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3 years ago

Why aren’t dominant genes ALWAYS more common?

Chemistry

1 answer:

Zolol [24]3 years ago

5 0

Answer:

selective breeding

Explanation:

Example: pretend that rabbits can only be brown or black and black is dominant. If only the brown rabbits reproduce it will eventually be more common Hope this helps

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What is the capacity to do work known as?

Jlenok [28]

Energy

Simply Energy.

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3 years ago

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Describe the structure of ammonium lauryl sulfate. Refer to the given diagram. Your answer should include the type of bonding, t

Anit [1.1K]

Answer:

The compound contains lauryl sulfate and ammoium ions. Lauryl sulfate contains lauric acid (in black and white) , the fatty acid formed by the covalent bonds between C-C attached to hydrogens, and sulfate ions attached to lauric acid (in red) with C-S covalent bond. Sulfer is attached to oxygen by covalent bonds. In Ammonium ions, N is surrounded by four hydrogen atoms.

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4 years ago

Which clouds are often associated with thunder and lightning?

FrozenT [24]

Cumulonimbus clouds, of course!. These clouds are known to carry rain, hail, and thunder. Bigger versions are known as supercells, deadly storms that can spew out tons of rain, hail, wind, and even tornadoes!

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3 years ago

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A solution contains 0.60 mg/ml mn2+. what minimum mass of kio4 must me added to 5.00 ml of the solution in order to completely o

azamat

The given solution of Mn²⁺ is 0.60 mg/mL.
Hence mass of Mn²⁺ in 5 mL of solution = 0.60 mg/mL x 5 mL = 3 mg

Molar mass of Mn = 54.9 g/mol
Hence, moles of Mn²⁺ = 3 x 10⁻³ g / 54.9 g/mol = 5.46 x 10⁻⁵ mol

The balanced equation for the reaction is,
2Mn²⁺ + 5KIO₄ + 3H₂O → 2MnO₄⁻ + 5KIO₃ + 6H⁺

The stoichiometric ratio between Mn²⁺ and KIO₄ is 2 : 5

Hence, moles of KIO₄ reacted = 5.46 x 10⁻⁵ mol x (5 / 2)
= 13.65 x 10⁻⁵ mol
Molar mass of KIO₄ = 230 g/mol

Hence needed mass of KIO₄ = 13.65 x 10⁻⁵ mol x 230 g/mol
= 0.031395 g
= 31.395 mg
≈ 31.4 mg

5 0

3 years ago

19.3 g of cadmium hydroxide reacted with 15.21 g of hydrobromic acid. How many grams of water can be made?

Alecsey [184]

Answer:

$m_{H_2O}=3.384gH_2O$

Explanation:

Hello,

In this case, the chemical reaction is:

$Cd(OH)_2+2HBr\rightarrow CdBr_2+2H_2O$

Thus, we first identify the limiting reactant by computing the yielded moles of water by both of the reactants:

$n_{H_2O}^{by\ Cd(OH)_2}=19.3gCd(OH)_2*\frac{1molCd(OH)_2}{146.4gCd(OH)_2}*\frac{2molH_2O}{1molCd(OH)_2}=0.264molH_2O\\\\n_{H_2O}^{by\ HBr}=15.21gHBr*\frac{1molHBr}{80.9gHBr}*\frac{2molH_2O}{2molHBr}=0.188molH_2O$

In such a way, since HBr yields less water than cadmium hydroxide, we infer that HBr is the limiting one, therefore, the yielded mass of water are:

$m_{H_2O}=0.188molH_2O*\frac{18gH_2O}{1molH_2O}\\ \\m_{H_2O}=3.384gH_2O$

Regards.

4 0

3 years ago