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garri49 [273]
3 years ago
11

0.25 mol of nitric acid is used to make a 0.10 M nitric acid solution. How many grams of solute was used?

Chemistry
1 answer:
Yanka [14]3 years ago
7 0

Answer:

15.75 grams of HNO3 was used and dissolved in 2.5 liters of solvent, to make a 0.10 M solution

Explanation:

Step 1: Data given

Nitric acid = HNO3

Molar mass of H = 1.01 g/mol

Molar mass of N = 14.0 g/mol

Molar mass O = 16.0 g/mol

Number of moles nitric acid (HNO3) = 0.25 moles

Molairty = 0.10 M

Step 2: Calculate molar mass of nitric acid

Molar mass HNO3 = Molar mass H + molar mass N + molar mass (3*O)

Molar mass HNO3 = 1.01 + 14.0 + 3*16.0

Molar mass HNO3 = 63.01 g/mol

Step 3: Calculate mass of solute use

Mass HNO3 = moles HNO3 * molar mass HNO3

Mass HNO3 = 0.25 moles * 63.01 g/mol

Mass HNO3 = 15.75 grams

15.75 grams of HNO3 was used and dissolved in 2.5 liters of solvent, to make a 0.10 M solution

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Mass of One Water Molecule? The answer says the mass of one water molecule is 2.99 X 10^-23
zmey [24]
The mass of one mole of water it is 18 amu, but you need to find the mass of a molecule of water, therefore you calculate the mass of one mole of water, which is 18 amu and you divided by Avogadro's number which is 6,022 x 10^23. The result is 2,989 x 10^-23. Hope I helped you. If you have any questions ask :) Good luck.
6 0
3 years ago
A 110.0-mL sample of a solution that is 3.0×10−3 M in AgNO3 is mixed with a 230.0- mL sample of a solution that is 0.10 M in NaC
almond37 [142]

Answer:

[Ag⁺] = 0.0666M

Explanation:

For the addition of Ag⁺ and CN⁻, the (Ag(CN)₂⁻ is produced, thus:

Ag⁺ + 2CN⁻  ⇄  Ag(CN)₂⁻

Kf = 1x10²¹ = [Ag(CN)₂⁻] / [CN⁻]² [Ag⁺]

As initial concentrations of Ag⁺ and CN⁻ are:

[Ag⁺] = 0.110L × (3.0x10⁻³mol / L) = 3.3x10⁻⁴mol / (0.110L + 0.230L) = 9.7x10⁻⁴M

[CN⁻] = 0.230L × (0.1mol / L) = 0.023mol / (0.110L + 0.230L) = 0.0676M

The equilibrium concentrations of each compound are:

[CN⁻] = 9.7x10⁻⁴M - x

[Ag⁺] = 0.0676M - x

[Ag(CN)₂⁻] = x

<em>Where x is reaction coordinate</em>

Replacing in Kf formula:

1x10²¹ = [x] / [9.7x10⁻⁴M - x]² [0.0676M - x]

1x10²¹ = [x] / 6.36048×10⁻⁸ - 0.000132085 x + 0.06954 x² - x³

-1x10²¹x³ + 6.954x10¹⁹x² - 1.32085x10¹⁷ x + 6.36x10¹³ = x

-1x10²¹x³ + 6.954x10¹⁹x² - 1.32085x10¹⁷ x + 6.36x10¹³ = 0

Solving for x:

X = 9.614x10⁻⁴M

Thus, equilibrium concentration of Ag⁺ is:

[Ag⁺] = 0.0676M - 9.614x10⁻⁴M = <em>0.0666M</em>

6 0
2 years ago
How many Valance electrons are in an atom of phosphorous
densk [106]

Answer:

Five

Explanation:

Valance electrons are the outermost electrons in an atom. There are five electrons in the third layer of the phosphorus atom. So your answer is five valance electrons.

7 0
3 years ago
In this experiment, 0.070 g of caffeine is dissolved in 4.0 ml of water. The caffeine is then extracted from the aqueous solutio
Damm [24]

2.0ml of methylene chloride solution is used each time to extract caffeine from the aqueous solution.  

Consider the concentration of caffeine obtained during each individual extraction from the aqueous solution to be C.  

The total amount of caffeine obtained during each extraction is calculated as

(Total volume of water used to make up the caffeine aqueous solution) x (concentration of caffeine obtained during each individual extraction from the aqueous solution) + (Volume of methylene chloride added during each extraction x distribution coefficient of caffeine x concentration of caffeine obtained during each individual extraction from the aqueous solution)  


Substituting these values we get                                                            

The total amount of caffeine obtained during each extraction                

 = (4.0×C )+ (2.0×4.6 × C)                                                                              

= 13.2 C


The amount of caffeine remaining in the aqueous solution is calculated as  

(Total volume of water used to make up the caffeine aqueous solution) x (concentration of caffeine obtained during each individual extraction from the aqueous solution)


Substituting these values we get                                                            

The amount of caffeine remaining in the aqueous solution = 4 × C                                                                                            

The fraction of caffeine remaining in aqueous solution is calculated as  

= (The total amount of caffeine obtained during each extraction)/ (The amount of caffeine remaining in the aqueous solution)                    

=4.0 C/13.2 C                                                                                                

= 1/3.3.  

Therefore the fraction of caffeine left in aqueous solution after 3 extractions is =(1/3.3)^3  =0.028

Therefore, the total amount of caffeine extracted                            

=0.070 × (1-(1/3.3)^3)                                                                                      

= 0.068 g


5 0
3 years ago
Read 2 more answers
Only a small fraction of a weak acid ionizes in aqueous solution. What is the percent ionization of a 0.100-M solution of acetic
Mademuasel [1]

Answer:

1.33%

Explanation:

In an aqueous solution, a weak acid such as acetic acid, will be in equilibrium with its conjugate base, acetate ion, thus:

CH₃CO₂H(aq) + H₂O(l) ⇌ H₃O⁺(aq) + CH₃CO₂⁻(aq )

Where dissociation constant, ka, is defined as the ratio of concentrations of products and reactants:

Ka = 1.8x10⁻⁵ = [H₃O⁺] [CH₃CO₂⁻] / [CH₃CO₂H]

<em>H₂O is not taken into account in the equilibrium because is a pure liquid</em>

<em />

When a solution of acetic acid becomes to equilibrium, the original concentration of the acid decreases producing more H₃O⁺ and CH₃CO₂⁻.

The concentrations at equilibrium when a 0.100M solution of acetic acid reaches this state, is:

[CH₃CO₂H] = 0.100M - X

[H₃O⁺] = X

[CH₃CO₂⁻] = X

<em>Where X is reaction coordinate.</em>

Replacing in Ka expression:

1.8x10⁻⁵ = [H₃O⁺] [CH₃CO₂⁻] / [CH₃CO₂H]

1.8x10⁻⁵ = [X] [X] / [0.100M - X]

1.8x10⁻⁶ - 1.8x10⁻⁵X = X²

1.8x10⁻⁶ - 1.8x10⁻⁵X - X² = 0

Solving for X:

X = -0.00135 → False solution. There is no negative concentrations.

X = 0.00133 → Right solution.

That means concentration of acetate ion is:

[CH₃CO₂⁻] = 0.00133M.

Now, percent ionization is defined as 100 times the ratio between weak acid that is ionizated, [CH₃CO₂⁻] = 0.00133M, per initial concentration of the acid, [CH₃CO₂H] = 0.100M. Replacing:

% Ionization = 0.00133M / 0.100M × 100 =

<h3>1.33%</h3>

<em />

<em />

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4 0
2 years ago
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