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NeX [460]
1 year ago
9

does adding boiled, deionized water to the titrating flask to wash the wall of the erlenmeyer flask and the buret tip increase,

decrease, or have no effect on the ksp value of the ca(oh)2? explain.
Chemistry
1 answer:
olga55 [171]1 year ago
4 0

Addition of  boiled, deionized water to the titrating flask to wash the wall of the erlenmeyer flask and the buret tip will have no effect on the Ksp value of ca(oh)2.

There will be no effect on the Ksp value as boiled deionised water is not able to alter the number of hydronium and hydroxide ions. As no change in the ions happen so there will be no change in Ksp value. The equilibrium constant for a solid material dissolving in an aqueous solution is the solubility product constant, Ksp. It stands for the degree of solute dissolution in solution. A substance's Ksp value increases with how soluble it is.

To know more about, solubility product constant, click here,

brainly.com/question/6960236

#SPJ4

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3 years ago
If a pork roast must absorb kJ to fully cook, and if only 10% of the heat produced by the barbecue is actually absorbed by the r
kolezko [41]

Answer:

1,033.56 grams of carbon dioxide was emitted into the atmosphere.

Explanation:

C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g),\Delta H_{rxn}=-2044 kJ/mol (assuming)

Energy absorbed by pork,E = 1.6\times 10^3 kJ (assuming)

Total energy produced by barbecue = Q

Percentage of energy absorbed by pork = 10%

10\%=\frac{E}{Q}\times 100

Q=\frac{E}{10}\times 100=\frac{1.6\times 10^3 kJ}{10}\times 100=1.6\times 10^4 kJ

Since, it is a energy produced in order to indicate the direction of heat produced we will use negative sign.

Q = -1.6\times 10^4 kJ

Moles of propane burnt to produce Q energy =n

n\times \Delta H_{rxn}=Q

n=\frac{Q}{\Delta H_{rxn}}=\frac{-1.6\times 10^4 kJ}{-2044 kJ/mol}=7.83 mol

According to reaction , 1 mol of propane gives 3 moles of carbon dioxide. then 7.83 moles of will give:

\frac{1}{3}\times 7.83 mol=23.49 mol carbon dioxide gas.

Mass of 23.49 moles of carbon dioxide gas:

23.49 mol × 44 g/mol =1,033.56 g

1,033.56 grams of carbon dioxide was emitted into the atmosphere.

8 0
2 years ago
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