Answer:
Barium<Strontium<Calcium <Magnesium< Beryllium
Explanation:
Electronegativity is defined as the ability of an atom in a bond to attract the shared electrons of the bond towards itself.
Electronegativity is a periodic trend that decreases down the group and increases across the period.
Hence, if i want to arrange Beryllium, Barium, Strontium, Magnesium, Calcium in order of increasing electronegativity, i will have;
Barium<Strontium<Calcium <Magnesium< Beryllium
The heat released by the water when it cools down by a temperature difference AT
is Q = mC,AT
where
m=432 g is the mass of the water
C, = 4.18J/gºC
is the specific heat capacity of water
AT = 71°C -18°C = 530
is the decrease of temperature of the water
Plugging the numbers into the equation, we find
Q = (4329)(4.18J/9°C)(53°C) = 9.57. 104J
and this is the amount of heat released by the water.
The average mass of an atom is calculated with the formula:
average mass = abundance of isotope (1) × mass of isotope (1) + abundance of isotope (2) × mass of isotope (2) + ... an so on
For the boron we have two isotopes, so the formula will become:
average mass of boron = abundance of isotope (1) × mass of isotope (1) + abundance of isotope (2) × mass of isotope (2)
We plug in the values:
10.81 = 0.1980 × 10.012938 + 0.8020 × mass of isotope (2)
10.81 = 1.98 + 0.8020 × mass of isotope (2)
10.81 - 1.98 = 0.8020 × mass of isotope (2)
8.83 = 0.8020 × mass of isotope (2)
mass of isotope (2) = 8.83 / 0.8020
mass of isotope (2) = 11.009975
mass of isotope (1) = 10.012938 (given by the question)
Answer:
2x+3y=-6
4x+3y=12
3x-y=5
5x+3y=1
Explanation:
standard form: ax + by = C
Explanation:
The two half equations are;
3e + HNO3 → NO
S→ H2SO4 + 6e
When balancing half equations, we have to make sure the number of electrons gained is equal to the number of electrons lost.
<em>Which factor will you use for the top equation?</em>
We multiply by 2 to make the number of electrons = 6e
<em>Which factor will you use for the bottom equation?</em>
We multiply by 1 to make the number of electrons = 6e