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lutik1710 [3]
3 years ago
12

Which is the most chemically active of all the elements?

Chemistry
2 answers:
liberstina [14]3 years ago
7 0
Answer
Fluorite
Explanation
Fluorine is a yellowish, poisonous, highly corrosive gas. It is the most chemically active nonmetallic element and is the most electronegative of all the elements. It is a member of Group 17 (the halogens ) of the periodic table .
SSSSS [86.1K]3 years ago
6 0

Answer:

its Fluorite

Explanation:

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In an unconformity between two layers of rock, how is the lower layer usually described?
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The following unbalanced equation illustrates the overall reaction by which the body utilizes glucose to produce energy: C6H12O6
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Answer:

the conversion factor is f= 6  mol of glucose/ mol of CO2

Explanation:

First we need to balance the equation:

C6H12O6(s) + O2(g) → CO2(g) + H2O(l) (unbalanced)

C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l) (balanced)

the conversion factor that allows to calculate the number of moles of CO2 based on moles of glucose is:

f = stoichiometric coefficient of CO2 in balanced reaction / stoichiometric coefficient of glucose in balanced reaction

f = 6 moles of CO2 / 1 mol of glucose = 6  mol of glucose/ mol of CO2

f = 6 mol of CO2/ mol of glucose

for example, for 2 moles of glucose the number of moles of CO2 produced are

n CO2 = f * n gluc = 6 moles of CO2/mol of glucose * 2 moles of glucose= 12 moles of CO2

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Compounds can be made by combining ________
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The equilibrium constant, Kc, for the following reaction is 0.967 at 650 K. 2NH3(g) N2(g) 3H2(g) When a sufficiently large sampl
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Answer: Concentration of NH_3 in the equilibrium mixture is 0.31 M

Explanation:

Equilibrium concentration of H_2 = 0.729 M

The given balanced equilibrium reaction is,

                 2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)

Initial conc.            x                0           0

At eqm. conc.     (x-2y) M     (y) M   (3y) M

The expression for equilibrium constant for this reaction will be:

3y = 0.729 M

y = 0.243 M

K_c=\frac{[y]\times [3y]^3}{[x-2y]^2}

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

K_c=\frac{0.243\times (0.729)^3}{(x-2\times 0.243)^2}

0.967=\frac{0.243\times (0.729)^3}{(x-2\times 0.243)^2}

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Thus concentration of NH_3 in the equilibrium mixture is 0.31 M

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