From the reaction above, the rate is given by the following formula:
r = -(1/2) dA / dt = - dB / dt = (1/3) dC/ dt
Note that A and B charge is negative due to they decrease with time
given dA / dt = -0.110 M/s
hence dB / dt = -0.110 / 2 = -0.055 M/s
dC / dt = (-3/2) (-0.110) = 0.165 M/s
The formula of the given compounds are as follows:
a. copper (II) phosphate : Cu₃(PO₄)₂
b. phosphorus trichloride : PCl₃
c. potassium sulfite : K₂SO₃
d. strontium nitride : Sr₃N₂
e. nitrous acid : HNO₂
<h3>What is the formula of a compound?</h3>
The formula of a compound represents the compound using the symbol of the component elements in the compound showing the ratio in which the atoms of the elements combine in the compound.
The formula of the given compounds are as follows:
a. copper (II) phosphate : Cu₃(PO₄)₂
b. phosphorus trichloride : PCl₃
c. potassium sulfite : K₂SO₃
d. strontium nitride : Sr₃N₂
e. nitrous acid : HNO₂
In conclusion, the formula of a compound represents the compound with symbols.
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Answer:
ΔU = −55.45 kJ
Explanation:
From first law of thermodynamics in chemistry, we have;
ΔU = Q + W
where;
ΔU is change in internal energy
Q is the net heat transfer
W is the net work done
We are given;
Q = 74.6 kJ
But Q will be negative since heat is released
Thus;
ΔU = -74.6 kJ + W
We are given;
Constant pressure; P = 35 atm = 35 × 101325 = 3546375 N/m²
Volume before reaction; Vi = 8.2 L = 0.0082 m³
Volume after reaction; V_f = 2.8 L = 0.0028 m³
Now,
W = -P(V_f - V_i)
W = - 3546375(0.0028 - 0.0082)
W = 19.15 KJ
Thus;
ΔU = Q + W
ΔU = -74.6 kJ + 19.15 KJ =
ΔU = −55.45 kJ
a. 48.6 is magnesium and 32.0 is oxygen
b. 80.6
c. also 80.6
d. yes, because the product has equal mass to the reactants
Answer:
The order is:
F >Be >Li >Ba
Explanation:
Electrons are held in atoms by their attraction to the nucleus which means that to remove an electron from the atom energy is needed.
The ionization energy is the minimum energy necessary to remove an electron from an atom in the gas phase and ground state, the electron removed being the outermost, that is, the furthest from the nucleus. The further away the electron is from the nucleus, the easier it is to remove it, that is, the less energy is needed.
By increasing the atomic number of the elements of the same group, the nuclear attraction on the outermost electron decreases, since the atomic radius increases. Then the ionization energy decreases. In other words, in a group it decreases from top to bottom because the size of the atom increases and it is easier to remove an external electron.
By increasing the atomic number of the elements of the same period, the nuclear attraction on the outermost electron increases, since the atomic radius decreases. Therefore, in a period, as the atomic number increases, the ionization energy increases. In summary, in a period it increases from left to right as the effective nuclear charge increases and it increases thanks to the decrease in the size of the atom.
Taking these considerations into account, the order is:
<u><em>F >Be >Li >Ba</em></u>
