The practical limit to ages that can be determined by radiocarbon dating is about 41000-yr-old sample, what percentage of the original 614c atoms remains? (for percentage you can either enter the percentage with a percent sign "65.1%" or the percentage as a fraction "0.651".)

In percentage, the sample of C-4 remains = 0.7015 %

Explanation:

The Half life  Carbon 14 =  5730 year

Where, k is rate constant

So,  

The rate constant, k = 0.000120968 year⁻¹

Time = 41000 years

Using integrated rate law for first order kinetics as:

Where,  

is the concentration at time t

is the initial concentration

So,  

In percentage, the sample of C-4 remains = 0.7015 %

answer: option (b) is the correct answer.

explanation:

a covalent bond is formed by sharing of electrons. whereas polarity arises due to difference in electronegativity of two atoms.

therefore, more is the electronegativity of an atom which combines with the titanium, more readily it will induce a partial positive and negative charge.

thus, out of the given options fluorine is most electronegative and has an incomplete octet. therefore, it will readily combine with titanium to share the electrons and complete its octet.  

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g