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3 years ago

Which equation represents the correct net ionic equation for the reaction between Ca(OH)2 and H2SO4?

Chemistry

1 answer:

Georgia [21]3 years ago

7 0

Answer: The correct net ionic equation for the reaction is $2H^{+}(aq)+2OH^{-}(aq)\rightarrow 2H_2O(l)$

Explanation:

Net ionic equation is defined as the equations in which spectator ions are not included.

Spectator ions are the ones that are present equally on the reactant and product sides. They do not participate in the reaction.

The balanced molecular equation is:

$Ca(OH)_2(aq)+H_2SO_4(aq)\rightarrow CaSO_4(aq)+2H_2O(l)$

The complete ionic equation follows:

$Ca^{2+}(aq)+2OH^-(aq)+2H^+(aq)+SO_4^{2-}(aq)\rightarrow 2Ca^{2+}(aq)+SO_4^{2-}(aq)+H_2O(l)$

As calcium and sulfate ions are present on both sides of the reaction. Thus, they are considered spectator ions.

The net ionic equation follows:

$2H^{+}(aq)+2OH^{-}(aq)\rightarrow 2H_2O(l)$

Hence, the correct net ionic equation for the reaction is $2H^{+}(aq)+2OH^{-}(aq)\rightarrow 2H_2O(l)$

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A sample of gas contains 6.25 × 10-3 mol in a 500.0 mL flask at 265°C. What is the pressure of the gas in kilopascals? Which var

RideAnS [48]

55.9 kPa; Variables given = volume (V), moles (n), temperature (T)

We must calculate p from V, n, and T, so we use the Ideal Gas Law:

pV = nRT

Solve for p: p = nRT/V

R = 8.314 kPa.L.K^(-1).mol^(-1)

T = (265 + 273.15) K = 538.15 K

V = 500.0 mL = 0.5000 L

∴ p = [6.25 x 10^(-3) mol x 8.314 kPa·L·K^(-1)·mol^(-1) x 538.15 K]/(0.5000 L) = 55.9 kPa

4 0

3 years ago

Read 2 more answers

One of the compounds used to increase the octane rating of gasoline is toluene (pictured). Suppose 43.3 mL of toluene (d = 0.867

Thepotemich [5.8K]

Answer:

(A)

Density = Mass / Volume

Mass = Density × Volume

$= 0.867 g/mL \times 43.3mL = 37.5411 g Toluene$

$1C_6 H_5 CH_3 + 9 O_2 > 7 CO_2 + 4 H_2 O$

Mole ratio of toluene : Oxygen is 1 : 9

$37.5411 g \text { Toluene } \times \frac{1 \text {mol} \text {toluene}}{92 g \text { toluene}} \times \frac{9 {mol} O_{2}}{1 \text {mol} \text { toluene }} \times \frac{32 g O_{2}}{1 {mol} O_{2}}=117 g O_{2}(\text {Answer})$

(B)

1 mole of Toluene produces 7 moles of $CO_2$ gas and 4 moles of $H_2 O$ Vapour

So the mole ratio is 1 : 11

$37.5411 g Toluene $\times \frac{1 \text { mol toluene }}{92 g \text { toluene }} \times \frac{11 \mathrm{mol} \text { gas }}{1 \text { mol toluene }} $$\\\\=4.49 \text { mol gaseous products (Answer) } $$

(C)

1mole contains $6.022\times10^{23}$ molecules

$37.5411 g Toluene $\times \frac{1 \text { mol toluene }}{92 g \text { toluene}} \times \frac{4 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}}{1 \mathrm{mol} \text { toluene }} \times \frac{6.022 \times 10^{23} \text { molecules } \mathrm{H}_{2} \mathrm{O}}{1 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}} $\\\\$=9.82 \times 10^{23} \text { molecules } \mathrm{H}_{2} \mathrm{O} \text { (Answer) } $$