Answer:
C. Combustion
Explanation:
Combustion- an organic compound containing carbon, hydrogen and sometimes oxygen reacts with oxygen gas to form carbon dioxide and water.
Answer: -
12.59
Explanation: -
Strength of NaOH = 0.0179 M
Volume of NaOH = 58.0 mL = 58.0/1000 = 0.058 L
Number of moles = 0.0179 M x 0.058 L
= 1.04 x 10⁻³ mol
Mol of [OH⁻] given by NaOH = 1.04 x 10⁻³ mol
Strength of Ba(OH)₂ = 0.0294 M
Volume of Ba(OH)₂ = 60.0 mL = 60.0/1000 = 0.060 L
Number of moles = 0.0294 M x 0.060 L
= 1.76 x 10⁻³ mol
Mol of [OH⁻] given by Ba(OH)₂ =2 x 1.76 x 10⁻³ mol
Total [OH⁻] = 1.04 x 10⁻³ mol + 2 x 1.76 x 10⁻³ mol
= 4.56 x 10⁻³ mol
Total volume of the mixture = 58.0 + 60.0
= 118.0 mL
118.0 mL of the solution has 4.56 x 10⁻³ mol [OH⁻]
1000 mL of the solution has 
= 0.0386 mol
Using the relation
pOH = - log [OH-]
= - log 0.0386
= 1.41
Using the relation
pH + pOH = 14
pH = 14 - 1.41
= 12.59
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.
Answer:
B is incorrect
Explanation:
Ocean are made of salt water which is undrinkable therefore they are not any type of drinking water source.