The genetic code from DNA is carried into the
cytoplasm by____

Answer:

i) pH = 0.6990

ii) pH = 2.389

Explanation:

i) Before adding aqueous NaOH, there are 25.00 mL of 0.2000 M HCl. HCl reacts with the water in the aqueous solution as follows:

HCl + H₂O ⇒ H₃O⁺ + Cl⁻

The HCl and H₃O⁺ are related to each other through a 1:1 molar ratio, so the concentration of H₃O⁺ is equal to the HCl concentration.

The pH is related to the hydronium ion concentration as follows:

pH = -log([H₃O⁺]) = -log(0.2000) = 0.699

ii) Addition of NaOH causes the following reaction:

H₃O⁺ + NaOH ⇒ 2H₂O + Na⁺

The H₃O⁺ and NaOH react in a 1:1 molar ratio. The amount of NaOH added is calculated:

n = CV = (0.2000 mol/L)(24.00 mL) = 4.800 mmol NaOH

Thus, 4.800 mmol of H₃O⁺ were neutralized.

The initial amount of H₃O⁺ present was:

n = CV = (0.2000 mol/L)(25.00 mL) = 5.000 mmol H₃O⁺

The amount of H₃O⁺ that remains after addition of NaOH is:

(5.000 mmol) - (4.800 mmol) = 0.2000 mmol

The concentration of H₃O⁺ is the amount of H₃O⁺ divided by the total volume. The total volume is (25.00 mL) + (24.00 mL ) = 49.00 mL

C = n/V = (0.2000 mmol) / (49.00 mL) = 0.004082 M

The pH is finally calculated:

pH = -log([H₃O⁺]) = -log(0.004082) = 2.389

Answer:

823.2 J

Explanation:

PE = mgh

= (7 kg) (9.8 m/s^2) (12 m)

= 823.2 J

Answer:

RequiredAnswer:-

Mass=m=7kg

Height=h=12m

Gravitational force=g=10m/s^2

• As we know that

• Substitute the values

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

Final answer:

The pH value in the titration flask after 25.00 mL of the 0.10 M KOH solution is added to 50.00 mL of 0.10 M HNO2 solution is 3.35.

Explanation:

The subject of this question is titration, which is a method used in chemistry to measure the concentration of an unknown solution. Given 50.00 mL of 0.10 M HNO2 (nitrous acid, Ka = 4.5 × 10-4), titrated with 0.10 M KOH (potassium hydroxide), we need to calculate the pH after 25.00 mL of the KOH solution is added.

First, we need to find the moles of the HNO2 and the KOH. Moles equals Molarity times Volume. So, for HNO2, it is 0.10 M * 0.050 L which equals 0.005 moles. For KOH, it is 0.10 M * 0.025 L which equals 0.0025 moles.

Then, subtract the moles of OH- from the moles of HNO2 to determine the concentration of HNO2 left, which is 0.005 moles - 0.0025 moles = 0.0025 moles. Divide this by the total volume of the solution (50.00 mL + 25.00 mL = 75.00 mL or 0.075 L to determine the new concentration of HNO2, 0.0025 moles / 0.075 L = 0.033 M. Then use the given Ka value with the equation [H+] = sqrt(Ka * [HNO2]) to get [H+].

To find acids' pH, we use the formula pH = -log[H+]. Use the calculated [H+] to find the pH.

Upon performing these calculations, the resulting pH value should be approximately 3.35 after 25.00 mL of the KOH solution is added, so the answer is (b) 3.35.

Learn more about titration here:

brainly.com/question/38139486

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Explanation:

CH2)3(g)CH3CH=CH2(g) [(CH2)3], M       time, min

0.128               0

6.40×10-2          14.4

3.20×10-2        28.8

1.60×10-2          43.2

(1) What is the half-life for the reaction starting at t=0 min? min

Half life is the amount of time required for a substance to decay by half of it's initial concentration.

Starting form 0, the initial concentration = 0.128

After 14.4 mins, the final concentration is now exactly half of the initial concentration. This means 14.4 min is the half life starting from t=0min

What is the half-life for the reaction starting at t=14.4 min?

Starting form 14.4min, the initial concentration = 6.40×10-2

After 14.4 mins (28.8 - 14.4), the final concentration is now exactly half of the initial concentration. This means 14.4 min is the half life starting from t=14.4min

Does the half-life increase, decrease or remain constant as the reaction proceeds?

The half life is a constant factor, hence it remains constant as the reaction proceeds.

(2) Is the reaction zero, first, or second order?

Because the half life is independent of the concentration, it is a first order reaction.

In a zero order reaction, the half life Decreases as the reaction progresses; as concentration decreases.

In a first order reaction, the half life Increases with decreasing concentration.

(3) Based on these data, what is the rate constant for the reaction? min-1

The realtionship between the half life and rate onstant is;

k = 0.693 / half life

k = 0.693 / 14.4

k = 0.048125 min-1