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3 years ago

Phosphoric acid is neutralized by potassium hydroxide according to the following reaction:

1 answer:

5 0

Balance the reaction first:

3KOH + H3PO4 —> K3PO4 + 3H2O

So for every mol of H3PO4, you need 3 mol of OH- to fully neutralize the acid, since H3PO4 is polyprotic.

0.0200 L KOH • (2.000 mol KOH / L KOH) • (1 mol H3PO4 / 3 mol KOH) = 0.0133 mol H3PO4

Divide this by the volume of H3PO4 to get the concentration.

0.0133 mol H3PO4 / 0.0250 L = 0.532 M H3PO4

You might be interested in

What is the empirical formula of a compound composed of 30.5 g potassium (k) and 6.24 g oxygen (o)?

VLD [36.1K]

K5O2

convert grams to moles, divide both by the smallest mole mass, multiply that until hole.

30.5 g K ÷ 39.10 = .78 mol
6.24 g O ÷ 16 = .39 mol
.78 mol ÷ .39 mol = 2.5
.39 mol ÷ .39 mol = 1
2.5 x 2 = 5
1 x 2 = 2
K5O2

5 0

3 years ago

Consider the following reaction at a high temperature. Br2(g) ⇆ 2Br(g) When 1.35 moles of Br2 are put in a 0.780−L flask, 3.60 p

UNO [17]

Answer : The equilibrium constant $K_c$ for the reaction is, 0.1133

Explanation :

First we have to calculate the concentration of $Br_2$ .

$\text{Concentration of }Br_2=\frac{\text{Moles of }Br_2}{\text{Volume of solution}}$

$\text{Concentration of }Br_2=\frac{1.35moles}{0.780L}=1.731M$

Now we have to calculate the dissociated concentration of $Br_2$ .

The balanced equilibrium reaction is,

$Br_2(g)\rightleftharpoons 2Br(aq)$

Initial conc. 1.731 M 0

At eqm. conc. (1.731-x) (2x) M

As we are given,

The percent of dissociation of $Br_2$ = $\alpha$ = 1.2 %

So, the dissociate concentration of $Br_2$ = $C\alpha=1.731M\times \frac{1.2}{100}=0.2077M$

The value of x = 0.2077 M

Now we have to calculate the concentration of $Br_2\text{ and }Br$ at equilibrium.

Concentration of $Br_2$ = 1.731 - x = 1.731 - 0.2077 = 1.5233 M

Concentration of $Br$ = 2x = 2 × 0.2077 = 0.4154 M

Now we have to calculate the equilibrium constant for the reaction.

The expression of equilibrium constant for the reaction will be :

$K_c=\frac{[Br]^2}{[Br_2]}$

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

$K_c=\frac{(0.4154)^2}{1.5233}=0.1133$

Therefore, the equilibrium constant $K_c$ for the reaction is, 0.1133

7 0

3 years ago

Which new diagnosis would prompt the provider to discontinue a prn order for magnesium hydroxide? group of answer choices renal

dusya [7]

Renal failure is the new diagnosis that the provider would prompt to discontinue a PRN order for magnesium hydroxide.

<h3>What is Renal failure?</h3>

Renal failure is a special condition where the person is unable to manage the function of the kidney organ.

The magnesium hydroxide may cause renal failure because this salt must be eliminated by the kidney and therefore it may trigger health complications.

In conclusion, renal failure is the new diagnosis that the provider would prompt to discontinue a PRN order for magnesium hydroxide.

Learn more about Renal failure here:

brainly.com/question/20404873

#SPJ1

6 0

2 years ago

Given the balanced equation below, calculate the moles of aluminum that are needed to react completely with 28.7 moles of FeO. Y

Alexxx [7]

Answer:

43.05 moles of Al needed to react with 28.7 moles of FeO.

Explanation:

Given data:

Moles of FeO = 28.7 mol

Moles of Al needed to react with FeO = ?

Solution:

Chemical equation:

2Al + 3FeO → 3Fe + Al₂O₃

Now we will compare the moles of Al with FeO.

FeO : Al

2 : 3

28.7 : 3/2×28.7 = 43.05 mol

Thus 43.05 moles of Al needed to react with 28.7 moles of FeO.

8 0

3 years ago

When considering metal complexation with EDTA, if you are comparing 2 metals, the metal with a higher ____________ will react wi

DENIUS [597]

If you are comparing 2 metals, the metal with a higher <u>Number of free ions</u> will react with EDTA first

EDTA is a type of chemical which binds certain metal ions such as calcium and magnesium. some of the functions of EDTA includes:

Preventing blood clotting of blood samples
prevention of the formation of Biofilm by bacterias

The EDTA will readily react with metals which have a hiogher number of free ions that it can bind with.

Hence we can conclude that If you are comparing 2 metals, the metal with a higher <u>Number of free ions</u> will react with EDTA first.

Learn more about EDTA : brainly.com/question/10818175

7 0

2 years ago