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kumpel [21]
3 years ago
6

When aqueous solutions of sodium cyanide and nitric acid are mixed, an aqueous solution of sodium nitrate and hydrocyanic acid r

esults. Write the net ionic equation for the reaction.
Chemistry
2 answers:
deff fn [24]3 years ago
8 0

Answer:

There is no net ionic equation

Explanation:

Step 1: Data given

sodium cyanide = NaCN

nitric acid = HNO3

sodium nitrate = NaNO3

hydrocyanic acid = HCN

Step 2: The unbalanced equation

NaCN(aq) + HNO3(aq) → NaNO3(aq) + HCN(aq)

The equation is already balanced.

Step 3: the net ionic equation

The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will , after canceling those spectator ions in both side, look like this:

Na+ + CN- + H+ NO3-   → Na+ + NO3- + H+ + CN-

There is no net ionic equation

If you don't have a net ionic equation to balance, that means you have what is usually called a molecular equation (or a complete molecular equation)

Viktor [21]3 years ago
6 0

Answer:

There is no net ionic equation

Explanation:

We identify the reactants and the products for the reaction

NaCN, HNO₃, NaNO₃, HCN

In this case, they are all soluble in water (aq)

We write the reaction, by the ionic form

Na⁺(aq) + CN⁻ (aq) + H⁺ (aq) + NO₃⁻(aq)  → Na⁺(aq) + NO₃⁻(aq)  + H⁺(aq) + CN⁻(aq)

All of the ions are spectators, they all react so there is no net ionic equation

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3 years ago
A scientist mixed two chemicals, observed the results, and concluded that an endothermic reaction took place. Which of the follo
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8 0
3 years ago
Given equilibrium partial pressures of PNO2= 0.247 atm, PNO = 0.0022atm, and PO2 = 0.0011 atm calculate the equilibrium constant
maxonik [38]
Answer 1:
Equilibrium constant (K) mathematically expressed as the ratio of the concentration of products to concentration of reactant. In case of gaseous system, partial pressure is used, instead to concentration.

In present case, following reaction is involved:

                        2NO2    ↔      2NO + O2

Here, K = \frac{[PNO]^2[O2]}{[PNO2]^2}

Given: At equilibrium, <span>PNO2= 0.247 atm, PNO = 0.0022atm, and PO2 = 0.0011 atm
</span>
Hence,  K = \frac{[0.0022]^2[0.0011]}{[0.247]^2}
                 = 8.727 X 10^-8

Thus, equilibrium constant of reaction = 8.727 X 10^-8
.......................................................................................................................

Answer 2:
Given: <span>PNO2= 0.192 atm, PNO = 0.021 atm, and PO2 = 0.037 atm.

Therefore, Reaction quotient = </span>\frac{[PNO]^2[O2]}{[PNO2]^2}
                                              = \frac{[0.021]^2[0.037]}{[0.192]^2}
                                              = 4.426 X 10^-4.

Here, Reaction quotient > Equilibrium constant.

Hence, <span>the reaction need to go to reverse direction to reattain equilibrium </span>
5 0
3 years ago
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