Answer is: mass of methanol is 528.32 grams.
N(CH₃OH) = 9.93·10²⁴; number of methanol molecules.
n(CH₃OH) = N(CH₃OH) ÷ Na (Avogadro constant).
n(CH₃OH) = 9.93·10²⁴ ÷ 6.022·10²³ 1/mol.
n(CH₃OH) = 16.49 mol; amount of substance.
m(CH₃OH) = n(CH₃OH) · M(CH₃OH).
m(CH₃OH) = 16.49 mol · 32.04 g/mol.
m(CH₃OH) = 528.32 g.
The answer should be A, the number of electrons increases by 1, from left to right.
Across the periods (horizontal direction) in periodic table, each element has one more electron then the one before it. The number of period determines the number of occupied electron shells of the elements in that period. Which means, period 1 elements have 1 occupied electron shell, period 2 has 2, period 3 has 3 etc.
Therefore, the answer is A, we always go from left to right, and plus since period 2 elements only have 2 occupied electrons so their 2nd electron shell electrons must increase throughout the period.
The correct answer is passive transport.
When molecules are moved from areas of high concentration to areas of low concentration in order to gain energy, that is considered to be passive transport. If the opposite happens, then that would be active transport.
You need the heat of fusion of ice to solve this problem. <span>the heat of fusion of water = 334 J/g </span><span>
the formula to solve for heat is </span>q = m·ΔH<span>f</span>
just plug the values
q= (50.0 g) (334 j/g)= 16700 joules or 16.7 Kj
Answer: A) Ar
Explanation: Argon [Ar] has an atomic no of 18 and the electronic configuration is:
Ar :
Chlorine [Cl] has atomic no of 17 and thus the electronic configuration is :
Cl :
Magnesium [Mg] has atomic no of 12 and thus the electronic configuration is :
Mg :
Sodium has atomic no of 11 and thus the electronic configuration is :
Cl :
As we move across a period the number of electrons are being added to the same shell but the number of protons increases, thus the electrons are being tightly held and thus the atomic radius keeps on decreasing.
Thus the smallest atom would have a tightly bound electron nearer to the nucleus. As argon is the last element of the period, it is the smallest in the period and thus largest amount of energy will be required to remove the outermost electron.
