Answer:
The component that dissolves the other component is called the solvent. Solute – The component that is dissolved in the solvent is called solute
Once the substance stops dissolving, the system is at equlibrium with the water and the undissolved salt now, if it is in the process of dissolving because it is completely soluble but has not been able to completely dissolve, it is not at equilibrium
Answer:
most liquids do expand when they freeze
Explanation:
water is actually one that expands and there are probably many more too
i hope this helps you
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>
Answer:
START and STOP, ReAcTanTS means START, PrOducTS means STOP
Explanation:
1. We find words 'STOP' and 'START'.
2. Let's take a look at the word 'ReAcTanTS'. Enumerate the capital letters from 1 to 5 from left to right, that is,
'1': R,
'2': A,
'3': T,
'4': T,
'5': S.
Notice that if we take a combination of 54213, we obtain a word 'start'. This has, of course, a chemical application. Reactants are the ones we start with in a chemical reaction.
3. Let's take a look at the word 'PrOducTS'. Enumerate the capital letters from 1 to 4 from left to right, that is:
'1': P,
'2': O,
'3': T,
'4': S.
Notice that if we take a combination of 4321, we obtain a word 'stop'. Similarly to the previous word, it also has a direct relationship to the chemical context. Products are formed in a chemical reaction when reactants combine. This is the point where reaction comes to and end and eventually stops if it's a typical one-sided reaction.
