<em>answer:</em><em> </em><em>option </em><em>d </em><em>(</em><em>2</em><em>×</em><em>m</em><em>o</em><em>l</em><em>a</em><em>r</em><em> </em><em>mass </em><em>of </em><em>H </em><em>+</em><em>2</em><em>×</em><em>m</em><em>o</em><em>l</em><em>a</em><em>r</em><em> </em><em>mass </em><em>of </em><em>O</em>
<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em>
VOLUME= 5cm*10cm*2cm =100cm^3
but density of iron=7.874g/cm^3
mass=7.874g*100 =787.4g
mass of that block = 787.4g
Using a calculator:
(2.568 x 5.8)/4.186 = 3.5581460…
= 3.56 (3sf)
You didn’t specify the correct number of significant figures needed.
<span>the balanced equation for the reaction is as follows
Na</span>₂<span>SO</span>₄<span> + BaCl</span>₂<span> ----> 2NaCl + BaSO</span>₄
<span>stoichiometry of Na</span>₂<span>SO</span>₄<span> to BaCl</span>₂<span> is 1:1
first we need to find out which the limiting reactant is
limiting reactant is fully used up in the reaction.
number of Na2So4 moles - 0.5 mol number of BaCl2 moles - 60 g / 208 g/mol = 0.288 mol
since molar ratio is 1:1 equal number of moles of both reactants should react with each other
therefore BaCl2 is the limiting reactant and Na2SO4 is in excess. amount of product formed depends on number of limiting reactant present.
stoichiometry of BaCl</span>₂<span> to BaSO</span>₄<span> is 1:1.
therefore number of BaSO4 moles formed - 0.288 mol</span>
Answer:
7.5 M
Explanation:
In order to find a solution's molar concentration, or molarity, you need to determine how many moles of solute, which in your case is sodium sulfate,
Na
2
SO
4
, you get in one liter of solution.
That is how molarity was defined -- the number of moles of solute in one liter of solution.
So, you know that you have
0.090
moles of solute in
12 mL
of solution. Your goal here will be to scale up this solution by using this information as a conversion factor to help you determine the number of moles of solute present in
