Answer:
396 g OF CO2 WILL BE PRODUCED BY 270 g OF GLUCOSE IN A RESPIRATION PROCESS.
Explanation:
To calculate the gram of CO2 produced by burning 270 g of gucose, we first write out the equation for the reaction and equate the two variables involved in the question;
C6H12O6 + 6O2 -------> 6CO2 + 6H2O
1 mole of C6H12O6 reacts to form 6 moles of CO2
Then, calculate the molar mass of the two variables;
Molar mass of glucose = ( 12 *6 + 1* 12 + 16* 6) g/mol = 180 g/mol
Molar mass of CO2 = (12 + 16 *2) g/mol = 44 g/mol
Next is to calculate the mass of glucose and CO2 involved in the reaction by multiplying the molar mass by the number of moles
1* 180 g of glucose yields 6 * 44 g of CO2
180 g of glucose = 264 g of CO2
If 270 g of glucose were to be used, how many grams of CO2 will be produced;
so therefore,
180 g of glucose = 264 g of CO2
270 g of glucose = x grams of CO2
x = 264 * 270 / 180
x = 71 280 / 180
x = 396 g of CO2.
In other words, 396 g of CO2 will be produced by respiration from 270 g of glucose.
CaCl2 and KCl are both salts which dissociate in water
when dissolved. Assuming that the dissolution of the two salts are 100 percent,
the half reactions are:
<span>CaCl2 ---> Ca2+ + 2 Cl-</span>
KCl ---> K+ + Cl-
Therefore the total Cl- ion concentration would be coming
from both salts. First, we calculate the Cl- from each salt by using stoichiometric
ratio:
Cl- from CaCl2 = (0.2 moles CaCl2/ L) (0.25 L) (2 moles
Cl / 1 mole CaCl2)
Cl- from CaCl2 = 0.1 moles
Cl- from KCl = (0.4 moles KCl/ L) (0.25 L) (1 mole Cl / 1
mole KCl)
Cl- from KCl = 0.1 moles
Therefore the final concentration of Cl- in the solution
mixture is:
Cl- = (0.1 moles + 0.1 moles) / (0.25 L + 0.25 L)
Cl- = 0.2 moles / 0.5 moles
<span>Cl- = 0.4 moles (ANSWER)</span>
Answer:
The easiest way to separate a mixture of sugar and water is to use <em><u>distillation</u></em>, a process that separates substances based on their different boiling points. 《☆☆☆☆☆》
Explanation:
Hope it helps JOIN 《Æ §QŮÅĐ》
