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____ [38]
3 years ago
8

When a solution of 0.1 M Mg(NO3)2 was mixed with a limited amount of aqueous ammonia, a light white, wispy solid was observed, i

ndicating a reaction took place. Write the net ionic equation for this reaction.
Chemistry
1 answer:
Ipatiy [6.2K]3 years ago
8 0

<u>Answer:</u> The net ionic equation is written below.

<u>Explanation:</u>

Net ionic equation of any reaction does not include any spectator ions.

Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form.

The chemical equation for the reaction of magnesium nitrate and aqueous ammonia (ammonium hydroxide) is given as:

Mg(NO_3)_2(aq.)+2NH_4OH(aq.)\rightarrow Mg(OH)_2(s)+2NH_4NO_3(aq.)

A white precipitate of magnesium hydroxide is formed in the above reaction.

Ionic form of the above equation follows:

Mg^{2+}(aq.)+2NO_3^-(aq.)+2NH_4^+(aq.)+2OH^-(aq.)\rightarrow Mg(OH)_2(s)+2NH_4^+(aq.)+2NO_3^-(aq.)

As, ammonium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

Mg^{2+}(aq.)+2OH^-(aq.)\rightarrow Mg(OH)_2(s)

Hence, the net ionic equation is written above.

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5 0
3 years ago
To find time, t, the age of the universe, you must first isolate the variable t. Rewrite the velocity equation v = d / t so that
Alex17521 [72]

Answer:

t = d/v

Explanation:

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3 years ago
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san4es73 [151]

Answer:

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6 0
3 years ago
Read 2 more answers
An experiment shows that a 250 ?ml gas sample has a mass of 0.430 g at a pressure of 736 mmhg and a temperature of 28 ?c.
dmitriy555 [2]
What we're looking for here is the gas sample's molar mass given its mass, pressure, volume, and temperature. Recalling the gas law, we have

PV = nRT or
n = \frac{PV}{RT}

where R is <span>0.08206 L atm / mol K, P is the given pressure, T is the temperature, and V is the volume.

Before applying the values given, it is important to make sure that they are to be converted to have consistent units with that of R. 
</span>
Thus, we have

P = 736/ 729 = 0.968 atm
T = 28 + 273.15 = 301.15 K
V = 250/1000 = 0.250 L

Now, applying these converted values into the gas law, we have

n = \frac{(0.968 atm)(0.250 L)}{(0.08206 L.atm/mol.K)(301.15 K)}
n = 0.00979 moles

Given that the mass of the sample is 0.430 g, we have

molar mass = \frac{mass}{number of moles} 
molar mass = \frac{0.430}{0.00979} = 43.9

Thus, the gas sample has a molar mass of 43.9 g/mol.
4 0
3 years ago
Calculate the theoretical yield of ammonia produced by the reaction of 100g of H2 gas and 200g of N2 gas
ipn [44]
To get the theoretical yield of ammonia NH3:
first, we should have the balanced equation of the reaction:
3H2(g) + N2(g) → 2NH3(g)
Second, we start to convert mass to moles
moles of N2 = N2 mass / N2 molar mass
                     = 200 / 28 = 7.14 moles
third, we start to compare the molar ratio from the balanced equation between N2 & NH3 we will find that N2: NH3 = 1:2 so when we use every mole of N2 we will get 2 times of that mole of NH3 so,
moles of NH3 = 7.14 * 2 = 14.28 moles 
finally, we convert the moles of NH3 to mass again to get the mass of ammonia:
mass of NH3 = no.moles * molar mass of ammonia
                      = 14.28 * 17 = 242.76 g
6 0
3 years ago
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