1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Luden [163]
3 years ago
5

4. DNA replication:

Chemistry
2 answers:
garri49 [273]3 years ago
7 0

Answer:

b

Explanation:

icang [17]3 years ago
5 0

Answer:

A.

Explanation:

You might be interested in
On the reaction below, label the BSA, BSB, CA, and CB. CH3COOH + H2O → CH3COO– + H3O+
Ostrovityanka [42]

Answer:

Acid(BSA) = CH₃COOH

Base (BSB) = H₂O

Conjugate base (CB) = CH₃COO⁻

Conjugate acid (CA) = H₃O⁺

Explanation:

Equation of reaction;

CH₃COOH + H₂O → CH₃COO⁻ + H₃O⁺

Hello,

From my understanding of the question, we are required to identify the

1) Acid

2) Base

3) conjugate acid

4) conjugate base in the reaction

Acid (BSA) = CH₃COOH

Base (BSB) = H₂O

CA = conjugate acid = H₃O⁺

CB = conjugate base = CH₃COO⁻

7 0
3 years ago
A buffer consists of 0.120 M HNO2 and 0.150 M NaNO2 at 25°C. pka of HNO2 is 3.40. a. What is the pH of the buffer? b. What is th
Mashcka [7]

Explanation:

It is known that K_{a} of HNO_{2} = 4.5 \times 10^{-4}.

(a)  Relation between K_{a} and pK_{a} is as follows.

                       pK_{a} = -log (K_{a})

Putting the values into the above formula as follows.

                      pK_{a} = -log (K_{a})

                                    = -log(4.5 \times 10^{-4})

                                     = 3.347

Also, relation between pH and  pK_{a} is as follows.

              pH = pK_{a} + log\frac{[conjugate base]}{[acid]}

                     = 3.347+ log \frac{0.15}{0.12}

                    = 3.44

Therefore, pH of the buffer is 3.44.

(b)   No. of moles of HCl added = Molarity \times volume

                                            = 11.6 M \times 0.001 L

                                             = 0.0116 mol

In the given reaction, NO^{-}_{2} will react with H^{+} to form HNO_{2}

Hence, before the reaction:

No. of moles of NO^{-}_{2} = 0.15 M \times 1.0 L

                                           = 0.15 mol

And, no. of moles of HNO_{2} = 0.12 M \times 1.0 L

                                               = 0.12 mol

On the other hand, after the reaction :  

No. of moles of NO^{-}_{2} = moles present initially - moles added

                                          = (0.15 - 0.0116) mol

                                          = 0.1384 mol

Moles of HNO_{2} = moles present initially + moles added

                               = (0.12 + 0.0116) mol

                                = 0.1316 mol

As, K_{a} = 4.5 \times 10^{-4}

           pK_{a} = -log (K_{a})

                         = -log(4.5 \times 10^{-4})

                         = 3.347

Since, volume is both in numerator and denominator, we can use mol instead of concentration.

As, pH = pK_{a} + log \frac{[conjugate base]}{[acid]}

            = 3.347+ log {0.1384/0.1316}

            = 3.369

            = 3.37 (approx)

Thus, we can conclude that pH after the addition of 1.00 mL of 11.6 M HCl to 1.00 L of the buffer solution is 3.37.

6 0
3 years ago
1. Carbon is important because:
nikklg [1K]

Answer:

B

Explanation:

All organic material has carbon.

3 0
3 years ago
Read 2 more answers
Name the following alkyne:
Lady bird [3.3K]

Answer:

There isn't any picture for me to see

have a good day :)

Explanation:

3 0
3 years ago
A 98.0°C piece of cadmium (c=.850J/g°C) is placed in 150.0g of 37.0°C water. After sitting for a few minutes, both have a temper
Eduardwww [97]
The answer is 19.9 grams cadmium. 
Assuming there was no heat leaked from the system, the heat q lost by cadmium would be equal to the heat gained by the water:
     heat lost by cadmium = heat gained by the water
     -qcadmium = qwater
Since q is equal to mcΔT, we can now calculate for the mass m of the cadmium sample:
     -qcadmium = qwater
     -(mcadmium)(0.850J/g°C)(38.6°C-98.0°C)) = 150.0g(4.18J/g°C)(38.6°C-37.0°C)
     mcadmium = 19.9 grams
3 0
3 years ago
Other questions:
  • What is sodium mixed with carbon dioxide
    11·1 answer
  • Cual de los estados de agregacion es mas abundante en el universo
    5·1 answer
  • What happens to the atoms in a scoop of ice cream as it melts?
    14·2 answers
  • What does the term chemistry mean?
    8·1 answer
  • What might you expect to observe from this double-displacement reaction: CoCl2(aq) + Na2SO4(aq) ---> CoSO4 + 2NaCl?
    12·1 answer
  • Which type of radiological technology is useful for viewing tumors?
    6·1 answer
  • What is the smallest halogen atom<br>​
    9·2 answers
  • What is the energy of the motion of objects that are attached together?
    14·1 answer
  • Which metalloids would behave more like metals? Which metalloids would behave more like nonmetals?.
    11·2 answers
  • Use a periodic table to help you answer these questions if needed.
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!