What is the possible formula of a compound formed when sodium and chlorine bond

Does temperature have an affect on a magnets strength

Vlad [161]

If magnets are heated to the Curie point, they lose their ability to be magnetic. ... Cooling causes the molecules in the magnet to have less kinetic energy. This means that there is less vibration in the magnet's molecules, allowing the magnetic field they create to be more consistently concentrated in a given direction.

8 0

3 years ago

Please help due tomorrow Thanks

natka813 [3]

I believe the answer is the poles of the magnet.

8 0

3 years ago

<img src="https://tex.z-dn.net/?f=H_2PO_4%5E-%28aq%29%20%5Crightarrow%20H%5E%2B%28aq%29%20%2B%20HPO_4%5E%7B2-%7D%28aq%29" id="Te

klasskru [66]

Answer:

The pH of the buffer solution = 8.05

Explanation:

Using the Henderson - Hasselbalch equation;

pH = pKa₂ + log ( [HPO₄²-]/[H₂PO4⁻]

where pKa₂ = -log (Ka₂) = -log ( 6.1 * 10⁻⁸) = 7.21

Concentration of OH⁻ added = 0.069 M (i.e. 0.069 mol/L)

[H₂PO4⁻] after addition of OH⁻ = 0.165 - 0.069 = 0.096 M

[HPO₄²-] after addition of OH⁻ = 0.594 + 0.069 = 0.663 M

Therefore,

pH = 7.21 + log (0.663 / 0.096)

pH = 7.21 + 0.84

pH = 8.05

4 0

3 years ago

What is a possible quantum number set for an electron in the 3s orbital of a magnesium atom

Alik [6]

n = 3
l = 0
$m_l$ = 0

$m_s$ = 1/2 or -1/2

<h3>Explanation</h3>

There are four quantum numbers in an electron that orbits the atom.

n, the principal quantum number.
l, the angular quantum number.
$m_l$ , the magnetic quantum number.
$m_s$ , the spin quantum number.

n is a positive integer. The value of n indicates the main shell of the electron. The electron in question is in the 3s orbital. As a result, n = 3.

l is a non-negative integer. The value of l indicates the type of subshell ("orbital") of the electron. The types of subshells possible depends on the main shell. For example, both s and p orbitals exist in the second main shell. However, only the s orbital exists in the first main shell. The value of l ranges from 0 to n - 1.

l = 0 indicates an s orbital.
l = 1 indicates a p orbital.
l = 2 indicates a d orbital.
l = 3 indicates an f orbital.

The electron in question is in an s orbital. As a result, l = 0.

$m_l$ is an integer. The value of $m_l$ indicates the position of the electron within the subshell. The range of $m_l$ depends on the value of l. $m_l$ ranges from -l to l (that's -l, ..., -1, 0, 1, ... l). Accordingly, there are 2 l + 1 orbitals in a l subshell. l = 0 for this 3s electron. There's only one orbital in the 3s subshell. The only $m_l$ value possible for this electron is 0.

The value of $m_s$ is either - 1/2 or 1/2. It indicates the position of an electron within a single orbital. The value of $m_s$ does not depend on that of n, l, or $m_l$ . However, by the Pauli Exclusion Principle, at least one of the four numbers must differ for two electrons in the same atom. In case all three of n, l, and $m_l$ are the same, the two electrons must differ in $m_s$ . However, this question asks only for the number of one single electron. Thus, giving either - 1/2 or 1/2 shall work.

<h3>Reference</h3>

Vitz et. al, "5.8 Quantum Numbers (Electronic)", ChemPRIME (Moore et al.), Chemistry Libretexts. 27 Oct 2017.

7 0

4 years ago

When 80.0 mL of a 0.812 M barium chloride solution is combined with 40 mL of a 1.52 M potassium sulfate solution, 10.8 g of bari

BabaBlast [244]

Answer:

76.1%

Explanation:

The reaction that takes place is:

BaCl₂ + K₂SO₄ → BaSO₄ + 2KCl

First we determine how many moles of each reactant were added:

BaCl₂ ⇒ 80 mL * 0.812 M = 64.96 mmol BaCl₂

K₂SO₄ ⇒ 40 mL * 1.52 M = 60.8 mmol K₂SO₄

Thus K₂SO₄ is the limiting reactant.

Using the moles of the limiting reactant we calculate how many moles of BaSO₄ would have been produced if the % yield was 100%:

60.8 mmol K₂SO₄ * $\frac{1mmolBaSO_4}{1mmolK_2SO_4}$ = 60.8 mmol BaSO₄

Then we convert that theoretical amount into grams, using the molar mass of BaSO₄:

60.8 mmol BaSO₄ * 233.38 mg/mmol = 14189.504 mg BaSO₄

14189.504 mg BaSO₄ / 1000 = 14.2 g BaSO₄

Finally we calculate the % yield:

% yield = 10.8 g / 14.2 g * 100 %

% yield = 76.1%

7 0

3 years ago