Answer:
We can make 10 percent solution by volume or by mass. A 10% of NaCl solution by mass has ten grams of sodium chloride dissolved in 100 ml of solution. Weigh 10g of sodium chloride. Pour it into a graduated cylinder or volumetric flask containing about 80ml of water.
Explanation:
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A sample of an ideal gas has a volume of 2.30 L at 281 K and 1.02 atm. 1.76 atm is the pressure when the volume is 1.41 L and the temperature is 298 K.
<h3>What is Combined Gas Law ?</h3>
This law combined the three gas laws that is (i) Charle's Law (ii) Gay-Lussac's Law and (iii) Boyle's law.
It is expressed as
where,
P₁ = first pressure
P₂ = second pressure
V₁ = first volume
V₂ = second volume
T₁ = first temperature
T₂ = second temperature
Now put the values in above expression we get
P₂ = 1.76 atm
Thus from the above conclusion we can say that A sample of an ideal gas has a volume of 2.30 L at 281 K and 1.02 atm. 1.76 atm is the pressure when the volume is 1.41 L and the temperature is 298 K.
Learn more about the Combined gas Law here: brainly.com/question/13538773
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A general equation for a combustion reaction would be expressed as follows:
CxHy + (x+y/2)O2 = xCO2 + y/2H2O
Propane would obviously would only have carbon and hydrogen in its structure. Assuming a complete combustion, all of the carbon atoms would go to carbon dioxide and all of the hydrogen atoms to water. To determine the empirical, we determine the number of carbon and hydrogen atoms present.
moles C = 2.461 g CO2 ( 1 mol / 44.01 g ) ( 1 mol C / 1 mol CO2 ) = 0.06 mol C
moles H = 1.442 g H2O ( 1 mol / 18.02 g ) ( 2 mol H / 1 mol H ) = 0.16 mol H
Then, we divide the smallest amount to the each mole of the atoms. We do as follows:
C = 0.06 / 0.06 = 1
H = 0.16 / 0.06 = 2.67
Then we multiply a number in order to obtain a whole number ratio between the atoms.
1 CH2.67
2 C2H5.34
3 C3H8 <-------- empirical formula
Answer:
Y = 92.5 %
Explanation:
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In this case, since the reaction between lead (II) nitrate and potassium bromide is:
Exhibits a 1:2 mole ratio of the former to the later, we can calculate the moles of lead (II) bromide product to figure out the limiting reactant:
Thus, the limiting reactant is the KBr as it yields the fewest moles of PbBr2 product. Afterwards, we calculate the mass of product by using its molar mass:
And the resulting percent yield:
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