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alexandr402 [8]
3 years ago
8

what is the maximum number of moles of h2o that can be produced when 2.0 moles of nh3 are completly reacted. Formula: 4NH3+5O2--

>4NO+6H2O

Chemistry
2 answers:
kondaur [170]3 years ago
5 0
4 mole NH3 forms 6 moles of water, so 2 mole NH3 will form 3 moles of water

Degger [83]3 years ago
3 0
I hope it cleared your doubt and ALL THE BEST!

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Rus_ich [418]

This is true because as thermal energy increases, molecules move faster and liquids are turned into gas.

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Learn more: brainly.com/question/18124975

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A loop of wire turns between two permanent magnets in a generator. What is
lorasvet [3.4K]

Answer:

a current flows through the loop of wire.

Explanation: my answers were worded diffrently

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3 years ago
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What are the main components in dna
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Adenine , guanine, cytosine and thymine
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2 years ago
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g A 500. mL solution contains 0.665 M NaC2H3O2 and 0.475 M HC2H3O2. What mass of HCl in grams needs to be added for the solution
77julia77 [94]

Answer:

7.38g HCl

Explanation:

Using H-H equation for acetic buffer:

pH = pKa + log [NaC2H3O2] / [HC2H3O2]

<em>Where pKa is -log Ka = 4.74 and [] could be taken as moles of each compound.</em>

The initial moles of each specie is:

[NaC2H3O2]:

0.500L * (0.665mol/L) = 0.3325moles

[HC2H3O2]:

0.500L * (0.475mol/L) = 0.2375 moles

That means total moles are:

[NaC2H3O2] + [HC2H3O2] = 0.57 moles <em>(1)</em>

And solving H-H equation for a pH of 4.21:

4.21 = 4.74 + log [NaC2H3O2] / [HC2H3O2]

0.29512 = [NaC2H3O2] / [HC2H3O2] <em>(2)</em>

Replacing (1) in (2):

0.29512 = 0.57mol - [HC2H3O2] / [HC2H3O2]

0.29512 [HC2H3O2] = 0.57mol - [HC2H3O2]

1.29512 [HC2H3O2] = 0.57mol

[HC2H3O2] = 0.44 moles

The HCl reacts with NaC2H3O2 producing HC2H3O2, that means you need to add:

0.44 moles - 0.2375 moles =

0.2025 moles of HCl

Using molar mass of HCl (36.45g/mol), to convert these moles to grams:

0.2025 moles * (36.45g/mol) =

<h3>7.38g HCl</h3>

8 0
3 years ago
Use the given half reactions to "construct" an electrolytic cell. Zn^2+ + 2 e^---------&gt;Zn E°cell = -0.76 V Cu^2+ + 2 e^-----
Neko [114]

Answer:

See explanation and image attached

Explanation:

The standard cell potential at 298 K is given by;

E°cathode - E°anode

Hence;

E°cell = 0.34 V - (-0.76 V)

E°cell = 0.34 V + 0.76 V

E°cell = 1.1 V

To reduce Zn^2+ to Zn then Zn must be the cathode, hence;

E°cell = (-0.76 V) - 0.34 V

E°cell = -1.1 V

5 0
3 years ago
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