Answer:
= 25 %
Explanation:
Given that :
the genotype attributed to trolls having one eye = (EE, Ee); that is homzygous dominant or heterozygote dominant for eye number
Likewise those that posses two eyes have (ee) ; that is homzygous recessive
Now; two heterozygous one eyed trolls are crossed ; we have :
Ee × Ee
The punnet square for this cross looks like what we have below:
E e
E EE Ee
e Ee ee
From the above cross; we have:
1 EE ---- (homozygous dominant ) indicating troll having one eye
2 Ee ---- ( heterozygote dominant ) indicating troll having one eye
1 ee ----- ( homozygous recessive ) indicating troll having two eyes
The genotypic ratio is 1:2:1
Thus; the expected genotype ratio of the two eyed offspring = 
= 25 %
Answer:
1. 0.662 M
2. 2.294 M
3. 0.357 M
Explanation:
The Molarity of a compound is simply defined as the mole of compound per unit litre of the solution. It can be obtained mathematically by the following equation:
Molarity = mole /Volume
With the equation above, let us calculate the molarity for the question given above.
1. Mole of CuSO4 = 0.49 mol Volume = 0.74 L.
Molarity of CuSO4 =?
Molarity = mole /Volume
Molarity of CuSO4 = 0.49/0.74
Molarity of CuSO4 = 0.662 M
2. Mole of Co(NO3)2 = 0.78 mol Volume = 0.34 L.
Molarity of Co(NO3)2 =?
Molarity = mole /Volume
Molarity of Co(NO3)2 = 0.78/0.34
Molarity of Co(NO3)2 = 2.294 M
3. Mole of K2Cr2O4 = 0.30 mol Volume = 0.84L.
Molarity of K2Cr2O4 =?
Molarity = mole /Volume
Molarity of K2Cr2O4 = 0.30/0.84
Molarity of K2Cr2O4 = 0.357 M
Answer:
N₂O₂
Explanation:
The empirical formula NO tells us that the number of Nitrogen atoms will always be equal to the number of Oxygen atoms.
The <em>molar mass of the simplest compound, NO</em>, would be:
- Molar Mass of NO = Molar Mass of N + Molar Mass of O = 30 g/mol
We <u>divide the molar mass of the problem compound by the molar mass of the simplest compound</u>:
This means we<u> multiply the empirical formula times </u><u>2</u>:
- The molecular formula is N₂O₂
<u>Answer:</u> The initial amount of Uranium-232 present is 11.3 grams.
<u>Explanation:</u>
All the radioactive reactions follows first order kinetics.
The equation used to calculate half life for first order kinetics:
We are given:
Putting values in above equation, we get:
Rate law expression for first order kinetics is given by the equation:
![k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B2.303%7D%7Bt%7D%5Clog%5Cfrac%7B%5BA_o%5D%7D%7B%5BA%5D%7D)
where,
k = rate constant = 
t = time taken for decay process = 206.7 yrs
![[A_o]](https://tex.z-dn.net/?f=%5BA_o%5D)
= initial amount of the reactant = ?
[A] = amount left after decay process = 1.40 g
Putting values in above equation, we get:
![0.0101yr^{-1}=\frac{2.303}{206.7yrs}\log\frac{[A_o]}{1.40}](https://tex.z-dn.net/?f=0.0101yr%5E%7B-1%7D%3D%5Cfrac%7B2.303%7D%7B206.7yrs%7D%5Clog%5Cfrac%7B%5BA_o%5D%7D%7B1.40%7D)
![[A_o]=11.3g](https://tex.z-dn.net/?f=%5BA_o%5D%3D11.3g)
Hence, the initial amount of Uranium-232 present is 11.3 grams.
No idea.. I think if you take angle (<) MNL then divide those...
