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MariettaO [177]
3 years ago
7

What must happen before a body cell can begin mitotic cell division

Chemistry
1 answer:
bixtya [17]3 years ago
5 0

Prophase is the first stage of Mitosis and also the beginning stage in which the cell prepares itself to begin to other stages of Mitosis.

Hope this helps!!!

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When an excited electron in a hydrogen atom falls from ????=5 to ????=2, a photon of blue light is emitted. If an excited electr
poizon [28]

Answer:

n = 3

Explanation:

Given the formula for the transition energy of an atom with 1 electron:

E=-13.6*Z^{2}*(\frac{1}{n_{f}^{2}}-\frac{1}{n_{i}^{2}} ) eV

For the H transition n=5 to n=2:

E=-13.6*(\frac{1}{4}-\frac{1}{25} ) eV=-2.856 eV

Then we solve for nf with Z=2 (Helium)

n_{f}=\sqrt{\frac{n_{i}^{2}*Z^{2}*13.6 eV }{2.856eV*n_{i}^{2}+13.6eV*Z^{2}} }

n_{f}=\sqrt{\frac{4^{2}*2^{2}*13.6 eV }{2.856eV*4^{2}+13.6eV*2^{2}} }=3

Is near 3, actually the energy of the transitions are:

H (5⇒2) = -2.85 eV = 434 nm (Dark blue)

He (4⇒3) = -2.64 eV = 469 nm (Light blue)

I thought it was cool to see the actual colors. Included them.

4 0
3 years ago
How many liters of H2O gas are produced when
anyanavicka [17]
1 mole C3H8 produces 4 moles H2O. So, first we convert 32 grams of propane to moles and then find moles of H2O. Then convert moles of H2O to grams of H2O
Moles of H2O produced = 32 g C3H8 x 1 mole/44 g x 4 moles H2O/mole C3H8 = 2.909 moles H2O
Grams H2O produced = 2.909 moles H2O x 18 g/mole = 52.36 g = 52 g H2O
8 0
2 years ago
Answer is C
Vladimir [108]

Answer:

c)

Explanation:

add water to the acid to decrease its molarity

5 0
3 years ago
6.25 g of HCl(g), a strong acid, is dissolved in water to make a 280-ml solution.
elena-14-01-66 [18.8K]

pH= -log[0,611]= 0,21

7 0
3 years ago
How many milliliters of 0.200 M NH4OH are needed to react with 12.0 mL of 0.550 M FeCl3?
trapecia [35]

Answer:

9.9 ml of 0.200M NH₄OH(aq)

Explanation:

3NH₄OH(Iaq) + FeCl₃(aq) => NH₄Cl(aq) + Fe(OH)₃(s)

?ml of 0.200M NH₄OH(aq) reacts completely with  12ml of 0.550M FeCl₃(aq)

1 x Molarity NH₄OH x Volume Am-OH Solution(L) = 2 x Molarity FeCl₃ x Volume FeCl₃ Solution

1(0.200M)(Vol Am-OH Soln) = 3(0.550M)(0.012L)

=> Vol Am-OH Soln = 3(0.550M)(0.012L)/1(0.200M) = 0.0099 Liter = 9.9 milliliters  

5 0
3 years ago
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