Help me which is it

What are the different types of energy used in matter?

ValentinkaMS [17]

Condensation I think

4 0

3 years ago

Calculate the concentration of acetic acid, HAc, and acetate ion, Ac−, in a 0.25M acetate buffer solution with pH = 5.36. "0.25M

Neporo4naja [7]

Answer:

[HAc] = 0.05M

[Ac⁻] = 0.20M

Explanation:

The Henderson-Hasselbalch formula for the acetic acid buffer is:

pH = pka + log₁₀ [Ac⁻] / [HAc]

Replacing:

5.36 = 4.76 + log₁₀ [Ac⁻] / [HAc]

3.981 = [Ac⁻] / [HAc] (1)

Also, as total concentration of buffer is 0.25M it is possible to write:

0.25M = [Ac⁻] + [HAc] (2)

Replacing (2) in (1)

3.981 = 0.25M - [HAc] / [HAc]

3.981 [HAc] = 0.25M - [HAc]

4.981 [HAc] = 0.25M

[HAc] = 0.05M

Replacing this value in (2):

0.25M = [Ac⁻] + 0.05M

[Ac⁻] = 0.20M

I hope it helps!

7 0

4 years ago

If 15.0 g of nitrogen reacts with 15.0 g of hydrogen, 10.5 g of ammonia is

Alexxx [7]

Answer:

Percent yield = 57%

Explanation:

Given data:

Mass of nitrogen = 15.0 g

Mass of hydrogen = 15.0 g

Mass of ammonia produced = 10.5 g

Percent yield = ?

Solution:

Chemical equation:

N₂+ 3H₂ → 2NH₃

Number of moles of hydrogen:

Number of moles = mass/molar mass

Number of moles = 15.0 g/ 2 g/mol

Number of moles = 7.5 mol

Number of moles of nitrogen:

Number of moles = mass/molar mass

Number of moles = 15.0 g/ 28 g/mol

Number of moles = 0.54 mol

Now we will compare the moles of ammonia with nitrogen and hydrogen from balance chemical equation.

N₂ : NH₃

1 : 2

0.54 : 2×0.54 = 1.08

H₂ : NH₃

3 : 2

7.5 : 2/3×7.5= 5 mol

Theoretical yield of ammonia:

Mass = number of moles × molar mass

Mass = 1.08 × 17 g/mol

Mass = 18.36 g

Percent yield:

Percent yield = (actual yield / theoretical yield) ×100

Percent yield = (10.5 g/ 18.36 g) ×100

Percent yield = 0.57 ×100

Percent yield = 57%

8 0

3 years ago

Matter that has a uniform and definite composition is called?

enot [183]

Matter that has a uniform and definite composition is called a vapor

8 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