HClO, perchloric acid is a weak acid. Unlike strong acids like HCl or H2SO4 it not dissociate completely but partially such that an equilibrium exists between the dissociated ions and the undissociated acid. The equilibrium is as shown below:
HClO + H2O ↔ H3O⁺ + ClO⁻
Since HClO is a weak acid, the reverse reaction is favored over the forward reaction. Thus apart from water, HClO will be present in large amounts.
Answer:
atomic mass 112.4 and Cadmium (Cd)
Explanation:
You have 73.35 g of MO.
After the reaction the O is removed and you only have M which the mass is 64.21 g.
With that you can calculate the mass of O removed:
Mass of O = 9.14 g
Mass = AM * moles ; (AM : Atomic Mass)
9.14 = 16 * moles
moles = 9.14 / 16
moles = 0.57125
The formula of the metal oxide is MO, meaning it has 1 mole of M per mole of O. 73.35 had 0.57125 moles of O, then it also had 0.57125 moles of M, and the remaining mass of 64.21 g represents those moles
Mass = AM * moles ; (AM : Atomic Mass)
64.21 = AM * 0.57125
AM = 64.21/0.57125 = 112.4 amu
The metal with an atomic mass of 112.4 is Cadmium (Cd), therefore the metal oxide is CdO
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.
Nobelium discovered in Berkeley California
