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Lina20 [59]
3 years ago
13

The pedigree represents how a certain sex-linked recessive trait is inherited.

Chemistry
2 answers:
Zina [86]3 years ago
6 0

Answer:c

Explanation:

i think that what is was for me

Phantasy [73]3 years ago
6 0
XDY i believe is the answer
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How would you make a 2.00L of 0. 500M sodium chloride solution. (assume you have a fully equipped lab with water); sketch, calc
Gwar [14]

Answer:

Sodium chloride solution:

First you need to calculate the mass of salt needed (done in the explanation), which is 58.44g. Then it have to be weighted in an analytical balance in a weighting boat and then transferred into a 2L volumetric flask that is going to be filled until the mark with distilled water.

Sulfuric acid dilution:

First you need to calculate the volume needed (done in the explanation), it is 16.6 mL. Using a graduated pipette one measures this volume and transfer it into a 2L volumetric flask that is already half filled with distilled water, and then one fills it until its mark.

Explanation:

Sodium chloride solution:

Each liter of a 0.500M solution has half mol, so 2L of said solution has 1 mol of salt. Sodium chloride molar mass is 58.44g/mol, so in 2L of solution there is 58.44g of salt. That`s the mass that`s going to be weighted and transferred to a 2L volumetric flask.

Sulfuric acid dilution:

This is the equation for dilution of solutions:

c_{1} v_{1} =c_{2} v_{2}

Where "c1" stands for the initial concentration (stock solution concentration), "v1" for the initial volume (volume of stock solution used), "c2" for the desired concentration and "v2" for the desired volume.

When we are diluting from a stock solution we want to know how much do we have to pipette from the stock solution into our volumetric flask. We do so by isolating the "v1" term from the dilution equation:

v_{1} =\frac{c_{2} v_{2} }{c_{1} }

in this case that would be:

v_{1} =\frac{0.100 x2.0 }{12.0 }=0.0166L=16.6mL

5 0
3 years ago
What common household element can, over time, reduce airflow, insulate components, reduce heat exchange or even cause the system
lina2011 [118]

Answer:

The correct answer is Dust

Explanation

Dust is a dry dirt in powder form usually found on surfaces of items in a building, it comprises of very small particles of soil, sand and sometimes includes toxic substances, skin cells,bacteria, soil particles, particles of clothing material, tiny pieces of dead insects and pollen

4 0
4 years ago
Read 2 more answers
Consider two gases, A and B, are in a container at room temperature. What effect will the following changes have on the rate of
Oksanka [162]

Answer:

B: increase.

Explanation:

When we are considering two gases A and B in a container at room temperature .

We have to find the change on  rate of reaction when the number of molecules of gases A is doubled

Let [A]=a and [B]=b

A+B\rightarrow product

Rate of reaction

R_1=k[A][B]=kab

We know that concentration is increases with increase in number of moles

When the number of molecules of gases A is doubled then concentration of gases A increases.

Therefore ,[A]=2a

Rate of reaction

R_2=k(2a)(b)=2kab

R_2=2R_1

Hence, the rate of reaction is  2 times the initial rate of reaction.Therefore, the rate of reaction will increase when the number of molecules of gases A is doubled.

Answer: B: increase.

4 0
4 years ago
What is a property of a solution with a high pH? Select all that apply. *
Lena [83]

Answer:

Bases (solutions with a high pH) feel slipper, have an -OH group, and are corrosive.

8 0
3 years ago
A 101.2 ml sample of 1.00 m naoh is mixed with 50.6 ml of 1.00 m h2so4 in a large styrofoam coffee cup; the cup is fitted with a
Murrr4er [49]

The enthalpy change of the reaction when sodium hydroxide and sulfuric acid react can be calculated using the mass of solution, temperature change, and specific heat of water.

The balanced chemical equation for the reaction can be represented as,

H_{2}SO_{4}(aq) + 2NaOH (aq) ----> Na_{2}SO_{4}(aq) + 2H_{2}O(l)

Given volume of the solution = 101.2 mL + 50.6 mL = 151.8 mL

Heat of the reaction, q = m C .ΔT

m is mass of the solution = 151.8 mL * \frac{1 g}{1 mL} = 151.8 g

C is the specific heat of solution = 4.18 \frac{J}{g. ^{0}C}

ΔT is the temperature change = 31.50^{0}C - 21.45^{0}C = 10.05^{0}C

q = 151.8 g (4.18 \frac{J}{g ^{0}C})(10.05^{0}C) = 6377 J

Moles of NaOH = 101.2 mL * \frac{1L}{1000 mL}*\frac{1.00 mol}{L} = 0.1012 mol NaOH

Moles of H_{2}SO_{4} = 50.6 mL * \frac{1 L}{1000 mL} * \frac{1.0 mol}{1 L} = 0.0506 mol H_{2}SO_{4}

Enthalpy of the reaction = \frac{6377 J*\frac{1kJ}{1000J}}{0.0506 mol} = 126 kJ/mol

5 0
3 years ago
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