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4 years ago

Which periodic group has the smallest atomic radius?

Chemistry

1 answer:

kobusy [5.1K]4 years ago

6 0

Answer:

Helium

Explanation:

Atomic radii vary in a predictable manner across the periodic table. As can be seen in the figures below, the atomic radius increases from top to bottom in a group and decreases from left to right over a period of time. Helium is therefore the smallest element, and francium is the largest.

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Potassium chlorate (kclo3) decomposes in a reaction described by this chemical equation:2kclo3(s) → 2kcl(s) + 3o2(g)if 36.0 gram

cricket20 [7]

<span>If 36 gm of potassium chlorate enter into the reaction, the total mass of the two products will still be 36 gm because if there is only one reactant, the mass of the compounds after the reaction will be same that reactant based on the law of conservation of matter.</span>

6 0

3 years ago

A sample of a gas occupies 2.76 L at 303K. What would the volume be if the temperature was increased to 404K? Assume constant

Artist 52 [7]

Answer:

If the temperature was increased to 404 K, its volume would be 3.68 L.

Explanation:

Charles' Law gives a relationship between the volume and the temperature of the gas at constant temperature. This law states that the volume of a given amount of gas held at constant pressure is directly proportional to the temperature.

$V\propto T$

$\dfrac{V_1}{V_2}=\dfrac{T_1}{T_2}$

Let

$V_1=2.76\ L\\\\T_1=303\ K\\\\T_2=404\ K$

Let $V_2$ is new volume. Using above formula we get :

$V_2=\dfrac{V_1T_2}{T_1}\\\\V_2=\dfrac{2.76\times 404}{303}\\\\V_2=3.68\ L$

If the temperature was increased to 404 K, its volume would be 3.68 L.

6 0

4 years ago

Calculate the pH at 25 degrees celsius of a 0.39 M solution of pyridinium chloride (c5h5nhcl) . Note that pyridine (c5h5n) is a

lutik1710 [3]

Answer:

$pH=2.3$

Explanation:

Hello!

In this case, since pyridinium chloride has a pKb of 8.77 which is a Kb of 1.70x10⁻⁹ and therefore a Ka of 5.89x10⁻⁵ which means it tends to be acidic, we write its ionization via:

$C_5H_5NHCl(aq)+H_2O(l)\rightleftharpoons C_5H_5NCl^-(aq)+H_3O^+(aq)$

Because it is a Bronsted base which donates one hydrogen ion to the water to produce hydronium. Thus, we write the equilibrium expression with the aqueous species only:

$Ka=\frac{[C_5H_5NCl^-][H_3O^+]}{[C_5H_5NHCl]}$

In terms of the reaction extent $x$ , we write:

$5.89x10^{-5}=\frac{x*x}{0.39-x}$

Thus, solving for $x$ we obtain:

$x_1=-0.0048M\\\\x_2=0.0048M$

Clearly the solution is 0.0048 M because to negative values are not allowed, therefore, since it equals the concentration of hydronium which defines the pH, we write:

$pH=-log([H_3O^+])=-log(0.0048)\\\\pH=2.3$