During MITOSIS, the parent, diploid (2n), cell is divided to create two identical, diploid (2n), daughter cells. ... After cytokinesis, the ploidy of the daughter cells remains the same because each daughter cell contains 4 chromatids, as the parent cell did.

4 0

3 years ago

Hello! I just need a little bit of help. I'm supposed to design an experiment on how reaction rates are determined and affected

alexandr1967 [171]

Get 3 cups of water at the exact same temperature, using the thermometer to check.
Label the cups as ‘whole’, ‘pieces’, and ‘crushed’
Next, get something to dissolve, in this case, polident. Take one of the polident tablets and break it into 4 pieces, and set it aside.
Take another polident tablet and this time put it into a different cup, and crush it. Set it aside.
Keep the last tablet whole.
Set up your stopwatch and drop the polident tablet that is whole in the cup labeled ‘whole’, starting the stopwatch at the same time.
Watch the cup and see when the tablet is fully dissolved, then stop the stopwatch.
Record the time in the table.
Repeat steps 6-8 for both the ‘pieces’ and ‘crushed’ tablets.

Hope this helps! Please let me know if you need more help, or if you think my answer is incorrect. Brainliest would be MUCH appreciated. Have a great day!

Stay Brainy!

− $xXheyoXx$

3 0

2 years ago

What is the mass in grams of one mole of many of any substance known as

OLEGan [10]

The molecular weight of the substance.

7 0

3 years ago

The rate constant k for a certain reaction is measured at two different temperatures temperature 376.0 °c 4.8 x 108 280.0 °C 2.3

9966 [12]

Answer:

The activation energy for this reaction = 23 kJ/mol.

Explanation:

Using the expression,

$\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )$

Where,

$k_1\ is\ the\ rate\ constant\ at\ T_1$

$k_2\ is\ the\ rate\ constant\ at\ T_2$

$E_a$ is the activation energy

R is Gas constant having value = 8.314×10⁻³ kJ / K mol

$k_2=2.3\times 10^8$

$T_2=280\ ^0C$

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (280 + 273.15) K = 553.15 K

$T_2=553.15\ K$

$k_1=4.8\times 10^8$

$T_1=376\ ^0C$

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (376 + 273.15) K = 649.15 K

$T_1=649.15\ K$

So,

$\left(\ln \left(\:\frac{4.8\times \:\:\:10^8}{2.3\times \:\:\:10^8}\right)\right)\:=-\frac{E_a}{8.314\times \:10^{-3}\ kJ/mol.K}\times \:\left(\frac{1}{649.15\ K}-\frac{1}{553.15\ K}\right)$

$E_a=-\frac{10^{-3}\times \:8.314\ln \left(\frac{10^8\times \:4.8}{10^8\times \:2.3}\right)}{-\frac{96}{359077.3225}}\ kJ/mol$

$E_a=-\frac{\frac{8.314\ln \left(\frac{4.8}{2.3}\right)}{1000}}{-\frac{96}{359077.3225}}\ kJ/mol$

$E_a=22.87\ kJ/mol$

<u>The activation energy for this reaction = 23 kJ/mol.</u>

6 0

3 years ago