Fill in the blanks in order w/ these numbers:
1. 2 1 2 1
2. 3 1 1
3. 1 2 1 2
4. 2 1 1
5. 4 3 2
6. 1 1 2
Hope this helps.
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.
According to Avogadro's law 1 mole contains 6.022 ×10^23 particles
1 mole of carbon = 44.01 g/ mol
Therefore;
44.01 g = 6.022 ×10^23 molecules
Hence, 1.68×10^26 molecules will have a mass of ;
(44.01 × 1.68×0^26) / 6.022×10^23
= 1.228 × 10^4 molecules
I think the correct answer is to use its electron dot structure. Fluorine is the most electronegative element among the halogens. It has 5 electrons in its 2p shell wherein it should have 6 electrons.
