1) Well, you need to logically think about which of the pecan trees would be better. Tree 3 may have more clusters but they taste bitter so this is out of the option. So Tree 4 and Tree 1 would be best because they both have rich and buttery tastes and they both have a lot of clusters.
2) Not quite sure about this one, sorry :(
Answer:
A
Explanation:
To label an element correctly using a combination of the symbol, mass number and atomic number furnishes some important information about the element.
We can obtain these information from the element provided that correct labeling of the element is presented. Firstly, after writing the symbol of the element, the atomic number is placed as a subscript on the left while the mass number of the atomic mass is placed as a superscript on the same left.
Looking at the question asked, we have the element symbol in the correct position as Ca, with 42 also in the correct position which is the mass number. The third number which is 20 is thus the atomic number of the element.
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.
87.8 , 31 Celsius=87.8(88) in Fahrenheit
Given buffer:
potassium hydrogen tartrate/dipotassium tartrate (KHC4H4O6/K2C4H4O6 )
[KHC4H4O6] = 0.0451 M
[K2C4H4O6] = 0.028 M
Ka1 = 9.2 *10^-4
Ka2 = 4.31*10^-5
Based on Henderson-Hasselbalch equation;
pH = pKa + log [conjugate base]/[acid]
where pka = -logKa
In this case we will use the ka corresponding to the deprotonation of the second proton i.e. ka2
pH = -log Ka2 + log [K2C4H4O6]/[KHC4H4O6]
= -log (4.31*10^-5) + log [0.0451]/[0.028]
pH = 4.15
