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2 years ago

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The nonmetals helium, neon, and argon and the metals copper, silver, and gold are among only a few elements that are found in na

ture in their pure forms. What can you infer about the chemical reactivity of these elements

1 answer:

Fynjy0 [20]2 years ago

5 0

Answer:

A pure element is a pure substance that has only one kind of atoms. If a substance is present in nature in elemental form, it means that the element is not very reactive as reactive elements exist in the form of their compounds with atoms of other elements. Element can be categorized as a metal or a non-metal based on some physical observable properties. Generally, metals are shiny lustrous solids and nonmetals are gases. Metals are good conductors of heat and electricity whereas nonmetals are not conductors.

Explanation:

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The value of Henry's law constant for oxygen in water at 242C is 1.66 x 1 ox M/torr. b. c. Calculate the solubility of oxygen in

DochEvi [55]

25ca an the oxygen gas is 2 atom ur mum

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3 years ago

When 0.42 g of a compound containing C, H, and O is burned completely, the products are 1.03 g CO2 and 0.14 g H2O. The molecular

Luda [366]

<u>Answer:</u> The empirical and molecular formula for the given organic compound is $C_2H_2O_4$ and $C_6H_4O_2$

<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 $CO_2=1.03g$

Mass of $H_2O=0.14g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

<u>For calculating the mass of carbon:</u>

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

So, in 1.03 g of carbon dioxide, $\frac{12}{44}\times 1.03=0.28g$ of carbon will be contained.

<u>For calculating the mass of hydrogen:</u>

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

So, in 0.14 g of water, $\frac{2}{18}\times 0.14=0.016g$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.42) - (0.28 + 0.016) = 0.124 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 = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.28g}{12g/mole}=0.023moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.016g}{1g/mole}=0.016moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.124g}{16g/mole}=0.00775moles$

<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 0.00775 moles.

For Carbon = $\frac{0.023}{0.00775}=2.96\approx 3$

For Hydrogen = $\frac{0.016}{0.00775}=2.06\approx 2$

For Oxygen = $\frac{0.00775}{0.00775}=1$

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

The ratio of C : H : O = 3 : 2 : 1

Hence, the empirical formula for the given compound is $C_3H_{2}O_1=C_3H_2O$

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 = 108.10 g/mol

Mass of empirical formula = 54 g/mol

Putting values in above equation, we get:

$n=\frac{108.10g/mol}{54g/mol}=2$

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

$C_{(3\times 2)}H_{(2\times 2)}O_{(1\times 2)}=C_6H_4O_2$

Thus, the empirical and molecular formula for the given organic compound is $C_3H_2O$ and $C_6H_4O_2$

6 0

3 years ago

A 52.0 g of Copper (specific heat=0.0923cal/gC) at 25.0C is warmed by the addition of 299 calories of energy. find the final tem

Leto [7]

Answer : The final temperature of the copper is, $87.29^oC$

Solution :

Formula used :

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

where,

Q = heat gained = 299 cal

m = mass of copper = 52 g

c = specific heat of copper = $0.0923cal/g^oC$

$\Delta T=\text{Change in temperature}$

$T_{final}$ = final temperature = ?

$T_{initial}$ = initial temperature = $25^oC$

Now put all the given values in the above formula, we get the final temperature of copper.

$299cal=52g\times 0.0923cal/g^oC\times (T_{final}-25^oC)$

$T_{final}=87.29^oC$

Therefore, the final temperature of the copper is, $87.29^oC$

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3 years ago

The equilibrium for the reaction between (ch3)2nh a weak base

nadezda [96]

The equilibrium for the dissolution of the weak base is ;(CH3)2NH(aq) + H2O(l) ⇄ (CH3)2NH3^+(aq) + OH^-(aq)

<h3>What is a weak base?</h3>

A weak base is one that does not ionize completely in solution. As such, a weak base will have a very low base dissociation constant Kb reflecting its minimal dissociation in solution.

The question is incomplete hence we are are unable to work out the equilibrium but in solution it will look like this;

(CH3)2NH(aq) + H2O(l) ⇄ (CH3)2NH3^+(aq) + OH^-(aq)

Learn more about weak base: brainly.com/question/4131966

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2 years ago

What is chromatography?

denpristay [2]

Its <span>c.chromatography is the process of separating solutions on the basis of their boiling points </span>

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3 years ago