All categories

2 years ago

How would technology affect the population growth rate?

Chemistry

2 answers:

Rasek [7]2 years ago

8 0

B. It would lead to rapid population increase with high birth rates and low death rates due to the medical advancements

nexus9112 [7]2 years ago

6 0

Answer:

B. It would lead to rapid population increase with high birthrates and lower death rates.

Explanation:

Technology will increase birth rate and decrease death rate. Advancement in medical care allows better prenatal care, facilitates intensive care for children born with critical diseases, improved medications and baby food is also critical in saving the life of babies born. Also several vaccinations are available to prevent children from succumbing to life threatening diseases. In vitro fertilization is facilitating birth rates.

Additionally, better healthcare facilities are available at old age such as stem cell therapy, early diagnosis of cancer, treatment for Alzheimer’s and more.

You might be interested in

Sulfonation of benzene has the following mechanism: (1) 2 H2SO4 ⇌ H3O+ + HSO4− + SO3 [fast] (2) SO3 + C6H6 → H(C6H5+)SO3− [slow]

ziro4ka [17]

Question is incomplete, complete question is as follows :

Complete Question : .Sulfonation of benzene has the following mechanism:

(1) 2 H2SO4 ⇌ H3O+ + HSO4− + SO3

[fast]

(2) SO3 + C6H6 → H(C6H5+)SO3−

[slow]

(3) H(C6H5+)SO3− + HSO4− → C6H5SO3− + H2SO4

[fast]

(4) C6H5SO3− + H3O+ → C6H5SO3H + H2O

[fast]

write the overall rate law for the initial rate of the reaction as a fraction.

Rate=k(________/_________)

Answer:

The overall rate law for the initial reaction is = $k_{overall} [H_{2}SO_{4}]^{2} [C_{6}H_{6}]$

Explanation :

Frist of all, all the common terms are cancelled out and written the overall reaction.

As we know that the rate depednant step is the slowest step of the reaction, rate law is :

rate = $k_{2} [SO_{3}][C_{6}H_{6}]$

But the problem is that SO3 cannot be written in the overall rate law because it is an intermediate.

Rate law for synthesis of S03 is as follows :

rate = $k_{1}[H_{2}SO_{4}]^{2}$

Hence when we substitute equation 2 in equation one,

Rate comes out to be = $k_{overall} [H_{2}SO_{4}]^{2} [C_{6}H_{6}]$

6 0

2 years ago

Joseph drives 25miles North in 15 minutes then he drives 40 miles east in 30 minutes, then he uses cruise control option for 1 h

Musya8 [376]

Answer: 1.25 miles per minute

Explanation:

Average speed is the rate of change of total distance covered per unit time.

i.e Average speed = (Total distance / Time taken)

Total distance covered = (25miles + 40 miles + 70 miles + 15 miles)

= 150 miles

Total time taken = ( 15 minutes + 30 minutes + 1 hour + 15 minutes) = 120 minutes

Since 60 minutes = 1 hour, the total time taken is 120 minutes

Now, apply Average speed = (Total distance / Time taken)

= (150 miles / 120 minutes)

= 1.25 miles per minutes

Thus, Joseph drove with an average speed of 1.25 miles per minute.

7 0

2 years ago

0.00105 liters is the same as how many cm and ml

Ainat [17]

The answer is 1.05 cubic centimeters and 1.05 mL (1 cubic centimeter is equal to 1 mL)

8 0

3 years ago

Help me please, will give brainliest if correct!

mafiozo [28]

The answer would be C

4 0

3 years ago

Read 2 more answers

Match each substance with the correct designation for the equation HSO3- + CH3NH2 <=> SO32- + CH3NH3+ HSO3- CH3NH2 SO32- C

Zanzabum

Answer:

$HSO_3^-$ : conjugate acid of $SO_3^{2-}$

$CH_3NH_2$ : conjugate base of $CH_3NH_3^+$

$SO_3^{2-}$ : conjugate base of $HSO_3^-$

$CH_3NH_3^+$ : conjugate acid of $CH_3NH_2$

Explanation:

According to the Bronsted-Lowry conjugate acid-base theory, an acid is defined as a substance which looses donates protons and thus forming conjugate base and a base is defined as a substance which accepts protons and thus forming conjugate acid.

$HSO_3^-+CH_3NH_2\rightleftharpoons SO_3^{2-}+CH_3NH_3^+$

Here in forward reaction $CH_3NH_2$ is accepting a proton, thus it is considered as a base and after accepting a proton, it forms $CH_3NH_3^+$ which is a conjugate acid.

And $HSO_3^-$ is losing a proton, thus it is considered as an acid and after loosing a proton, it forms $SO_3^{2-}$ which is a conjugate base.

Similarly in the backward reaction, $CH_3NH_3^+$ is loosing a proton, thus it is considered as a acid and after loosing a proton, it forms $CH_3NH_2$ which is a conjugate base.

And $SO_3^{2-}$ is accepting a proton, thus it is considered as a base and after accepting a proton, it forms $HSO_3^{-}$ which is a conjugate acid.

4 0

3 years ago