Answer:
2 Atm; 2.016 g
Explanation:
Changing the volume without changing the temperature or mass only changes the pressure. Volume and pressure are inversely proportional so halving the volume will double the pressure.
P = 1 Atm, T = 0 °C are "standard" temperature and pressure (STP). The volume of 1 mole of gas is 22.4 L under these conditions. That means the amount of hydrogen gas in the cylinder is 1 mole, so has a mass of 2.016 g.
After the volume reduction, the pressure is 2 Atm, and the mass remains 2.016 g.
Answer:
incorporates both ionic bonding and covalent bonding.
Explanation:
A covalent bond is formed when an element shares its valence electron with another element. This bond is formed between two non metals.
An ionic bond is formed when an element completely transfers its valence electron to another element. The element which donates the electron is known as electropositive element and the element which accepts the electrons is known as electronegative element. This bond is formed between a metal and an non-metal.
For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.
Here potassium is having an oxidation state of +1 called as
cation and nitrate
is an anion with oxidation state of -1. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral
.
is formed by sharing of electrons between two non metals nitrogen and oxygen.
Thus
incorporates both ionic bonding and covalent bonding.
Answer:
a) After the balloon inflated after 440 uL of dropwise due to the reaction of 1-Decene and the solution in the conical vial. b)
⇒ 16
c) No
was not the limiting reactant.
Explanation:
Generally, hydrogenation is the chemical reaction between a compound or element and molecular hydrogen in the presence of catalysts such as platinum.
a) After the balloon inflated after 440 uL of dropwise 1-Decene solution was added due to the reaction between 1-Decene and the solution in the conical vial.
b)
⇒ 16
c)
was not the limiting reactant based on the mol to mol ratio of
and decane which is 1:1. Therefore, if 0.8 mol of decane was produced then 0.8 mol of
would also be produced.