Answer:
92.5925925926
Explanation:
8 million divided by the seconds in a day witch is 86400
The balanced chemical reaction is:
<span>2Na + 2H2O → 2NaOH + H2
</span><span>
We first use the amount of hydrogen gas to be produced and the molar mass of the hydrogen gas to determine the amount in moles to be produced. Then, we use the relation from the reaction to relate H2 to Na.
53.2 g H2 ( 1 mol / 2.02 g ) ( 2 mol Na / 1 mol H2 ) ( 22.99 g / 1 mol ) = 1210.96 g Na
1210.96 g Na ( 1 mL / 0.97 g ) = 1248.41 mL Na needed</span>
Answer:
C₃H₄O₄
Explanation:
In order to get the empirical formula of a compound, we have to follow a series of steps.
Step 1: Divide the percent by mass of each element by its atomic mass.
C: 34.6/12.01 = 2.88
H: 3.9/1.01 = 3.86
O: 61.5/16.00 = 3.84
Step 2: Divide all the numbers by the smallest one, i.e., 2.88
C: 2.88/2.88 = 1
H: 3.86/2.88 ≈ 1.34
O: 3.84/2.88 ≈ 1.33
Step 3: Multiply all the numbers by a number that makes all of them integer
C: 1 × 3 = 3
H: 1.34 × 3 = 4
O: 1.33 × 3 = 4
The empirical formula is C₃H₄O₄.
Answer:
K = 3.37
Explanation:
2 NH₃(g) → N₂(g) + 3H₂(g)
Initially we have 4 mol of ammonia, and in equilibrium we have 2 moles, so we have to think, that 2 moles have been reacted (4-2).
2 NH₃(g) → N₂(g) + 3H₂(g)
Initally 4moles - -
React 2moles 2m + 3m
Eq 2 moles 2m 3m
We had produced 2 moles of nitrogen and 3 mol of H₂ (ratio is 2:3)
The expression for K is: ( [H₂]³ . [N₂] ) / [NH₃]²
We have to divide the concentration /2L, cause we need MOLARITY to calculate K (mol/L)
K = ( (2m/2L) . (3m/2L)³ ) / (2m/2L)²
K = 27/8 / 1 → 3.37
