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Lapatulllka [165]
4 years ago
10

What is the mass of one mole of nitrogen atoms?

Chemistry
2 answers:
OLEGan [10]4 years ago
4 0

Answer:

28.0

Explanation:

Effectus [21]4 years ago
3 0

Answer:

14.01g

Explanation:

This depends on what periodic table you are using however if you use the official one used on the AP exam, it is 14.01g

Stoichiometry

1 mol N*14.01g/1 mol N

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Given the NaOH density =1.0698g mL^-1, How many mL of .71M HCl would be required to neutralize 11.33g of 2.03 M NaOH?
SSSSS [86.1K]

 The volume in ml of 0.71 M HCl  that would  be  required to neutralize  11.33 g   of 2.03 M NaOH  is 30.3 ml

<u><em>calculation</em></u>

Step 1 : write equation  for reaction

NaOH + HCl →NaCl +H₂O

Step 2:find the volume of NaOH

volume=mass/ density

= 11.33 g/ 1.0698 g/ml  =10.59 ml

Step 3: find the  moles of NaOH

moles  = molarity  x volume  in liters

molarity= 2.03 M=2.03 mol/l

volume  in liters =10.59/1000 =0.0106 L

moles  = 2.03 M  x 0.0106 L =0.0215  moles

step 4:  use the mole ratio to determine the moles of HCl

from  equation in step 1 , NaOH:HCl  is 1:1 therefore the moles of HCl is also 0.0215  moles

Step 5: find volume of  HCl

volume= moles/ molarity

molarity  =0.71 M =0.71 mol/l

=0.0215 moles /0.71  mol/l=0.0303 L

into ml =0.0303 x 1000=30.3 ml

3 0
3 years ago
Account for the change when NO2Cl is added using the reaction quotient Qc. Match the words in the left column to the appropriate
ryzh [129]

Answer:

A. Disturbing the equilibrium by adding NO2Cl decreases Qc to a value less than Kc.

B. To reach a new state of equilibrium, Qc therefore increases which means that the denominator of the expression for Qc decreases.

C. To accomplish this, the concentration of reagents decreases, and the concentration of products increases.

Explanation:

Hello,

In this case, for the equilibrium reaction:

NO_2Cl(g)+NO(g)\rightleftharpoons NOCl(g)+NO_2(g)

Whose equilibrium expression is:

Kc=\frac{[NO_2][NOCl]}{[NO_2Cl][NO]}

The proper matching is:

A. Disturbing the equilibrium by adding NO2Cl decreases Qc to a value less than Kc, since the denominator becomes greater, therefore, Qc decreases.

B. To reach a new state of equilibrium, Qc therefore increases which means that the denominator of the expression for Qc decreases, since the lower the denominator, the higher Qc as it has the concentration of reactants.

C. To accomplish this, the concentration of reagents decreases, and the concentration of products increases, since the reactants must be consumed in order to reestablish equilibrium by shifting the reaction towards the products.

Best regards.

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