Answer:
prophase I
Explanation:
Homologous chromosomes pair up during prophase I of meiosis.
Answer:
This reaction has infinite ways to be balanced
Explanation:
To balance this equation we can use the algebraic method:
N2O4(g) + CO → NO(g) + CO2(g)+NO2(g)
Where we write each molecule as a letter:
A + B → C + D + E
Then, we write the equations according the number of atoms of each molecule. That is:
Oxygen → 4A + B = C + 2D + 2E <em>(1)</em>
Nitrogen → 2A = C + E <em>(2)</em>
Carbon → B = D <em>(3)</em>
Then, we have to give 1 arbitral number for a letter. For example:
B = 1; D = 1
<em>(1) </em>4A + 1 = C + 2 + 2E
4A = C + 2E + 1
2A = C + E <em>(2) </em>Twice <em>(2):</em>
4A = 2C + 2E
Subtracting <em>(1) </em>in <em>(2)</em>
C + 2E + 1 = 2C + 2E
C + 1 = 2C
1 = C
Si 1 = C:
4A + 1 = 1 + 2 + 2E
4A = 2 + 2E <em>(1)</em>
y:
2A = 1 + E <em>(2)</em>
Twice:
4A = 2 + 2E
As <em>(1) </em>and <em>(2) </em>are the same equation:
<h3>This reaction has infinite ways to be balanced</h3><h3 />
For example:
N2O4(g) + CO → NO(g) + CO2(g)+NO2(g)
Answer:
<h3>the equilibrium constant of the decomposition of hydrogen bromide is 0.084</h3>
Explanation:
Amount of HBr dissociated
2HBr(g) ⇆ H2(g) + Br2(g)
Initial Changes 2.15 0 0 (mol)
- 0.789 + 0.395 + 0.395 (mol)
At equilibrium 1.361 0.395 0.395 (mole)
Concentration 1.361 / 1 0.395 / 1 0.395 / 1
at equilibrium (mole/L)
![K_c=\frac{[H_2][Br_2]}{[HBr]^2} \\\\=\frac{(0.395)(0.395)}{(1.361)^2} \\\\=\frac{0.156025}{1.852321} \\\\=0.084](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BH_2%5D%5BBr_2%5D%7D%7B%5BHBr%5D%5E2%7D%20%5C%5C%5C%5C%3D%5Cfrac%7B%280.395%29%280.395%29%7D%7B%281.361%29%5E2%7D%20%5C%5C%5C%5C%3D%5Cfrac%7B0.156025%7D%7B1.852321%7D%20%5C%5C%5C%5C%3D0.084)
<h3>Therefore, the equilibrium constant of the decomposition of hydrogen bromide is 0.084</h3>
Answer:
i think its Environmental and mutagens
Explanation:
