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tensa zangetsu [6.8K]
3 years ago
15

A sample of wood that originally contained 100 grams of carbon-14 now contains only 25 grams of carbon-14. approximately how man

y years ago was this sample part of a living tree?
Chemistry
1 answer:
BARSIC [14]3 years ago
6 0
In order for 100 grams of Carbon 14 to decay to 25 grams, 2 half lives worth of time would have passed. 

<span>log(100/25) / log(2) = 2 half lives </span>

<span>2 * 5730 years = 11,460 years</span>
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Breathing equipment used by rescue workers needs to capture the CO2 the humans breath out and produce O2 for them to breath in,
const2013 [10]

Answer: 4KO_2+2CO_2\rightarrow 2K_2CO_3+3O_2

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

The balanced chemical equation for reaction of potassium superoxide  with carbon dioxide to produce oxygen and potassium carbonate will be:

4KO_2+2CO_2\rightarrow 2K_2CO_3+3O_2

8 0
3 years ago
A mixture of 10 cm3 of methane and 10 cm3 of ethane was sparked with an excess of oxygen. After cooling to room temperature, thr
Tpy6a [65]
In the combustion process using excess oxygen, each mole of methane results to 1 mole of co2 while ethane produces 2 moles of Co2. Under same conditions, these can be translated to volume. Hence the total volume absorbed is 10 cm3 + 20 cm3 = 30 cm3.
5 0
3 years ago
If a car can go from 0 to 60 mi/hr in 8.0 seconds, what would be its final speed after 5.0 seconds if its starting speed were 50
tangare [24]

Answer:

87.5 mi/hr

Explanation:

Because a = Δv / Δt (a = vf - vi/ Δt), we need to find the acceleration first to know the change in velocity so we can determine the final velocity.

vf = 60 mi/hr

vi = 0 mi/hr

Δt = 8 secs

a = vf - vi/ Δt

= 60 mi/hr - 0 mi/hr/ 8 secs

= 60 mi/hr / 8 secs

= 7.5 mi/hr^2

Now that we know the acceleration of the car is 7. 5 mi/hr^2, we can substitute it in the acceleration formula to find the final velocity when the initial velocity is 50 mi/hr after 5 secs.

vi = 50 mi/ hr

Δt = 5 secs

a = 7.5 mi/ hr^2

a = vf - vi/ Δt

7.5 = vf - 50 mi/hr / 5 secs

37.5 = vf - 50

87.5 mi/ hr = vf

7 0
3 years ago
In bohr's atomic theory, when an electron moves from one energy level to another energy level more distant from the nucleus,
labwork [276]

Answer is: energy is absorbed.

According to the Bohr model of the atom:

1. Electrons orbit the nucleus in orbits that have a set size and energy.

2. Energy levels of electrons are discrete (certain discrete values of energy).

3. Electrons can jump from one energy level to another, absorbing or emitting electromagnetic radiation with a frequency ν (energy difference of the levels).

5 0
3 years ago
What is the total energy change for the following reaction:CO+H2O-CO2+H2
Alekssandra [29.7K]

Answer:

\large \boxed{\text{-41.2 kJ/mol}}

Explanation:

Balanced equation:    CO(g) + H₂O(g) ⟶ CO₂(g) + H₂(g)

We can calculate the enthalpy change of a reaction by using the enthalpies of formation of reactants and products

\Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)

(a) Enthalpies of formation of reactants and products

\begin{array}{cc}\textbf{Substance} & \textbf{$\Delta_{\text{f}}$H/(kJ/mol}) \\\text{CO(g)} & -110.5 \\\text{H$_{2}$O} & -241.8\\\text{CO$_{2}$(g)} & -393.5 \\\text{H$_{2}$(g)} & 0 \\\end{array}

(b) Total enthalpies of reactants and products

\begin{array}{ccr}\textbf{Substance} & \textbf{Contribution)/(kJ/mol})&\textbf{Sum} \\\text{CO(g)} & -110.5& -110.5 \\\text{H$_{2}$O(g)} &-241.8& -241.8\\\textbf{Total}&\textbf{for reactants} &\mathbf{ -352.3}\\&&\\\text{CO}_{2}(g) & -393.5&-393.5 \\\text{H}_{2} & 0 & 0\\\textbf{Total}&\textbf{for products} & \mathbf{-393.5}\end{array}

(c) Enthalpy of reaction \Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)= \text{-393.5 kJ/mol - (-352.3 kJ/mol}\\= \text{-393.5 kJ/mol + 352.3 kJ/mol} = \textbf{-41.2 kJ/mol}\\ \text{The total enthalpy change is $\large \boxed{\textbf{-41.2 kJ/mol}}$}

4 0
3 years ago
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