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3 years ago

The half-life of carbon-14 is about ____________ years. Fill in the blank.

Chemistry

2 answers:

erma4kov [3.2K]3 years ago

8 0

Answer:

5730 years

Explanation: for more on radiocarbon dating.) and hence explain why the amount of carbon-14 in the sample decays by half over any period of 5730 years.

katrin2010 [14]3 years ago

5 0

The half-life of carbon-14 is about 5730 years

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Which process(es) can be used to separate mixtures but not compounds?

hodyreva [135]

Think it would be distillation

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3 years ago

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How many moles of Mg are present in 62.23 grams of Mg? (Show your calculations for full credit.)

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3 years ago

The pKa of the α‑carboxyl group of serine is 2.21 , and the pKa of its α‑amino group is 9.15 . Calculate the average net charge

sleet_krkn [62]

Answer:

Net charge in serine at pH equal to 8.30 is "0"

Explanation:

At pH > $pK_{a}$ , carboxyl group exists as $-CO_{2}^{-}$ (charged)

At pH < $pK_{a}$ , carboxyl group exists as -COOH (neutral)
At pH > $pK_{a}$ , amino group exists as $-NH_{2}$ (neutral)
At pH < $pK_{a}$ , amino group exists as $-NH_{3}^{+}$ (charged)
So, at pH = 8.30, both carboxyl and amino group exists in charged state.
Net charge in serine at pH equal to 8.30 is "0".
Structure of serine at pH equal to 8.30 has been shown below.

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4 years ago

Assuming that all the energy given off in the reaction goes to heating up only the air in the house, determine the mass of metha

nirvana33 [79]

Answer:

0.92 kg

Explanation:

The volume occupied by the air is:

$35.0m\times 35.0m \times 3.2m \times \frac{10^{3}L }{1m^{3} } =3.9 \times 10^{6} L$

The moles of air are:

$3.9 \times 10^{6} L \times \frac{1.00mol}{22.4L} =1.7 \times 10^{5}mol$

The heat required to heat the air by 10.0 °C (or 10.0 K) is:

$1.7 \times 10^{5}mol \times \frac{30J}{K.mol} \times 10.0 K = 5.1 \times 10^{7}J$

Methane's heat of combustion is 55.5 MJ/kg. The mass of methane required to heat the air is:

$5.1 \times 10^{7}J \times \frac{1kgCH_{4}}{55.5 \times 10^{6} J } =0.92kgCH_{4}$

3 0

3 years ago