Answer:
2Ag(s) + 2H²S(g) + O2(g) ➡️ Ag2S(s) + 2H2O(g)
Explanation:
Sorry for my typo. but you understand
Radon-220 undergoes alpha decay with a half-life of 55.6 s.?
Assume there are 16,000 atoms present initially and calculate how many atoms will be present at 0 s, 55.6 s, 111.2 s, 166.8 s, 222.4 s, and 278.0 s (all multiples of the half-life). Express your answers as integers separated by commas.
Calculate how many atoms are present at 50 s, 100 s, and 200 s (not multiples of the half-life).
The half life of a radioactive isotope refers to the time taken for half of the number of original number of atoms present in the sample to decay.
The equation below gives the number of atoms present at time t
[tex]N=Noe^-kt[/tex]
N = Number of atoms present at time t
No = Number of atoms initially present
k = decay constant
t = time taken
Given that;
t1/2 = 0.693/k
where t1/2 = half life
k = 0.693/t1/2
k = 0.693/ 55.6 s
k = 0.0125 s-1
Substituting values;
N = 16,000 e^-0.0125(0)
N = 16,000 atoms
At 50 s
N = 16,000 e^-0.0125(50)
= 8564 atoms
At 100 s
N = 16,000 e^-0.0125(100)
= 4584 atoms
At 200 s
N = 16,000 e^-0.0125(200)
= 1313 atoms
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Dylan has a coworker who is always showing up late and then not finishing his work on time . It's frustrating the other members of the team . What can he do that might help the situation ? a ) Complain about the coworker to other team members b ) Ask his coworker if he understands his job responsibilities c ) Tell his boss that the coworker is slacking off d ) Complete his coworker's work for him
In the reoxidation of QH2 by purified ubiquinone-cytochrome c reductase (Complex III) from heart muscle, the overall stoichiometry of the reaction requires 2 mol of cytochrome c per mole of QH2 because:
Answer: Options related to your question is missing below are the missing options
a. cytochrome c is a one-electron acceptor, whereas QH2 is a two-electron donor.
b. cytochrome c is a two-electron acceptor, whereas QH2 is a one-electron donor.
c. cytochrome c is water soluble and operates between the inner and outer mitochondrial membranes
d. heart muscle has a high rate of oxidative metabolism, and therefore requires twice as much cytochrome c as QH2 for electron transfer to proceed normally.
e. two molecules of cytochrome c must first combine physically before they are catalytically active.
answer:
cytochrome c is a one-electron acceptor, whereas QH2 is a two-electron donor. ( A )
Explanation:
The overall stoichiometry of the reaction requires 2 mol of cytochrome per mole of QH2 because a cytochrome is simply a one-electron acceptor while QH2 is not a one-electron donor ( i.e. it is a two-electron donor )
An electron donor in a reaction is considered a reducing agent because it donates its electrons to another compound thereby self oxidizing itself in the process.
What type of oxide is each of the following? NO2, CO, Fe2O3, Al2O3, P20,
Nitrogen Dioxide
Carbon Monoxide
Iron (II) Oxide
Aluminum Oxide
DiPhosphorus Monoxide
Write the balanced equation for the hydration of CuSO4CuSO4. Indicate the physical states using the abbreviations (ss), (ll), or (gg) for solid, liquid, or gas, respectively. Use (aqaq) to indicate the aqueous phase. Indicate appropriate charges on negative and positive ions if they are formed.
Answer:
CuSO4(s) + 5H2O(l) ----> CuSO4.5H2O(s)
Explanation:
Hydration is the process by which anhydrous CuSO4 acquires molecules of water of crystalization to form the pentahydrate.
The water of crystalization becomes attached go the crystals of the CuSO4 to form the hydrated salt.
Beginning with solid anhydrous CuSO4 we have;
CuSO4(s) + 5H2O(l) ----> CuSO4.5H2O(s)
there is 3.5 g of fat in a granola bar. You determine the fat content to be 4.0 G in the lab. What is the percent error
Answer:
[tex]error = 4.0 - 3.5 = 0.5 \\ \\ percent \: error = \frac{0.5}{3.5} \times 100 \\ \\ = 14.29\% [/tex]
Identify the compound that possesses a permanent dipole. Please choose the correct answer from the following choices, and then select the submit answer button. Answer choices acetone, (CH3)2CO cyclohexane, C6H12 pentane, C5H12 methane, CH4.
