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KengaRu [80]
2 years ago
9

What changes within the star A lead to the red giant B?

Chemistry
1 answer:
Lady_Fox [76]2 years ago
8 0

Answer: The fusion of hydrogen to form helium

Explanation:

You might be interested in
50.00 mL of 0.10 M HNO 2 (nitrous acid, K a = 4.5 × 10 −4) is titrated with a 0.10 M KOH solution. After 25.00 mL of the KOH sol
boyakko [2]

Answer:

b. 3.35

Explanation:

To calculate the pH of a solution containing both acid and its salt (produced as a result of titration) we need to use Henderson’s equation i.e.

pH = pKa + log ([salt]/[acid])     (Eq. 01)

Where  

pKa = -log(Ka)        (Eq. 02)

[salt] = Molar concentration of salt produced as a result of titration

[acid] = Molar concentration of acid left in the solution after titration

Let’s now calculate the molar concentration of HNO2 and KOH considering following chemical reaction:

HNO2 + KOH ⇆ H2O + KNO2    (Eq. 03)

This shows that 01 mole of HNO2 and 01 mole of KOH are required to produce 01 mole of KNO2 (salt). And if any one of them (HNO2 and KOH) is present in lower amount then that will be considered the limiting reactant and amount of salt produced will be in accordance to that reactant.

Moles of HNO2 in 50 mL of 0.01 M HNO2 solution = 50/1000x0.01 = 0.005 Moles

Moles of KOH in 25 mL of 0.01 M KOH solution = 25/1000x0.01 = 0.0025 Moles

As it can be seen that we have 0.0025 Moles of KOH therefore considering Eq. 03 we can see that 0.0025 Moles of KOH will react with only 0.0025 Moles of HNO2 and will produce 0.0025 Moles of KNO2.

Therefore

Amount of salt produced i.e [salt] = 0.0025 moles       (Eq. 04)

Amount of acid left in the solution [acid] = 0.005 - 0.0025 = 0.0025 moles (Eq.05)

Putting the values in (Eq. 01) from (Eq.02), (Eq. 04) and (Eq. 05) we will get the following expression:

pH= -log(4.5x10 -4) + log (0.0025/0.0025)

Solving above we get  

pH = 3.35

5 0
3 years ago
A buffer is a solution that is a mixture of either a weak acid and its conjugate base or a weak base and its conjugate acid. Whe
leva [86]

Answer:

a. The conjugate base of an acidic buffer will accept hydrogen protons when a strong acid is added to the solution.

b. An acidic buffer solution is a mixture of a weak acid and its conjugate base.

e. The weak acid of an acidic buffer will donate hydrogen protons when a strong base is added to the solution.

Explanation:

<em>Which of the statements correctly describe the properties of a buffer?</em>

a. The conjugate base of an acidic buffer will accept hydrogen protons when a strong acid is added to the solution. TRUE. The conjugate base neutralizes the excess of hydrogen protons.

b. An acidic buffer solution is a mixture of a weak acid and its conjugate base. TRUE.

c. An acidic buffer solution is a mixture of a weak base and its conjugate acid. FALSE. This is a basic buffer solution.

d. The weak acid of an acidic buffer will accept hydrogen protons when a strong base is added to the solution. FALSE. The weak acid will react with the hydroxyl ions from the added base.

e. The weak acid of an acidic buffer will donate hydrogen protons when a strong base is added to the solution. TRUE. These hydrogen protons will form water.

f. The conjugate base of an acidic buffer will donate hydrogen protons when a strong acid is added to the solution. FALSE. It will accept hydrogen protons.

7 0
3 years ago
Given that a for HBrO is 2. 8×10^−9 at 25°C. What is the value of b for BrO− at 25°C?
lara [203]

If Ka for HBrO is 2. 8×10^−9 at 25°C, then the value of Kb for BrO− at 25°C is 3.5× 10^(-6).

<h3>What is base dissociation constant? </h3>

The base dissociation constant (Kb) is defined as the measurement of the ions which base can dissociate or dissolve in the aqueous solution. The greater the value of base dissociation constant greater will be its basicity an strength.

The dissociation reaction of hydrogen cyanide can be given as

HCN --- (H+) + (CN-)

Given,

The value of Ka for HCN is 2.8× 10^(-9)

The correlation between base dissociation constant and acid dissociation constant is

Kw = Ka × Kb

Kw = 10^(-14)

Substituting values of Ka and Kw,

Kb = 10^(-14) /{2.8×10^(-9) }

= 3.5× 10^(-6)

Thus, we find that if Ka for HBrO is 2. 8×10^−9 at 25°C, then the value of Kb for BrO− at 25°C is 3.5× 10^(-6).

DISCLAIMER: The above question have mistake. The correct question is given as

Question:

Given that Ka for HBrO is 2. 8×10^−9 at 25°C. What is the value of Kb for BrO− at 25°C?

learn more about base dissociation constant:

brainly.com/question/9234362

#SPJ4

7 0
1 year ago
The product of the nitration reaction will have a nitro group at which position with respect to the methyl group? Group of answe
zimovet [89]

Answer:

Mostly Para

Explanation:

First, let's assume that the molecule is the toluene (A benzene with a methyl group as radical).

Now the nitration reaction is a reaction in which the nitric acid in presence of sulfuric acid, react with the benzene molecule, to introduce the nitro group into the molecule. The nitro group is a relative strong deactiviting group and is metha director, so, further reactions that occur will be in the metha position.

Now, in this case, the methyl group is a weak activating group in the molecule of benzene, and is always ortho and para director for the simple fact that this molecule (The methyl group) is a donor of electrons instead of atracting group of electrons. Therefore for these two reasons, when the nitration occurs,it will go to the ortho or para position.

Now which position will prefer to go? it's true it can go either ortho or para, however, let's use the steric hindrance principle. Although the methyl group it's not a very voluminous and big molecule, it still exerts a little steric hindrance, and the nitro group would rather go to a position where no molecule is present so it can attach easily. It's like you have two doors that lead to the same place, but in one door you have a kid in the middle and the other door is free to go, you'll rather pass by the door which is free instead of the door with the kid in the middle even though you can pass for that door too. Same thing happens here. Therefore the correct option will be mostly para.

4 0
3 years ago
10. Which of the following mirror is used to form smaller image
vitfil [10]

Answer:

Correct answer is b) convex mirror

Explanation:

A convex mirror, which bulges outward, reflects at a wider angle near its edges than at its center, creating a slightly distorted image that's smaller than actual size. Convex mirrors have many uses. The smaller size of the images means that you can see more with these surfaces, hence their use in safety mirrors.

3 0
3 years ago
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