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kolbaska11 [484]
3 years ago
10

Suppose there is 0.63 g of HNO3 per 100 mL of a particular solution. What is the concentration of the HNO3 solution in moles per

liter?
Chemistry
1 answer:
Vadim26 [7]3 years ago
4 0

Answer:

There are 0.09996826 moles per liter of the solution.

Explanation:

Molar mass of HNO3: 63.02

Convert grams to moles

0.63 grams/ 63.02= 0.009996826

Convert mL to L and place under moles (mol/L)

100mL=0.1 L

0.009996826/0.1= 0.09996826 mol/L

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Which of these molecules is nonpolar?<br> O A. PF 3 <br>O B. 02 <br>O C. CO <br>O D. CHCI​
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3 years ago
A 10.0 g pat of butter raise the water level in a graduated cylinder by 11.6 mL. What is the density of the butter
Nina [5.8K]

- formula for density is mass divided by volume

therefore density of butter = 10.0g divided by 11.6ml = 0.8620689 g/cm³ ≈ 0.862 g/cm³ (3sf)

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3 years ago
What is the ph of a 0.25 m solution of c6h5nh2 given that its kb is 1.8 x 10-6?
IrinaK [193]

The pH a 0.25 m solution of C₆H₅NH₂ is equal to 3.13.

<h3>How do we calculate pH of weak base?</h3>

pH of the weak base will be calculate by using the Henderson Hasselbalch equation as:

pH = pKb + log([HB⁺]/[B])

pKb = -log(1.8×10⁻⁶) = 5.7

Chemical reaction for C₆H₅NH₂ is:

                          C₆H₅NH₂ + H₂O → C₆H₅NH₃⁺ + OH⁻

Initial:                     0.25                           0            0

Change:                    -x                             x             x

Equilibrium:        0.25-x                           x             x

Base dissociation constant will be calculated as:
Kb = [C₆H₅NH₃⁺][OH⁻] / [C₆H₅NH₂]

Kb = x² / 0.25 - x

x is very small as compared to 0.25, so we neglect x from that term and by putting value of Kb, then the equation becomes:

1.8×10⁻⁶ = x² / 0.25

x² = (1.8×10⁻⁶)(0.25)

x = 0.67×10⁻³ M = [C₆H₅NH₃⁺]

On putting all these values on the above equation of pH, we get

pH = 5.7 + log(0.67×10⁻³/0.25)

pH = 3.13

Hence pH of the solution is 3.13.

To know more about Henderson Hasselbalch equation, visit the below link:
brainly.com/question/13651361

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5 0
2 years ago
When the following oxidation–reduction reaction in acidic solution is balanced, what is the lowest whole-number coefficient for
ruslelena [56]

Answer:

b. 16, reactant side

Explanation:

Let's consider the following redox reaction.

MnO₄⁻(aq) + I⁻(aq) → Mn²⁺(aq) + I₂(s)

We can balance it using the ion-electron method.

Step 1: Identify both half-reactions

Reduction: MnO₄⁻(aq) → Mn²⁺(aq)

Oxidation: I⁻(aq) → I₂(s)

Step 2: Perform the mass balance, adding H⁺(aq) and H₂O(l) where appropriate

MnO₄⁻(aq) + 8 H⁺(aq) → Mn²⁺(aq) + 4 H₂O(l)

2 I⁻(aq) → I₂(s)

Step 3: Perform the charge balance, adding electrons where appropriate

MnO₄⁻(aq) + 8 H⁺(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l)

2 I⁻(aq) → I₂(s)  + 2 e⁻

Step 4: Multiply both half-reactions by numbers so that the number of electrons gained and lost are equal

2 × (MnO₄⁻(aq) + 8 H⁺(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l))

5 × (2 I⁻(aq) → I₂(s)  + 2 e⁻)

Step 5: Add both half-reactions and cancel what is repeated on both sides

2 MnO₄⁻(aq) + 16 H⁺(aq) + 10 e⁻ + 10 I⁻(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 I₂(s)  + 10 e⁻

The balanced reaction is:

2 MnO₄⁻(aq) + 16 H⁺(aq) + 10 I⁻(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 I₂(s)

5 0
3 years ago
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