Answer:
acetone, (CH3)2CO cyclohexane are the compound that possesses a permanent dipole
Explanation:
Permanent dipole describes the partial charge separation that can occur within a molecule along with the bond dat form between 2 different atoms
Chlorine radicals perform the first propagation step:
a. in comparison to bromine radicals.
b. radicals form easily in the presence of chlorine radicals.
c. Subsequently, the resulting radicals can react with bromine in a second propagation step to yield monobrominated products.
Answer:
b. radicals form easily in the presence of chlorine radicals.
Explanation:
Chlorine radicals perform the first propagation step: because "radicals form easily in the presence of chlorine radicals."
This is because the first propagation step consumes a CHLORINE RADICAL while the second propagation step regenerates a CHLORINE RADICAL. In this way, a chain reaction occurs, whereby one CHLORINE RADICAL can ultimately cause thousands of molecules of methane to be converted into chloromethane with C12 present.
Hence, in this case, the correct answer is that "radicals form easily in the presence of chlorine radicals."
A quantity of ideal gas requires 800 kJ to raise the temperature of the gas by 10.0 K when the gas is maintained at constant volume. The same quantity of gas requires 900 kJ to raise the temperature of the gas by 10.0 K when the gas is maintained at constant pressure. What is the adiabatic gas constant of this gas
Answer:
[tex]\gamma=1.125[/tex]
Explanation:
From the question we are told that:
Initial Heat [tex]Q_1=800kJ[/tex]
initial Temperature [tex]T_1=10.0K[/tex]
Final Heat [tex]Q_2=800kJ[/tex]
Final Temperature [tex]T_2=10.0K[/tex]
Generally the equation for Adiabatic constant is mathematically given by
[tex]\gamma=\frac{Cp}{Cv}[/tex]
Since
Equation for Heat [tex]dQ=nCdT[/tex]
Where
[tex]n_1=n_2\\\\T_1=T_2[/tex]
Therefore
[tex]Q_1=Cv\\\\Cv=800[/tex]
And
[tex]Cp=900[/tex]
Therefore
[tex]\gamma=\frac{900}{800}\\\\\gamma=\frac{9}{8}[/tex]
[tex]\gamma=1.125[/tex]
Calculate the numerical value of the equilibrium constant, Kc, for the reaction below if the equilibrium concentrations for CO, H2 , CH4 and H2O are 0.989 M, 0.993 M, 1.078 M and 0.878 M, respectively. (calculate your answer to three sig figs)
CO(g) + 3 H2(g) ⇌ CH4(g) + H2O(g)
Kc = [CH4]×[H2O] / [CO]×[H2]^3
Kc = 1.078×0.878 / (0.989×0.933^3)
Kc = 0.977
The numerical value of the equilibrium constant, Kc, for the given reaction is found to be 0.977.
What is Equilibrium constant?The Equilibrium constant may be defined as the numerical value that significantly indicated the correlation between the amounts of products and reactants present at equilibrium in a reversible chemical reaction at a definite temperature.
According to the question, the reaction is as follows:
[tex]CO +3H_2[/tex] ↔ [tex]CH_4+ H_2O[/tex].
The equilibrium concentrations are 0.989 M, 0.993 M, 1.078 M and 0.878 M, respectively.
Now, the equilibrium constant is calculated by the following formula:
Kc = [CH4]×[H2O] / [CO]×[tex][H_2]^3[/tex]= 1.078×0.878 / (0.989×0.93[tex]3^3[/tex]).
= 0.9464/(0.989 × 0.8121)
= 0.977.
Therefore, the numerical value of the equilibrium constant, Kc, for the given reaction is found to be 0.977.
To learn more about Equilibrium constant, refer to the link:
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What is the oxidation state of rubidium (Rb)?
A. +1
B. -2
C. +2
D. -1
Answer:
The answer is A. +1
Explanation:
What trends were seen in medeleevs periodic table
Answer:
groups are based on how many electrons to become stable
Explanation:
I NEED A ANSWER FOR THIS QUESTION, ASAP
Predict whether reactants or products will be favored at equilibrium for the below reaction.
Kp= 2.2 x 10^6 at 298K
2COF2 (g) + ⇌ CO2(g) + CF4(g)
Answer:
The products will be favored at equilibrium.
Explanation:
The balanced chemical equation for the reaction is the following:
2 COF₂ (g) + ⇌ CO₂(g) + CF₄(g)
The reactant is COF₂ (left side) and the products are CO₂ and CF₄ (right side).
The equilibrium constant is given by the ratio between the partial pressures (P) of products and reactants, because they are in the gas phase. Thus, the expression of the equilibrium constant is the following:
[tex]Kp = \frac{P(CO_{2}) P(CF_{4}) }{P(COF_{2} )^{2} } = 2.2 x 10^{6}[/tex]
Since Kp>>>>1 ⇒ (P(CO₂) x P(CF₄)) > (P(COF₂))²
So, the partial pressures of the products (CO₂ and CF₄) are higher than the partial pressure of the reactant (COF₂).
