Answer: a ) NaCl
b) No compound.
Explanation:
Atomic number of A is 11 , thus it is sodium. It has electronic configuration of 2,8,1 and thus has 1 valence electron and can form .
Atomic number of B is 18 , thus it is argon. It has electronic configuration of 2,8,8 and thus it is an inert gas.
Atomic number of C is 1 , thus it is hydrogen. It has electronic configuration of 1. It can only share electrons.
Atomic number of D is 17 , thus it is chlorine. It has electronic configuration of 2,8,7 and thus has 7 valence electron and can form .
a) A and D :
Here Sodium is having an oxidation state of +1 called as cation and anion is . Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral
b) B and C
As B is an inert gas , it wont combine with C.
For conversion of kPa and atm, the relationship is 1 atm = 101.325 kPa. So for 20000 kPa, the atm number is 20000/101.325=197.4 atm.
Might be D bc it’s the only one that makes sense to me
Answer:
3.46*10⁻²² kJ
Explanation:
By the Avogadro's number, 1 mole of electrons at 1 mole of atoms correspond to 6.02x10²³ electrons. So it's necessary 208.4 kJ to remove 6.02x10²³ electrons. To remove a single electron:
6.02*10²³ electrons ---------------- 208.4 kJ
1 electron ---------------- x
By a simple direct three rule:
6.02*10²³ x = 208.4
x = 3.46*10⁻²² kJ
CO(g) +2H2--->CH3OH
2.50g H2*1mol/2g=1.25 mol H2
30.0L CO*1mol/22.4L=1.34 mol CO,
according to reaction 1 mol CO needs 2 mol H2,so 1.34 mol CO need 2.68 mol H2, so 1) limiting teactant is H2 (H)
2)1.25 mol CH3OH will be produced, 1.25 mol*32g/mol=40.0 g CH3OH
3) 1.25 mol H2 needs 0.625 g CO
1.34-0.625=0.715 g CO leftover