Answer:
1. C + O₂ → CO₂
2. C + CO₂ → 2 CO
3. Fe₂O₃ + 3 CO → 2 Fe + 3 CO₂
Answer:
2) 0.4 mol
Explanation:
Step 1: Given data
- Volume of the solution (V): 500 mL
- Molar concentration of the solution (M): 0.8 M = 0.8 mol/L
Step 2: Convert "V" to L
We will use the conversion factor 1 L = 1000 mL.
500 mL × 1 L/1000 mL = 0.500 L
Step 3: Calculate the moles of KBr (solute)
The molarity is the quotient between the moles of solute (n) and the liters of solution.
M = n/V
n = M × V
n = 0.8 mol/L × 0.500 L = 0.4 mol
The answer is B. neutrons have no charge
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).
The answer is the second choice
