All categories

3 years ago

PLS HELP ASAP I NEED NUMBER 8

Chemistry

2 answers:

marishachu [46]3 years ago

8 0

Explanation:

determine the genotypes of the parent organisms.

write down your "cross" (mating)

draw a p-square.

4. " ...

determine the possible genotypes of the offspring by filling in the p-square.

summarize results (genotypes & phenotypes of offspring)

mr Goodwill [35]3 years ago

5 0

A punnet square can be used to predict and see the possibilities of the genotypes.

You might be interested in

How many milligrams of sodium sulfide are needed to completely react with 25.00 ml of a 0.0100 m aqueous solution of cadmium nit

NARA [144]

Na₂S(aq) + Cd(NO₃)₂(aq) = CdS(s) + 2NaNO₃(aq)

v=25.00 mL
c=0.0100 mmol/mL
M(Na₂S)=78.046 mg/mmol

n(Na₂S)=n{Cd(NO₃)₂}=cv

m(Na₂S)=M(Na₂S)n(Na₂S)=M(Na₂S)cv

m(Na₂S)=78.046*0.0100*25.00≈19.5 mg

5 0

3 years ago

Read 2 more answers

An aqueous CsCl solution is 8.00 wt% CsCl and has a density of 1.0643 g/mL at 20°C. What is the boiling point of this solution?

umka2103 [35]

Answer: The boiling point of solution is 100.53

Explanation:

We are given:

8.00 wt % of CsCl

This means that 8.00 grams of CsCl is present in 100 grams of solution

Mass of solvent = (100 - 8) g = 92 grams

The equation used to calculate elevation in boiling point follows:

$\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of pure solution}$

To calculate the elevation in boiling point, we use the equation:

$\Delta T_b=iK_bm$

Or,

$\text{Boiling point of solution}-\text{Boiling point of pure solution}=i\times K_b\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Boiling point of pure solution = 100°C

i = Vant hoff factor = 2 (For CsCl)

$K_b$ = molal boiling point elevation constant = 0.51°C/m

$m_{solute}$ = Given mass of solute (CsCl) = 8.00 g

$M_{solute}$ = Molar mass of solute (CsCl) = 168.4 g/mol

$W_{solvent}$ = Mass of solvent (water) = 92 g

Putting values in above equation, we get:

$\text{Boiling point of solution}-100=2\times 0.51^oC/m\times \frac{8.00\times 1000}{168.4g/mol\times 92}\\\\\text{Boiling point of solution}=100.53^oC$

Hence, the boiling point of solution is 100.53

6 0

3 years ago

If all the water in 430.0 mL of a 0.45 M NaCI solution evaporates what is the mass of NaCI will remain

skad [1K]

Answer:

11.31 g.

Explanation:

Molarity is defined as the no. of moles of a solute per 1.0 L of the solution.

M = (no. of moles of solute)/(V of the solution (L)).

∴ M = (mass/molar mass)of NaCl/(V of the solution (L)).

∴ mass of NaCl remained after evaporation of water = (M)(V of the solution (L))(molar mass) = (0.45 M)(0.43 L)(58.44 g/mol) = 11.31 g.

6 0

3 years ago

Read 2 more answers

How might the government make sure scientific research is done in an ethical way?

antiseptic1488 [7]

Answer:

They could possibly make daily check ups to the research facility?

7 0

3 years ago

Explain how stagnant sulfureted water may greatly accelerate metallic corrosion. (See the Email Link in Moodle) | (7 marks)

slavikrds [6]

Answer:

sulfur promotes oxide-reduction reactions.

Explanation:

In stagnant water, some solutes tend to precipitate. When Sulfur precipitate and touch a metal, Sulfur is being reduced and the metal is oxidated. This depends of potential redox of each element.

6 0

3 years ago