Answer:
Plants normally turn dark green and look stunted (both leaves and stalks). Older leaves are first affected and may become violet. Sometimes the brown tips of the leaves remain fragile and their maturity seems to be delayed.
Explanation:
it is what it is booiiii
Reaction of dissociation: Ag₂SO₄ → 2Ag⁺ + SO₄²⁻.
m(Ag₂SO₄) = 4 g.
V(Ag₂SO₄) = 1 l.
n(Ag₂SO₄) = m(Ag₂SO₄) ÷ M(Ag₂SO₄).
n(Ag₂SO₄) = 4 g ÷ 311,8 g/mol.
n(Ag₂SO₄) = 0,0128 mol.
n(Ag⁺) = 2 · 0,0128 mol = 0,0256 mol.
n(Ag₂SO₄) = n(SO₄²⁻) = 0,0128 mol.
c(Ag⁺) = n ÷ V = 0,0256 mol ÷ 1 l = 0,0256 mol/l.
Ksp = c(Ag⁺)² · c(SO₄²⁻).
Ksp = (0,0256 mol/l)² · 0,0128 mol/l.
Ksp = 8,3·10⁻⁶.
Answer: 404.04 kJ.
Explanation:
To calculate the moles, we use the equation:
moles of 
According to stoichiometry :
2 moles of on burning produces = 1036 kJ
Thus 0.78 moles of on burning produces =
Thus the enthalpy change when burning 26.7 g of hydrogen sulfide is 404.04 kJ.
Answer:
4
Explanation:
your chosen answer is right as unsaturated mean contains double bond and 1st and 3rd is Wrong as Carbon only can have 4 bonds and 2nd is Wrong as it is saturated.
hope this helps :)
Answer:
1528.3L
Explanation:
To solve this problem we should know this formula:
V₁ / T₁ = V₂ / T₂
We must convert the values of T° to Absolute T° (T° in K)
21°C + 273 = 294K
70°C + 273 = 343K
Now we can replace the data
1310L / 294K = V₂ / 343K
V₂ = (1310L / 294K) . 343K → 1528.3L
If the pressure keeps on constant, volume is modified directly proportional to absolute temperature. As T° has increased, the volume increased too
