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3 years ago

Enter the balanced NET IONIC equation for the potentially unbalanced equation AgNO3(aq)+NaCl(aq)→AgCl(s)+NaNO3(aq).

Chemistry

1 answer:

Zigmanuir [339]3 years ago

4 0

Answer:

$Ag^+(aq)+Cl^-(aq)\rightarrow AgCl(s)$

Explanation:

Hello!

In this case, since the net ionic equations are ionic representations of the molecular equation in which the spectator ions (those at both reactants and products sides) are cancelled out, we first write the complete ionic equation for this reaction, considering that the solid silver chloride is not ionized due to its precipitation:

$Ag^+(aq)+NO_3^-(aq)+Na^+(aq)+Cl^-(aq)\rightarrow AgCl(s)+Na^+(aq)+NO_3^-(aq)$

Whereas the nitrate and sodium ions are cancelled out for the aforementioned reason as they are the spectator ions, to obtain:

$Ag^+(aq)+Cl^-(aq)\rightarrow AgCl(s)$

Which is the required net ionic equation.

Best regards!

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A sample of the sugar d-ribose (C5H10O5) of mass 0.727 g was placed in a calorimeter and then ignited in the presence of excess

alexandr1967 [171]

Answer:

The internal energy of combustion of d-ribose = -2127 kJ/mol

The enthalpy of formation of d-ribose = -1269.65 kJ/mol

Explanation:

Step 1: Data given

Mass of d-ribose = 0.727 grams

The temperature rose by 0.910 K

In a separate experiment in the same calorimeter, the combustion of 0.825 g of benzoic acid, for which the internal energy of combustion is −3251 kJ mol−1, gave a temperature rise of 1.940 K.

Molar mass of benzoic acid = 122.12 g/mol

Step 2: Calculate ΔU for benzoic acid

The calorimeter is a constant-volume instrument so:

ΔU = q

ΔU = (0.825 g/ 122.12 g/mol) * (−3251 kJ /mol)

ΔU = -21.96 kJ

Step 3: Calculate ΔU for d-ribose

c = |q| / ΔT

⇒ with ΔT = 1.940 K

c = 21.96 kJ / 1.940 K

c = 11.32 kJ /K

For d-ribose: ΔU = -cΔT

ΔU = -11.32 kJ/K * 0.910 K

ΔU = - 10.3 kJ

Step 4: Calculate moles of d-ribose

moles ribose = 0.727 grams / 150.13 g/mol

moles ribose = 0.00484 moles

Step 5: Calculate the internal energy of combustion for d-ribose

ΔrU = ΔU / n

ΔrU = -10.3 kJ / 0.004842 moles

ΔrU = -2127 kJ/mol

Step 6: Calculate The enthalpy of formation of d-ribose

The combustion of ribose is:

C5H10O5(s) + 5O2(g) → 5CO2(g) + 5H20(l)

Since there is no change in the number of moles of gas, ΔrH = ΔU

For the combustion of ribose, we consider the following reactions:

5CO2(g) + 5H2O(l) → C5H10O5(s) +5O2(g) ΔH = -2127 kJ/mol

C(s) + O2(g) → CO2(g) ΔH = -393.5 kJ/mol

H2(g) + 1/2 O2(g) → H2O(l) ΔH = -285.83 kJ/mol

ΔH = 2127 kJ/mol + 5(-393.5 kJ/mol) + 5(-285.83 kJ/mol)

ΔH = 2127 kJ/mol - -1967.5 kJ/mol - 1429.15 kJ/mol

ΔH = -1269.65 kJ/mol

The internal energy of combustion of d-ribose = -2127 kJ/mol

The enthalpy of formation of d-ribose = -1269.65 kJ/mol

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Why does a change in tempurature indicate that a chemical reaction may have occured?

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Explanation:

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