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2 years ago

Which one of the following combinations cannot function as a buffer solution?

Chemistry

1 answer:

ANEK [815]2 years ago

3 0

Answer:

c.) HNO₃ & NaNO₃

Explanation:

A buffer system has 2 components:

A weak acid and its conjugate base or
A weak base and its conjugate acid

Which one of the following combinations cannot function as a buffer solution?

a.) HCN & KCN. YES. HCN is a weak acid and CN⁻ (coming from KCN) its conjugate base.

b.) NH₃ & (NH₄)₂SO₄. YES. NH₃ is a weak base and NH₄⁺ (coming from (NH₄)₂SO₄) its conjugate acid.

c.) HNO₃ & NaNO₃. NO. HNO₃ is a strong acid.

d.) HF & NaF. YES. HF is a weak acid and F⁻ (coming from NaF) its conjugate base.

e.) HNO₂ & NaNO₂. YES. HNO₂ is a weak acid and NO₂⁻ (coming from NaNO₂) its conjugate base.

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Aluminum has a face-centered cubic unit cell. How many atoms of Al are present in each unit cell

Serjik [45]

Answer:4

Explanation:

As shown in the image attached, a face-centred cubic structure has 8 atoms at the corners and 6 face center atoms.

Each corner atom contributes to eight cell, so per unit cell 1/8 ×8 =1atom

Face center atoms contributes to two unit cells 1/2 × 6=3atoms

Total atoms =3+1=4atoms

Therefore the atoms in Al FCC per unit cell is 4

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3 years ago

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Answer:

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3 years ago

Barium sulfate is made by the following reaction.

ipn [44]

Answer:

%age Yield = 96 %

Explanation:

The balance chemical equation for given double replacement reaction is,

Ba(NO₃)₂ + Na₂SO₄ → BaSO₄ + 2 NaNO₃

Step 1: Calculate moles of Ba(NO₃)₂:

Moles = Mass / M.Mass

Moles = 75.1 g / 261.33 g/mol

Moles = 0.2873 moles of Ba(NO₃)₂

Step 2: Find out moles of BaSO₄ formed:

According to balance chemical equation,

1 mole of Ba(NO₃)₂ produced = 1 mole of BaSO₄

So,

0.2873 moles of Ba(NO₃)₂ will produce = X moles of BaSO₄

Solving for X,

X = 0.2873 mol × 1 mol / 1 mol

X = 0.2873 moles of BaSO₄

Step 3: Calculate Theoretical Mass of BaSO₄:

Mass = Moles × M.Mass

Mass = 0.2873 mol × 233.38 g/mol

Mass = 67.07 g of BaSO₄

Step 4: Calculate %age Yield as:

Theoretical Yield = 67.07 g

Actual Yield = 64.4 g

%age Yield = ???

Formula Used:

%age Yield = (Actual Yield ÷ Theoretical Yield) × 100

Putting Values,

%age Yield = (64.4 g ÷ 67.07 g) × 100

%age Yield = 96.01 % ≈ 96 %

6 0

2 years ago

PLEASE HELP I HAVE TO DO THIS BENCHAMRK QUESTION!!!

GuDViN [60]

PLEASE HELP I HAVE TO DO THIS BENCHAMRK QUESTION!!!

The chart shows parts of a plant and an animal, at different levels of organization. Consider the plant images. If we are referring to levels of organization in a plant which picture can BEST be used to fill in box Y?

first picture (A

8 0

3 years ago

Find the percent composition of OXYGEN in Manganese (III) nitrate, Mn(NO3)3.

BaLLatris [955]

Answer:

59.8%

Explanation:

First find the Mr of manganese (III) nitrate.

Mr of Mn(NO₃)₃ = 54.9 + (14 × 3) + (16 × 3 × 3) = 240.9

Since we have to find the percentage composition of oxygen, we need to find the Mr of oxygen in the compound, which is:

Mr of (O₃)₃ = (16 × 3) × 3 = 144

Now we can find percentage composition / percentage by mass of oxygen.

% composition = $\frac{Mr\ of\ oxygen\ in\ compound}{Mr\ of\ compound}$ × 100

% composition = $\frac{144}{240.9}$ × 100 = 59.776%

∴ % compostion of oxygen in maganese(III)nitrate is 59.8% (to 3 significant figures).

8 0

2 years ago