Answer:
Approximately 
.
Explanation:
Note that both figures in the question come with four significant figures. Therefore, the answer should also be rounded to four significant figures. Intermediate results should have more significant figures than that.
<h3>Formula mass of strontium hydroxide</h3>
Look up the relative atomic mass of 
, 
, and 
on a modern periodic table. Keep at least four significant figures in each of these atomic mass data.
Calculate the formula mass of 
:
.
<h3>Number of moles of strontium hydroxide in the solution</h3>
means that each mole of formula units have a mass of 
.
The question states that there are 
of in this solution.
How many moles of formula units would that be?
.
<h3>Molarity of this strontium hydroxide solution</h3>
There are 
of formula units in this 
solution. Convert the unit of volume to liter:
.
The molarity of a solution measures its molar concentration. For this solution:
.
(Rounded to four significant figures.)
Answer:
1. C + O₂ → CO₂
2. C + CO₂ → 2 CO
3. Fe₂O₃ + 3 CO → 2 Fe + 3 CO₂
Answer:
I think its a double reaction
Explanation:
Answer:
0.166M
Explanation:
In a neutralization, the acid, H₂SO₄, reacts with a base, KOH, to produce a salt, K₂SO₄ and water. The reaction is:
H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O
To solve this problem, we need to determine moles of H2SO4 and moles of KOH that reacts to find the moles of sulfuric acid that remains after the reaction:
<em>Moles H2SO4:</em>
0.650L * (0.430mol /L) = 0.2795moles H2SO4
<em>Moles KOH:</em>
0.600L * (0.240mol / L) = 0.144 moles KOH
Moles of sulfuric acid that reacts with 0.144 moles of KOH are:
0.144 moles KOH * (1mol H2SO4 / 2 mol KOH) = 0.072 moles of H2SO4 react.
And remain:
0.2795moles H2SO4 - 0.072moles H2SO4 = 0.2075 moles of H2SO4 reamains.
In 0.650L + 0.600L = 1.25L:
Molar concentration of sulfuric acid:
0.2075 moles of H2SO4 / 1.25L =
<h3>0.166M</h3>
Answer:
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