Full net ionic equation of nitric acid and solid aluminum hydroxide please

1 Point

Mila [183]

Answer:

Option C. 296K

Explanation:

T°C = 23°C

To covert the above to Kelvin temperature, use the following equation

T(K) = T°C + 273

T(K) = 23 + 273

T(K) = 296K

7 0

3 years ago

A student prepared an equilibrium solution by mixing the following solutions:A.2.00 mL of 0.00250 M Fe(NO3)3B.5.00 mL of 0.00250

Crank

Explanation:

$Molarity=\frac{\text{Moles of solute}}{\text{Volume of solution Liter}}$

A. 2.00 mL of 0.00250 M $Fe(NO_3)_3$

Moles of ferric nitrate = n

Volume of ferric nitrate = 2.00 ml = 0.002 L ( 1 mL=0.001 L)

Molarity of ferric nitrate = 0.00250 M

$n=0.00250 M\times 0.002 L=0.000005 mol$

B. 5.00 mL of 0.00250 M $KSCN$

Moles of KSCN = n'

Volume of KSCN = 5.00 ml = 0.005 L ( 1 mL=0.001 L)

Molarity of KSCN = 0.00250 M

$n'=0.00250 M\times 0.005 L=0.0000125 mol$

C. 3.00 mL of 0.050 M $HNO_3$

Moles of nitric acid = n''

Volume of nitric acid = 3.00 ml = 0.003 L ( 1 mL=0.001 L)

Molarity of nitric acid = 0.050 M

$n=0.050 M\times 0.003 L=0.00015 mol$

After mixing A, B and C together and their respective initial concentration before reaction.

After mixing A, B and C together the volume of the solution becomes = V

V = 0.002 L=0.005 L+0.003 L= 0.010 L

Concentration of ferric nitrate :

$[Fe(NO_3)_3]=\frac{0.000005 mol}{0.010 L}=0.0005 M$

Concentration of ferric ions :

$[Fe^{3+}]=1\times [Fe(NO_3)_3]=0.0005 M$

Concentration of nitrate ions from ferric nitrate:

$[NO_3^{-}]=3\times [Fe(NO_3)_3]=0.0015 M$

Concentration of KSCN :

$[KSCN]=\frac{0.0000125 mol}{0.010 L}=0.00125 M$

Concentration of $SCN^-$ ions:

$[SCN^-]=1\times [KSCN]=0.00125 M$

Concentration of potassium ions:

$[K^+]=1\times [KSCN]=0.00125 M$

Concentration of nitric acid :

$[HNO_3]=\frac{0.00015 mol}{0.010 L}=0.015 M$

Concentration of hydrogen ion :

$[H^+]=1\times [HNO_3]=0.015 M$

Concentration of nitrate ions from nitric acid :

$[NO_3^{-}]=1\times [HNO_3]=0.0015 M$

Concentration of nitrate ion in mixture = 0.0015 M + 0.0015 M = 0.0030 M

$Fe^{3+}+SCN^-\rightleftharpoons Fe(NCS)^{2+}$

given concentration of $Fe(NCS)^{2+}$ at equilbrium = $3.6\times 10^{-5} M = 0.000036 M$

initially :

0.0005 M 0.00125 M 0

At equilibrium

(0.0005-0.000036) M (0.00125-0.000036) M 0.000036 M

0.000464 M 0.001214 M 0.000036 M

The expression of an equilibrium constant will be given as;

$K_c=\frac{[Fe(NCS)^{2+}]}{[Fe^{3+}][SCN^{-}]}$

$=\frac{0.000036 M}{0.000464 M\times 0.001214 M}=63.91$

The value for the equilibrium constant is 63.91.

6 0

3 years ago

A test tube can be used to hold

Andrej [43]

What grade is this? I forgot but if u tell me the grade i can prob remember

5 0

3 years ago

A barium hydroxide solution is prepared by dissolving 2.81 g of Ba(OH)2 in water to make 25.6 mL of solution. What is the concen

chubhunter [2.5K]

The concentration of the solution in units of molarity is 0.64 M and the concentration of the solution in terms of g/mL is 0.10976 g/mL.

<h3>What is the concentration of the solution?</h3>

The measure of the amount of the solute dissolved in the given amount of the solvent or solution is said to be the concentration of the solution.

It is given by the expression,

Concentration of the solution = mass of the solute/total volume of the solution

Units: g/mL

<h3>What is the concentration of the solution in units of molarity?</h3>

The concentration of the solution in units of molarity is given by the expression,

Molarity = (no. of moles of the solute)/(Volume in L)

Units: moles/L

<h3>Calculation:</h3>

It is given that the solute is barium hydroxide - Ba(OH)₂

The mass of Ba(OH)₂ = 2.81 g

When in dissolved in water, the total volume of the solution = 25.6 mL

The molar mass of Ba(OH)₂ = 171.34 g/mol

Then,

No. of moles = (mass of Ba(OH)₂)/(molar mass of Ba(OH)₂)

i.e.,

No. of moles = $\frac{2.81}{171.34}$ = 0.0164 mole

Concentration = mass of the solute/total volume

I.e., C = $\frac{2.81}{25.6}$ = 0.10976 g/mL

Then the concentration of the solution in units of molarity is,

Molarity = (no. of moles)/(Volume of the solution in L)

⇒ Molarity = $\frac{0.0164}{25.6/1000}$ = 0.64 M

Therefore, the concentration of the solution in units of molarity is 0.64 M.

Learn more about molarity here:

brainly.com/question/26873446

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8 0

2 years ago

What is the difference b/w synthesis reaction and combination reaction

suter [353]

Direct Combination Reaction : The reaction in which two or more substances combine to form a single substance is called direct combination reaction. Synthesis Reaction: The reaction in which two or more elements combine to form a compound is called direct combination reaction.

HOPED I HELP BRIANEST PLZZ

4 0

3 years ago

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