Answer:
120g
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
Sn + 2HF —> SnF2 + H2
Next, we shall determine the number of mole of HF needed to react with 3 moles of Sn.
From the balanced equation above, 1 mole of Sn reacted with 2 moles of HF.
Therefore, 3 moles of Sn will react with = 3 x 2 = 6 moles of HF.
Finally, we shall convert 6moles of HF to grams
This is illustrated below:
Number of mole of HF = 6moles
Molar Mass of HF = 1 + 19 = 20g/mol
Mass of HF =..?
Mass = number of mole x molar Mass
Mass of HF = 6 x 20
Mass of HF = 120g
Therefore, 120g of HF is needed to react with 3 moles of Sn
Answer:
C2H4O
Explanation:
C H O
54.54/12 9.09/1 36.37/16
4.545 9.09 2.27
4.545/2.27 9.09/2.27 2.27/2.27
2.00 4.00 1
Empirical formula is C2H4O
Smog
formed by mixture of smoke and fog
Let us suppose that we have 1 mol of FeCr2O4. I'm going to use approximate masses because I have no idea what your periodic table will say. Just put in the exact masses from your periodic table.
Fe = 56
Cr*2 = 2*52 = 104
O4 = 4*16 = 64
===========
Total = 56 + 104 + 64 = 224
The % oxygen = (64 / 224) * 100 = 28.5 % but none of your answers match this. Perhaps you are talking about Fe2(Cr2O4)3 The brackets make all the difference in the world.
Without going through all the detail I did before, The molecular mass is
Fe * 2 = 112
Cr * 6 = 312
O * 12 = 192
The total molecular mass = 616
The % Oxygen = (192 / 616) * 100 = 31% roughly. That answer isn't there either.
Let's wait and see who else answers.
Answer:
d) A - 70% B - 30%
Explanation:
If x is the abundance of A, and 1−x is the abundance of B, then:
x (32.0) + (1−x) (33.0) = 32.3
32x + 33 − 33x = 32.3
33 − x = 32.3
x = 0.7
The abundance of A is 70%, and the abundance of B is 30%.
