Answer:
Potassium iodide increases the decomposition rate of hydrogen peroxide.
Explanation:
Potassium iodide increases the decomposition rate of hydrogen peroxide because potassium iodide act as a catalyst. A catalyst speed up the process of chemical reaction without reacting with the molecules present in reaction. If the potassium iodide is not present as a catalyst for the decomposition of hydrogen peroxide then the decomposition of hydrogen peroxide takes too much time because the catalyst is absent that speed up the reaction.
Answer:
518 mL
Explanation:
We can solve this using Boyle's Law Formula
P1V1 = P2V2
where p1 = initial pressure, p2 = final pressure, v1 = initial volume and v2 = final volume
here , the initial pressure is 1 atm and the initial volume is 725mL
we are given the final pressure 1.4 and we need to find the final volume
so we have p1v1 = p2v2
==> plug in p1 = 1 , v1 = 725 mL and p2 = 1.4
(1)(725) = (1.4)v2
==> multiply 1 and 725
725 = (1.4)(v2)
==> divide both sides by 1.4
v2 = 518
N2 would have a volume of 518mL at 1.4atm
Please, observe that it is not right to say that a substance content heat.
Heat is not something that a body or substance content. Heat is the transmission of energy due to difference of temperatures.
An endothermic reactions is that where the reactants abosorb energy from the surroundings to occur. The products, then, will be higher in energy than the reactants while the surroundings get colder.
Answer:
1.28 mol
Explanation:
mole = mass/molar mass
n = v/v/cm³
mass = 0. 075g
v = 1dm³ =1000cm³
n= m/MV=0.075/58.44(1000)
n =1.28 mol
Answer:
12.32 L.
Explanation:
The following data were obtained from the question:
Mass of CH4 = 8.80 g
Volume of CH4 =?
Next, we shall determine the number of mole in 8.80 g of CH4. This can be obtained as follow:
Mass of CH4 = 8.80 g
Molar mass of CH4 = 12 + (1×4) = 12 + 4 = 16 g/mol
Mole of CH4 =?
Mole = mass/Molar mass
Mole of CH4 = 8.80 / 16
Mole of CH4 = 0.55 mole.
Finally, we shall determine the volume of the gas at stp as illustrated below:
1 mole of a gas occupies 22.4 L at stp.
Therefore, 0.55 mole of CH4 will occupy = 0.55 × 22.4 = 12.32 L.
Thus, 8.80 g of CH4 occupies 12.32 L at STP.
