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2 years ago

What percentage of radioactive substance remains after two half-lives

Chemistry

1 answer:

Ksju [112]2 years ago

3 0

Answer:

After 2 half-lives there will be 25% of the original isotope, and 75% of the decay product. After 3 half-lives there will be 12.5% of the original isotope, and 87.5% of the decay product. After 4 half-lives there will be 6.25% of the original isotope, and 93.75% of the decay product.

Explanation:

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What is the molarity of a solution that has 2.52 grams of NaCO3 dissolved to

sammy [17]

Answer:

Explanation:

2.52g/ 0.125L= 20.16M

5 0

2 years ago

For the reaction KClO2⟶KCl+O2 KClO2⟶KCl+O2 assign oxidation numbers to each element on each side of the equation. K in KClO2:K i

frosja888 [35]

Answer:

Explanation:

The formula of the reaction:

KClO₂ → KCl + O₂

To assign oxidation numbers, we have to obey some rules:

Elements in an uncombined state or one whose atoms combine with one another to form molecules have an oxidation number of zero.
The charge on simple ions signifies their oxidation number.
The algebraic sum of all the oxidation number of all atoms in a neutral compound is zero. For radicals with charges, their oxidation number is the charge.

The oxidation number of K in KClO₂:

K + (-1) + 2(-2) = 0

K-5 = 0

K = +5

The oxidation number of K in KCl:

K + (-1) = 0

K = +1

The oxidation number Cl in KClO₂ is -1

For Cl in KCl, the oxidation number is -1

For O in KClO₂, the oxidation number is (2 x -2) = -4

For O in O₂, the oxidation number is 0

K moves from an oxidation state of +5 to +1. This is a gain of electrons and K has undergone reduction. We then say K is reduced.

O moves from an oxidation state of -4 to 0. This is a loss of electrons and O has undergone oxidation. We say O is oxidized.

7 0

3 years ago

Read 2 more answers

Strigol is an important plant hormone that is released by crops such as rice and sugarcane. Unfortunately for these plants, stri

Len [333]

The hydrogen deficiency index( HDI) of strigol is = 10

<h3>How to calculate HDI:</h3>

The hydrogen deficiency index is used to measure the number of degree of unsaturation of an organic compound.

Strigol is an example of an organic compound because it contains carbons and hydrogen.

To calculate the HDI using the molecular formula given (C19H20O6) the formula for HDI is used which is:

$hdi = \frac{1}{2} (2c + 2 + n - h - x)$

where C = number of carbon atoms = 19

n= number of nitrogen atoms = 0

h= number of hydrogen atoms = 20

X = number of halogen atoms = 0

Note that oxygen was not considered because it forms two bonds and has no impact.

There for HDI =

$\frac{1}{2} (2 \times 19 + 2 + 0 - 20 - 0)$

HDI=

$\frac{1}{2} (40 - 20)$

HDI =

$\frac{1}{2} \times 20$

HDI = 10

Therefore, the hydrogen deficiency index of strigol is = 10

Learn more about unsaturated compounds here:

brainly.com/question/490531

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2 years ago

Arrange the following H atom electron transitions in order of decreasing wavelength of the photon absorbed or emitted:

Katyanochek1 [597]

The decreasing order of wavelengths of the photons emitted or absorbed by the H atom is : b → c → a → d

Rydberg's formula :

$\frac{1}{\lambda} = R_h (\frac{1}{n_1^2}-\frac{1}{n_2^2} )$ ,

where λ is the wavelength of the photon emitted or absorbed from an H atom electron transition from $n_1$ to $n_2$ and $R_h$ = 109677 is the Rydberg Constant. Here $n_1$ and $n_2$ represents the transitions.

(a) $n_1$ =2 to $n_2$ = infinity

$\frac{1}{\lambda} = 109677\times (\frac{1}{2^2}-\frac{1}{\infty^2} )$ = 109677/4 [since 1/infinity = 0] Therefore, $\lambda$ = 4 / 109677 = 0.00003647 m

(b) $n_1$ =4 to $n_2$ = 20

$\frac{1}{\lambda} = 109677\times (\frac{1}{4^2}-\frac{1}{20^2} )$ = 6580.62

Therefore, $\lambda$ = 1 / 6580.62 = 0.000152 m

(c) $n_1$ =3 to $n_2$ = 10

$\frac{1}{\lambda} = 109677\times (\frac{1}{3^2}-\frac{1}{10^2} )$ = 11089.56

Therefore, $\lambda$ = 1 / 11089.56 = 0.00009 m

(d) $n_1$ =2 to $n_2$ = 1

$\frac{1}{\lambda} = 109677\times (\frac{1}{2^2}-\frac{1}{1^2} )$ = - 82257.75

Therefore, $\lambda$ = 1 /82257.75 = - 0.0000121 m

[Even though there is a negative sign, the magnitude is only considered because the sign denotes that energy is emitted.]

So the decreasing order of wavelength of the photon absorbed or emitted is b → c → a → d.

Learn more about the Rydberg's formula athttps://brainly.com/question/14649374

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1 year ago

Which of the following chemical reactions is an oxidation-reduction reaction? CuSO4 + BaCl2 yields BaSO4 + CuCl2 Pb(NO3)2 + 2NaC

Ratling [72]

The answer is N2 + 3H2 yields 2NH3. The oxidation-reduction reaction means that there is electrons transfer during the reaction which means that the valence changed.

6 0

3 years ago

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