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3 years ago

Make the equation equal Fe(NO3)3+KSCN —> Fe(SCN)3 + K(NO3)3

Chemistry

1 answer:

Advocard [28]3 years ago

5 0

Answer:

Fe(NO₃)₃ + 3KSCN → Fe(SCN)₃ + 3KNO₃

Explanation:

Chemical equation:

Fe(NO₃)₃ + KSCN → Fe(SCN)₃ + KNO₃

Balanced Chemical equation:

Fe(NO₃)₃ + 3KSCN → Fe(SCN)₃ + 3KNO₃

Type of reaction:

It is double displacement reaction.

In this reaction the anion or cation of both reactants exchange with each other. In given reaction the cation Fe⁺³ exchange with cation K⁺.

The given reaction equation is balanced so there are equal number of atoms of each elements are present on both side of equation and completely hold the law of conservation of mass.

Double replacement:

It is the reaction in which two compound exchange their ions and form new compounds.

AB + CD → AC +BD

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vagabundo [1.1K]

Answer:

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3 years ago

C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l)

stiv31 [10]

¹/3 C3H8(g) + ⁵/3 O2(g)

Explanation:

The coefficient before every molecule is representative of the number of moles. We can represent it in ration form so as to calculate the question;

C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(l) means;

For every 1 mole of C₃H₈(g) and 5 moles of O₂(g) produces 3 moles of CO₂(g) and 4 moles of H₂O(l).

Therefore to produce 1.00 mole of CO₂(g);

We represent it in ratio;

C₃H₈(g) : CO₂(g)

1 : 3

What about ;

? (x) : 1

We cross multiply;

3x = 1 * 1

X = 1/3

We evaluate the same for O₂;

O₂(g) : CO₂(g)

5 : 3

What about

? (x) : 1

3x = 5 * 1

x = 5/3

Learn More:

For more on evaluating moles in chemical reactions check out;

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7 0

3 years ago

Calculate the ph of a 0.20 m solution of iodic acid (hio3, ka = 0.17).

saveliy_v [14]

Iodic acid partially dissociates into H+ and IO3-
Assuming that x is the concentration of H+ at equilibrium, and sine the equation says the same amount of IO3- will be released as that of H+, its concentration is also X. The formation of H+ and IO3- results from the loss of HIO3 so its concentration at equilibrium is 0.20 M - x
Ka = [H+] [IO3-] / [HIO3];
Initially, [H+] ≈ [IO3-] = 0 and [HIO3] = 0.20; 
At equilibrium [H+] ≈ [IO3-] = x and [HIO3] = 0.20 - x; 
so 0.17 = x² / (0.20 - x); 
Solving for x using the quadratic formula: 
x = [H+] = 0.063 M or pH = - log [H+] = 1.2.

7 0

4 years ago

Calculate the values of ΔU, ΔH, and ΔS for the following process:

ladessa [460]

Answer:

ΔU = 45.814 KJ

ΔH = 46.4375 KJ

ΔS = 18.76 J/K

Explanation:

H2O(l) → H2O(l) → H2O(steam)

298.15K, 1atm ΔHp 373.15K,1atm ΔHv 373.15K,1 atm

∴ ΔHp = Qp = nCpΔT

∴ n H2O = 1 mol

∴ Cp,n = 75.3 J/mol.K

∴ ΔT = 373.25 - 298.15 = 75 K

⇒ Qp = (1 mol)*(75.3 J/mol.K)*(75K) = 5647.5 J

⇒ ΔHp = 5647.5 J = 5.6475 KJ

⇒ ΔH = ΔHp + ΔHv

∴ ΔHv = 40.79 KJ/mol * 1 mol = 40.79 KJ

⇒ ΔH = 5.6475 KJ + 40.79 KJ = 46.4375 KJ

ideal gas:

∴ ΔH = ΔU + PΔV

∴ V1 = nRT1/P1 = ((1)*(0.082)*(298.15))/1 = 24.45 L

∴ V2 = nRT2/P2 = ((1)*(0.082)*(373.15))/ 1 = 30.59 L

⇒ ΔV = V2 - V1 = 6.15 L * (m³/1000L) = 6.15 E-3 m³

∴ P = 1 atm * (Pa/ 9.86923 E-6 atm) = 101325.027 Pa

⇒ ΔU = ΔH - PΔV = 46.4375 KJ - ((101325.027 Pa*6.15 E-3m³)*(KJ/1000J))

⇒ ΔU = 46.4375 KJ - 0.623 KJ

⇒ ΔU = 45.814 KJ

∴ ΔS = Cv,n Ln (T2/T1) + nR Ln (V2/V1)

⇒ ΔS = (75.3) Ln(373.15/298.15) + (1)*(8.314) Ln (30.59/24.45)

⇒ ΔS = 16.896 J/K + 1.863 J/K

⇒ ΔS = 18.76 J/K

3 0

4 years ago

11 A student combined 1.2 g of H2 gas and 5.4 g of He gas in a 2.5 L closed container. What was the pressure exerted by the H2 g

blagie [28]

The correct answer is 5.8 atm.

7 0

3 years ago

Read 2 more answers