All categories

3 years ago

When might a dominant trait appear?

Chemistry

1 answer:

daser333 [38]3 years ago

7 0

Answer:

A dominant allele will always allow a specific trait to show up no matter if we have two dominant copies (BB) or just one (Bb). A trait from a recessive allele will only appear if it is paired with another recessive allele

Explanation:

(•_•)

<) )╯all the single ladies

/ \

(•_•)

\( (> all the single ladies

/ \

(•_•)

<) )╯oh oh oh

You might be interested in

Assume that you have developed a tracer with first-order kinetics and k = 0.1 s − 1 . For any given mass, how long will it take

anygoal [31]

Answer: It will take 6.93 sec for tracer concentration to drop by 50% and 13.9 sec for tracer concentration to drop by 75%

Explanation:

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant = $0.1s^{-1}$

t = age of sample

a = let initial amount of the reactant = 100 g

a - x = amount left after decay process

a) for tracer concentration to drop by 50%

a - x = amount left after decay process = 50

$t=\frac{2.303}{k}\log\frac{100}{50}$

$t=\frac{0.693}{0.1}$

$t=6.93sec$

It will take 6.93 sec for tracer concentration to drop by 50%

b) for tracer concentration to drop by 75 %

a - x = amount left after decay process = 25

$t=\frac{2.303}{k}\log\frac{100}{100-75}$

$t=\frac{2.303}{0.1}\times \log\frac{100}{25}$

$t=13.9sec$

It will take 13.9 sec for tracer concentration to drop by 75%.

5 0

3 years ago

Fe(s) + CuSO4(aq) <===> Cu(s) + FeSO4(aq)

Ludmilka [50]

Answer : The original concentration of copper (II) sulfate in the sample is, $5.6\times 10^{-1}g/L$

Explanation :

Molar mass of Cu = 63.5 g/mol

First we have to calculate the number of moles of Cu.

Number of moles of Cu = $\frac{\text{Mass of Cu}}{\text{Molar mass of Cu}}=\frac{89\times 10^{-3}g}{63.5g/mol}=1.40\times 10^{-3}mole$

Now we have to calculate the number of moles of $CuSO_4$

Number of moles of Cu = Number of moles of $CuSO_4$

Number of moles of $CuSO_4$ = $1.40\times 10^{-3}mole$

Now we have to calculate the molarity of $CuSO_4$

$\text{Molarity}=\frac{\text{Moles of }CuSO_4\times 1000}{\text{Volume of solution (in mL)}}$

Now put all the given values in this formula, we get:

$\text{Molarity}=\frac{1.40\times 10^{-3}mole\times 1000}{400.mL}=0.0035M$

To change mol/L into g/L, we need to multiply it with molar mass of $CuSO_4$

Molar mass of $CuSO_4$ = 159.609 g/mL

Concentration in g/L = $0.0035M\times 159.609g/mol=0.5586g/L\approx 5.6\times 10^{-1}g/L$

Thus, the original concentration of copper (II) sulfate in the sample is, $5.6\times 10^{-1}g/L$

3 0

3 years ago

Polonium crystallizes in a simple cubic unit cell with an edge length of 3.36 Å.

Diano4ka-milaya [45]

a. The unit cell is the smallest group of atom which have overall symmetry of a crystal, and from which is the entire letters can be buled built up by repetition in 3 dimensions.

b. The volume(v) of the unit cell is equal to the cell edge length (a)cubed.

c. density of polonium is 9.32g/cm3.

8 0

3 years ago

Describe two environmental problems associated with long chain molecules

Leya [2.2K]

do you mean polymers or organic compounds?

8 0

3 years ago

What happens when a current is applied to two electrodes immersed in pure water?

frosja888 [35]

Nothing at all happens because pure water cannot conduct electricity

7 0

3 years ago