QUICK ANSWER
J.J. Thomson's cathode ray experiment was a set of three experiments that assisted in discovering electrons. He did this using a cathode ray tube or CRT. It is a vacuum sealed tube with a cathode and anode on one side.
Answer : Any natural sources of CFC's are not known only the major sources like aerosols, propellants, refrigerants,etc are known. So, if any natural sources are given then it cannot be called as a major source for emitting CFC into environment.
An atom of carbon has 4 electrons in its outermost shell, which means that
<span>its ionic charge is 4+ or 4-
</span>Si is in same group as carbon so its also 4+ or 4-
Germanium is 4+.
Sn is also 2+ or 4+
Pb is usually +2
Answer:
As with the hydrogen-ion concentration, the concentration of the hydroxide ion can be expressed logarithmically by the pOH. The pOH of a solution is the negative logarithm of the hydroxide-ion concentration. pOH=−log[OH−] The pH of a solution can be related to the pOH.
Answer:
87.9%
Explanation:
Balanced Chemical Equation:
HCl + NaOH = NaCl + H2O
We are Given:
Mass of H2O = 9.17 g
Mass of HCl = 21.1 g
Mass of NaOH = 43.6 g
First, calculate the moles of both HCl and NaOH:
Moles of HCl: 21.1 g of HCl x 1 mole of HCl/36.46 g of HCl = 0.579 moles
Moles of NaOH: 43.6 g of NaOH x 1 mole of NaOH/40.00 g of NaOH = 1.09 moles
Here you calculate the mole of H2O from the moles of both HCl and NaOH using the balanced chemical equation:
Moles of H2O from the moles of HCl: 0.579 moles of HCl x 1 mole of H2O/1 mole of HCl = 0.579 moles
Moles of H2O from the moles of NaOH: 1.09 moles of HCl x 1 mole of H2O/1 mole of NaOH = 1.09 moles
From the calculations above, we can see that the limiting reagent is HCl because it produced the lower amount of moles of H2O. Therefore, we use 0.579 moles and NOT 1.09 moles to calculate the mass of H2O:
Mass of H2O: 0.579 moles of H2O x 18.02 g of H2O/1 mole of H2O = 10.43 g
% yield of H2O = actual yield/theoretical yield x 100= 9.17 g/10.43 g x 100 = 87.9%
