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ludmilkaskok [199]

2 years ago

7

What is a property of a solution with a high pH? Select all that apply. *

1 answer:

Lena [83]2 years ago

8 0

Answer:

Bases (solutions with a high pH) feel slipper, have an -OH group, and are corrosive.

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A cell contains the same DNA as its parent cell. This is a result of which process? mitosis meiosis gamete formation fertilizati

lesya692 [45]

Answer:

Mitosis

Explanation:

A cell that contains the same DNA as its parent cell is a product of mitosis.

Mitosis is a cell division process that takes place in somatic cells.
In meiosis a single cell divides to produce two daughter cells each having the same DNA as the parent
Mitosis ensures that the same DNA is found between the parent
Mitosis is used for repairing and production of new cells.

3 0

2 years ago

Read 2 more answers

Please Help 8-9!!!!!!!

seropon [69]

The labeled diagram is given in the image attached.

As it can be seen from the image that freezing is when energy is removed from the system at 0 ⁰ while melting is when energy is added at 0⁰.

Also when energy is added at 100⁰C, it causes boiling while when it is removed at 100⁰C, it causes condensation.

Melting point of water is 0⁰C while boiling point is 100⁰C

3 0

2 years ago

(6.022*10^23)*2.58=<br> what’s the answer

Sladkaya [172]

The answer would be
$1.553676 \times {10}^{24}$

8 0

3 years ago

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5x^2 + 3×^2-x+7x<br>Help plz

PSYCHO15rus [73]

Answer:

5x^2 + 3x^2 -x +7x

$5x^2+3x^2+6x = 8x^2+6x$

factorize it to get

2x(4x+3)

3 0

3 years ago

The rate constant for a certain reaction is measured at two different temperatures:

Talja [164]

Answer: The activation energy Ea for this reaction is 22689.8 J/mol

Explanation:

According to Arrhenius equation with change in temperature, the formula is as follows.

$ln \frac{k_{2}}{k_{1}} = \frac{-E_{a}}{R}[\frac{1}{T_{2}} - \frac{1}{T_{1}}]$

$k_1$ = rate constant at temperature $T_1$ = $2.3\times 10^8$

$k_2$ = rate constant at temperature $T_2$ = $4.8\times 10^8$

$E_a$ = activation energy = ?

R= gas constant = 8.314 J/kmol

$T_1$ = temperature = $280.0^0C=(273+280)=553K$

$T_2$ = temperature = $376.0^0C=(273+376)=649K$

Putting in the values ::

$ln \frac{4.8\times 10^8}{2.3\times 10^8} = \frac{-E_{a}}{8.314}[\frac{1}{649} - \frac{1}{553}]$

$E_a=22689.8J/mol$

The activation energy Ea for this reaction is 22689.8 J/mol

3 0

2 years ago