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4 years ago

Which is a property that Mendeleev predicted for gallium (Ga)?

2 answers:

8 0

The correct option is B.
Mendeleev was the one who originated the idea of arranging elements in the periodic table according to their chemical and physical properties. He left spaces in the periodic table and predicted the discovery of those elements that had not been discovered then. One of these elements is Gallium. He predicted that gallium is going to be a metal and he gave the properties that the element will possess. He also predicted that the element gallium will be placed under aluminium in the periodic table.

notsponge [240]4 years ago

7 0

the correct answer is b it is a metal

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A new potential heart medicine, code-named X-281, is being tested by a pharmaceutical company, Pharma-pill. As a research techni

torisob [31]

Answer:

a. pka = 3,73.

b. pkb = 10,27.

Explanation:

a. Supposing the chemical formula of X-281 is HX, the dissociation in water is:

HX + H₂O ⇄ H₃O⁺ + X⁻

Where ka is defined as:

$ka = \frac{[H_3O^+][X^-]}{[HX]}$

In equilibrium, molar concentrations are:

[HX] = 0,089M - x

[H₃O⁺] = x

[X⁻] = x

pH is defined as -log[H₃O⁺]], thus, [H₃O⁺] is:

$[H_3O^+]} = 10^{-2,40}$

[H₃O⁺] = <em>0,004M</em>

Thus:

[X⁻] = 0,004M

And:

[HX] = 0,089M - 0,004M = <em>0,085M</em>

$ka = \frac{[0,004][0,004]}{[0,085]}$

ka = 1,88x10⁻⁴

And <em>pka = 3,73</em>

b. As pka + pkb = 14,00

pkb = 14,00 - 3,73

I hope it helps!

4 0

4 years ago

A 21.82 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 21.33 grams of CO2 and 4

morpeh [17]

<u>Answer:</u> The molecular formula for the given organic compound is $C_2H_2O_4$

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=21.33g$

Mass of $H_2O=4.366g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 21.33 g of carbon dioxide, $\frac{12}{44}\times 21.33=5.82g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 4.366 g of water, $\frac{2}{18}\times 4.366=0.485g$ of hydrogen will be contained.

Mass of oxygen in the compound = (21.82) - (5.82 + 0.485) = 15.515 g

To formulate the empirical formula, we need to follow some steps:

<u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{5.82g}{12g/mole}=0.485moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.485g}{1g/mole}=0.485moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{15.515g}{16g/mole}=0.969moles$

<u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.485 moles.

For Carbon = $\frac{0.485}{0.485}=1$

For Hydrogen = $\frac{0.485}{0.485}=1$

For Oxygen = $\frac{0.969}{0.485}=1.99\approx 2$

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 1 : 2

Hence, the empirical formula for the given compound is $C_1H_{1}O_2=CHO_2$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{molecular mass}}{\text{empirical mass}}$

We are given:

Mass of molecular formula = 90.04 g/mol

Mass of empirical formula = 45 g/mol

Putting values in above equation, we get:

$n=\frac{90.04g/mol}{45g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(1\times 2)}H_{(1\times 2)}O_{(2\times 2)}=C_2H_2O_4$

Thus, the molecular formula for the given organic compound is $C_2H_2O_4$

4 0

3 years ago

In order to perform a chemical reaction, 225 mL of 0.500 M lead (II) nitrate is required. How much 5.00 M stock solution is requ

vladimir1956 [14]

Answer:

what

Explanation:

what ````

4 0

3 years ago

Calculate the second volumes. <br> 7.03 Liters at 31 C and 111 Torr to STP

DIA [1.3K]

The second volume : V₂= 0.922 L

<h3> Further explanation </h3><h3>Given </h3>

7.03 Liters at 31 C and 111 Torr

Required

The second volume

Solution

T₁ = 31 + 273 = 304 K

P₁ = 111 torr = 0,146 atm

V₁ = 7.03 L

At STP :

P₂ = 1 atm

T₂ = 273 K

Use combine gas law :

P₁V₁/T₁ = P₂V₂/T₂

Input the value :

0.146 x 7.03 / 304 = 1 x V₂/273

V₂= 0.922 L

4 0

3 years ago

If a solution has a pOH of 8.71, what is the [H+]?

TEA [102]

Answer: $5.1\times 10^{-6}M$

Explanation:

pH or pOH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration.

$pH=-\log [H^+]$

$pH+pOH=14$

Putting in the values:

$pH=14-8.71=5.29$

$5.29=-log[H^+]$

$[H^+]=5.1\times 10^{-6}$

Thus $[H^+]$ is $5.1\times 10^{-6}M$

6 0

4 years ago