Carbon = 12.010. Oxygen = 15.999 x 2 15.999 x 2 = 31.998 + 12.010 = 44.008 \frac{37.15 grams * 1 mole CO2}{44.008 grams}
Answer:
A.
Explanation:
The <u>tertiary structure </u>of proteins is related to the interactions between the amino acids of the <u>primary structure</u>. Thus, these interactions give it a specific three-dimensional configuration which is very sensitive to <u>functionality</u>.
For example, <u>allosteric inhibitions</u> are related to this concept. When the <u>inhibitor</u> changes the tertiary structure of the protein it loses all <u>activity</u> and for the catalysis of the reaction.
Thus, the primary structure (which is related to the specific <u>sequence of amino acids</u>) will determine the tertiary structure since the chain folds will be a consequence of<u> intra-amino acid interactions</u>.
Yes, anything with carbonate, hydrogen carbonate (bicarbonate) at the end is a carbonate.
Examples:NaHCO3 (Sodium hydrogen carbonate or Sodium bicarbonate)Na2CO3 (Sodium carbonate)
Answer:
8.934 g
Step-by-step explanation:
We know we will need a balanced equation with masses and molar masses, so let’s gather all the information in one place.
M_r: 192.12 44.01
H₃C₆H₅O₇ + 3NaHCO₃ ⟶ Na₃C₆H₅O₇ + 3H₂O + 3CO₂
m/g: 13.00
For ease of writing, let's write H₃C₆H₅O₇ as H₃Cit.
(a) Calculate the <em>moles of H₃Cit
</em>
n = 13.00 g × (1 mol H₃Cit /192.12 g H₃Cit)
n = 0.067 67 mol H₃Cit
(b) Calculate the <em>moles of CO₂
</em>
The molar ratio is (3 mol CO₂/1 mol H₃Cit)
n = 0.067 67 mol H₃Cit × (3 mol CO₂/1 mol H₃Cit)
n = 0.2030 mol CO₂
(c) Calculate the <em>mass of CO₂
</em>
m = 0.2030 mol CO₂ × (44.01 g CO₂/1 mol CO₂)
m = 8.934 g CO₂
The inner core us the densest
