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3 years ago

Which of these is an example of sexual reproduction, involving meiosis?

Chemistry

2 answers:

dsp733 years ago

7 0

A bacterium splits in half, forming two new bacteria.

lesya692 [45]3 years ago

5 0

Answer:

A bacterium splits in half, forming two new bacteria

Explanation:

Meiosis is a process where a single cell divides twice to produce four cells containing half the original amount of genetic information

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If m< CAB = 80 degrees, then find m< BAD

eimsori [14]

Answer:

Explanation:

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7 0

3 years ago

Determine the rate law, including the values of the orders and rate law constant, for the following reaction using the experimen

olga55 [171]

Answer: Rate law= $k[A]^1[B]^2$ , order with respect to A is 1, order with respect to B is 2 and total order is 3. Rate law constant is $3L^2mol^{-2}s^{-1}$

Explanation: Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

$Rate=k[A]^x[B]^y$

k= rate constant

x = order with respect to A

y = order with respect to A

n = x+y = Total order

a) From trial 1: $1.2\times 10^{-2}=k[0.10]^x[0.20]^y$ (1)

From trial 2: $4.8\times 10^{-2}=k[0.10]^x[0.40]^y$ (2)

Dividing 2 by 1 : $\frac{4.8\times 10^{-2}}{1.2\times 10^{-2}}=\frac{k[0.10]^x[0.40]^y}{k[0.10]^x[0.20]^y}$

$4=2^y,2^2=2^y$ therefore y=2.

b) From trial 2: $4.8\times 10^{-2}=k[0.10]^x[0.40]^y$ (3)

From trial 3: $9.6\times 10^{-2}=k[0.20]^x[0.40]^y$ (4)

Dividing 4 by 3: $\frac{9.6\times 10^{-2}}{4.8\times 10^{-2}}=\frac{k[0.20]^x[0.40]^y}{k[0.10]^x[0.40]^y}$

$2=2^x,2=2^1$ , x=1

Thus rate law is $Rate=k[A]^1[B]^2$

Thus order with respect to A is 1 , order with respect to B is 2 and total order is 1+2=3.

c) For calculating k:

Using trial 1: $1.2\times 10^{-2}=k[0.10]^1[0.20]^2$

$k=3 L^2mol^{-2}s^{-1}$ .

6 0

3 years ago

When developing an experimental design, which action would improve the

11111nata11111 [884]

Answer:

Explanation:

I'm right I took the test

4 0

2 years ago

How does the arrangement of elements in periods relate to electron configuration

VashaNatasha [74]

Electronic Configuration of elements in a period is same because If you see the electronic Configuration of elements in a period you will notice that the valence shell electrons for all elements are present in the same Shell. For example, in first period consisting of Hydrogen and Helium, both the elements' valence electrons are present in the same Shell.
Electronic Configuration of Hydrogen,
1s^1
Electronic Configuration of Helium,
1s^2

Both elements' valance electrons are present in the 1st shell

(This is just a small example to understand the concept because other periods are long but the first period is short that's why I gave the example of the first period)

4 0

3 years ago

Which is a higher concentration, 10-9 M or 10-8 M? Explain.

andreyandreev [35.5K]

Answer:

10-9 Millimeters/liters

Explanation:

Because 10-9 M Is more than 10-8 M

I hope this is correct

3 0

3 years ago