triangle

6 0

3 years ago

Read 2 more answers

A second- order reaction of the type A + B -->P was carried out in a solution that was initially 0.075 mol dm^-3 in A and 0.0

andriy [413]

Answer:

a) 16.2 dm^3/mol*h

b) 6.1 × 10^3 s, 2.5 × 10^3 s (it is different to the hint)

Explanation:

We can use the integrated rate equation in order to obtain k.

For the reaction A+ B --> P the reaction rate is written as

Rate = $-\frac{dC_A}{dt} = -\frac{dC_B}{dt} = \frac{dC_P}{dt} = kC_AC_B$

If $C_{A0}$ and $C_{B0}$ are the inital concentrations and x the concentration reacted at time t, so $C_A=C_{A0} -x$ and $C_B=C_{B0} -x$ and the rate at time t is written as:

$-\frac{dx}{dt} =-k(C_{A0} -x)(C_{B0}-x)$

Separating variables and integrating

$\int\limits^x_0 {\frac{1}{(C_{A0}-x)(C_{B0}-x)} } \, dx = \int\limits^t_0 {k} \, dt$

The integral in left side is solved by partial fractions, it can be used integral tables

$\frac{1}{C_{B0}-C_{A0}}(ln\frac{C_{A0}}{C_{A0}-x}-ln\frac{C_{B0}}{C_{B0}-x}) =kt$

Using logarithm properties (ln x - ln y = ln(x/y))

$\frac{1}{C_{B0}-C_{A0}}(ln\frac{C_{A0}C_{B}}{C_{A}C_{B0}}) =kt$

Using the given values k can be calculated. But the data seems inconsistent since if the concentration of A changes from 0.075 to 0.02 mol dm^-3 it implies that 0.055 mol dm^-3 of A have reacted after 1 h, so according to the reaction given the same quantity of B should react, and we only have a $C_{B0}$ of 0.05 mol dm^-3.

Assuming that the concentration of B fall to 0.02 mol dm^-3 (and not the concentration the A). So we arrive to the answer given in the Hint.

So, the values given are t= 1, $C_{A0}=0.075$ , $C_{B0}=0.05$ , $C_{B}=0.02$ , it implies that the quantity reacted, x, is 0.03 and $C_{A}=0.075-x = 0.045$ . Then, the value of k would be

$kt = \frac{1}{0.05-0.075}(ln\frac{0.075*0.02}{0.045*0.05})$

$k = 16.21 \frac{dm^3}{mol*1h}$

b) the question b requires calculate the time when the concentration of the specie is half of the initial concentration.

For reactant A, It is solved with the same equation

$\frac{1}{C_{B0}-C_{A0}}(ln\frac{C_{A0}C_{B}}{C_{A}C_{B0}}) =kt$

but suppossing that $C_A= C_{A0}/2=0.0375$ so $C_B=C_{B0}- C_{A0}/2=0.0125$ , k=16.2 and the same initial concentrations. Replacing in the equation

$t=\frac{1}{16.2(0.05-0.075)}(ln\frac{0.075*0.0125}{0.0375*0.05})$

$t=1.71 h = 1.71*3600 s = 6.1*10^3 s$

For reactant B, $C_B= C_{B0}/2=0.025$ so $C_A=C_{A0}- C_{B0}/2=0.05$ , k=16.2 and the same initial concentrations. Replacing in the equation

$t=\frac{1}{16.2(0.05-0.075)}(ln\frac{0.075*0.025}{0.05*0.05})$

$t=0.71 h = 0.7*3600 s = 2.5*10^3 s$

Note: The procedure presented is correct, despite of the answer be something different to the given in the hint, I obtain that result if the k is 19.2... (maybe an error in calculation of given numbers)

3 0

3 years ago

MIn photosynthesis, what are the two major reactions that take place?

Alla [95]

Answer:

photosyntheisis

Explanation:

The process by which plants and some other organisms capture enrgy in sunlight and use it to make food (Btw ik this is right becuase i did the same thing and i have the definition of what it is and other definitions!

5 0

3 years ago

For Gay-Lussac's law, temperature is measured in _____.

Thepotemich [5.8K]

For Scientist Joseph Louis Gay-Lussac's law, temperature is measured on an absolute scale, specifically Kelvins or k. The formula may be expressed in P/T=k (P over T equals k). The answer to this question is 'kelvins'. I hope this helps.

7 0

3 years ago