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3 years ago

Which of the following elements has the largest first ionization energy? PLEASE HELP

Chemistry

2 answers:

arlik [135]3 years ago

3 0

Answer:

Boron

Explanation:

Hope it helps you......

BigorU [14]3 years ago

3 0

Answer:

B. Nitrogen

D . Boron

Explanation:

both are the element has the largest first ionization energy

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According to the periodic table, carbon has an atomic mass of 12.011 u. This indicates that the most abundant isotope of carbon

Eduardwww [97]

The most abundant of all of the isotopes of an element will be the one who's mass the mass of element is closest to. In this case, the mass of atomic carbon is closest to the mass of carbon-12.
Thus, Carbon-12 is the most abundant isotope.

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4 years ago

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The diagram shows an animal cell.

LenKa [72]

Answer:

Explanation:

Assume your diagram is like the one below.

X represents a mitochondrion.

That's where the Tricarboxylic Acid Cycle converts a single glucose molecule into six molecules of CO₂.

W is wrong. It represents a vacuole, which can store both nutrients and waste products for later elimination.

Y is wrong. It represents the nucleolus, which plays a critical role in the synthesis of ribosomes.

Z is wrong. It represents the cytoplasm, which is where cell processes like glycolysis and protein synthesis take place.

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3 years ago

The decomposition of hydrogen peroxide was studied, and the following data were obtained at a particular temperature: Time (s) (

harina [27]

Answer:

Part a: The rate of the equation for 1st order reaction is given as $Rate=k[H_2O_2]$

Part b: The integrated Rate Law is given as $[H_2O_2]=[H_2O_2]_0 e^{-kt}$

Part c: The value of rate constant is $7.8592 \times 10^{-4} s^{-1}$

Part d: Concentration after 4000 s is 0.043 M.

Explanation:

By plotting the relation between the natural log of concentration of $H_2O_2$ , the graph forms a straight line as indicated in the figure attached. This indicates that the reaction is of 1st order.

Part a

Rate Law

The rate of the equation for 1st order reaction is given as

$Rate=k[H_2O_2]$

Part b

Integrated Rate Law

The integrated Rate Law is given as

$[H_2O_2]=[H_2O_2]_0 e^{-kt}$

Part c

Value of the Rate Constant

Value of the rate constant is given by using the relation between 1st two observations i.e.

t1=0, M1=1.00

t2=120 s , M2=0.91

So k is calculated as

$-k(t_2-t_1)=ln{\frac{M_2}{M_1}}\\-k(120-0)=ln{\frac{0.91}{1.00}}\\k=\frac{-0.09431}{-120}\\k=7.8592 \times 10^{-4} s^{-1}$

The value of rate constant is $7.8592 \times 10^{-4} s^{-1}$

Part d

Concentration after 4000 s is given as

$-k(t_2-t_1)=ln{\frac{M_2}{1.0}}\\-7.8592 \times 10^{-4}(4000-0)=ln{\frac{M_2}{1.00}}\\-3.1437=ln{\frac{M_2}{1.00}}\\M_2=e^{-3.1437}\\M_2=0.043 M$

Concentration after 4000 s is 0.043 M.

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3 years ago

Calculate the frequency of light having wavelength of 456 nm.

Lubov Fominskaja [6]

Answer:

Calculate the frequency of light having wavelength of 456 nm.

1.22*10^8nm

Explanation:

7 0

2 years ago

PLEASE HELP!!! (Look at photo)

pychu [463]

Answer

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Explanation:

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3 years ago