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Triss [41]
4 years ago
7

Write orbital diagrams (boxes with arrows in them) to represent the electron configurations of carbon before and after sp hybrid

ization.

Chemistry
1 answer:
NemiM [27]4 years ago
4 0
Carbon has an electron configuration of 1s^2 2s^2 2p^2. During sp hybridization, one s and one p orbital of carbon combine to form two sp hybrid orbitals.


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A forensic anthropologist they discovers the bones of a human whose tibia has stopped growing but who's a clavicle appears to ha
yulyashka [42]

Explanation:

  • <em><u>the</u></em><em><u> </u></em><em><u>person</u></em><em><u> </u></em><em><u>was</u></em><em><u> </u></em><em><u>between</u></em><em><u> </u></em><em><u>1</u></em><em><u>3</u></em><em><u> </u></em><em><u>and</u></em><em><u> </u></em><em><u>16</u></em><em><u> </u></em><em><u>years</u></em><em><u> </u></em><em><u>old</u></em>
6 0
3 years ago
8. A sample of potassium chlorate (KCIO,) was heated in a test tube and decomposed 2KC?(s) 302 (g) + 2KCIO, (s) The oxygen was c
dangina [55]

Answer:

Partial pressure of O_{2} in the gas was 733 torr and mass of KClO_{3} in the sample was 2.12 g.

Explanation:

a) Total pressure of gas = (partial pressure of water vapour)+(partial pressure of O_{2})

Here partial pressure of water vapour is 21 torr and total pressure of gas is 754 torr.

So, partial pressure of O_{2}= (total pressure of gas)-(partial pressure of water vapour) = (754 torr) - (21 torr) = 733 torr

b) Lets assume that O_{2} behaves ideally. Hence-

                                            PV=nRT

where P is pressure of O_{2}, V is volume of O_{2} , n is number of moles of O_{2} , R is gas constant and T is temperature in kelvin

here P = 733 torr = (733\times 0.001316)atm = 0.9646 atm

        V = 0.65 L, R = 0.082 L.atm/(mol.K), T=(273+22)K = 295 K

   So, n=\frac{PV}{RT}

                   = \frac{(0.9646 atm)\times (0.65 L)}{(0.082 L.atm/(mol.K))\times (295 K)}

                   = 0.0259 moles

As 3 moles of O_{2} are produced from 2 moles of KClO_{3} therefore 0.0259 moles of O_{2} are produced from (\frac{2\times 0.0259}{3}) moles or 0.0173 moles of KClO_{3}.

Molar mass of KClO_{3}= 122.55 g

So mass of KClO_{3} in sample = (0.0173\times 122.55)g

                                                                    = 2.12 g

7 0
3 years ago
PLEASE ANSWER
RoseWind [281]

It is a radio wave

Explanation:

The energy of an electromagnetic waves is related to its frequency by the equation:

E=hf

where

h is the Planck constant

f is the frequency of the wave

We notice that the energy of an electromagnetic wave is proportional to its frequency. This means that waves with higher frequencies, such as gamma rays, x-rays, are more energetic, while waves with lower frequencies, such as microwave and radio waves, are the less energetic.

In this case, we want to find the frequency of a wave with energy

E = 0.000001 eV

Converting into Joules,

E=1\cdot 10^{-6} eV \cdot 1.6\cdot 10^{-19} J/eV=1.6\cdot 10^{-25} J

Solving for the frequency,

f=\frac{E}{h}=\frac{1.6\cdot 10^{-25}}{6.63\cdot 10^{-34}}=2.4\cdot 10^8 Hz = 240 MHz

Which falls in the range of frequency of radio waves.

Learn more about electromagnetic waves:

brainly.com/question/9184100

brainly.com/question/12450147

#LearnwithBrainly

4 0
3 years ago
Which form of asexual reproduction describes the process of a cell dividing to
stepan [7]

Answer:A

Explanation:Binary Fission, meaning ‘getting divided into half’ is a type of asexual reproduction where a single living cell grows twice its size and then splits to form two identical daughter cells, each carrying a copy of the parent cell’s genetic material. Examples of cells that use binary fission for division

3 0
3 years ago
Read 2 more answers
An object with a pre-weighed mass of exactly (and correctly) 0.54 g is given to 2 students. One student obtains a weight of 0.59
Alina [70]

Answer:

Both student have same percent error.

Explanation:

Given data:

Actual mass of object = 0.54 g

Measured value by one student = 0.59 g

Measured value by second student = 0.49 g

Percent error = ?

Solution:

Formula

Percent error = (measured value - actual value / actual value) × 100

Percent error of first student:

percent error = (0.59 g - 0.54 / 0.54 ) ×100

percent error =  9.3 %

Percent error of second student:

Percent error = (0.49 g - 0.54 / 0.54 ) ×100

Percent error = - 9.3 %

Both student have same percent error. The only difference is that first student measure the greater value then actual value and second student measure the less value then actual, however difference was same and gives same percent error.

3 0
4 years ago
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