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4 years ago

How do you determine the correct subscripts in a chemical formula

1 answer:

4 0

BY CHECKING THE REACTIVITY OF AN ELEMENT WHICH IS MOST REACTIVE OR NOT AND YOU STUDY TYPE OF CHEMICAL REACTION IN 1 CH AND YOU CHECK THE REACTIVITY OF ELEMENTS IN 3 CH METALS AND NON METALS PAGE NO 45 IN NCERT BOOK

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For a molecule to be considered inorganic it must not contain ?

horsena [70]

Answer:

It must not contain protein.

8 0

4 years ago

34. Given the fact that the concentrations of reactants and

Nina [5.8K]

Answer:

The word dynamic is used to describe chemical equilibrium because even though the concentration of reactants and products remains constant, the reaction does not stop, but still continues with the rate of forward reaction equal to the rate of backward reaction. This is opposed to static equilibrium where there is no reaction at all.

Hope this helps you out!

3 0

3 years ago

Hydrogen sulfide decomposes according to the following reaction, for which Kc = 9.30 10-8 at 700°C. 2 H2S(g) 2 H2(g) + S2(g) If

Vikentia [17]

Answer:

The equilibrium concentration of hydrogen gas is 0.0010 M.

Explanation:

The equilibrium constant of the reaction = $K_c=9.30\times 10^{-8$ }

Moles of hydrogen sulfide = 0.31 mol

Volume of the container = 4.1 L

$[concentration]=\frac{moles}{volume (L)}$

$[H_2S]=\frac{0.31 mol}{4.1 L}=0.076 M$

$2H_2S(g)\rightleftharpoons 2H_2(g)+S_2(g)$

Initially

0.076 M

At equilibrium

(0.076-2x) 2x x

The expression of an equilibrium constant :

$K_c=\frac{[H_2]^2[S_2]}{[H_2S]^2}$

$9.30\times 10^{-8}=\frac{(2x)^2\times x}{(0.076-x)^2}$

Solving for x:

x = 0.00051

The equilibrium concentration of hydrogen gas:

$[H_2]=2x=2\times 0.00051 M=0.0010 M$

4 0

4 years ago

How many moles of gas occupy 98 l at a pressure of 2.8 atmospheres and a temperature of 292 k?

Furkat [3]

From the ideal gas equation, we know that PV = nRT
where, P = pressure, V = volume occupied, n = number of moles of gas, R = universal gas constant=0.082L atm mol-1 K-1 and T = temperature.
Given: P = 2.8 atm, V = 98 l and T = 292 k.
Therefore n = $\frac{PV}{RT}$ = $\frac{2.8 X 98}{0.082 X 292}$ = 11.46

6 0

3 years ago

For the following reaction, 5.78 grams of chlorine gas are mixed with excess iron. The reaction yields 6.79 grams of iron(III) c

NeX [460]

Answer:

a. Theoretical yield = 8.81 g

b. 77 %

Explanation:

We begin from the reaction:

2Fe (s) + 3Cl₂ (g) → 2FeCl₃ (s)

and the excess is the iron, so the limiting reagent is the chlorine.

We convert mass to moles → 5.78 g / 70.90 g/mol = 0.0815 mol

Ratio is 3:2 so we think this:

From 3 moles of chlorine, I can produce 2 moles of Iron (III) chloride

Then 0.0815 moles would produce (0.0815 . 2) /3 = 0.0543 moles

This is the theoretical yield. We convert the moles to mass:

0.0543 mol . 162.2 g /mol = 8.81 g

If the reaction yields 6.79 g of FeCl₃ and the theoretical yield is 8.81 g ,we can obtain the percent yield: (6.79 / 8.81) . 100 = 77 %

5 0

3 years ago