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DENIUS [597]
3 years ago
10

how many grams of nahco3 (fm84.007) must be added ro 4.00g of k2co3 (fm 138.206) to give a ph of 10.80 in 500ml of water​

Chemistry
1 answer:
Anna11 [10]3 years ago
6 0

Answer:

0.804g of NaHCO₃ you must add

Explanation:

pKa of HCO₃⁻/CO₃²⁻ is 10.32.

It is possible to find pH of a buffer by using H-H equation, thus:

pH = pka + log [A⁻] / [HA]

<em>Where [HA] is concentration of acid (HCO₃⁻) and [A⁻] is concentration of conjugate acid (CO₃²⁻).</em>

Moles of CO₃²⁻ = K₂CO₃ are:

4.00g ₓ (1mol / 138.206g) = 0.0289 moles CO₃²⁻

Replacing:

10.80 = 10.32 + log [0.0289] / [HCO₃⁻]

[HCO₃⁻] = 0.009570 moles you need to add to obtain the desire pH

As molar mass of NaHCO₃ is 84.007g/mol, mass of NaHCO₃ is:

0.009570 moles ₓ (84.007g / mol) =

<h3>0.804g of NaHCO₃ you must add</h3>

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kipiarov [429]

Answer:

The length of an edge of this unit cell is 407.294 pm

Explanation:

Face centered cubic structure contains 4 atoms in each unit cell and 12 coordination number, occupying about 74% volume of the total cell. Face centered cubic structure is known for efficient use of space for atom packing.

To determine the edge length, a relationship between the radius of the atom and edge length is used.

X = R√8

Where;

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Therefore, the length of an edge of this unit cell is 407.294 pm

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3 years ago
Valance shell example..​
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Answer:

<h2>Oxygen has six valence electrons, two in the 2s subshell and four in the 2p subshell.</h2>

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3 0
2 years ago
A sample of nitrogen occupies 10.0 liters at 25°C what would be the new volume at 20°C? 7.9 L Ob 9.8 L 10.2 L 10.6 L
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Answer:

  9.4 liter

Explanation:

1) Data:

V₁ = 10.0 L

T₁ = 25°C = 25 + 273.15 K = 298.15 K

 P₁ = 98.7 Kpa

 T₂ = 20°C = 20 + 273.15 K = 293.15 K

  P₂ = 102.7 KPa

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2) Formula:

Used combined law of gases:

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3) Solution:

Solve the equation for V₂:

  V₂ = P₁V₁ T₂ / (P₂ T₁)

Substitute and compuite:

V₂ = P₁V₁ T₂ / (P₂ T₁)

V₂ = 98.7 KPa × 10.0 L × 293.15 K / (102.7 KPa × 298.15 K)

V₂ =  9.4 liter ← answer

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