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nydimaria [60]
3 years ago
7

If the volume occupied by 0.500 mol of nitrogen gas at 0°C is 11.2 L, then the volume occupied by 2.00 mol of nitrogen gas at th

e same temperature and pressure will be:
A. the data given is not sufficient to determine the volume
B. 22.4 L
C. 44.8 L
D. the same as that occupied by 0.500 mol of nitrogen gas
Chemistry
1 answer:
kiruha [24]3 years ago
8 0
The  volume  occupied  by 2.00  moles of nitrogen  gas at  the same  temperature  and pressure  will be

 0.500  moles = 11.2 Liters
 what about 2 moles =? liters

by cross  multiplication

= 11.2 liters  x   2moles/ 0.500 moles  =  44.8  liters 
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In this nuclear reaction, a neutron gets converted into proton and electron. The electron released is the beta-minus particle.

The charge on these particles is -1 and mass of these particles is negligible.

5 0
3 years ago
Read 2 more answers
What is the boiling point of a solution produced by adding 610 g of cane sugar (molar mass 342.3 g/mol) to 1.4 kg of water? For
wel

Answer:

Boiling point of solution is 100.65^{0}\textrm{C}

Explanation:

Cane sugar is a non-volatile solute.

According to Raoult's law for a non-volatile solute dissolved in a solution-

                              \Delta T_{b}=K_{b}.m

Where, \Delta T_{b} is elivation in boiling point of solution, K_{b} is ebbulioscopic constant of solvent (how much temperature is raised for dissolution of 1 mol of non-volatile solute) and m is molality of solution.

Here, K_{b}=0.51^{0}\textrm{C}.kg.mol^{-1}

610 g of cane sugar = \frac{610}{342.3} moles of cane sugar

                                  = 1.78 moles of cane sugar

So, molality of solution (m) = \frac{1.78}{1.4}mol.kg^{-1}=1.27mol.kg^{-1}

Plug in all the values in the above equation, we get-

\Delta T_{b}=0.51^{0}\textrm{C}.kg.mol^{-1}\times 1.27mol.kg^{-1}=0.65^{0}\textrm{C}

So, boiling point of solution = (100+0.65)^{0}\textrm{C}=100.65^{0}\textrm{C}              

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3 years ago
Help science!! *10 points*
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Answer:

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What is the difference aldehyde and ketone in organic chemistry​
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2 years ago
How many grams of ammonia are produced when 1.0 mole of nitrogen reacts
Varvara68 [4.7K]

Answer: N2(g) + 3H2-> 2NH3(g)  This is the balanced equation

Note the mole ratio between N2, H2 and NH3.  It is 1 : 3 : 2  This will be important.

 

moles N2 present = 28.0 g N2 x 1 mole N2/28 g = 1 mole N2 present

moles H2 present = 25.0 g H2 x 1 mole H2/2 g = 12.5 moles H2 present

Based on mole ratio, N2 is limiting in this situation because there is more than enough H2 but not enough N2.

 

moles NH3 that can be produced = 1 mole N2 x 2 moles NH3/mole N2 = 2 moles NH3 can be produced  

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NOTE:  The key to this problem is recognizing that N2 is limiting, and therefore limits how much NH3 can be produced.

Explanation: here you go!! good luck! hope this helped

7 0
3 years ago
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