Answer:
I think D but dont count on me
In a reaction involving sodium metal and chlorine gas, the amount of sodium metal required to react with 0.950 grams of chlorine gas would be 0.62 grams
<h3>Stoichiometric calculation</h3>
From the balanced equation of the reaction:
2Na (s) + Cl2 (g) ---------> 2NaCl (s)
The mole ratio of Na to Cl2 is 2:1.
Mole of 0.950 grams of chlorine gas = 0.950/70.906
= 0.0134 moles
Equivalent mole of Na = 0.0134 x 2 = 0.0268 moles
Mass of 0.0268 moles Na = 0.0268 x 22.99
= 0.62 grams
More on stoichiometric calculations can be found here: brainly.com/question/8062886
Answer:
5.12 L
Explanation:
P1V1/T1 = P2V2/T2
1.71 atm *10.0L /100 K = 9.70 atm *x/290.29 K
x = 1.71 atm *10.0L*290.29K /(100 K * 9.70 atm)= 5.12 L
Hey there!
Cr₄(P₂O₇)₃
Cr: 4 x 51.996 = 207.984
P: 6 x 30.97 = 185.82
O: 21 x 16 = 336
-------------------------------------
729.804 g/mol
The molar mass of Cr₄(P₂O₇)₃ is 729.804 g/mol.
Hope this helps!