That which flows onward uniformly

What is the basis for the classification of changes in matter into physical and chemical changes quizlet

laiz [17]

Explanation:

The basis for classifying changes in matter into physical and chemical change is to see if a material is altered after going through a process.

In a physical change only the state of matter is usually altered especially its form or shape.

Chemical change alters a substance by producing new kinds that are totally different from it.

In a chemical change, new kinds of matter are usually formed.
Such changes are usually irreversible.
Change of state is a reversible process.

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Chemical change brainly.com/question/9388643

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6 0

3 years ago

Why it is so hard to find life on other planets.

STatiana [176]

Answer:

Because people on earth may not have the correct technology.

Explanation:

5 0

3 years ago

What is the percent yield of CuS for the following reaction given that you start with 15.5 g of Na2S and 12.1 g CuSO4? The actua

notka56 [123]

Answer:

(B) 42.1%

Explanation:

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Given: For $Na_2S$

Given mass = 15.5 g

Molar mass of $Na_2S$ = 78.0452 g/mol

Moles of $Na_2S$ = 15.5 g / 78.0452 g/mol = 0.1986 moles

Given: For $CuSO_4$

Given mass = 12.1 g

Molar mass of $CuSO_4$ = 159.609 g/mol

Moles of $CuSO_4$ = 12.1 g / 159.609 g/mol = 0.0758 moles

According to the given reaction:

$Na_2S+CuSO_4\rightarrow Na_2SO_4+CuS$

1 mole of $Na_2S$ react with 1 mole of $CuSO_4$

0.1986 mole of $Na_2S$ react with 0.1986 mole of $CuSO_4$

Available moles of $CuSO_4$ = 0.0758 moles

Limiting reagent is the one which is present in small amount. Thus, $CuSO_4$ is limiting reagent. (0.0758 < 0.1986)

The formation of the product is governed by the limiting reagent. So,

1 mole of $CuSO_4$ gives 1 mole of $CuS$

0.0758 mole of $CuSO_4$ gives 0.0758 mole of $CuS$

Molar mass of $CuS$ = 95.611 g/mol

Mass of $CuS$ = Moles × Molar mass = 0.0758 × 95.611 g = 7.2473 g

Theoretical yield = 7.2473 g

Given experimental yield = 3.05 g

% yield = (Experimental yield / Theoretical yield) × 100 = (3.05/7.2473) × 100 = 42.1 %

Option B is correct.

6 0

3 years ago

When an electron drops from a higher energy level to a lower one, energy is emitted. in comparison, how much energy is required

NNADVOKAT [17]

Energy is released when an electron transitions from one energy level to another. In contrast, the same amount of energy is needed to carry out the process, the other way around, from the bottom elevation to the upper one.

What occurs when an electron transitions from one energy level to another?

The energy of the electron drops when it changes levels, and the atom releases photons. The electron emits a photon when it transitions from a greater to a lower energy level. The energy emitted is precisely the energy that is lost when an electron moves to a level with less energy.
An atom's electrons have negative energy. The electron must be given energy in order to be removed from the hydrogen atom, as shown by the negative sign. The quantity of energy in the atom will rise by supplying the electron with energy. Similar to how a ball on Earth chooses to rest in valleys rather than hills, the electron wants to spend the majority of its time at a lower energy level.
For a brief period of time, the electron remains in an excited state. The energy required to bring the electron to its lower-energy state will be released when the electron transitions between excited and unexcited states.

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7 0

1 year ago

A sample of gas occupies 40.0 mL at -123°C. What volume does the sample occupy at 27°C if the pressure is held constant? V1T2 =

Brums [2.3K]

Answer:

80mL

Explanation:

Step 1:

Data obtained from the question.

Initial Volume (V1) = 40mL

Initial temperature (T1) = –123°C

Final temperature (T2) = 27°C

Final volume (V2) =..?

Step 2:

Conversion of celsius temperature to Kelvin temperature.

T(K) = T(°C) + 273

Initial temperature (T1) = –123°C =

–123°C + 273 = 150K

Final temperature (T2) = 27°C = 27°C + 273 = 300K

Step 3:

Determination of the final volume.

This can be obtained as follow:

V1/T1 = V2/T2

Initial Volume (V1) = 40mL

Initial temperature (T1) = 150K

Final temperature (T2) = 300

Final volume (V2) =..?

V1/T1 = V2 /T2

40/150 = V2 /300

Cross multiply

150 x V2 = 40 x 300

Divide both side by 150

V2 = (40 x 300) /150

V2 = 80mL

Therefore, the new volume of the gas is 80mL

6 0

3 years ago