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max2010maxim [7]
3 years ago
9

How many grams of nitrogen gas are needed to produce 34 g of ammonia

Physics
1 answer:
scZoUnD [109]3 years ago
7 0

Answer:

28 grams

Explanation:

The equation for the reaction is

3H(2) + N(2) -> 2NH(3)

Then we have.

The molar mass, M of ammonia is 17 g/mol.

34 grams of ammonia, NH3 then would be

34 g / 17 g/mol

= 2 moles

2 moles of ammonia will be obtained from

(2 * 1) / 2

​= 1 mole of nitrogen

The molar masses of nitrogen is 28 g/mol

2 moles of nitrogen corresponds to 1 * 28 = 28 grams.

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Answer:

True, overtone occurs when two or more sounds are produced at the same time.

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The formation of condensation on a glass of ice water causes the ice to melt faster than it would otherwise. If 8.55 g of conden
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Answer:

m = 62.14 g

Explanation:

Energy used to melt the ice is the energy released by the condensation of the water forms on the glass

so here we have

energy for the condensation of water is given as

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now the energy of vaporization is given as

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3 years ago
a liquid with a specific heat of 1900J/Kg° C has 4750 joules of heat energy is added to it. if the temperature increases from 20
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E=mcθ
where E is the energy added,m is the mass,c is the specfic heat capacity and θ is the change in temprature.
making m subject u get E/c<span>θ=m
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plugging the values  we get :
</span>\frac{4750}{1900*10}
<span>
solving it we get the answer that is 0.25kg or 250 grams

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Which is a front in which cold air is replacing warm air at the surface?
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In a simple circuit a 6-volt dry cell pushes charge through a single lamp which has a resistance of 3 Ω. According to Ohms law t
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Answer:

The answers and workings is in the Explanation section

Explanation:

<em>In a simple circuit a 6-volt dry cell pushes charge through a single lamp which has a resistance of 3 Ω. According to Ohms law the current through the circuit is _____________ Amps. </em>

According to Ohm’s law V =I*R  

Where V = Voltage, I = Current and R = Resistance

I = V/R =6/3 =2 Amps of current

Answer = 2 Amps

<em> If a second identical lamp is connected in series, the 6-volt battery must push a charge through a total resistance of ________ Ω</em>.  <em>The current in the circuit is then __________ Amps. </em>

Since the second lamp is connected in series with the first, the total resistance will R₂ = 3 + 3 = 6Ω

2 resistance in series  R₂ =  6Ω

The current in the circuit with the two lamps connected in series is I₂ =V/R₂ =6/6 = 1 Amps

The current is 1 Amps

Answer =  6Ω  and 1 Amps

<em>If a third identical lamp is connected in series, the total resistance is now _________Ω.  The current through all three lamps in series is now _________ Amps.  </em>

Since the third lamp is connected in series with the first and second, the total resistance will R₃ = 3 + 3 + 3 = 9Ω

total resistance of the 3 lamps R₃ =  9Ω

The current in the circuit with the three lamps connected in series is

I =V/R₃ =6/9 = 0.67 Amps

The current through the 3 lamps I₃ = 0.67 Amps

Answer =  9Ω  and 0.67 Amps

<em>The current through each individual lamp is __________ Amps.  </em>

Since all 3 lamps are connected in series, the same current will flow through each of the  3 lamps, and that current is I₃  

The current through each individual lamp is 0.67 Amp

Answer = 0.67 Amp

<em>What is the power when a voltage of 120 volts drives a current of 3 amps through a device?  </em>

The formula for power P = I*V =120*3 = 360 Watts

power P  = 360 Watts

Answer = 360 Watts

<em>What is the current when a 90-W light bulb is connected to 120 V? </em>

From P =I*V, make I the subject of the formula, I = P/V =90/120 = 0.75

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Answer =  0.75 Amps

<em>How much current does a 75-W light bulb draw when connected to 120 V?</em>  

Current I =P/V = 75/120 = 0.625 Amps.

Answer = 0.625 Amps

<em>If part of an electric circuit dissipates energy at 6 W when it draws a current of 3 A, what voltage is impressed across it? </em>

Voltage V =P/I =6/3 =2 Volts

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<em> If a 60 W light bulb at 120 V is left on in your house to prevent burglary, and the power company charges 10 cents per kilowatt-hour, how much will it cost to leave the bulb on for 30 days? Show your work. </em>

24 hours make 1 day, so the number of hours the bulb was left on = 24 *30 = 720 hours

The power rating of the bulb is 60W = 60/1000 = 0.06 KiloWatt

Total power consumed in kilowatt-hour = 0.06 * 720 = 43.2 kilowatt-hour

Cost for 30 days = 0.1*43.2 = $4.32 ( note that 10 Cents = $0.1)

Answer =  $4.32

4 0
3 years ago
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