Answer:
m = 180 g
Explanation:
Given data:
Energy absorbed = 108 J
Mas of gold = ?
Initial temperature = 25°C
Final temperature = 29.7 °C
Specific heat capacity of gold = 0.128 J/g.°C
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT =29.7 °C - 25°C
ΔT = 4.7 °C
108 J = m ×0.128 J/g.°C ×4.7 °C
108 J = m ×0.60 J/g
m = 108 J/0.60 J/g
m = 180 g
Answer:
The answer to your question is letter B, 2-methylhexane.
Explanation:
Remember that for naming organic compounds first, we need to look for the largest chain of carbons.
In your example, the largest chain is horizontal and has 6 carbons.
Later, we need to circle all the branches, in your example there is only one branch located close to the left side
After that, we number the carbons of the main chain, starting in the corner with more branches, in your example we start from the first carbon on the left.
Finally, start naming the number of the carbon branch, later hte name of the branch and finally the name of the main chain.
Answer:
V₂ → 106.6 mL
Explanation:
We apply the Ideal Gases Law to solve the problem. For the two situations:
P . V = n . R . T
Moles are still the same so → P. V / R. T = n
As R is a constant, the formula to solve this is: P . V / T
P₁ . V₁ / T₁ = P₂ .V₂ / T₂ Let's replace data:
(1.20 atm . 73mL) / 112°C = (0.55 atm . V₂) / 75°C
((87.6 mL.atm) / 112°C) . 75°C = 0.55 atm . V₂
58.66 mL.atm = 0.55 atm . V₂
58.66 mL.atm / 0.55 atm = V₂ → 106.6 mL
Answer is: n<span>o, because the ion product is less than the Ksp of lead iodide. </span>
Chemical dissociation 1: KI(s) → K⁺(aq) + I⁻(aq).
Chemical dissociation 2: Pb(NO₃)₂(s) → Pb²⁺(aq) + 2NO₃⁻(aq).
Chemical reaction: Pb²⁺(aq) + 2I⁻(aq) → PbI₂(s).
Ksp(PbI₂) = 7.1·10⁻⁹.
V = 500 mL ÷ 1000 mL/L = 0.5 L.
c(KI) = c(I⁻) = 0.0025 mol ÷ 0.5 L.
c(I⁻) = 0.005 M.
c(Pb(NO₃)₂) = c(Pb²⁺) = 0.00004 mol ÷ 0.5 L.
c(Pb²⁺) = 0.00008 M.
Q = c(Pb²⁺) · c(I⁻)².
Q = 8·10⁻⁵ M · (5·10⁻³ M)².
Q = 2·10⁻⁹; <span> the ion product.</span>
