Answer:
V = 34.55 L
Explanation:
Given that,
No of moles, n = 1.4
Temperature, T = 20°C = 20 + 273 = 293 K
Pressure, P = 0.974 atm
We need to find the volume of the gas. It can be calculated using Ideal gas equation which is :
PV=nRT
R is gas constant, 
Finding for V,
So, the volume of the gas is 34.55 L.
the particles of solids move but very slowly.
the particles of liquids move moderately fast
the particles of gas move very fast.
The reasons for this movement is the space the particles are together. Since there is not move space between solids, the particles move slowly, water there is moderate space, and air there is a lot of space.
Answer: The empirical formula for the given compound is 
Explanation : Given,
Percentage of C = 38.8 %
Percentage of H = 16.2 %
Percentage of N = 45.1 %
Let the mass of compound be 100 g. So, percentages given are taken as mass.
Mass of C = 38.8 g
Mass of H = 16.2 g
Mass of N = 45.4 g
To formulate the empirical formula, we need to follow some steps:
Step 1: Converting the given masses into moles.
Moles of Carbon =
Moles of Hydrogen = 
Moles of Nitrogen = 
Step 2: Calculating the mole ratio of the given elements.
For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 3.23 moles.
For Carbon = 
For Hydrogen = 
For Oxygen = 
Step 3: Taking the mole ratio as their subscripts.
The ratio of C : H : N = 1 : 5 : 1
Hence, the empirical formula for the given compound is 
Because ionic compounds' strong bonds form network structures, which have a stronger attraction than the covalent compounds which are molecules.
A foreign DNA molecule can be incorporated into a bacterial plasmid during a transformation reaction.
<h3>How to explain the reaction?</h3>
With the aid of two enzymes, ligase and restriction enzymes, a foreign DNA molecule can be incorporated into a bacterial plasmid during a transformation reaction. Each enzyme detects a target DNA sequence and cuts it nearby, while ligase aids in connecting the DNA. When two bits of DNA have complimentary bases, it facilitates their joining.
Plasmid and the insert fragment are both present in the microfuge tube, and they both have compatible sticky ends. However, the ligase has been denatured and is no longer active because the prior student left it outside rather than freezing it; despite this, we had already put the ligase into the tube. Ligase aids in binding the plasmid and insert fragments together, but because it is denatured in this instance, it will no longer be able to do so. As a result, no transformation process will take place. And since ligase links DNA fragments together by catalyzing the development of connections between the nearby nucleotides, the two fragments will not be able to unite.
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