Answer:
1. All red calves i.e. RR
2. All roan calves i.e RW
3. 2 red calves (RR) and two roan calves (RW)
Explanation:
According to this question, a gene coding for fur colour in cattle is involved. Red alleles (R) and white alleles (W) are co-dominant to produce a roan cattle (RW). The possible traits of the following crosses are (see attached punnet square):
1) A red bull (RR) is mated to a red (RR) cow: All red calves i.e. RR
2) A red (RR) bullis mated with white (WW) cow: All roan calves i.e RW
3) A roan bull (RW) is mated with red (RR) cow: 2 red calves (RR) and two roan calves (RW).
Orbital
All atoms have the same number of electrons as protons. Negative electrons are attracted to the positive nucleus. This force of attraction keeps electrons constantly moving around the nucleus. The region where an electron is most likely to be found is called an orbital.
Here you go :)
Answer:
1.00 × 10¹⁸
Explanation:
1. Calculate the <em>energy of one photon</em>
The formula for the energy of a photon is
<em>E</em> = <em>hc</em>/λ
<em>h</em> = 6.626 × 10⁻³⁴ J·s; <em>c</em> = 2.998 × 10⁸ m·s⁻¹
λ = 477 nm = 477 × 10⁻⁹ m Insert the values
<em>E</em> = (6.626 × 10⁻³⁴ × 2.998× 10⁸)/(477 × 10⁻⁹)
<em>E</em> = 4.165× 10⁻¹⁹ J
2. Calculate the <em>number of photons</em>
Divide the total energy by the energy of one photon.
No. of photons = 0.418 × 1/4.165 × 10⁻¹⁹
No. of photons = 1.00 × 10¹⁸
3. 9
20. 60 (1hour)
To get to 60 from 20 we multiply by three. 3 times 3 is 9.
A is true
B is true
C is true
D is true
E is false
Answer : The half life of 28-Mg in hours is, 6.94
Explanation :
First we have to calculate the rate constant.
Expression for rate law for first order kinetics is given by:
where,
k = rate constant
t = time passed by the sample = 48.0 hr
a = initial amount of the reactant disintegrate = 53500
a - x = amount left after decay process disintegrate = 53500 - 10980 = 42520
Now put all the given values in above equation, we get
Now we have to calculate the half-life.
Therefore, the half life of 28-Mg in hours is, 6.94
