All categories

3 years ago

Why is it important for organelles to double during the G2 phase of the cell cycle?

Chemistry

1 answer:

boyakko [2]3 years ago

6 0

Explanation:

so then we would know about the reproduction system and how we are born and what happens with the cells inside the womens womb and how the sperm cell meets the egg cell and creates the egg

You might be interested in

What is true about the speed of molecules in a sample of gas held at a constant temperature?

vovikov84 [41]

At the constant temperatures the lighter the gas molecules, the faster the average velocity of the gas molecules.

3 0

3 years ago

Number of

Karolina [17]

Mass plus atomic and number

8 0

3 years ago

4.2g of sodium bicarbonate is equivalent to how many moles of sodium bicarbonate

Mnenie [13.5K]

<span>The mass of one mole of sodium bicarbonate (aka NaHCO3) is equal to 1 * 22.99g/mol + 1 * 1.00g/mol + 1 * 12.01g/mol + 3 * 16.00g/mol = 83.91g/mol. From this, we can convert 4.2g of NaHCO3 to moles by dividing by 83.91g/mol, to get 0.050 moles of sodium bicarbonate.</span>

7 0

3 years ago

Read 2 more answers

I don't know which the correct answer it is ???

Virty [35]

When I answered i got D

4 0

3 years ago

A 0.4272 g sample of an element contains 2.241 x 10 ^21 atoms . what is the symbol of the element?

grin007 [14]

Answer:

Likely $\rm In$ (indium.)

Explanation:

Number of atoms: $N = 2.241 \times 10^{21}$ .

Dividing, $N$ , the number of atoms by the Avogadro constant, $N_A \approx 6.023 \times 10^{23} \; \rm mol^{-1}$ , would give the number of moles of atoms in this sample:

$\begin{aligned} n &= \frac{N}{N_{A}} \\ &\approx \frac{2.241 \times 10^{21}}{6.023 \times 10^{23}\; \rm mol^{-1}} \approx 3.72 \times 10^{-3}\; \rm mol \end{aligned}$ .

The mass of that many atom is $m = 0.4272\; \rm g$ . Estimate the average mass of one mole of atoms in this sample:

$\begin{aligned}M &= \frac{m}{n} \\ &\approx \frac{0.4272\; \rm g}{3.72 \times 10^{-3}\; \rm mol} \approx 114.82\; \rm g \cdot mol^{-1}\end{aligned}$ .

The average mass of one mole of atoms of an element ( $114.82\; \rm g \cdot mol^{-1}$ in this example) is numerically equal to the average atomic mass of that element. Refer to a modern periodic table and look for the element with average atomic mass $114.82$ . Indium, $\rm In$ , is the closest match.

5 0

3 years ago