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3 years ago

NO GUESSING!ONLY RIGHT ANSWER

Chemistry

1 answer:

lutik1710 [3]3 years ago

7 0

<h3> <u>ANSWER</u></h3>

2. Neon, only

<h3><u>EXPLANATION</u></h3>

When Na and F combine they form an electron configuration of 2-8. Na electron configuration is 2-8-1 while F is 2-7, so when they form an ionic bond F will gain Na outermost electron to complete its valence shell due to having a higher negativity. Neon has an electron configuration of 2-8 but argon has 2-8-8.

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Lithium has two stable isotopes with masses of 6.01512 amu and 7.01600 amu. The average molar mass of Li is 6.941 amu. What is t

Dvinal [7]

Answer : The percent abundance of Li isotope-1 and Li isotope-2 is, 6.94 % and 93.1 % respectively.

Explanation :

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

Let the fractional abundance of Li isotope-1 be 'x' and the fractional abundance of Li isotope-2 will be '100-x'

For Li isotope-1 :

Mass of Li isotope-1 = 6.01512 amu

Fractional abundance of Li isotope-1 = x

For Li isotope-2 :

Mass of Li isotope-2 = 7.01600 amu

Fractional abundance of Li isotope-2 = 100-x

Average atomic mass of Li = 6.941 amu

Putting values in equation 1, we get:

$6.941=[(6.01512\times x)+(7.01600\times (100-x))]$

By solving the term 'x', we get:

$x=694.048$

Percent abundance of Li isotope-1 = $\frac{694.048}{100}=6.94\%$

Percent abundance of Li isotope-2 = 100 - x = 100-6.94 = 93.1 %

6 0

3 years ago

Consider the following reaction at a high temperature. Br2(g) ⇆ 2Br(g) When 1.35 moles of Br2 are put in a 0.780−L flask, 3.60 p

UNO [17]

Answer : The equilibrium constant $K_c$ for the reaction is, 0.1133

Explanation :

First we have to calculate the concentration of $Br_2$ .

$\text{Concentration of }Br_2=\frac{\text{Moles of }Br_2}{\text{Volume of solution}}$

$\text{Concentration of }Br_2=\frac{1.35moles}{0.780L}=1.731M$

Now we have to calculate the dissociated concentration of $Br_2$ .

The balanced equilibrium reaction is,

$Br_2(g)\rightleftharpoons 2Br(aq)$

Initial conc. 1.731 M 0

At eqm. conc. (1.731-x) (2x) M

As we are given,

The percent of dissociation of $Br_2$ = $\alpha$ = 1.2 %

So, the dissociate concentration of $Br_2$ = $C\alpha=1.731M\times \frac{1.2}{100}=0.2077M$

The value of x = 0.2077 M

Now we have to calculate the concentration of $Br_2\text{ and }Br$ at equilibrium.

Concentration of $Br_2$ = 1.731 - x = 1.731 - 0.2077 = 1.5233 M

Concentration of $Br$ = 2x = 2 × 0.2077 = 0.4154 M

Now we have to calculate the equilibrium constant for the reaction.

The expression of equilibrium constant for the reaction will be :

$K_c=\frac{[Br]^2}{[Br_2]}$

Now put all the values in this expression, we get :

$K_c=\frac{(0.4154)^2}{1.5233}=0.1133$

Therefore, the equilibrium constant $K_c$ for the reaction is, 0.1133

7 0

3 years ago

Which statement describes the light reaction of photosynthesis?

cupoosta [38]

Answer:

Explanation:

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3 years ago

Covert from scientific notation<br> 1.3 x 10¹²=

Lady_Fox [76]

Answer:

1,300,000,000,000

Explanation:

1.3 x 10^12

We want to convert this from scientific notation.

Tip: in scientific notation the exponent tells you how many place you move the decimal point over. If the exponent is negative you move the decimal point to the left. Ex. For, 4.1 x 10^-8, we would move the decimal point over 8 times to the left to get .00000041. When the exponent is positive we move over to the right. Ex. For, 7.6 x 10^7 we would move the decimal point over 7 times to the right to get 76,000,000

So to convert 1.3 x 10^12 we simply move over the decimal point over 12 times to the right.

1.3 x 10^12 ------> 1,300,000,000,000

Our answer is 1,300,000,000,000

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2 years ago

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Ainat [17]

Answer:

HCL, NACL and CO are the examples of chemical formulas

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3 years ago