Answer:
Ribosomes preform biological synthesis. (mRNA Translation). They also link amino acids together to form polypeptide chain.
Answer:
0,1 mol
Explanation:
We know that the formula of concentration is C= moles of solute/ volume
0,4 M= moles of solute/ 250 mL
Convert mL to L 250 mL =0,25 L
0,4 M x 0,25 L= moles of solute
0,1 moles= moles of solute
Taking into account the reaction stoichiometry, you can observe that:
- one mole of Ca₃P₂ produces 2 mol of PH₃.
- the mole ratio between phosphine and calcium phosphide is 2 mol PH₃ over 1 mol Ca₃P₂.
<h3>Reaction stoichiometry</h3>
In first place, the balanced reaction is:
Ca₃P₂ + 6 H₂O → 3 Ca(OH)₂ + 2 PH₃
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- Ca₃P₂:1 mole
- H₂O: 6 moles
- Ca(OH)₂: 3 moles
- PH₃: 2 moles
The molar mass of the compounds is:
- Ca₃P₂: 182 g/mole
- H₂O: 18 g/mole
- Ca(OH)₂: 74 g/mole
- PH₃: 34 g/mole
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- Ca₃P₂: 1 mole ×182 g/mole= 182 grams
- H₂O: 6 moles× 18 g/mole= 108 grams
- Ca(OH)₂: 3 moles ×74 g/mole= 222 grams
- PH₃: 2 moles ×34 g/mole= 68 grams
<h3>Correct statements</h3>
Then, by reaction stoichiometry, you can observe that:
- one mole of Ca₃P₂ produces 2 mol of PH₃.
- the mole ratio between phosphine and calcium phosphide is 2 mol PH₃ over 1 mol Ca₃P₂.
Learn more about the reaction stoichiometry:
<u>brainly.com/question/24741074</u>
<u>brainly.com/question/24653699</u>
The volume in liters occupied by 22.6 g of I₂ gas at STP is 1.99 L (answer A)
<u><em>calculation</em></u>
Step: find the moles of I₂
moles= mass÷ molar mass
from periodic table the molar mass of I₂ is 253.8 g/mol
moles = 22.6 g÷253.8 g/mol =0.089 moles
Step 2:find the volume of I₂ at STP
At STP 1 moles =22.4 L
0.089 moles= ? L
<em>by cross multiplication</em>
={ (0.089 moles x 22.4 L) /1 mole} = 1.99 L
Answer: Ions are formed by the addition of electrons to, or the removal of electrons from,
Explanation: