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Andrej [43]
2 years ago
5

You have an unknown compound of Potassium derivatives (KIxOy). This sample contains 17.0g of potassium, and 27.8g of oxygen and

55.2g of iodine. If you prepared another sample, but this time it had 460.0g of potassium periodate, how much oxygen and iodine would you have in your sample?
Chemistry
1 answer:
Alex777 [14]2 years ago
7 0

Answer:

The body temperatures in degrees Fahrenheit of a sample of adults in onle small town are 96.6 99.6 96. 3 97.3 99.8 97.7 97,6 97.9 99.2 Assume body temperatures of adults are normally distributed Based on this data find the 9900 confidence Interval of the mean body temperature of adults in the town Enter Your answer as an Open-interval (i.e. parentheses) accurate to 3 decimal places. Assume the data is from normally distributed population_ 9900 CI = Previewv Tp Enter 4euleie er using intewval notation. Example: [2,5) Uge U fcr unicr to combine interval:, Example: (-00,2] U [4,00} Entet each Faluce nunber (like 33,2.2172) Or 33 calculztion (like 5/3,243 5-4) Enter DNE for ar empty set, Use 00 to enter Infinity. Enter each valze accurate to decial place: Get Help: Points possible: 10 This attempt of 3 Me ye ingtmctor absutthis questin Poelis quesivto 6 EpIC e 6 DeA

Explanation:

(Unsure if this helped but hope it did!)

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Calculate the molarity of a solution consisting of 25.0 g of KOH in 3.00 L of solution.
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0.15M

Explanation:

The equation for molarity is M= n/L. Where "M" is Molarity, "n" is the number of moles of solute, and "L" is the total liters in solution.

You need to calculate the number of moles from the given grams. The molar mass of KOH is (39.098+ 16 +1.008)= 56.106g. To calculate the mols of KOH, \frac{25.0g}{1} × \frac{1 mol}{56.106g} = 0.44558... mol, you see that the grams unit cancel out leaving you with mol as the unit.

The volume is given in L already so no need to do any conversion. M= \frac{0.4558mol}{3.00L} = 0.1485M ≈ 0.15M

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