Answer:
C12H22O11(aq) + H2O(l) —> 4C2H5OH(aq) + 4CO2(g)
Explanation:
When aqueous sugar (sucrose) react with water in the presence of yeast, the following products are obtained as shown in the equation below:
C12H22O11(aq) + H2O(l) —> C2H5OH(aq) + CO2(g)
Now, we shall balance the equation as follow:
There are a total of 24 atoms of H on the left side and 6 atoms on the right side. It can be balance by putting 4 in front of C2H5OH as shown below:
C12H22O11(aq) + H2O(l) —> 4C2H5OH(aq) + CO2(g)
There are a total of 9 atoms of C on the right side and 12 atoms on the left side. It can be balance by putting 4 in front of CO2 as shown below:
C12H22O11(aq) + H2O(l) —> 4C2H5OH(aq) + 4CO2(g)
Now the equation is balanced.
Answer:
C4H6
Explanation:
See attached table
Convert each of the masses into moles by dividing the mass by the molar mass of that element. That yields 3.83 moles of C and 6 moles of O. I rounded up the C to 4 moles to result in an empirical formula of C4H6
Answer:
The three blanks for this answer, are
1. volumen
2. moles
3. Temperature and pressure.
So, Avogadro's law states that the volume of a gas is directly proportional to the moles of the gas when temperature and pressure stay the same
Explanation:
Imagine you have 10 moles of a gas which is contained in 50 L. How many moles of that gas, you will have if the volumen has been reduced to 10 L. (Of course, don't forget that T° and pressure are the same)
There is an equation like this, initial moles /initial volume = moles at the end/volume at the end, (Avogadro law for gases), so 10/50 =moles at the end/10. When u operate, moles at the end = (10 x 10) / 50.
Moles at the end are 2. Did u get it?. Volumen has been reduced, also the moles.
Color.
Density.
Hardness.
freezing point.
Length.
Location.
Smell.
Temperature.
Volume.
Brittleness.
Hope this helps. :)
Let's eliminate these one by one.
The first pair would not be the same, as X would most likely be in group IA, and Y would be in group VIIA, because of their tendency to gain and lose electrons.
The second pair would also violate the same rule, but X would most likely be in group IIA, and Y would most likely be in group VIA.
The third pair would not be the same, as X is most likely in group VIIA, and since Y has eight valence electrons, it is most likely a noble gas.
The final pair has X with atomic number 15, making it phosphorous. Phosphorous wants to gain 3 electrons to have a full octet of 8 outer "valence" electrons, and Y would also like to gain 3 electrons. This means it is possible that the final pair would be in the same group.
