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2 years ago

Difference between Physical and chemical changes At least six examples for each type of change

Chemistry

2 answers:

Aleksandr [31]2 years ago

7 0

Physical changes involve a new form or shape of matter

Tju [1.3M]2 years ago

4 0

Physical changes involve a new form or shape of matter,


Chemical change is any change that results in the formation of new chemical substances.

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4.4273 x 10 to the 5th divided by 5.3668 x 10 to the 3rd

Rasek [7]

(4.4273 × 10¹⁰)/(5.3668 × 10³) = 0.824 94 × 10⁷ = 8.2494 × 10⁶

6 0

3 years ago

You wish to measure the iron content of the well water on the new property you are about to buy. You prepare a reference standar

djverab [1.8K]

1.04 ⨯ 10 $^{-4}$ M

<h3>Explanation</h3>

A = ε $\cdot l \cdot c$ by the Beer-Lambert law, where

A the absorbance,
$l$ the path length,
ε the molar absorptivity of the solute, and
$c$ concentration of the solution.

A and ε are the same for both solutions. Therefore, $l \cdot c$ is constant; $l$ is inversely proportional to $c$ . The 100 mL sample would have a concentration 1/4.78 times that of the 45.0 mL reference.

The 13.0 mL standard solution has a concentration of 5.17 ⨯ $10^{-4}$ M. Diluting it to 45.0 mL results in a concentration of $5.17 \times 10^{-4} \times \frac{13.0}{45.0} =$ 1.494 M.

$c$ is inversely related to $l$ for the two solutions. As a result, c₂ = $c_1 \cdot \frac{l_1}{l_2} = 1.494 \times 10^{-4} \times \frac{1}{4.78} =$ 3.126 M.

The 30.0 mL sample has to be diluted by 30.0 / 100.0 times to produce the 100.0 mL solution being tested. The 100.0 mL solution has a concentration of 3.126 M. Therefore, the 30.0 mL solution has a concentration of $3.126 \times \frac{100.0}{30.0} =$ 1.04 ⨯ $10^{-4}$ M.

6 0

3 years ago

Suppose we have a compound that is 4.330 % Li, 22.10 % Cl, 39.89 % O, and 33.69 % H2O. What is the compounds formula?

amid [387]

The formula of compound is LiClO4.3H2O

calculation

find the mole of each element

that is moles for Li,Cl,O and that of H2O

moles = % composition/ molar mass

For Li = 4.330/ 6.94 g/mol= 0.624 moles

Cl=22.10/35.5=0.623 moles

39.89/16 g/mol =2.493 moles

H20= 33.69/18 g/mol= 1.872 moles

find the mole ratio by dividing each moles by smallest number of mole ( 0.624 moles)

that is for Li= 0.624/0.623= 1

Cl= 0.623/0.623=1

O = 2.493/0.623 =4

H2O= 1.872/0.623=3

<h3>Therefore the formula=LiClO4.3H2O</h3><h3 />

8 0

3 years ago

which atoms attains a stable valence electron configuration by bonding with another atom 1) neon 2) radon 3) helium 4) hyrdogen

Sergeu [11.5K]

Well, when an atom attains a stable valence electron, it means that the outer electrons are complete and so cannot attain any more electrons. For the first shell, it is complete when it has 2 electrons, the second shell is complete when it has 8 electrons, all the other shells also have a particular number when complete. Anyway, i believe the answer is HYDROGEN because when HYDROGEN combines with another atom of HYDROGEN, the outer shell is completed. This is because HYDROGEN has only 1 electron. If the two HYDROGENS, which both have 1 electron combine, they make the electrons 2, which is complete for the first shell, HYDROGEN ends in the first shell. Since the electrons become 2, the shell is at stable valence. In all the other options, this happens;
NEON- It has 10 electrons, 2 in the first shell and 8 in the second. So the the shells are already complete, so it can't bond with any thing, which is completely against the question.
RADON- Radon has 86 electrons.
HELIUM- Helium has 2 electrons, so the shell is already full, and cannot bond, so it goes against the question. The question says BY BONDING.
So the answer is definitely 4) HYDROGEN
Hope i helped. Have a nice day, by the way, i'm very sure it's hydrogen.

3 0

3 years ago

You are instructed to create 900. mL of a 0.29 M phosphate buffer with a pH of 7.8. You have phosphoric acid and the sodium salt

Aleksandr [31]

Answer:

B and C

Explanation:

When we have to do a buffer solution we always have to choose the reaction that has the pKa closer to the desired pH value. When we find the pKa values we will obtain:

$pKa_1=-Log[6.9x10^-^3]=2.16$

$pKa_2=-Log[6.2x10^-^8]=7.20$

$pKa_3=-Log[4.8x10^-^13]=12.31$

The closer value is pKa2 with a value of 7.2. Therefore we have to use the second reaction. In which $H_2PO_4^-^1$ is the acid and $HPO_4^-^2$ is the base. Therefore the answer for the first question is B and the answer for the second question is C.

8 0

2 years ago