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3 years ago

Could anyone give me the answer to this plz 20 points

Chemistry

1 answer:

svetoff [14.1K]3 years ago

6 0

13 - Periodic table
14 - Dimitri mandeleev
15 - groups

Mark me brainiest pls it right answer

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atroni [7]

The last one would be false

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4 years ago

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Andru [333]

Co2 = two covalent bonds
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3 years ago

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Can someone pls help me with this question

artcher [175]

Answer:

The answer to your question is 24.325

Explanation:

Data

Magnesium-24 Abundance = 78.70%

Magnesium-25 Abundance = 10.13%

Magnesium-26 Abundance = 11.17%

Process

1.- Convert the abundance to decimals

Magnesium-24 Abundance = 78.70/100 = 0.787

Magnesium-25 Abundance = 10.13/100 = 0.1013

Magnesium-26 Abundance = 11.17/100 = 0.1117

2.- Write an equation

Average atomic mass = (Atomic mass-1 x Abundance 1) + (Atomic mass 2 x

Abundance-2) + (Atomic mass 3 x Abundance 3)

3.- Substitution

Average atomic mass = (24 x 0.787) + (25 x 0.1013) + (26 x 0.1117)

4.- Simplification

Average atomic mass = 18.888 + 2.533 + 2.904

5.- Result

Average atomic mass = 24.325

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4 years ago

Duncan knows that it takes 36400 cal of energy to heat a pint of water from room temperature to boiling. However, Duncan has pre

antiseptic1488 [7]

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3 years ago

The equilibrium constant for the reaction 2NO(g)+Br2(g)⥫⥬==2NOBr(g) is Kc=1.3×10−2 at 1000 K. At this temperature does the equil

o-na [289]

Answer :

The equilibrium favors NO and $Br_2$ .

(1) The value of equilibrium constant for this reaction is, 76.9

(2) The value of equilibrium constant for this reaction is, 8.77

Explanation:

The given chemical equation is:

$2NO(g)+Br_2(g)\rightarrow 2NOBr(g)$

The value of equilibrium constant for the above equation is $K_c=1.3\times 10^{-2}$ .

The value of $K_c$ that means equilibrium lies to the left side. Thus, the equilibrium favors NO and $Br_2$ .

We need to calculate the equilibrium constant for the given equation of above chemical equation, which is:

(1) $2NOBr(g)\rightarrow 2NO(g)+Br_2(g)$

The equilibrium constant for the reverse reaction will be the reciprocal of the initial reaction.

The value of equilibrium constant for reverse reaction is:

$K_{c_1}=\frac{1}{K_c}$

$K_{c_1}=\frac{1}{1.3\times 10^{-2}}=76.9$

Thus, the value of equilibrium constant for this reaction is, 76.9

(2) $NOBr(g)\rightarrow NO(g)+\frac{1}{2}Br_2(g)$

The equilibrium constant for the reverse reaction will be the reciprocal of the initial reaction.

If the equation is multiplied by a factor of '1/2', the equilibrium constant of the reverse reaction will be the 1/2 power of the equilibrium constant of initial reaction.

The value of equilibrium constant for reverse reaction is:

$K_{c_2}=(\frac{1}{K_c})^{1/2}$

$K_{c_2}=(\frac{1}{1.3\times 10^{-2}})^{1/2}=8.77$

Thus, the value of equilibrium constant for this reaction is, 8.77

7 0

3 years ago

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