Independent variables don't change
Dependent variables change as a result of the alteration made during the experiment
Answer:
Hmm, what can I help you with?
Explanation:
The answers is 3.0 g/cm^3
Explanation:
You use the formula
Density= mass/volume
When they give you mass and volume.
The mass is 15 grams and the volume would be 5 cm^3.
So the density would be
15/5= 3
Answer:
C15H24O
Explanation:
TO GET THE EMPIRICAL FORMULA, WE NEED TO KNOW THE MASSES AND CONSEQUENTLY THE NUMBER OF MOLES OF EACH OF THE INDIVIDUAL CONSTITUENT ELEMENTS.
FIRSTLY, WE CAN GET THE MASS OF THE CARBON FROM THAT OF THE CARBON IV OXIDE. WE NEED TO KNOW THE NUMBER OF MOLES OF CARBON IV OXIDE GIVEN OFF. THIS CAN BE CALCULATED BY DIVIVDING THE MASS BY THE MOLAR MASS OF CARBON IV OXIDE. THE MOLAR MASS OF CARBON IV OXIDE IS 44G/MOL
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HENCE, THE NUMBER OF MOLES OF CARBON IV OXIDE IS 4.122/44 WHICH EQUALS 0.094. SINCE THERE IS ONLY ONE ATOM OF CARBON IN CO2 THEN THEY HAVE EQUAL NUMBER OF MOLES AND THUS THE NUMBER OF MOLES OF CARBON IS 0.094. WE CAN THEN PROCEED TO CALCULATE THE MASS OF CO2 PRESENT. THIS CAN BE CALCULATED BY MULTIPLYING THE NUMBER OF MOLES BY THE ATOMIC MASS UNIT. THE ATOMIC MASS UNIT OF CARBON IS 12. HENCE, THE MASS OF CO2 PRESENT IS 12 * 0.094 = 1.128g</h3><h3>
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WE CAN NOW GET THE MASS OF THE HYDROGEN BY MULTIPLYING THE NUMBER OF MOLES OF WATER BY 2 AND ALSO ITS ATOMIC MASS UNIT</h3><h3>
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TO GET THE NUMBER OF MOLES OF WATER, WE SIMPLY DIVIDE THE MASS BY THE MOLAR MASS. THE MOLAR MASS OF WATER IS 18g/mol. The NUMBER OF MOLES IS THUS 1.350/18 = 0.075</h3><h3>
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THE NUMBER OF MOLES OF HYDROGEN IS TWICE THAT OF WATER SINCE IT CONTAINS 2 ATOMS PER MOLECULE OF WATER. ITS NUMBER OF MOLES IS THUS 0.075*2 = 0.15 MOLE</h3><h3>
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THE MASS OF HYDROGEN IS THUS 0.075 * 2 * 1 = 0.15g</h3><h3>
</h3><h3>WE CAN NOW FIND THE MASS OF OXYGEN BY SUBTRACTING THE MASSES OF HYDROGEN AND CARBON FROM THE TOTAL MASS.</h3><h3 /><h3>MASS OF OXYGEN = 1.376-0.15-1.128 = 0.098g</h3><h3 /><h3>THE NUMBER OF MOLES OF OXYGEN IS THUS 0.098/16 = 0.006125</h3><h3 /><h3>WE CAN NOW USE THE NUMBER OF MOLES TO OBTAIN THE EMPIRICAL FORMULA.</h3><h3 /><h3>WE DO THIS BY DIVIDING EACH BY THE SMALLEST NUMBER OF MOLES WHICH IS THAT OF THE OXYGEN.</h3><h3 /><h3>C = 0.094/0.006125 = 15</h3><h3>H = 0.15/0.006125 = 24</h3><h3>O = 1</h3><h3 /><h3>THE EMPIRICAL FORMULA IS THUS C15H24O</h3>
