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3 years ago

What happens to the concentration of both H3O+ and OH- ions as water is added to a base?

Chemistry

1 answer:

DanielleElmas [232]3 years ago

6 0

Answer:

The concentration of H₃O⁺ ions increase whereas the concentration of the OH⁻ decreases.

Explanation:

The water, H₂O, has its own equilibrium thus:

2H₂O(l) ⇄ OH⁻(aq) + H₃O⁺(aq)

Where the water equilibrium constant, Kw, is defined as:

Kw = 1x10⁻¹⁴ = [OH⁻] [H₃O⁺]

The addition of a base represents the increasing of [OH⁻]. that means the concentration of OH⁻ ions.

Based on LeCh principle, in an equilibrium, the addition of a product produce the decreasing in concentration of the other products trying to counteract the effect, that means the concentration of H₃O⁺ ions decreases.

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Food contains chemical potential energy that the body uses. How do you think that the body gains this energy

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By digesting the food

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3 years ago

Molarity to percent by mass. Convert 1.672 mol/L MgCl2(aq) solution to percent by mass of MgCl2 in the solution. The solution de

nignag [31]

Answer:

$\%m/m=14\%$

Explanation:

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In this case, since the molarity of magnesium chloride (molar mass = 95.211 g/mol) is 1.672 mol/L and we know the density of the solution, we can first compute the concentration in g/L as shown below:

$[MgCl_2]=1.672\frac{molMgCl_2}{L}*\frac{95.211gMgCl_2}{1molMgCl_2}=159.2\frac{gMgCl_2}{L}$

Next, since the density of the solution is 1.137 g/mL, we can compute the concentration in g/g as shown below:

$[MgCl_2]=159.2\frac{gMgCl_2}{L}*\frac{1L}{1000mL}*\frac{1mL}{1.137g}=0.14$

Which is also the by-mass fraction and in percent it turns out:

$\%m/m=0.14*100\%\\\\\%m/m=14\%$

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6 0

3 years ago

The momentum of an object depends on which two quantities?

Maurinko [17]

The answer is D; Mass and Velocity

6 0

3 years ago

Read 2 more answers

Air at 7S°F and14.6 though a cfm(cubic ft per minute) and the The flow rate is 48000 psia flows though a rectangular duct of Ix2

IgorLugansk [536]

Explanation:

The given data is as follows.

Air is at $75^{o}F$ and 14.6 psia.

$\varepsilon$ = 0.00015 ft, Flow rate, (Q) = 48000 $ft^{3}/m$

(a) Formula to calculate hydraulic radius $(r_{H})$ is as follows.

$r_{H} = \frac{\text{free flow area}}{\text{wet perimeter}}$

= $\frac{2 \times 1}{2(1) + 2(2)}$

= $\frac{1}{3}$ ft

Formula for equivalent diameter is as follows.

$D_{eq} = 4 \times r_{H}$

= $4 \times \frac{1}{3} ft$

= $\frac{4}{3}$ ft

(b) Formula for velocity floe is as follows.

Q = VA

V = $\frac{Q}{A}$

= $\frac{48000}{2 \times 1}$ ft/min

= 24000 ft/min

(c) Formula to calculate Reynold's number is as follows.

$R_{e}$ = $\frac{D \times V \times \rho}{\mu}$

= $\frac{\frac{4}{3} \times 24000 \times 0.0744}{0.0443}$ (as $\rho = 0.0744 lb/ft^{3}$ and $\mu$ = 0.0443 lb/ft. hr)

= 53742.66 hr/min

As 1 hr = 10 min. So, $53742.66 hr/min \times \frac{60 min}{1 hr}$

= 3224559.6

(d) Formula to calculate pressure drop $(\Delta P)$ is as follows.

$\frac{\Delta P}{L} = \frac{4f \rho V^{2}}{2Dg_{c}}$

Putting the given values into the above formula as follows.

$\frac{\Delta P}{L} = \frac{4f \rho V^{2}}{2Dg_{c}}$

= $\frac{4 \times 0.00015 \times 100 \times 0.0744 \times (24000)^{2}}{2 \times \frac{4}{3} \times {4}{3}}$

= 6.238 $lb/ft^{2}$

5 0

3 years ago

An order is given to administer 57 g of lactulose syrup, often used for treating complications of liver disease. The suspension

Angelina_Jolie [31]

Answer : The volume given to the patient should be, 85.5 mL

Explanation :

As we are given that the suspension contains 10g/15mL.

Now we have to determine the volume should be given to the patient.

As, 10 grams of lactulose syrup needed 15 mL volume of solution

So, 57 grams of lactulose syrup needed $\frac{57g}{10g}\times 15mL=85.5mL$ volume of solution

Thus, the volume given to the patient should be, 85.5 mL

8 0

3 years ago