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3 years ago

10

Validate how valence electrons determine the chemical reactivity of an element: give 2 examples of elements with high reactivity

and justify your response.

1 answer:

Ostrovityanka [42]3 years ago

8 0

Nitrogen is more reactive than oxygen and oxygen than chlorine

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Write down the formula for B<br> example:<br> Hydrogen + Fluorine = Hydrogen Fluorine <br><br> help

Tom [10]

Answer:

P³⁻ + Cl⁻ --> PCl₃

Explanation:

PCl₃: phosphorus trichloride. prefix in front of chloride is "tri"–meaning three.

5 0

3 years ago

Read 2 more answers

What is the final volume?

zaharov [31]

Answer:

Option A. 9.4 L

Explanation:

From the question given above, the following data were obtained:

Initial volume (V₁) = 8 L

Initial temperature (T₁) = 293 K

Final temperature (T₂) = 343 K

Final volume (V₂) =?

V₁ / T₁ = V₂ / T₂

8 / 293 = V₂ / 343

Cross multiply

293 × V₂ = 8 × 343

293 × V₂ = 2744

Divide both side by 293

V₂ = 2744 / 293

V₂ = 9.4 L

Therefore, the final volume of the gas is 9.4 L

6 0

3 years ago

(5.625 + 8.15) x 2.34 + 3.2

dalvyx [7]

The answer is 35.4335

Hope this helped! (Plz mark me brainliest!)

3 0

3 years ago

Read 2 more answers

The relative atomic mass of Chlorine is 35.45. Calculate the percentage abundance of the two isotopes of Chlorine, 35Cl and 37Cl

nata0808 [166]

Answer:

35Cl = 75.9 %

37Cl = 24.1 %

Explanation:

Step 1: Data given

The relative atomic mass of Chlorine = 35.45 amu

Mass of the isotopes:

35Cl = 34.96885269 amu

37Cl = 36.96590258 amu

Step 2: Calculate percentage abundance

35.45 = x*34.96885269 + y*36.96590258

x+y = 1 x = 1-y

35.45 = (1-y)*34.96885269 + y*36.96590258

35.45 = 34.96885269 - 34.96885269y +36.96590258y

0.48114731 = 1,99704989‬y

y = 0.241 = 24.1 %

35Cl = 34.96885269 amu = 75.9 %

37Cl = 36.96590258 amu = 24.1 %

3 0

3 years ago

Given the following information, what is the concentration of H2O(g) at equilibrium? [H2S](eq) = 0.671 M [O2](eq) = 0.587 M Kc =

MAVERICK [17]

<u>Answer:</u> The equilibrium concentration of water is 0.597 M

<u>Explanation:</u>

Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_{c}$

For a general chemical reaction:

$aA+bB\rightleftharpoons cC+dD$

The expression for $K_{eq}$ is written as:

$K_{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b}$

The concentration of pure solids and pure liquids are taken as 1 in the expression.

For the given chemical reaction:

$2H_2S(g)+O_2(g)\rightleftharpoons 2S(s)+2H_2O(g)$

The expression of $K_c$ for above equation is:

$K_c=\frac{[H_2O]^2}{[H_2S]^2\times [O_2]}$

We are given:

$[H_2S]_{eq}=0.671M$

$[O_2]_{eq}=0.587M$

$K_c=1.35$

Putting values in above expression, we get:

$1.35=\frac{[H_2O]^2}{(0.671)^2\times 0.587}$

$[H_2O]=\sqrt{(1.35\times 0.671\times 0.671\times 0.587)}=0.597M$

Hence, the equilibrium concentration of water is 0.597 M

8 0

4 years ago