Ask question

Ask question

All categories

2 years ago

8

Find the age t of a sample, if the total mass of carbon in the sample is mc, the activity of the sample is a, the current ratio

of the mass of 14 6c to the total mass of carbon in the atmosphere is r, and the decay constant of 14 6c is ?. assume that, at any time, 14 6c is a negligible fraction of the total mass of carbon and that the measured activity of the sample is purely due to 14 6c. also assume that the ratio of mass of 14 6c to total carbon mass in the atmosphere (the source of the carbon in the sample is the same at present and on the day when the number of 14 6c atoms in the sample was set.express your answer in terms of the mass ma of a 14 6c atom, mc, a, r, and ?.

1 answer:

kkurt [141]2 years ago

4 0

should be 7.57*10^3 years

You might be interested in

What would be the products for the reaction Br2 + KI →? (Just identify the correct products for the reaction. You do not need to

Savatey [412]

<u>Answer:</u> The products of the reaction will be $I_2\text{ and }KBr$

<u>Explanation:</u>

Single displacement reaction is defined as the reaction in which more reactive element displaces a less reactive element from its chemical reaction.

The general chemical equation for the single displacement reaction follows:

$A+BC\rightarrow AC+B$

The given chemical equation follows:

$Br_2+2KI\rightarrow I_2+2KBr$

Bromine element is more reactive than iodine element. Thus, can easily replace iodine from its chemical reaction.

Hence, the products of the reaction will be $I_2\text{ and }KBr$

5 0

2 years ago

6. Un volumen de 1.0 mL de agua de mar contiene casi 4 x 10-12 g de Au. El volumen total de agua en los océanos es de 1.5 x 1021

Aleks [24]

Answer:

The total amount of Au is $ $2.0\times10^{24}$

Explanation:

Given that,

Mass of 1.0 ml of Au $m=4\times10^{-2}\ g$

Total volume of water in oceans $V=1.5\times10^{21}\ L$

We need to calculate the volume in ml

Using given volume

$V=1.5\times10^{21}\times1000\ mL$

$V=1.5\times10^{24}\ mL$

We need to calculate the total mass of Au

Using given data

$1\ ml\ volume = 4\times10^{-2}\ g$

$1.5\times10^{24}\ ml=4\times10^{-2}\times1.5\times10^{24}$

So, The total mass of Au is $6\times10^{22}\ g$

The mass will be in ounce,

$Mass=0.035274\times6\times10^{22}$

$Mass=2.12\times10^{21}\ ounce$

The total amount of the Au Will be

$Total\ amount=2.12\times10^{21}\times948$

$Total\ amount=2.0\times10^{24}$

Hence, The total amount of Au is $ $2.0\times10^{24}$

3 0

1 year ago

When a known quantity of compound, at a known concentration, is added to a known volume of another compound to determine the con

Vladimir [108]

Answer:

A titration

Explanation:

A common example of a titration is when we have an acid of unknown concentration, so we add a known volume of a base of known concentration. This process lets us determine the concentration of the acid.

By definition, a titration is a quantitative analysis, as we determine how much of an analyte is there in a sample. However, <u>there are quantitative analyzes which are not titrations</u>. This is why the most appropiate answer is<em> a titration</em>.

5 0

1 year ago

The volume of a gas at 7.00°c is 49.0 ml. if the volume increases to 74.0 ml and the pressure is constant, what will the tempera

marshall27 [118]

Using charles law
v1/t1=v2/t2
v1=49ml
v2=74
t1=7+273=280k
t2=?
49/280=74/t2
0.175=74/t2 cross multiply
0.175t2=74
t2=74/0.175
t2=422k or 149celcius

8 0

1 year ago

Octane (C8H18) undergoes combustion according to the following thermochemical equation. 2C8H18(l) + 25O2(g) → 16CO2(g) + 18H2O(l

Zepler [3.9K]

Answer: The standard enthalpy of formation of liquid octane is -250.2 kJ/mol

Explanation:

The given balanced chemical reaction is,

$2C_8H_{18}(l)+25O_2(g)\rightarrow 16CO_2(g)+18H_2O(l)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{O_2}\times \Delta H_f^0_{(O_2)}+n_{H_2O}\times \Delta H_f^0_{(H_2O)}]-[n_{C_8H_{18}}\times \Delta H_f^0_{(C_8H_{18})+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

We are given:

$\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(C_8H_{18}(l))}=?kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-285.8kJ/mol$

Putting values in above equation, we get:

$-1.0940\times 10^4=[(16\times -393.5)+(18\times -285.8)]-[(25\times 0)+(2\times \Delat H_f{C_8H_{18}(l)}]$

$\Delta H^o_f_{(C_8H_{18}(l))}=-250.2kJ/mol$

Thus the standard enthalpy of formation of liquid octane is -250.2 kJ/mol

4 0

2 years ago