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3 years ago

An element is made up of only one kind of _

Chemistry

1 answer:

Afina-wow [57]3 years ago

3 0

Answer:

A chemical element is a substance that contains only one type of atom. If a substance contains more than one type of atom, it is a compound. An element can be a solid, liquid or gas. The smallest particle of such an element is an atom.

Explanation:

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I am a nonmetal.I am in the Oxygen family and in row 3.I have 6 valence electrons.I am yellow and have a stinky smell.Who am i

Dimas [21]

Answer:

sulfur

Explanation:

In oxygen family sulfur has yellow color and also having stinky smell. Thus given statements are about sulfur.

It is present in oxygen family.

It has six valance electrons.

Its atomic number is 16.

Its atomic weight is 32 amu.

The electronic configuration of sulfur is given below,

S₁₆ = 1s² 2s² 2p⁶ 3s² 3p⁴

We can see the valance shell is third shell and it have six electrons thus sulfur have six valance electrons. (3s² 3p⁴ )

Sulfur is used in vulcanisation process.

It is used in bleach and also as a preservative for many food.

it is used to making gun powder.

8 0

2 years ago

Consider the reaction: CH4 + 2O2 = CO2 + 2H2O

crimeas [40]

Answer:

The answer to your question is:

a) 80 g of O2

b) O2, 15.13 g of CO2

c) It's not posible to know which is the limiting reactant.

Explanation:

Reaction CH4 + 2O2 ⇒ CO2 + 2H2O

a. Calculate the grams of O2 needed to react with 20.00 grams of CH4. _____________

MW CH4 = 16 g

MW O2 = 32 g

16 g of CH4 ---------------- 2(32) g of O2

20 g -------------- x

x = (20 x 64) / 16 = 80 g of O2

b. Given 15.00 g. of CH4 and 22.00 g. of O2, identify the limiting reactant and calculate the grams of CO2 that can be produced. LR _________ grams CO2 _________ .

CH4 + 2O2 ⇒ CO2 + 2H2O

15 g 22 g

16 g of CH4 ---------------- 64 g of O2

15 g of CH4 --------------- x

x = (15 x 64) / 16 = 60 g of O2

The Limiting reactant is O2 because it is necessary 60g of O2 for 16 g of CH4 and there are only 22.

CH4 + 2O2 ⇒ CO2 + 2H2O

64 g of O2 ------------------ 44 g of CO2

22 g of O2 ------------------ x

x = (22 x 44)/ 64 = 15. 13 g of CO2

c. For the reaction CH4 + 2O2 = CO2 + 2H2O, if you have 10.31 g. of CH4 and an unknown amount of oxygen, and form 20.00 g. of CO2, i. Identify if there is a limiting reactant ______________ ii. Calculate the number of grams of the limiting reactant present if there is one. ______________

CH4 + 2O2 ⇒ CO2 + 2H2O

10.31 g 20 g

We can identify the limiting reactant if we know the quantity of the reactants, if we only know the quantity of one it is not posible to which is the limiting reactant.

4 0

3 years ago

On the basis of the information above, what is the approximate percent ionization of HNO2 in a 1.0 M HNO2 (aq) solution?

enot [183]

Answer:

The answer is "2%"

Explanation:

Equation:

$HNO_2\ (aq) \leftrightharpoons H^{+} \ (aq) + NO_2^{-}\ (aq) \\\\\ K_a = 4.0\times \ 10^{-4}$

$H^{+}=?$

Formula:

$Ka = \frac{[H^{+}][NO_2^{-}]}{[HNO_2]}$

Let

$[H^{+}] = [NO_2^{-}] = x$ at equilibrium

$x^2 = (4.0\times 10^{-4})\times 1.0\\\\x = ((4.0\times 10^{-4})\times 1.0)^{0.5} = 2.0 \times 10^{-2} \ M\\\\$

therefore,

$[H^{+}] = 2.0\times 10^{-2} \ M = 0.02 \ M$

Calculating the % ionization:

$= \frac{([H^{+}]}{[HNO_2])} \times 100 \\\\= \frac{0.02}{1}\times 100 \\\\= 2\%\\\\$

6 0

3 years ago

In a spontaneous fusion reaction, the total mass of the products is ________________________ the mass of the original elements.

Mice21 [21]

I believe the correct term that would fit the statement would be greater than. In a spontaneous fusion reaction, the total mass of the products is greater than the mass of the original elements. This nuclear reaction involves at least two nuclei that fuses to form one nuclei having larger mass than that of the reactant.

4 0

3 years ago

What is the [OH-] of a solution prepared by dissolving 0.0912 g of hydrogen chloride in sufficient pure water to prepare 250.0 m

velikii [3]

Answer: The $[OH^-]$ of a solution is $10^{-12}$ M

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

$Molarity=\frac{n\times 1000}{V_s}$

where,

n = moles of solute

$V_s$ = volume of solution in ml

moles of $HCl$ = $\frac{\text {given mass}}{\text {Molar mass}}=\frac{0.0912g}{36.5g/mol}=0.0025mol$

Now put all the given values in the formula of molality, we get

$Molarity=\frac{0.0025\times 1000}{250}=0.01$

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

$HCl\rightarrow H^++Cl^{-}$

According to stoichiometry,

1 mole of $HCl$ gives 1 mole of $H^+$

Thus $0.01$ moles of $HCl$ gives = $\frac{1}{1}\times 0.01=0.01$ moles of $H^+$

Putting in the values:

$[H^+][OH^-]=10^{-14}$

$[0.01][OH^-]=10^{-14}$

$[OH^-]=10^{-12}$

Thus the $[OH^-]$ of a solution prepared by dissolving 0.0912 g of hydrogen chloride in sufficient pure water to prepare 250.0 ml of solution is $10^{-12}$ M

8 0

3 years ago