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3 years ago

In a genotype what does the lower case letter represent

Chemistry

2 answers:

AlladinOne [14]3 years ago

6 0

The lower case letter represents the recessive allele.

vredina [299]3 years ago

3 0

Answer:

The lower case letter represents the recessive allele.

Explanation:

I just learned abt this today so im confident that this is correct

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What is the difference between a mixture and compound

bonufazy [111]

A compound is made up of a single type of molecule. A mixture, on the other hand, is made up of different substances that can be distinguished from one another.
As an example:
Water (pure water) is a compound since it's only made up of water molecules
Seawater is a mixture since it is made up of water molecules and soluble salts

3 0

3 years ago

Determine the number of valence electrons for magnesium

12345 [234]

Answer:

Magnesium has 2 valence electrons.

Explanation:

The atomic number of magnesium is 12. This means that it has 12 protons and 12 electrons. The first electron shell can hold 2 electrons, the second can hold 8, so the third will have 2 electrons. The outermost shell of the atom has two electrons. Therefore, magnesium has 2 valence electrons.

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4 years ago

Which move will produce light with the longer wavelength

mylen [45]

Answer:

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3 years ago

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marissa [1.9K]

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3 years ago

How many liters of gas will be in the closed reaction flask when 36.0L of ethane (C2H6) is allowed to react with 105.0L of oxyge

Ivan

Answer:- Volume of the gas in the flask after the reaction is 156.0 L.

Solution:- The balanced equation for the combustion of ethane is:

$2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(g)$

From the balanced equation, ethane and oxygen react in 2:7 mol ratio or 2:7 volume ratio as we are assuming ideal behavior.

Let's see if any one of them is limiting by calculating the required volume of one for the other. Let's say we calculate required volume of oxygen for given 36.0 L of ethane as:

$36.0LC_2H_6(\frac{7LO_2}{2LC_2H_6})$

= 126 L $O_2$

126 L of oxygen are required to react completely with 36.0 L of ethane but only 105.0 L of oxygen are available, It means oxygen is limiting reactant.

let's calculate the volumes of each product gas formed for 105.0 L of oxygen as:

$105.0LO_2(\frac{4LCO_2}{7L O_2})$

= 60.0 L $CO_2$

Similarly, let's calculate the volume of water vapors formed:

$105.0L O_2(\frac{6L H_2O}{7L O_2})$

= 90.0 L $H_2O$

Since ethane is present in excess, the remaining volume of it would also be present in the flask.

Let's first calculate how many liters of it were used to react with 105.0 L of oxygen and then subtract them from given volume of ethane to know it's remaining volume:

$105.0LO_2(\frac{2LC_2H_6}{7LO_2})$

= 30.0 L $C_2H_6$

Excess volume of ethane = 36.0 L - 30.0 L = 6.0 L

Total volume of gas in the flask after reaction = 6.0 L + 60.0 L + 90.0 L = 156.0 L

Hence. the answer is 156.0 L.

5 0

3 years ago