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AlekseyPX
3 years ago
9

If 33.6 grams of KCl are dissolved in 192 grams of water, what is the concentration of the solution in percent by mass? (3 point

s)
Chemistry
1 answer:
Brums [2.3K]3 years ago
4 0
The answer here would be 14.9 %KCl and here is how I can explain why:
33.5g / 225.6g x 100% = 14.9% 
<span>you are looking for % of KCl in the solution, you have to add the mass of the KCl and water to get the total mass of the solution</span>
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How to write a balanced chemical equation from empirical formula?
netineya [11]
Let's use the example: H2O --->  H2 + O2
We find how many elements of a product are on one side and how many elements on the other side.
Reactant: H=2 O=1
Product:   H=2 O=2
We need to make the same amount of hydrogen and oxegyn atoms on each side, regardless of how high the numbers are, and we do this by adding coefficients to the compounds.

Reactant: H=4 O=2
Product  : H=4 O=2
2 H2O--->   2 H2 + O2
8 0
3 years ago
Draw it draw lewis dot structures for each hypothetical molecule shown below, using the correct number of valence electrons for
soldi70 [24.7K]

The first molecule is a sensible molecule having complete octet of each atom such as C, H and O whereas the second molecule having hydrogen present between the aldehyde and methyl group and thus showing hydrogen is making bond with aldehyde and methyl as well which is not possible because hydrogen only having one electron in its octet due to which it can only form a single bond by sharing its valence electron.

6 0
3 years ago
The answer to (3.540)x(0.0065)x(401) should have ____.
miv72 [106K]
B. Two significant figures


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3 0
3 years ago
For the reaction 2kclo3(s)→2kcl(s)+3o2(g) calculate how many grams of oxygen form when each quantity of reactant completely reac
barxatty [35]
First, we need to get the molar mass of:

KClO3 = 39.1 + 35.5 + 3*16 = 122.6 g/mol

KCl =39.1 + 35.5 = 74.6 g/mol

O2 = 16*2 = 32 g/mol

From the given equation we can see that:

every 2 moles of KClO3 gives 3 moles of O2

when mass = moles * molar mass

∴ the mass of KClO3 = (2mol of KClO3*122.6g/mol) = 245.2 g

and the mass of O2 then = 3 mol * 32g/mol = 96 g

so, 245.2 g of KClO3 gives 96 g of O2

A) 2.72 g of KClO3: 

when 245.2 KClO3 gives → 96 g  O2

   2.72 g KClO3 gives →  X

X = 2.72 g KClO3 * 96 g O2/245.2 KClO3

    = 1.06 g of O2

B) 0.361 g KClO3:

when 245.2 g KClO3 gives → 96 g O2

     0.361 g KClO3 gives → X

∴ X = 0.361g KClO3 * 96 g / 245.2 g

       = 0.141 g of O2

C) 83.6 Kg KClO3:

when 245.2 g KClO3 gives → 96 g O2

       83.6 Kg KClO3 gives  →  X

∴X = 83.6 Kg* 96 g O2 /245.2 g KClO3

     = 32.7 Kg of O2

D) 22.4 mg of KClO3:

when 245.2 g KClO3 gives → 96 g O2

        22.4 mg KClO3 gives → X

∴X = 22.4 mg * 96 g O2 / 245.2 g KClO3

      = 8.8 mg of O2

     

 


7 0
3 years ago
Baking soda (NaHCO3) can be added to a fruit mix solution to create a carbonated drink. An example is the reaction between bakin
DanielleElmas [232]

Answer:

74.4 ml

Explanation:

          C₆H₈O₇(aq) + 3NaHCO₃(s) => Na₃C₆H₅O₃(aq + 3CO₂(g) + 3H₂O(l)

Given     15g = 15g/84g/mol = 0.1786mole Sodium Bicarbonate

From equation stoichiometry 3moles NaHCO₃ is needed for each mole citric acid or, moles of citric acid needed is 1/3 of moles sodium bicarbonate used.

Therefore, for complete reaction of 0.1786 mole NaHCO₃ one would need 1/3 of 0.1786 mole citric acid or 0.0595 mole H-citrate.

The question is now what volume of 0.8M H-citrate solution would contain 0.0595mole of the H-citrate? This can be determined from the equation defining molarity. That is => Molarity = moles solute / Liters of solution

=> Volume (Liters) = moles citric acid / Molarity of citric acid solution

=> Volume needed in liters = 0.0.0595 mole/0.80M = 0.0744 Liters or 74.4 ml

6 0
3 years ago
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