Write the word and balanced chemical equations for the reaction between:

What is another way that diamonds are useful

natali 33 [55]

Diamonds are common in the making of jewelry, but they are also used industrially because of their hardness. They are extremely useful in cutting, grinding, and drilling other materials.

7 0

3 years ago

Which types of atomic orbitals of the central atom mix to form hybrid orbitals in:<br> (b) CS₂;

Anettt [7]

<u>One s orbital</u> and <u>one p orbital</u> are the exact types of atomic orbitals of the central atom mix to form hybrid orbitals in CS₂

<h3>What is atomic orbital?</h3>

An atomic orbital is a function used in atomic theory and quantum mechanics to explain the position and wave-like behaviour of an electron in an atom. This function can be used to determine the likelihood of discovering any atom's electron in any particular area surrounding the nucleus.

The physical area or space where the electron may be calculated to be present, as predicted by the specific mathematical shape of the orbital, is referred to as an atomic orbital.

The three quantum numbers n, l, and $m_l$ which correspond to the electron's energy, angular momentum, and an angular momentum vector component, are used to describe all orbitals in an atom (magnetic quantum number).

Learn more about Atomic Orbital

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6 0

1 year ago

To standardize an H2SO4 solution, you put 20.00 mL of it in a flask with a few drops of indicator and put 0.450 M NaOH in a bure

hram777 [196]

To standardize 20.00 mL of 0.495 M H₂SO₄ are used 44.10 mL of 0.450 M NaOH in a neutralization reaction.

We want to standardize a H₂SO₄ solution with NaOH. The neutralization reaction is:

H₂SO₄ + 2 NaOH ⇒ Na₂SO₄ + H₂O

The buret, which contains NaOH, reads initially 0.63 mL, and 44.73 mL at the endpoint. The used volume of NaOH is:

$V = 44.73 mL - 0.63 mL = 44.10 mL$

44.10 mL of 0.450 M NaOH are used for the titration. The reacting moles of NaOH are:

$0.04410 L \times \frac{0.450mol}{L} = 0.0198 mol$

The molar ratio of H₂SO₄ to NaOH is 1:2. The moles of H₂SO₄ that react with 0.0198 moles of NaOH are:

$0.0198 mol NaOH \times \frac{1molH_2SO_4}{2molNaOH} = 0.00990 molH_2SO_4$

0.00990 moles of H₂SO₄ are in 20.00 mL of solution. The molarity of H₂SO₄ is:

$[H_2SO_4] = \frac{0.00990mol}{0.02000} = 0.495 M$

To standardize 20.00 mL of 0.495 M H₂SO₄ are used 44.10 mL of 0.450 M NaOH in a neutralization reaction.

Learn more: brainly.com/question/2728613

7 0

2 years ago

What represents the identity of an element

ololo11 [35]

Answer:

The identity of an element is its atomic number, which is the number of protons within the nucleus of one of its atoms.

Explanation:

5 0

3 years ago

Calculate the mass of ammonium sulfide (nh4)2s in 3.00 l of a 0.0200 m solution. g (nh4)2s

mezya [45]

Taking into account the definition of molarity, 4.08 grams of ammonium sulfide are present in 3.00 L of a 0.0200 M solution.

<h3>Definition of molarity</h3>

Molar concentration or molarity is a measure of the concentration of a solute in a solution and indicates the number of moles of solute that are dissolved in a given volume.

The molarity of a solution is calculated by dividing the moles of solute by the volume of the solution:

$Molarity=\frac{number of moles}{volume}$

Molarity is expressed in units $\frac{moles}{liter}$ .

<h3>Mass of ammonium sulfide</h3>

In this case, you know:

Molarity: 0.0200 M
number of moles: ?
volume: 3 L

Replacing in the definition of molarity:

$0.0200 \frac{moles}{liter}=\frac{number of moles}{3 L}$

Solving:

0.0200 $\frac{moles}{liter}$ × 3 L= number of moles

<u><em>0.06 moles= number of moles</em></u>

On the other side, the molar mass is the amount of mass that a substance contains in one mole. In this case, the molar mass of ammonium sulfide is 68 g/mole.

Then, the amount of mass contained in 0.06 moles is calculated as:

0.06 moles× $\frac{68 grams}{1 mole}$ = <u><em>4.08 grams</em></u>