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Blababa [14]
3 years ago
13

Determine the molar mass for ammonia (NH3) in g/mol.

Chemistry
1 answer:
navik [9.2K]3 years ago
5 0

Answer:

17.031 g/mol

Explanation:

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Calculate the answer. Express it in scientific notation and include the correct number of significant figures. (12 x 104 ) x (5
ipn [44]

Answer:

Explanation:

(12 x 104 ) x (5 x 10-²) = 6 x 10 ³ 6 x 105 6.0x10²

1.  (12 x 104 ) = 1248 or 1.248 x 10^3

2.  (1.248 x 10^3)(5 x 10^-2) = 6.240 x 10^1 or 60 rounded to 1 sig fig

I don't understand "= 6 x 10 ³ 6 x 105 6.0x10² "

6 0
1 year ago
Please I need this help me​
BaLLatris [955]

Answer:

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3 0
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A Carnot cycle operates between the temperatures limits of 400 K and 1600 K, and produces 3600 kW of net power. The rate of entr
TiliK225 [7]

The rate of entropy change:

The rate of entropy change of the working fluid during the heat addition process is 3 kW/K

What is the Carnot cycle?

  • The Carnot Cycle is a thermodynamic cycle made up of reversible isothermal expansion, adiabatic expansion, isothermal compression, and adiabatic compression processes in succession.
  • The ratio of the heat absorbed to the temperature at which the heat was absorbed determines the change in entropy.

The entropy of a system:

The rate of heat addition is expressed as,

Q = \frac{WT_{H}}{T_{H}- T_{L}}

The entropy of a system is a measure of how disorderly a system is getting. The rate of entropy generation during heat addition is,

S_{gen} = \frac{Q}{T_{H}} = \frac{W}{T_{H} - T_{L}}

Calculation:

<u>Given:</u>

T_{L} = 400K

T_{H} = 1600K

W = 3600 kW

Put all the values in the above equation, and we get,

S_{gen} = \frac{W}{T_{H} - T_{L}} = \frac{3600}{1600-400} = 3 kW/K

The rate of entropy change is 3 kW/K

Learn more about the Carnot cycle here,

brainly.com/question/13002075

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2 years ago
Which formula contains two non- metals?<br><br>BaO<br>NaBr<br>KI<br>Si O2​
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3 years ago
: Please predict products, label the states and write ionic and net ionic equations for each. These are all double replacement r
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Unmmmm i’m not quite understanding brhajsnsbfbfhrndnsnsjsmsbgbtieoou that but yes
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