Answer:
607 ppm
Explanation:
In this case we can start with the <u>ppm formula</u>:
If we have a solution of <u>0.0320 M</u>, we can say that in 1 L we have 0.032 mol of , because the molarity formula is:
In other words:
If we use the <u>atomic mass</u> of (19 g/mol) we can convert from mol to g:
Now we can <u>convert from g to mg</u> (1 g= 1000 mg), so:
Finally we can <u>divide by 1 L</u> to find the ppm:
<u>We will have a concentration of 607 ppm.</u>
I hope it helps!
B the color of the metal . Hope this helps
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Because its true our solar system is big
First, it is important to list the molar mass of the relevant substances.
Molar mass of magnesium = 24.305 g/mol
Molar mass of water = 18.0153 g/mol
Molar mass of H2 = 2.0159 g/mol
Second, we need to determine the limiting reactant for the chemical reaction. We take 4.73 g of Mg and determine the stoichiometric amount of water needed for it to be completely consumed. This is shown in the following equation:
4.73 g Mg x mol Mg/24.305 g x 2 mol H2O/1 mol Mg x 18.0153 g/mol H2O = 7.0119 g H2O
Thus, we have determined that 7.0119 g H2O is needed to completely react 4.73 g Mg. The given amount of 1.83 g H2O is insufficient which then indicates that water is the limiting reactant and should be the basis of our calculations.
Next, given 1.83 g H20, we calculate the theoretical yield of hydrogen gas using stoichiometry. The equation is then:
1.83 g H2O x mol H20/18.0153 g x 1 mol H2/2 mol H2O x 2.0159 g/mol H2 = 0.1024 g H2
However, the reaction yield was given to be 94.4%. The reaction yield is given by the formula percent yield = actual yield/ theoretical yield x 100%. Thus, the actual yield of hydrogen gas can be determined using the formula.
Actual yield of H2 = 0.94*0.1024 g H2
Thus, the amount of hydrogen gas produced is 0.0963 g.
and .
Assuming complete decomposition of both samples,
First compound:
; of the first compound would contain
Oxygen and mercury atoms seemingly exist in the first compound at a ratio; thus the empirical formula for this compound would be where the subscript "1" is omitted.
Similarly, for the second compound
; of the first compound would contain
and therefore the empirical formula
.