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inna [77]
3 years ago
8

5f2 (g) + 2nh3 (g) → n2f4 (g) + 6hf (g) if you have 58.5g nh3, how many grams of f2 are required for a complete reaction? how ma

ny grams of nh3 are required to produce 3.89g hf? how many grams of n2f4 can be produced from 217g f2?
Chemistry
1 answer:
xxTIMURxx [149]3 years ago
6 0
5F2 +2NH3 = N2f4 + 6HF

moles = mass/molar mass
mass =molar mass x mole

a)   moles of Nh3 = 58.5/17 = 3.44 moles
the mole  ratio of F2:NH3 = 5:2
the moles of F2 =3.44x5/2=8.6 moles
mass= 8.6 x 38 = 326.8 grams  of F2

B) moles of Hf = 3.89/20 = 0.1945 moles
mole ratio NH3:Hf =  2:6
moles of  NH3 = 0.1945 x2/6 = 0.0648 moles
mass =0.0648 x17 = 1.102 grams of NH3

C) moles of f2 = 217/38 =5.711 moles
mole ratio N2F6:F2 = 5:1
moles of N2F4 =5.711 x1/5 =1.142 moles
mass = 1.142 x104 = 118.77 grams of N2F4
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3 0
3 years ago
Calculate [H3O+] and [OH−] for each of the following solutions at 25 ∘C given the pH.<br> pH= 2.89
dimaraw [331]

Answer: The value of [H_{3}O^{+}] is 0.0012 M and [OH^{-}] is 1.02 \times 10^{-14}.

Explanation:

pH is the negative logarithm of concentration of hydrogen ion.

It is given that pH is 2.89. So, the value of concentration of hydrogen ions is calculated as follows.

pH = - log [H^{+}]\\2.89 = - log [H^{+}]\\conc. H^{+} = 0.0012 M

The relation between pH and pOH value is as follows.

pH + pOH = 14

0.0012 + pOH = 14

pOH = 14 - 0.0012 = 13.99

Now, pOH is the negative logarithm of concentration of hydroxide ions.

Hence, [OH^{-}] is calculated as follows.

pOH = - log [OH^{-}]\\13.99 = - log [OH^{-}]\\conc. OH^{-} = 1.02 \times 10^{-14} M

Thus, we can conclude that the value of [H_{3}O^{+}] is 0.0012 M and [OH^{-}] is 1.02 \times 10^{-14}.

6 0
3 years ago
Name three properties of solids that are different from those of liquids. Explain the differences for each.
olga55 [171]
1) Solids have a fixed shape
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3) Solids may fracture
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4 0
3 years ago
What is the ph of a buffer consisting of 0.200 m hc2h3o2 and 0.200 m kc2h3o2? the k a for hc2h3o2 is 1.8×10−5. view available hi
AURORKA [14]
PH = pKa + log \frac{[base]}{[Acid]}
Acid  is HC₂H₃O₂ and conjugate base is KC₂H₃O₂,
pKa = - log Ka = - log (1.8 x 10⁻⁵) = 4.74
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6 0
3 years ago
A concentration cell is built based on the following half reactions by using two pieces of zinc as electrodes, two Zn2+ solution
Lana71 [14]

Explanation:

The given reaction at cathode will be as follows.

At cathode: Zn^{2+} + 2e^{-} \rightarrow Zn,     E_{o} = -0.761 V

At anode: Zn \rightarrow Zn^{2+} + 2e^{-},       E_{o} = 0.761

Therefore, net reaction equation will be as follows.

                 Zn^{2+} + Zn \rightarrow Zn + Zn^{2+}

Initial:     0.129         -            -       0.427

Change:  -0.047      -            -     -0.047

Equilibrium: (0.129 - 0.047)      (0.427 - 0.047)

                = 0.082                       = 0.38

As E^{o}_{cell} for the given reaction is zero.

Hence, equation for calculating new cell potential will be as follows.

                   E_{cell} = E^{o}_{cell} - \frac{RT}{nF} ln \frac{[Zn^{2+}]_{products}}{[Zn^{2+}]_{reactants}}

                              = 0 - \frac{8.314 atm L/mol K \times 291 K}{2 \times 96500} ln \frac{0.38}{0.082}

                              = 0.019

Thus, we can conclude that the cell potential of the given cell is 0.019.

3 0
4 years ago
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