Removing chloride ion is the reaction away
Answer:
Rb<K<Ga<As<Se<S
Explanation:
We must remember that first ionization energy decreases down the group and increases across the period.
First ionization energy decreases down the group because of the addition of more shells which increases the distance between the nucleus and the outermost electron. Hence, Rb has a lower ionization energy that K.
Across the period, increase in the size of the nuclear charge causes the pull of the nucleus on the outermost electrons to increase thereby increasing the ionization energy. Hence ionization energy increases across the period. For this reason, the ionization energy of Ga<As<Se as shown.
Answer:
if you drop a water balloon onto the ground, its kinetic energy is converted mostly to thermal energy. If the balloon weighs 1 kilogram and you drop it from about 2 meters, it will heat up by less.
Explanation:
As you say, kinetic energy of large objects can be converted into this thermal energy. For example, if you drop a water balloon onto the ground, its kinetic energy is converted mostly to thermal energy. If the balloon weighs 1 kilogram and you drop it from about 2 meters, it will heat up by less than.
There are 67.2 liters of CO2 at STP
<h3>Further explanation</h3>
Given
3 mole of CO2
Required
Volume of CO2 at STP
Solution
Standard conditions for temperature and pressure are used as a reference in certain calculations or conditions
Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, Vm is 22.4 liters / mol.
So for 3 moles :
= 3 x 22.4 L
= 67.2 L
Answer:
Percentage mass of copper in the sample = 32%
Explanation:
Equation of the reaction producing Cu(NO₃) is given below:
Cu(s)+ 4HNO₃(aq) ---> Cu(NO₃)(aq) + 2NO₂(g) + 2H₂O(l)
From the equation of reaction, 1 mole of Cu(NO₃) is produced from 1 mole of copper. Therefore, 0.010 moles of Cu(NO₃) will be produced from 0.010 mole of copper.
Molar mass of copper = 64 g/mol
mass of copper = number of moles * molar mass
mass of copper = 0.01 mol * 64 g/mol = 0.64 g
Percentage by mass of copper in the 2.00 g sample = (0.64/2.00) * 100%
Percentage mass of copper in the sample = 32%
