For the answer to the question above asking w<span>hat volume of concentrated hydrochloric acid (12.0 M HCl) is required to make 2.0 liters of a 3.0 M HCl solution?</span> The answer is 3.0/12.0 * 2 L = So it's C
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Answer: C. 146 g
Explanation:
To calculate the moles, we use the equation:
Thus
For sodium nitrate
moles = 1.72
Molar mass of sodium nitrate = 85g/mol
Putting values in above equation, we get:
I can’t see the question :/
Number 1: b) single replacement
Number 2: a) double replacement
Number 3 d) decomposition
Number 4: c) synthesis
The equilibrium concentration of N₂ : 1.992
<h3>Further explanation</h3>
Given
Kc = 6.0 x 10⁻² at 500°C
0.253 M H₂ and 0.044 M NH₃
Reaction
3H₂(g) + N₂(g) = 2NH₃(g)
Required
The equilibrium concentration of N₂
Solution
Kc for the reaction :