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3 years ago

14

Solid Al(NO3)3 is added to distilled water to produce a solution in which the concentration on nitrate, [NO3^-], is 0.10 M. What

is the concentration of aluminum ion, [Al^3+] in this solution?

1 answer:

VikaD [51]3 years ago

5 0

Answer:

The concentration of nitrate ion $NO_{3}^{-}$ is 0.033 M.

Explanation:

$Al(NO_{3})_{3} \rightarrow Al^{3+} + NO_{3}^{-}$

1 mole of $[Al^{3+}]$ = 3 mole $NO_{3}^{-}$

Let $[NO_{3}^{-}]$ = 3y

then according to above reaction,

$[Al^{3+}]$ = y

$[NO_{3}^{-}]$ = 0.10 M (given)

$[Al^{3+}]$ = ?

3y = 0.10 M

$y = \frac{0.1}{3}$

y = 0.033 M

So ,concentration of $[Al^{3+}]$ = y = 0.033 M

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Hydrogen gas and aqueous zinc chloride are produced by the reaction of aqueous hydrochloric acid and solid zinc . Write a balanc

Ymorist [56]

<em>1. Write out all the ions/elements and indicate whether they are reactants or products. </em>

<u>Reactants:</u>

Aqueous Hydrochloric Acid: HCl (H has a charge of +1 and Cl has a charge of 1-, so when an ionic bond forms they will form HCl).

Solid Zinc: Zn (since it is a solid zinc with no other forms of elements, the zinc is pure).

<u>Products:</u>

Hydrogen gas: H2 (it has to be H2 because Hydrogen is a diatomic element, and when by itself during reactions, will be H2 instead of H).

Aqueous Zinc Chloride: ZnCl2 (Zn has a charge of 2+ and Cl has a charge of 1-, so through an ionic bond, they will form ZnCl2).

<em>2. Using this information, you can then write out the unbalanced equation.</em>

HCl + Zn -> H2 + ZnCl2

<em>3. Next, indicate their states of matter and any aqueous solutions. </em>

HCl (aq) + Zn (s) -> H2 (g) + ZnCl2 (aq)

<em>4. Try different numbers until you get the same numbers of H, Cl, and Zn on both sides. </em>

2 HCl (aq) + Zn (s) -> H2 (g) + ZnCl2 (aq)

As you can see, the equation contains 2 Hydrogen atoms on both sides, 2 Chloride atoms on both sides, and 1 Zinc atom on both sides. The equation is chemically balanced.

5 0

4 years ago

ale4655 [162]

Answer: Option (b) is the correct answer.

Explanation:

Since energy of reactants is less than the energy of products. Therefore, it means energy is absorbed during the reaction.

As the energy required to break the bonds in the reactants is greater than the energy released when products are formed.

Therefore, it is an endothermic reaction.

Thus, we can conclude that the statement, it is endothermic because the energy required to break bonds in the reactants is greater than the energy released when the products are formed is correct.

5 0

3 years ago

Read 2 more answers

Consider the reaction below for which K = 78.2 atm-1. A(g) + B(g) ↔ C(g) Assume that 0.386 mol C(g) is placed in the cylinder re

borishaifa [10]

Answer:

1.65 L

Explanation:

The equation for the reaction is given as:

A + B ⇄ C

where;

numbers of moles = 0.386 mol C (g)

Volume = 7.29 L

Molar concentration of C = $\frac{0.386}{7.29}$

= 0.053 M

A + B ⇄ C

Initial 0 0 0.530

Change +x +x - x

Equilibrium x x (0.0530 - x)

$K = \frac{[C]}{[A][B]}$

where

K is given as ; 78.2 atm-1.

So, we have:

$78.2=\frac{[0.0530-x]}{[x][x]}$

$78.2= \frac{(0.0530-x)}{(x^2)}$

$78.2x^2= 0.0530-x$

$78.2x^2+x-0.0530=0$

Using quadratic formula;

$\frac{-b+/-\sqrt{b^2-4ac} }{2a}$

where; a = 78.2 ; b = 1 ; c= - 0.0530

= $\frac{-b+\sqrt{b^2-4ac} }{2a}$ or $\frac{-b-\sqrt{b^2-4ac} }{2a}$

= $\frac{-(1)+\sqrt{(1)^2-4(78.2)(-0.0530)} }{2(78.2)}$ or $\frac{-(1)-\sqrt{(1)^2-4(78.2)(-0.0530)} }{2(78.2)}$

= 0.0204 or -0.0332

Going by the positive value; we have:

x = 0.0204

[A] = 0.0204

[B] = 0.0204

[C] = 0.0530 - x

= 0.0530 - 0.0204

= 0.0326

Total number of moles at equilibrium = 0.0204 + 0.0204 + 0.0326

= 0.0734

Finally, we can calculate the volume of the cylinder at equilibrium using the ideal gas; PV =nRT

if we make V the subject of the formula; we have:

$V = \frac{nRT}{P}$

where;

P (pressure) = 1 atm

n (number of moles) = 0.0734 mole

R (rate constant) = 0.0821 L-atm/mol-K

T = 273.15 K (fixed constant temperature )

V (volume) = ???

$V=\frac{(0.0734*0.0821*273.15)}{(1.00)}$

V = 1.64604

V ≅ 1.65 L

3 0

3 years ago

Phosphoric acid is a triprotic acid ( K a1 = 6.9 × 10 − 3 Ka1=6.9×10−3, K a2 = 6.2 × 10 − 8 Ka2=6.2×10−8, and K a3 = 4.8 × 10 −

irina1246 [14]

<u>Answer:</u> To calculate the pH of the buffer composed of $H_2PO_4^-\text{ and }HPO_4^{-2}$ , we use the $K_a2$

<u>Explanation:</u>

Phosphoric acid is a triprotic acid and it will undergo three dissociation reaction each having their respective dissociation constants.

The chemical equation for the first dissociation reaction follows:

$H_3PO_4\rightleftharpoons H_2PO_4^-+H^+;K_a1=6.9\times 10^{-3}$

The chemical equation for the second dissociation reaction follows:

$H_2PO_4^-\rightleftharpoons HPO_4^{2-}+H^+;K_a2=6.2\times 10^{-8}$

The chemical equation for the third dissociation reaction follows:

$HPO_4^{2-}\rightleftharpoons PO_4^{3-}+H^+;K_a3=4.8\times 10^{-13}$

To form a buffer composed of $H_2PO_4^-\text{ and }HPO_4^{-2}$ , we use the $K_a$ of second dissociation process

To calculate the $pK_a$ , we use the equation:

$pK_a=-\log (K_a)\\\\pK_a=-\log(6.2\times 10^{-8})\\\\pK_a=7.21$

To calculate the pH of buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a2+\log(\frac{[\text{conjugate base}]}{[\text{weak acid}]})$

$pH=pK_a2+\log(\frac{[HPO_4^{2-}]}{[H_2PO_4^-]})$

We are given:

$pK_a2$ = negative logarithm of second acid dissociation constant of phosphoric acid = 7.21

$[HPO_4^{2-}]$ = concentration of conjugate base

$[H_2PO_4^{-}]$ = concentration of weak acid

Hence, to calculate the pH of the buffer composed of $H_2PO_4^-\text{ and }HPO_4^{-2}$ , we use the $K_a2$

3 0

4 years ago

An object weighing 90 n has a volume of 45 ml and a mass of 9.0 g what is its density

UNO [17]

D= mass over volume

So, density equals 9.0g divided by 45ml.

SO, D=.25 g/ml

I'm not 100% sure this is correct, though. Best if luck. ♡

7 0

3 years ago