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garik1379 [7]
3 years ago
7

Which is the correct net ionic equation for the reaction of aqueous ammonia with nitric acid? Which is the correct net ionic equ

ation for the reaction of aqueous ammonia with nitric acid?NH3(aq) + NO3−(aq) → NH2−(aq) + HNO3(aq)NH4+(aq) + H+(aq) → NH52+(aq)NH2−(aq) + H+(aq) → NH3(aq)NH4+(aq) + NO3−(aq) → NH4NO3(aq)NH3(aq) + H+(aq) → NH4+(aq)
Chemistry
1 answer:
Svet_ta [14]3 years ago
4 0

Answer:

The last option:

  • NH₃ (aq) +  H⁺ (aq)  →  NH₄⁺ (aq)

Explanation:

1) Word equation

  • Aqueous ammonia + nitric acid → aqueous ammonium nitrate

2) Chemical (molecular) equation

  • NH₃ (aq) + HNO₃ (aq)  → NH₄ NO₃

3) Ionization reactions

Write the dissociation of the soluble ionic compounds:

  • HNO₃ → H⁺ + NO₃⁻
  • NH₄ NO₃ → NH₄⁺ + NO₃⁻

4) Total ionic equation:

  • NH₃ (aq) +  H⁺ (aq) + NO₃⁻ (aq) →  NH₄⁺ (aq) + NO₃⁻ (aq)

5) Net ionic equation

You must cancel the spectator ions, which are those ions that are repeated in both reactant and product sides, i.e. NO₃⁻. They are name spectator because they do not participate (change) during the reaction.

  • NH₃ (aq) +  H⁺ (aq)  →  NH₄⁺ (aq)

And that is the last choice of the list.

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Law Incorporation [45]
According to research, yes sulphur can penetrate glass.
7 0
3 years ago
Which of the following is true?
Tomtit [17]

Answer:

a. A reaction in which the entropy of the system increases can be spontaneous only if it is endothermic.

Explanation:

The change in free energy (ΔG) that is, the <u>energy available to do work</u>, of a system for a constant-temperature process is:

ΔG = ΔH - TΔS

  • When ΔG < 0 the reaction is spontaneous in the forward direction.
  • When ΔG > 0 the reaction is nonspontaneous. The reaction is

spontaneous in the opposite direction.

  • When ΔG = 0 the system is at equilibrium.

If <u>both ΔH and ΔS are positive</u>, then ΔG will be negative only when the TΔS  term is greater in magnitude than ΔH. This condition is met when T is large.

3 0
3 years ago
What is the volume (in liters) of a 5.98 gram sample of O2 at STP?
borishaifa [10]

Answer:

4.186 L

Explanation:

Using the pv=nrt equation and converting the grams of O2 into mols. After finding the number of mols by dividing 5.98 by 32 (2*the atomic weight of O) you plug that into the equation. So then you have 1*V=.186875*.08206*273 then you rearrange the equation to solve for v and get 4.186 L

4 0
3 years ago
Consider the reaction 2CO * O2 —&gt; 2 CO2 what is the percent yield of carbon dioxide (MW= 44g/mol) of the reaction of 10g of c
Arturiano [62]

Answer:

Y = 62.5%

Explanation:

Hello there!

In this case, for the given chemical reaction whereby carbon dioxide is produced in excess oxygen, it is firstly necessary to calculate the theoretical yield of the former throughout the reacted 10 grams of carbon monoxide:

m_{CO_2}^{theoretical}=10gCO*\frac{1molCO}{28gCO}*\frac{2molCO_2}{2molCO}  *\frac{44gCO_2}{1molCO_2}\\\\ m_{CO_2}^{theoretical}=16gCO_2

Finally, given the actual yield of the CO2-product, we can calculate the percent yield as shown below:

Y=\frac{10g}{16g} *100\%\\\\Y=62.5\%

Best regards!

8 0
3 years ago
A 20.0 g piece of aluminum at 5.00 C is dropped into 20.2 g of water at 90.00 C. The final temperature is 75.00 C. Use the First
bekas [8.4K]

Answer:

The specific heat of aluminium is 0.906 J/g°C

Explanation:

Step 1: data given

Mass of aluminium = 20.0 grams

Temperature = 5.00 °C

Mass of water = 20.2 grams

Temperature of water = 90.00 °C

The final temperature = 75.00 °C

Specific heat of water = 4.184 J/g°C

Step 2: calculate the specific heat of aluminium

heat won = heat lost

Qaluminium = -Qwater

Q = m*c* ΔT

m(aluminium * c(aluminium) *ΔT(aluminium = -m(water) * c(water) *ΔT(water)

⇒with m(aluminium) = mass of aluminium = 20.0 grams

⇒with c(aluminium) = the specific heat of aluminium = TO BE DETERMINED

⇒with ΔT(aluminium) = the change of temperature = T2 - T1 = 75.00 °C - 5.00 °C = 70.00 °C

⇒with m(water) = the mass of water = 20.2 grams

⇒with c(water) = the specific heat of water = 4.184 J/g°C

⇒with ΔT(water) = T2 - T1 = 75.00°C - 90.00 °C = -15.00 °C

20.0 * c(aluminium) * 70.00 = -20.2 * 4.184 * -15.00

c(aluminium) = 0.906 J/g°C

The specific heat of aluminium is 0.906 J/g°C

7 0
3 years ago
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