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3 years ago

Mendel crossed tall pea plants with short pea plants. the offspring were all tall plants.

Chemistry

1 answer:

ad-work [718]3 years ago

6 0

Answer:

1. Allele for shortness (t)

2. Allele for tallness (T)

Explanation:

Gregor Mendel performed numerous experiments using pea plants (Pisium sativum). He tested the height traits of pea plants by crossing tall pea plants (TT) and short pea plants (tt) to obtain offsprings that were all tall (Tt) in the first filial generation (F1).

This is possible because the allele for tallness is DOMINANT while the allele for shortness is RECESSIVE. This means that tallness allele (T) will mask the phenotypic expression of shortness allele (t) when in an heterozygous state (Tt) as in the F1 generation.

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Determine the mass of CaCO3 required to produce 40.0 mL CO2 at STP. Hint use molar volume of an ideal gas (22.4 L)

cupoosta [38]

Answer:

$m_{CaCO_3}=0.179gCaCO_3$

Explanation:

Hello,

In this case, since the undergoing chemical reaction is:

$CaCO_3(s)\rightarrow CaO(s)+CO_2(g)$

The corresponding moles of carbon dioxide occupying 40.0 mL (0.0400 L) are computed by using the ideal gas equation at 273.15 K and 1.00 atm (STP) as follows:

$PV=nRT\\\\n=\frac{PV}{RT}=\frac{1.00 atm*0.0400L}{0.082\frac{atm*L}{mol*K}*273.15 K})=1.79x10^{-3} mol CO_2$

Then, since the mole ratio between carbon dioxide and calcium carbonate is 1:1 and the molar mass of the reactant is 100 g/mol, the mass that yields such volume turns out:

$m_{CaCO_3}=1.79x10^{-3}molCO_2*\frac{1molCaCO_3}{1molCO_2} *\frac{100g CaCO_3}{1molCaCO_3}\\ \\m_{CaCO_3}=0.179gCaCO_3$

Regards.

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3 years ago

Question 1

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How many neutrons are found in a stable isotope of uranium-235?

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The combining of the nuclei of atoms is known as nuclear _____.

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A solution is made by adding 35.5 mL of concentrated hydrochloric acid ( 37.3 wt% , density 1.19 g/mL1.19 g/mL ) to some water i

erastova [34]

Answer:

1.73 M

Explanation:

We must first obtain the concentration of the concentrated acid from the formula;

Co= 10pd/M

Where

Co= concentration of concentrated acid = (the unknown)

p= percentage concentration of concentrated acid= 37.3%

d= density of concentrated acid = 1.19 g/ml

M= Molar mass of the anhydrous acid

Molar mass of anhydrous HCl= 1 +35.5= 36.5 gmol-1

Substituting values;

Co= 10 × 37.3 × 1.19/36.5

Co= 443.87/36.6

Co= 12.16 M

We can now use the dilution formula

CoVo= CdVd

Where;

Co= concentration of concentrated acid= 12.16 M

Vo= volume of concentrated acid = 35.5 ml

Cd= concentration of dilute acid =(the unknown)

Vd= volume of dilute acid = 250ml

Substituting values and making Cd the subject of the formula;

Cd= CoVo/Vd

Cd= 12.16 × 35.5/250

Cd= 1.73 M

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