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Elena L [17]
4 years ago
7

What observations led to the periodic law? what observations led to the periodic law? observation that certain elements possess

such property as radioactivity observation that the all halogen elements have similar properties observation that the all metals have similar properties observation that the properties of elements recur and certain elements have similar properties?
Chemistry
1 answer:
kvv77 [185]4 years ago
7 0
HISTORICAL INTRODUCTION.1<span> EVER since the establishment of the atomic theory by Dalton and Berzelius it was felt among chemists that there must be some relation between the atomic weights of the different elements and their properties. It was recognized very early that there exist groups of elements possessing related chemical and physical properties, and one of the earliest attempts to bring out this point is due to Dobereiner. In 1829 he tried to show that “many elements may be arranged in groups ()f three, in each of which the middle element has an atomic weight equal or approximately equal to the mean of the atomic weights of the two extremes.” As illustrations of this method of arrangement may be mentioned the following groups: Li, Na, K; Ca,Sr,</span>
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Data Table 3: Polystyrene Test Tube, 12x75mm
Lera25 [3.4K]

Answer:

Experiment 8 E Data Table 3 fl Data Table 4 fl Data Table 5 fl Data Table 6 Data Table 3: Polystyrene Test Tube, 12x75mm Volume of water at room temperature (V1 in mL) Volume of gas in polystyrene tube at boil (V2 in mL) Temperature of gas at boil inside polystyrene tube (°C) Volume of gas in polystyrenetube at room temperature (V3 in mL) Temperature of gas.

Explanation:

Hope this helps

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4 0
3 years ago
Gravity on Earth is 9.8 m/s2, and gravity on the Moon is 1.6 m/s2.
Ivanshal [37]

Answer:

Mass is the same but it weights 64 Newtons

Explanation:

First of Mass is the same in any sort of gravity. Now let's calculate weight

W = MG

where W = Weight

M = Mass

G = Gravity

W = (40kg)(1.6)

W = 64

Sorry for the spelling mistakes, hope this helps

7 0
3 years ago
Read 2 more answers
A solution of pH 5 is diluted 100 times. Find the pH of the resulting solution.
liq [111]
First, we calculate of the concentration of the H+ ions in the solution from the pH given. Then, calculate the new concentration after dilution. Calculation are as follows:

pH = -log[H+]
5 = -log[H+]
[H+] = 1 x 10^-5 M

M1V1 = M2V2
<span>1 x 10^-5 M (V1) = M2(100V1)
</span>M2 = 1 x 10^-7

pH =  -log[<span>1 x 10^-7</span>]
pH = 7

3 0
3 years ago
Read 2 more answers
How the calculation of the [OH-], pH and % ionization for 0.619 M ammonia (NH3) NH3 + H2O (liq) rightwards harpoon over leftward
fomenos

Answer:

[OH⁻] = 3.34x10⁻³M; Percent ionization = 0.54%; pH = 11.52

Explanation:

Kb of the reaction:

NH3 + H2O(l) ⇄ NH4+ + OH-

Is:

Kb = 1.8x10⁻⁵ = [NH₄⁺] [OH⁻] / [NH₃]

<em>As all NH₄⁺ and OH⁻ comes from the same source we can write: </em>

<em>[NH₄⁺] = [OH⁻] = X</em>

<em>And as </em>[NH₃] = 0.619M

1.8x10⁻⁵ = [X] [X] / [0.619M]

1.11x10⁻⁵ = X²

3.34x10⁻³ = X = [NH₄⁺] = [OH⁻]

<h3>[OH⁻] = 3.34x10⁻³M</h3><h3 />

% ionization:

[NH₄⁺] / [NH₃] * 100 = 3.34x10⁻³M / 0.619M * 100 = 0.54%

pH:

As pOH = -log [OH-]

pOH = 2.48

pH = 14 - pOH

<h3>pH = 11.52</h3>
5 0
3 years ago
"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"
Salsk061 [2.6K]

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 .

8 0
3 years ago
Read 2 more answers
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