What is material that will sink in Liquid water @ 4c

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

The specific heat capacity of lead is 0.13 J/g-K. How much heat (in J) is required to raise the temperature of 15 g of lead from

Alona [7]

Answer:

c. 29 J

Explanation:

Step 1: Given data

Specific heat capacity of Pb (c): 0.13 J/g.K (= 0.13 J/g.°C)

Mass of Pb (m): 15 g

Initial temperature: 22 °C

Final temperature: 37 °C

Step 2: Calculate the temperature change

ΔT = 37 °C - 22 °C = 15 °C

Step 3: Calculate the heat (Q) required to raise the temperature of the lead piece

We will use the following expression.

Q = c × m × ΔT

Q = 0.13 J/g.°C × 15 g × 15 °C = 29 J

6 0

3 years ago

Which household items would be considered a base? Select all that apply. *

mrs_skeptik [129]

Answer:

Drain Cleaner, soap, bleach

i think thats it, sorry if i got something wrong

6 0

3 years ago

All animals need oxygen gas (O2) for their primary cellular-level functioning. Inside the cell, O2 is split apart into oxygen at

il63 [147K]

Answer:

H₂O (water)

Explanation:

The reaction given is the glucose combustion inside the cells. During the combustion, it will be formed carbonic gas, water and energy.

The oxygen molecule is split in their two atoms of O. The element which has 1 proton and 1 electron is the hydrogen (H). So, when hydrogen reacts with oxygen, they form water (H₂O).

6 0

3 years ago

The two isotopes of chlorine are LaTeX: \begin{matrix}35\\17\end{matrix}Cl35 17 C l and LaTeX: \begin{matrix}37\\17\end{matrix}C

Kay [80]

Answer: The percentage abundance of $_{17}^{35}\textrm{Cl}$ and $_{17}^{37}\textrm{Cl}$ isotopes are 77.5% and 22.5% respectively.

Explanation:

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

Let the fractional abundance of $_{17}^{35}\textrm{Cl}$ isotope be 'x'. So, fractional abundance of $_{17}^{37}\textrm{Cl}$ isotope will be '1 - x'

For $_{17}^{35}\textrm{Cl}$ isotope:

Mass of $_{17}^{35}\textrm{Cl}$ isotope = 35 amu

Fractional abundance of $_{17}^{35}\textrm{Cl}$ isotope = x

For $_{17}^{37}\textrm{Cl}$ isotope:

Mass of $_{17}^{37}\textrm{Cl}$ isotope = 37 amu

Fractional abundance of $_{17}^{37}\textrm{Cl}$ isotope = 1 - x

Average atomic mass of chlorine = 35.45 amu

Putting values in equation 1, we get:

$35.45=[(35\times x)+(37\times (1-x))]\\\\x=0.775$

Percentage abundance of $_{17}^{35}\textrm{Cl}$ isotope = $0.775\times 100=77.5\%$

Percentage abundance of $_{17}^{37}\textrm{Cl}$ isotope = $(1-0.775)=0.225\times 100=22.5\%$

Hence, the percentage abundance of $_{17}^{35}\textrm{Cl}$ and $_{17}^{37}\textrm{Cl}$ isotopes are 77.5% and 22.5% respectively.

6 0

3 years ago