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4 years ago

[OH-] for a solution is

Chemistry

2 answers:

solong [7]4 years ago

8 0

Answer:

B = basic

Explanation:

Given data:

[OH⁻] = 5.35×10⁻⁴M

pH = ?

Solution:

pOH = -log[OH⁻]

pOH = - [5.35×10⁻⁴]

pOH = 3.272

it is known that,

pH + pOH = 14

pH = 14- pOH

pH = 14 - 3.272

pH = 10.728

The acidic pH is range from zero to less than 7 while 7 pH is neutral and above 7 the pH is basic. So, the given solution is basic.

ohaa [14]4 years ago

5 0

Answer:

grgggfggf

Explanation:

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What is the strongest evidence for hydrogen

Mumz [18]

Explanation:

4.hydrogen is able to accept or donate electrons,so it is the most versatile storm I the periodic chart

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3 years ago

How does the energy of the activated complex compare with the energies of reactants and products? select one:

Free_Kalibri [48]

Answer:

Option d. it is higher than the energy of both reactants and products

Explanation:

Activated complex, also known as transition state, is the intermediate structure formed in the course of a chemical reaction.

The activated complex is very unstable and of short life: it is at the peak of the potential chemical diagram, and can transform either into the reactants (backward) or the products (forward).

The activation energy of the reaction is the energy needed to reach the activated complex, then both reactants and products are lower in potential chemical energy than the activated complex, which is what explains why the activated complex can transform into one or another, reactants or products.

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4 years ago

A 1.50 L buffer solution is 0.250 M in HF and 0.250 M in NaF. Calculate the pH of the solution after the addition of 0.100 moles

alexgriva [62]

Answer : The pH of the solution is, 3.41

Explanation :

First we have to calculate the moles of $HF$ .

$\text{Moles of HF}=\text{Concentration of HF}\times \text{Volume of solution}$

$\text{Moles of HF}=0.250M\times 1.50L=0.375mol$

Now we have to calculate the value of $pK_a$ .

The expression used for the calculation of $pK_a$ is,

$pK_a=-\log (K_a)$

Now put the value of $K_a$ in this expression, we get:

$pK_a=-\log (6.8\times 10^{-4})$

$pK_a=4-\log (6.8)$

$pK_a=3.17$

The reaction will be:

$HF+OH^-\rightleftharpoons F^-+H_2O$

Initial moles 0.375 0.100 0.375

At eqm. (0.375-0.100) 0 (0.375+0.100)

= 0.275 = 0.475

Now we have to calculate the pH of solution.

Using Henderson Hesselbach equation :

$pH=pK_a+\log \frac{[Salt]}{[Acid]}$

$pH=pK_a+\log \frac{[F^-]}{[HF]}$

Now put all the given values in this expression, we get:

$pH=3.17+\log [\frac{(\frac{0.475}{1.50})}{(\frac{0.275}{1.50})}]$

$pH=3.41$

Thus, the pH of the solution is, 3.41

8 0

3 years ago

How many moles are in 68.5 liters of oxygen gas at STP?

valina [46]

3.05 moles of oxygen gas

6 0

3 years ago

Reynolds number, for an aorta of 0.9-centimeter diameter, calculate the blood flow in liters per minute when the flow regime in

olga55 [171]

Answer:

Flow in liters per minute is 4 L/min

Explanation:

For this case, we have an aorta of 0.9 cm of diameter (D). Let's suppose an uniform and constant diameter for calculation purposes.

D = (0.9 cm)(1m/100cm)

D = 0.009 m

It is required to calculate the blood flow in liters per minute when the flow regime changes from laminar to turbulent. Laminar flow is usually less than 2500 for Reynolds value, and Turbulent flow when Re is higher than 2500. So, we need to study the phenomenon for

Re = 2500.

Using the definition of Reynolds we can find the velocity average of the blood, and use it to find flow. Where blood density is $\rho$ , aorta diameter is D, average velocity is v and blood viscosity is $\mu$

$Re = \frac{\rho v D}{\mu } \\v = \frac{Re*\mu}{\rho D}$

From data problem, we have Re, D values. As we need blood density and blood viscosity we can find them in medical studies. For example: in this online document: Blood flow analysis of the aortic arch using computational fluid dynamics †

Satoshi Numata, Keiichi Itatani, Keiichi Kanda, Kiyoshi Doi, Sachiko Yamazaki, Kazuki Morimoto, Kaichiro Manabe, Koki Ikemoto, Hitoshi Yaku Author Notes

European Journal of Cardio-Thoracic Surgery, Volume 49, Issue 6, June 2016, Pages 1578–1585, https://doi.org/10.1093/ejcts/ezv459

Published: 20 January 2016

$\rho = 1060 kg/m^3\\\mu = 0.0004 kg/(ms)\\$

$v = \frac{Re*\mu}{\rho D}\\v = \frac{2500 * 0.004 kg/(ms)}{(1060 kg/m3)(0.009 m)}\\v = 1.04 m/s\\$

Average blood velocity is 1.04 m/s. After that, we can calculate the flow (Q) using the flow are (Ao) of aorta.

Q = v*Ao

$Q = (1.4m/s)*(\pi*(0.009m/2)^2)\\Q = 6.67 * 10^-5 \frac{m^3}{s}\\Q = (6.67 * 10^-5 \frac{m^3}{s})(\frac{1000 L}{1 m^3} ) (\frac{60 s}{1 min} )\\Q = 4 L/min$

Finally, the blood flow in liters per minute is 4 L/min when the flow regime in the aorta changes from laminar to turbulent.

7 0

3 years ago