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3 years ago

In the compounds formed by li, na, k, rb, and cs with chlorine, how many atoms of cl are there per atom of the metal?

1 answer:

5 0

To determine how many chlorine atoms for the compound made from the metals given, we need to know the ionic forms of the metals above. Lithium when in ionic form would have a charge of +1. Sodium metal would have a +1 ionic charge. Potassium would have a charge of +1 as well. Rubidium would have a charge of +1. And Cesium would have +1 also. These metals are located at group 1A so they would all have a +1 charge in the ionic form. To balance this charge, -1 should be paired which would be the charge of a chloride ion. Therefore, for each metal, only 1 chlorine atom would bond.

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"acid is responsible for the odor in rancid butter. a solution of 0.25 m butyric acid has a ph of 2.71. what is the ka for"

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Answer:- The Ka for the acid is $1.53*10^-^5$ .

Solution:- In general, monoprotic acid could be represented by HA. The dissociation equation for the ionization of HA is written as:

HA(aq)\rightarrow H^+(aq) + A^-(aq)

Now, we make the ice table for this equation as:

HA(aq)\rightarrow H^+(aq) + A^-(aq)

I 0.25 0 0

C -X +X +X

E (0.25 - X) X X

where, I stands for initial concentration, C stands for change in concentration and E stands for equilibrium concentration.

X is the change in concentration and from ice table it's same as the concentration of hydrogen ion that is calculated from given pH.

$Ka = [H^+][A^-]\frac{1}{HA}$

Where, Ka is the acid ionization constant. Let's plug in the values.

$Ka = \frac{X^2}{0.25-X}$

Let's calculate the value of X first using the equation:

$pH = -log[H^+]$ [/tex]

on taking antilog ob above equation we get:

$[H^+]=10^-^p^H$

$[H^+]=10^-^2^.^7^1$

$[H^+]$ = 0.00195

So, X = 0.001195

Let's plug in this value of X in the equation:-

$Ka=\frac{(0.00195)^2}{0.25-0.00195}$

$Ka=1.53*10^-^5$

So, the value of Ka for butyric acid is $1.53*10^-^5$ .

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: If a 250. mL sample of the above buffer solution initially has 0.0800 mol H2C6H5O7 - and 0.0600 mol HC6H5O7 2- , what would be

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Answer: New concentration of $HC_{6}H_{5}O^{2-}_{7}$ is 0.23 M.

Explanation:

The given data is as follows.

Moles of $HC_{6}H_{5}O^{2-}_{7}$ = 0.06 mol

Moles of $H_{2}C_{6}H_{5}O_{7}$ = 0.08 mol

Therefore, moles of $OH^{-}$ added are as follows.

Moles of $OH^{-}$ = $0.125 \times \frac{25}{1000}$

= 0.003125 mol

Now, new moles of $HC_{6}H_{5}O^{2-}_{7}$ = 0.06 + 0.003125

= 0.063125

Therefore, new concentration of $HC_{6}H_{5}O^{2-}_{7}$ will be calculated as follows.

Concentration = $\frac{0.063125}{0.275}$

= 0.23 M

Thus, we can conclude that new concentration of $HC_{6}H_{5}O^{2-}_{7}$ is 0.23 M.

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Answer:

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Which branch of chemistry specifically deals with quantitative analysis?

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Answer:

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