There are many important laws and theories in science. Which of the following is a law of science?

A 1000.0 ml sample of lake water in titrated using 0.100 ml of a 0.100 M base solution. What is the molarity (M) of the acid in

Fittoniya [83]

The molarity (M) of the acid in the lake water is $0.00001M$ .

Explanation:

In order to estimate the concentration of a solution in molarity, then the total number of moles of the solute is divided by the total volume of the solution.

According to the given information, the formula will be applied for calculating molarity (M) of the acid in the lake water is :

$M_1V_1=M_2V_2$

Here;

$M_1,M_2$ are molarity of acid in the lake water and base solution respectively. $V_1,V_2$ are volume of sample in the lake water and base solution respectively.

Given values are as follows:

$M_1=?\\M_2=0.100M\\V_1=1000ml\\V_2=0.100ml$

Putting these values in above equation :

$M_1V_1=M_2V_2$

$M_1(1000)=(0.100)(0.100)$

$M_1=\frac{(0.100)(0.100)}{1000}$

$M_1=0.00001M$

Therefore, the molarity (M) of the acid in the lake water is $0.00001M$ .

5 0

1 year ago

Ion arrangements of three general salts are represented below. h5sil917 h5sil917021 h5sil91703 which ion arrangement has the hig

vovangra [49]

This question is incomplete. Luckily, I found the same problem which is shown in the attached picture. To answer the question, we must know how the size and charge affect the lattice energy. The answer is: lattice energy increases with the increasing charge of the ions, and decreasing radius of the atoms. Therefore, the ranking would be: A .

4 0

2 years ago

1 2 3 4 5 6 7 8 9 10 Suna passes an electric current through a sample of clear, colorless, and odorless liquid. As the experimen

BaLLatris [955]

The question is incomplete, the complete question is;

Suna passes an electric current through a sample of clear, colorless, and odorless liquid. As the experiment continues, bubbles form, and the volume of liquid decreases. Suna collects samples of two colorless, odorless gases that bubble out of the liquid. One of the gases burns. Neither the original liquid nor the other gas burns. Which is the best explanation of her results? The electric current changed some of the sample to gas even though the sample was not breaking down. Therefore, the original liquid is a compound. The electric current released a gas that was odorless and colorless, like the original sample. Therefore, the original liquid is an element. The sample was broken down by the electric current and formed a new substance that could burn. Therefore, the original liquid is a compound. The sample lost some of its volume, but the gas still had the same chemical makeup as the original sample. Therefore, the original liquid is an element.

Answer:

The sample was broken down by the electric current and formed a new substance that could burn. Therefore, the original liquid is a compound.

Explanation:

When electric current is passed through a compound, the compound may become broken down to release its constituents. We refer to this phenomenon as electrolysis. We can now say that the substance has been 'decomposed' electrolytically.

Since the original sample was decomposed to yield a gas that could burn and one that couldn't burn even though the original sample couldn't burn, then the original sample is a compound.

8 0

2 years ago

Copper(II) sulfide, CuS, is used in the development of aniline black dye in textile printing. What is the maximum mass of CuS wh

Naya [18.7K]

Answer:

1.82 g is the maximum mass of CuS.

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Or,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Given :

For $CuCl_2$ :

Molarity = 0.500 M

Volume = 38.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 38.0×10⁻³ L

Thus, moles of $CuCl_2$ :

$Moles=0.500 \times {38.0\times 10^{-3}}\ moles$

Moles of $CuCl_2$ = 0.019 moles

For $(NH_4)_2S$ :

Molarity = 0.600 M

Volume = 42.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 42.0×10⁻³ L

Thus, moles of $(NH_4)_2S$ :

$Moles=0.600 \times {42.0\times 10^{-3}}\ moles$

Moles of $(NH_4)_2S$ = 0.0252 moles

According to the given reaction:

$CuCl_2_{(aq)}+(NH_4)_2S_{(aq)}\rightarrow CuS_{(s)}+2NH_4Cl_{(aq)}$

1 mole of $CuCl_2$ reacts with 1 mole of $(NH_4)_2S$

So,

0.019 mole of $CuCl_2$ reacts with 0.019 mole of $(NH_4)_2S$

Moles of $(NH_4)_2S$ = 0.019 mole

Available moles of $(NH_4)_2S$ = 0.0252 mole

Limiting reagent is the one which is present in small amount. Thus, $CuCl_2$ is limiting reagent.

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

1 mole of $CuCl_2$ gives 1 mole of $CuS$

0.019 mole of $CuCl_2$ gives 0.019 mole of $CuS$

Moles of $CuS$ formed = 0.019 moles

Molar mass of $CuS$ = 95.611 g/mol

Mass of $CuS$ = Moles × Molar mass = 0.019 × 95.611 g = 1.82 g

1.82 g is the maximum mass of CuS.

5 0

2 years ago

The maximum number of electrons that are possible at a given energy level depends on the number of ____ at the energy level.

VMariaS [17]

Answer:

 orbitals

Explanation:

Each orbital may contain a maximum of two electrons.

The energy level is the principal quantum number: 1, 2, 3, 4, 5, 6, 7.

The principal quantum number limits the kind of orbital (ℓ ) and the number of orbitals (mℓ ).

I. Principal quantum number, n = 1

Angular or orbital quantum number : ℓ = 0 ⇒ ortibal s

Magnetic quantum number: mℓ = 0

⇒ number of orbitals 1 ⇒ number of electrons: 2

II. Principal quantun number, n = 2

Angular or orbital quantum number : ℓ = 0 and 1 ⇒ ortibal s and p

Magnetic quantum number:

for ℓ: mℓ = 0: 1 s ortibal

for ℓ = 1: mℓ = -1, 0, and +1: 3 p ortitals

⇒ number of orbitals 4 ⇒ number of electrons: 8

III. Principal quantun number, n = 3

Angular or orbital quantum number : ℓ = 0, 1, and 2 ⇒ ortibal s, p and d

Magnetic quantum number:

for ℓ: mℓ = 0: 1 s ortibal

for ℓ = 1: mℓ = -1, 0, and +1: 3 p ortitals

for ℓ = 2: mℓ = -2, -1, 0, +1, and +2: 5 d ortitals

⇒ number of orbitals 9 ⇒ number of electrons: 18

IV. Principal quantun number, n = 4

Angular or orbital quantum number : ℓ = 0, 1, 2, and 3 ⇒ ortibal s, p, d and f

Magnetic quantum number:

for ℓ: mℓ = 0: 1 s ortibal

for ℓ = 1: mℓ = -1, 0, and +1: 3 p ortitals

for ℓ = 2: mℓ = -2, -1, 0, +1, and +2: 5 d ortitals

for ℓ = 3: mℓ = ±3, ±2, ±1, and 0: 7 f ortitals

⇒ number of orbitals 16 ⇒ number of electrons: 32.

Thus, you have:

energy level, n maximum number of maximum number of

orbitals electrons

1 1 2

2 4 8

3 9 18

4 16 32

Those numbers follow a rule: n² and 2n². You can verify that the previous numbers are in accordance with those formulas.

3 0

2 years ago