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sergey [27]
3 years ago
12

How many moles are in 1.2x10^3 grams of ammonia, NH3?

Chemistry
1 answer:
KonstantinChe [14]3 years ago
6 0
I don't have a calculator with me right now, but that mass would be 1200 grams. Divide the given amount of grams by the molar mass of NH3, which is 17.031g/mol. (Nitrogen + 3(hydrogen)). Again, sorry I didn't have a calculator. But all you would need to do is divide 1200 by 17.031. If you need to use sig figs, your answer should have 2 because the 1.2 x 10^3 limits your amount of sig figs.
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28. Which set of coefficients will balance this chemical equation?
Makovka662 [10]
2021 hoped I helped you god bless you
5 0
3 years ago
Lead(II) nitrate and ammonium iodide react to form lead(II) iodide and ammonium nitrate according to the reaction Pb(NO3)2(aq)+2
Setler79 [48]

Answer:

a) volume of ammonium iodide required =349 mL

b) the moles of lead iodide formed = 0.0436 mol

Explanation:

The reaction is:

Pb(NO_{3})_{2}+2NH_{4}I -->PbI_{2}+2NH_{4}NO_{3}

It shows that one mole of lead nitrate will react with two moles of ammonium iodide to give one mole of lead iodide.

Let us calculate the moles of lead nitrate taken in the solution.

Moles=molarityX volume (L)

Moles of lead nitrate = 0.360 X 0.121 =0.0436 mol

the moles of ammonium iodide required = 2 X0.0436 = 0.0872 mol

The volume of ammonium iodide required will be:

volume=\frac{moles}{molarity}=\frac{0.0872}{0.250}=0.349L=349mL

the moles of lead iodide formed = moles of lead nitrate taken = 0.0436 mol

7 0
3 years ago
Consider the reaction of A(g) + B(g) + C(g) => D(g) for whichthe following data were obtained:
AveGali [126]

Answer:

rate= k[A]²[B]²[C]

Explanation:

When concentration of A is increased two times ,keeping other's concentration constant , rate of reaction becomes 4 times .

So rate is proportional to [A]²

When concentration of B is increased two times , keeping other's concentration constant,rate of reaction becomes 4 times.

So rate is proportional to [B]²

When concentration of C is increased two times , keeping other's concentration constant, rate of reaction becomes 2 times.

So rate is proportional to [C]

So rate= k[A]²[B]²[C]

5 0
3 years ago
Suppose that in an ionic compound, "m" represents a metal that could form more than one type of ion. in the formula mf2 , the ch
geniusboy [140]
F (Fluorine) is in column (group/family) VIIA, or the "halogens". When you see the halogens (Fluorine, Chlorine, Bromine, and Iodine) in combination with a metal, each halogen atom present will carry a -1 charge. We can see that the atom has no charge, so the metal must cancel out the negative charges brought by the two fluorine atoms.
(Charge on m) + 2*(charge on fluorine) = 0
(Charge on m) + 2*(-1) = 0
(Charge on m) - 2 = 0
Charge on m ion = +2
3 0
3 years ago
A galvanic (voltaic) cell consists of an electrode composed of nickel in a 1.0 M nickel(II) ion solution and another electrode c
Andre45 [30]

<u>Answer:</u> The standard potential of the cell is 0.77 V

<u>Explanation:</u>

We know that:

E^o_{Ni^{2+}/Ni}=-0.25V\\E^o_{Cu^{+}/Cu}=0.52V

The substance having highest positive E^o reduction potential will always get reduced and will undergo reduction reaction.

The half reaction follows:

<u>Oxidation half reaction:</u> Ni(s)\rightarrow Ni^{2+}(aq)+2e^-

<u>Reduction half reaction:</u> Cu^{+}(aq)+e^-\rightarrow Cu(s)       ( × 2)

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

Putting values in above equation follows:

E^o_{cell}=0.52-(-0.25)=0.77V

Hence, the standard potential of the cell is 0.77 V

5 0
3 years ago
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