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kati45 [8]
4 years ago
13

Please can someone help I will mark brainiest!

Chemistry
1 answer:
vazorg [7]4 years ago
6 0

Answer:

2Cl- ⇒ Cl ↓2+ 2e

Explanation: sorry if this is not what you were looking for.

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During dehydration synthesis, an enzyme removes a ____________ functional group from the monomers of one organic molecule and a
Reika [66]

Linking monomers together to form a polymer .This chemical reaction also forms water molecules.

<h3>What is Polymerization?</h3>

This is a type of reaction which involves the linking of two or more monomers to form a polymer.

Dehydration reaction forms water molecules as part of the product thereby making it the most appropriate choice.

Read more about Dehydration here brainly.com/question/1301665

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3 0
2 years ago
The half life for the decay of carbon- is years. Suppose the activity due to the radioactive decay of the carbon- in a tiny samp
EleoNora [17]

The question is incomplete. The complete question is:

The half-life for the decay of carbon-14 is 5.73x10^3 years. Suppose the activity due to the radioactive decay of the carbon-14 in a tiny sample of an artifact made of woodfrom an archeological dig is measured to be 2.8x10^3 Bq. The activity in a similiar-sized sample of fresh wood is measured to be 3.0x10^3 Bq. Calculate the age of the artifact. Round your answer to 2 significant digits.

Answer:

570 years

Explanation:

The activity of the fresh sample is taken as the initial activity of the wood sample while the activity measured at a time t is the present activity of the wood artifact. The time taken for the wood to attain its current activity can be calculated from the formula shown in the image attached. The activity at a time t must always be less than the activity of a fresh wood sample. Detailed solution is found in the image attached.

6 0
3 years ago
PLZZZZZZZZZZZZZZZZZZ15. As much as 90 percent of the oxygen in our atmosphere is the result of
vladimir2022 [97]

Answer:

The correct answer is - option D. photosynthesis.

Explanation:

It is shown by the study that most of the atmospheric oxygen comes from the photosynthesis by plants as oxygen is the byproduct of the photosynthesis. Photosynthesis is the process that uses light energy, carbon dioxide, and water to produce food or glucose/sugar and release oxygen as the byproduct.

Many scientists believe that oceanic phytoplankton that releases oxygen by the photosynthesis process makes 80 to 85% of the total oxygen of the atmospheric oxygen.

7 0
3 years ago
Read 2 more answers
The metal zirconium becomes superconducting at temperatures below 3.4000K.
dybincka [34]

Answer:

-269.75 degrees Celsius

Explanation:

7 0
3 years ago
The composition of a liquid-phase reaction 2A - B was monitored spectrophotometrically. The following data was obtained: t/min 0
o-na [289]

Answer:

1) The order of the reaction is of FIRST ORDER

2)   Rate constant k = 5.667 × 10 ⁻⁴

Explanation:

From the given information:

The composition of a liquid-phase reaction 2A - B was monitored spectrophotometrically.

liquid-phase reaction 2A - B signifies that the reaction is of FIRST ORDER where the rate of this reaction is directly proportional to the concentration of A.

The following data was obtained:

t/min                    0         10         20          30             40          ∞

conc B/(mol/L)    0       0.089    0.153     0.200       0.230    0.312

For  a first order reaction:

K = \dfrac{1}{t} \ In ( \dfrac{C_{\infty} - C_o}{C_{\infty} - C_t})

where :

K = proportionality  constant or the rate constant for the specific reaction rate

t = time of reaction

C_o = initial concentration at time t

C _{\infty} = final concentration at time t

C_t = concentration at time t

To start with the value of t when t = 10 mins

K_1 = \dfrac{1}{10} \ In ( \dfrac{0.312 - 0}{0.312 - 0.089})

K_1 = \dfrac{1}{10} \ In ( \dfrac{0.312 }{0.223})

K_1 =0.03358 \  min^{-1}

K_1 \simeq 0.034 \  min^{-1}

When t = 20

K_2= \dfrac{1}{20} \ In ( \dfrac{0.312 - 0}{0.312 - 0.153})

K_2= 0.05 \times  \ In ( 1.9623)

K_2=0.03371 \ min^{-1}

K_2 \simeq 0.034 \ min^{-1}

When t = 30

K_3= \dfrac{1}{30} \ In ( \dfrac{0.312 - 0}{0.312 - 0.200})

K_3= 0.0333 \times  \ In ( \dfrac{0.312}{0.112})

K_3= 0.0333 \times  \ 1.0245

K_3 = 0.03412 \ min^{-1}

K_3 = 0.034 \ min^{-1}

When t = 40

K_4= \dfrac{1}{40} \ In ( \dfrac{0.312 - 0}{0.312 - 0.230})

K_4=0.025 \times  \ In ( \dfrac{0.312}{0.082})

K_4=0.025 \times  \ In ( 3.8048)

K_4=0.03340 \ min^{-1}

We can see that at the different time rates, the rate constant of k_1, k_2, k_3, and k_4 all have similar constant values

As such :

Rate constant k = 0.034 min⁻¹

Converting it to seconds ; we have :

60 seconds = 1 min

∴

0.034 min⁻¹ =(0.034/60) seconds

= 5.667 × 10 ⁻⁴ seconds

Rate constant k = 5.667 × 10 ⁻⁴

4 0
3 years ago
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