Answer:
At equilibrium, the concentration of the reactants will be greater than the concentration of the products. This does not depend on the initial concentrations of the reactants and products.
Explanation:
The value of Kc gives us an idea of the extent of the reaction. A big Kc (Kc > 1) means that in the equilibrium there are more products than reactants, and the opposite happens for a small Kc (Kc < 1). The equilibrium is reached no matter what the initial concentrations are.
The value of the equilibrium constant is relatively SMALL; therefore, the concentration of reactants will be GREATER THAN the concentration of products. This result is INDEPENDENT OF the initial concentration of the reactants and products.
The sun's rays hit the Earth at a extreme angle, which causes the days to become shorter, so yes, the sun's rays do hit the Earth at a extreme angle.
Answer:
c. 20.0332 g to 20,0 g
Explanation:
A significant figure is each of the digits of a number that are used to express it to the required degree of accuracy, starting from the first non-zero digit, with the exception of the trailing zeros.
<em>Which of the following examples illustrates a number that is correctly rounded to three significant figures?
</em>
a. 109 526 g to 109 500 g. NO. The rounded number has 4 significant figures: 109 500.
b. 0.03954 g to 0.040 g. NO. The rounded number has 2 significant figures: 0.040.
c. 20.0332 g to 20.0 g. YES. The rounded number has 3 significant figures: 20.0.
d. 04.05438 g to 4.054 g. NO. The rounded number has 4 significant figures: 4.054.
e. 103.692 g to 103.7g. NO. The rounded number has 4 significant figures: 103.7.
Answer:
91.16% has decayed & 8.84% remains
Explanation:
A = A₀e⁻ᵏᵗ => ln(A/A₀) = ln(e⁻ᵏᵗ) => lnA - lnA₀ = -kt => lnA = lnA₀ - kt
Rate Constant (k) = 0.693/half-life = 0.693/10³yrs = 6.93 x 10ˉ⁴yrsˉ¹
Time (t) = 1000yrs
A = fraction of nuclide remaining after 1000yrs
A₀ = original amount of nuclide = 1.00 (= 100%)
lnA = lnA₀ - kt
lnA = ln(1) – (6.93 x 10ˉ⁴yrsˉ¹)(3500yrs) = -2.426
A = eˉ²∙⁴²⁶ = 0.0884 = fraction of nuclide remaining after 3500 years
Amount of nuclide decayed = 1 – 0.0884 = 0.9116 or 91.16% has decayed.
