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2 years ago

12

Which statement correctly describes a difference between aesexual and sexual reproduction?

1 answer:

mote1985 [20]2 years ago

6 0

Answer:

(3rd option) Asexually reproduced offsprings have the same amount of genetic information as the parent, while sexually reproduced offsprings have half as much genetic information.

Explanation:

This is variation is caused as a result of both gamete production and fertilisation.

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Give the characteristic of a first order reaction having only one reactant.a. The rate of the reaction is not proportional to th

kicyunya [14]

Answer:

E) The rate of the reaction is directly proportional to the concentration of the reactant.

Explanation:

Give the characteristic of a first order reaction having only one reactant.

A) The rate of the reaction is not proportional to the concentration of the reactant.

B) The rate of the reaction is proportional to the square of the concentration of the reactant.

C) The rate of the reaction is proportional to the square root of the concentration of the reactant.

D) The rate of the reaction is proportional to the natural logarithm of the concentration of the reactant.

E) The rate of the reaction is directly proportional to the concentration of the reactant.

3 0

2 years ago

We want to calculate the concentrations of all species in a 0.58 M Na 2 SO 3 (sodium sulfite) solution. The ionization constants

NeX [460]

Explanation:

Reaction equation is as follows.

$Na_{2}SO_{3}(s) \rightarrow 2Na^{+}(aq) + SO^{2-}_{3}(aq)$

Here, 1 mole of $Na_{2}SO_{3}$ produces 2 moles of cations.

$[Na^{+}] = 2[Na_{2}SO_{3}] = 2 \times 0.58$

= 1.16 M

$[SO^{2-}_{3}] = [Na_{2}SO_{3}]$ = 0.58 M

The sulphite anion will act as a base and react with $H_{2}O$ to form $HSO^{-}_{3}$ and $OH^{-}$ .

As, $K_{b} = \frac{K_{w}}{K_{a_{2}}}$

= $\frac{10^{-14}}{6.3 \times 10^{-8}}$

= $1.59 \times 10^{-7}$

According to the ICE table for the given reaction,

$SO^{2-}_{3} + H_{2}O \rightleftharpoons HSO^{-}_{3} + OH^{-}$

Initial: 0.58 0 0

Change: -x +x +x

Equilibrium: 0.58 - x x x

So,

$K_{b} = \frac{[HSO^{-}_{3}][OH^{-}]}{[SO^{2-}_{3}]}$

$1.59 \times 10^{-7} = \frac{x^{2}}{0.58 - x}$

$x^{2} = 1.59 \times 10^{-7} \times (0.58 - x)$

x = 0.0003 M

So, x = $[HSO^{-}_{3}] = [OH^{-}]$ = 0.0003 M

$[SO^{2-}_{3}]$ = 0.58 - 0.0003

= 0.579 M

Now, we will use $[HSO^{-}_{3}]$ = 0.0003 M

The reaction will be as follows.

$HSO^{2-}_{3} + H_{2}O \rightleftharpoons H_{2}SO_{3} + OH^{-}$

Initial: 0.0003

Equilibrium: 0.0003 - x x x

$K_{b} = \frac{x^{2}}{0.0003 - x}$

$K_{b} = \frac{K_{w}}{K_{a_{1}}}$

= $\frac{10^{-14}}{1.4 \times 10^{-2}}$

= $7.14 \times 10^{-13}$

Therefore, $7.14 \times 10^{-13} = \frac{x^{2}}{0.0003 - x}$

As, x <<<< 0.0003. So, we can neglect x.

Therefore, $x^{2} = 7.14 \times 10^{-13} \times 0.0003$

= $0.00214 \times 10^{-13}$

x = $0.0146 \times 10^{-6}$

x = $[OH^{-}] = [H_{2}SO_{3}]$ = $1.46 \times 10^{-8}$

$[H^{+}] = \frac{10^{-14}}{[OH^{-}]}$

= $\frac{10^{-14}}{0.0003}$

= $3.33 \times 10^{-11}$ M

Thus, we can conclude that the concentration of spectator ion is $3.33 \times 10^{-11}$ M.

5 0

2 years ago

Which of the following is a true statement?

Ede4ka [16]

Energy can accomplish work

3 0

2 years ago

Read 2 more answers

Which of these outdoor activities would you most likely participate in during the winter?

Elza [17]

Answer:

D.

Explanation:

Translation:

A. make a barbecue

B. organize a picnic

C. swimming in the pool

D. to go skiing

Barbecue is usually a summer activity, so we eliminate that.

It would be too cold for a picnic!

It would be too cold to go swimming, and that's more of a summer thing!

Skiing is a snow activity, and since it is winter, it is likely to know.

4 0

2 years ago

Question 5 (1 point)

Marianna [84]

Answer:

P=12.16 atm

Explanation:

Using the formula of ideal gas law:

PV = nRT

P= nRT/V

n= number of moles

R= Avogadro constant = 0.0821

T= Temperature in K => ºC + 273.15 K

P= (1.50 moles)(0.0821)( 296.15 K)/ 3.00L

P= 12.15

7 0

2 years ago