Answer:
+2
Explanation:
10 e will 'balance ' 10 p then there is two + charges leftover
First, we need to find the atomic mass of

.
According to the periodic table:
The atomic mass of Carbon = C = 12.01
The atomic mass of Hydrogen = H = 1.008
The atomic mass of Oxygen = O = 16
As there are 6 Carbons, 12 Hydrogens and 6 Oxygens, therefore:
The
molar mass of

= 6 * 12.01 + 12 * 1.008 + 6 * 16
The
molar mass of

= 180.156
grams/moleNow that we have the molar mass of

, we can find the grams of glucose by using:
mass(of glucose in grams) = moles(of glucose given in moles) * molar mass(in grams/mole)
Therefore,
mass(of glucose in grams) = 2.47 * 180.156
mass(of glucose in grams = 444.99 grams
Ans: Mass of glucose in grams in 2.47 moles =
444.99 grams
-i
Correct answer: B
Cooling curve is the plot of temperature versus time as the sample is allowed to cool. In a cooling curve, we start at a temperature greater than the boiling point. At this temperature, the sample is in gaseous state. At the boiling point, there is no change in temperature as the gaseous and liquid states are in equilibrium. As the temperature reduces further, the liquid starts to condense and at the melting point of the sample the liquid undergoes phase transition to solid state. At the melting temperature, a second plateau is observed as the temperature remains unchanged. At temperatures below the melting point, the sample exists as a solid.
So from the curve, the second plateau is observed at around -111
. This point represents the phase transition from liquid to solid state.
A3B9 represents a molecular formula. The representation of the empirical formula for this compound is AB3. This is so because the empirical formula is the simplest ratio of the atoms present in the molecule. You get AB3 when you divide the subscripts of A3B9, this is 3 and 9, by the greatest common factor, which is 3. 3/3 = 1 and 9/3 = 3, so the subscripts for the empirical formula are 1 and 3, which is what AB3 represents. <span>Answer: AB3.</span>
Answer:
Higher than 59 °C because dipole-dipole interactions in iodine monochloride are stronger than dispersion forces in bromine.
Explanation:
I just took the test and i got it right