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olga nikolaevna [1]
3 years ago
9

What is the pH of 1.00 L of a buffer that is 0.110 M nitrous acid (HNO2) and 0.200 M NaNO2? (pKa of HNO2 = 3.40)

Chemistry
1 answer:
Andrei [34K]3 years ago
4 0

Answer:

pH = 3.65    

Explanation:

given data

pKa of HNO2 = 3.40

nitrous acid (HNO2) = 0.110 M

NaNO2 = 0.200 M

to find out

What is the pH

solution

we get here ph for acidic buffer  that is express as

pH = pKa + log(salt÷acid)      ........................1

put here value and we get

pH = 3.40 + log(0.200÷0.110)

pH = 3.65    

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A student constructs a coffee cup calorimeter and places 50.0 mL of water into it. After a brief period of stabilization, the te
Grace [21]

Answer:

The calorimeter constant is  = 447 J/°C

Explanation:

The heat absorbed or released (Q) by water can be calculated with the following expression:

Q = c × m × ΔT

where,

c is the specific heat

m is the mass

ΔT is the change in temperature

The water that is initially in the calorimeter (w₁) absorbs heat while the water that is added (w₂) later releases heat. The calorimeter also absorbs heat.

The heat absorbed by the calorimeter (Q) can be calculated with the following expression:

Q = C × ΔT

where,

C is the calorimeter constant

The density of water is 1.00 g/mL so 50.0 mL = 50.0 g. The sum of the heat absorbed and the heat released is equal to zero (conservation of energy).

Qabs + Qrel = 0

Qabs = - Qrel

Qcal + Qw₁ = - Qw₂

Qcal = - (Qw₂ + Qw₁)

Ccal . ΔTcal = - (cw . mw₁ . ΔTw₁ + cw . mw₂ . ΔTw₂)

Ccal . (30.31°C - 22.6°C) = - [(4.184 J/g.°C) × 50.0 g × (30.31°C - 22.6°C) +  (4.184 J/g.°C) × 50.0 g × (30.31°C - 54.5°C)]

Ccal  = 447 J/°C

5 0
3 years ago
The reaction converting glycerol to glycerol-3-phosphate can be represented as: glycerol + hpo42− ⟶ glycerol-3-phosphate + h2o.
Volgvan

Answer;

Yes; this reaction be spontaneous if coupled with the hydrolysis of ATP.

Explanation;

The reaction converting glycerol to glycerol-3-phosphate (energetically unfavorable) can be coupled with the conversion of ATP to ADP (energetically favorable):

Glycerol + HPO42 ⟶glycerol-3-phosphate+H2O

ATP + H2O⟶ ADP + HPO42− + H+

6 0
3 years ago
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All of the deer, grass, maple trees, and owls in a forest would be considered a
Alexxx [7]
A habitat because it has animals and trees
4 0
3 years ago
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For the following example, list the given and unknown information (including gratis or moles)
Setler79 [48]

Answer:

9.6 moles O2

Explanation:

I'll assume it is 345 grams, not gratis, of water.  Hydrogen's molar mass is 1.01, not 101.

The molar mass of water is 18.0 grams/mole.

Therefore:  (345g)/(18.0 g/mole) = 19.17 or 19.2 moles water (3 sig figs).

The balanced equation states that:  2H20 ⇒ 2H2 +02

It promises that we'll get 1 mole of oxygen for every 2 moles of H2O, a molar ratio of 1/2.

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6 0
2 years ago
Which shows the rational expression written using the least common denominator?
Alexandra [31]

Answer:

The first one.

\frac{x + 1}{4x^{2} } and \frac{4(x + 1)}{4x^{2} }

Explanation:

When comparing two fractions with variables like this, it's important to get to the same denominator in order to compare apples with apples and then be able to do not only comparisons but also perform additions/subtractions.

Question is which denominator to use and how to reach it.

In this case, the question and the answer choices do the work for you.  The question asks which one is the LEAST common denominator, and the answers show denominators x² and 4x².  The smallest of these is x², however, we can't simplify the first fraction to get to the x² denominator, so we'll go for the 4x².

So, the first fraction has already the correct denominator (4x²), we just have to transform the second one.

We multiply it by 1, expressed in a different way.  Since we're multiplying by one, we're not affecting the value, just the way it looks.

Let's do it!, to get the denominator to go from x² to 4x², we need to multiply it by 4... so we'll multiply by 4/4 (which is 1, neutral for the multiplication).

\frac{x + 1}{x^{2} } * \frac{4}{4} = \frac{4(x + 1)}{4x^{2} }

And now you have both fractions on the same denominator, without having changed their value, just their looks

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