Answer: The derivative of a constant term is always 0. So the acceleration of the body would be zero. Hence, the acceleration of a body moving with uniform velocity will always be zero.
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Answer:
1. C- Three.
2. A- Methionine
3. D- Translocation.
4. C- OH.
5. A - 5'
6. A - 3' carbon
7. A. adenine and guanine
Explanation:
1. A codon is a group of three nucleotide sequence that encodes or specifies an amino acid. This means that, during translation (second stage of gene expression), when a CODON is read, an amino acid is added to the growing peptide chain.
2. The codon that initiates the translation process is called a start codon. It has a sequence: AUG and it specifies Methionine amino acid. Hence, during translation where a tRNA binds to the mRNA codon to read it and add its corresponding amino acid, a tRNA with a complementary sequence of AUG (start codon) binds to it and carries Methionine amino acid.
3. Translocation is a process during translation whereby the mRNA-tRNA moeity moves forward in the ribosome to allow another codon to move into the vacant site for translation process to continue.
4. The sugar component of a nucelotide that makes up the nucleic acid (DNA or RNA) i.e. ribose or deoxyribose, contains an hydroxyll functional group (-OH).
5. A nucleotide consists of a pentose (five carbon) sugar, phosphate group and a nitrogenous base. The phosphate group (PO43-) is attached to the 5' carbon of the sugar molecule.
6. The free hydroxyll group (-OH) of the five carbon sugar molecule in DNA is attached to its 3' carbon.
7. Nitrogenous bases are the third component of a nucleotide, the other two being pentose sugar and phosphate group. The nitrogenous bases are four viz: Adenine, Guanine, Cytosine, and Thymine. These bases are classified into Purines and Pyrimidines based on the similarity in their structure. Adenine (A) and Guanine (G) are Purines because they possess have two carbon-nitrogen rings, as opposed to one possessed by Pyrimidines (Thymine and Cytosine).
Answer:
207.03°C
Explanation:
The following data were obtained from the question:
V1 (initial volume) = 6.80 L
T1 (initial temperature) = 52.0°C = 52 + 273 = 325K
P1 (initial pressure) = 1.05 atm
V2 (final volume) = 7.87 L
P2 (final pressure) = 1.34 atm
T2(final temperature) =?
Using the general gas equation P1V1/T1 = P2V2/T2, the final temperature of the gas sample can be obtained as follow:
P1V1/T1 = P2V2/T2
1.05 x 6.8/325 = 1.34 x 7.87/T2
Cross multiply to express in linear form as shown below:
1.05 x 6.8 x T2 = 325 x 1.34 x 7.87
Divide both side by 1.05 x 6.8
T2 = (325 x 1.34 x 7.87) /(1.05 x 6.8)
T2 = 480.03K
Now, let us convert 480.03K to a number in celsius scale. This is illustrated below:
°C = K - 273
°C = 480.03 - 273
°C = 207.03°C
Therefore, the final temperature of the gas will be 207.03°C
Answer:Butane > ethane > methane, because between bigger molecules there are stronger van der Waals forces and also higher molar mass means they need to be given more energy to have enough kinetic energy to move quickly, freely in gas.
There are multiple butene isomers (Butene) and some (2-Butenes - cis and trans) actually have higher boiling point than n-Butane (there is also Isobutane, of course, with quite much lower boiling point than all of them) and some (1-Butene, Isobutylene) have lower, so this isn't really a fair or simple question. But on simplest level, it can again be said that 1-butene has lower boiling point because it has very similar shape but slightly lower molar mass (2H less) than n-butane.
Explanation:
Answer:
Macromolecules. A very large organic molecule composed of many smaller molecules, 1)Carbohydrates, 2)proteins, 3)lipids, 4)nucleic acids. Three of the four classes of macromolecules that are polymers. 1.Carbohydrates.