Therefore, products will be favored at equilibrium at 298 K.
Compare the total number of modes for 4 moles of a monoatomic gas and 1 mole of a gas consisting of linear triatomic molecules (such as CO2 gas). If these two gases, initially at difference temperatures, were placed in the same container and allowed to reach equilibrium, which gas (if any) would have a greater change in temperature
Answer:
23
Explanation:
How many grams of NaCl (MM = 58.44g/mol) are in 250mL of a 0.75 molar solution?
Answer:
[tex]\boxed {\boxed {\sf 11 \ grams \ NaCl}}[/tex]
Explanation:
We are asked to find how many grams of sodium chloride are in a solution.
1. Moles of SoluteMolarity is a measure of concentration in moles per liter.
[tex]molarity= \frac{ moles \ of \ solute}{liters \ of \ solution}[/tex]
We know the molarity is 0.75 molar. 1 molar is the same as 1 mole per liter, so the solution contains 0.75 moles of sodium chloride per liter.
There are 250 milliliters of solution but molarity uses liters for volume. We must convert milliliters to liters. Remember that 1 liter contains 1000 milliliters. Set up a ratio and use dimensional analysis to convert.
[tex]250 \ mL * \frac{1 \ L} {1000\ mL} = \frac{ 250}{1000} \ L = 0.250 \ L[/tex]
Now we know the molarity and the liters of solution, but the moles of solute are unknown.
molarity = 0.75 mol NaCl/L moles of solute =x liters of solution = 0.25 LSubstitute the values into the formula.
[tex]0.75 \ mol \ NaCl/L = \frac{x}{0.250 \ L}[/tex]
We are solving for the moles of solute, so we must isolate the variable x. It is being divided by 0.250 liters. The inverse of division is multiplication, so multiply both sides of the equation by 0.250 L.
[tex]0.250 \ L *0.75 \ mol \ NaCl/L = \frac{x}{0.250 \ L} * 0.250 \ L[/tex]
[tex]0.250 \ L *0.75 \ mol \ NaCl/L = x[/tex]
The units of liters cancel.
[tex]0.250 * 0.75 \ mol \ NaCl[/tex]
[tex]\bold {0.1875 \ mol \ NaCl}[/tex]
2. Grams of SoluteNow that we have calculated the moles of solute, we must convert this to grams. We use the molar mass or the mass of 1 mole of a substance. Sodium chloride's molar mass is given and it is 58.44 grams per mole. This means there are 58.44 grams of sodium chloride in 1 mole of sodium chloride.
Set up a ratio so we can convert using dimensional analysis.
[tex]\frac {58.44 \ g \ NaCl}{1 \ mol \ NaCl}[/tex]
Multiply by the number of moles we calculated.
[tex]0.1875 \ mol \ NaCl *\frac {58.44 \ g \ NaCl}{1 \ mol \ NaCl}[/tex]
The units of moles of sodium chloride cancel.
[tex]0.1875 *\frac {58.44 \ g \ NaCl}{1}[/tex]
[tex]\bold {10.9575 \ g \ NaCl}[/tex]
3. Round using Significant FiguresThe original measurements of molarity and volume have 2 significant figures, so our answer must have the same. For the number we calculated, that is the ones place. The 9 in the tenths place tells us to round the 0 up to a 1.
[tex]11 \ g \ NaCl[/tex]
There are approximately 11 grams of sodium chloride in 250 mL of a 0.75 molar solution.
An equilibrium mixture of PCl5(g), PCl3(g), and Cl2(g) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 Torr, respectively. A quantity of Cl2(g) is injected into the mixture, and the total pressure jumps to 263.0 Torr at the moment of mixing. The system then re-equilibrates. The chemical equation for this reaction is
Answer:
p'PCl3 = 6.8 torr
p'Cl2 =26.4 torr
p'PCl5 =223.4 torr
Explanation:
An equilibrium mixture of PCl5(g), PCl3(g), and Cl2(g) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 Torr, respectively. A quantity of Cl2(g) is injected into the mixture, and the total pressure jumps to 263.0 Torr at the moment of mixing. The system then re-equilibrates. The chemical equation for this reaction is
PCl3(g) + Cl2(g) ---> PCl5(g)
Calculate the new partial pressures after equilibrium is reestablished. [in torr]
pPCl3
pCl2
pPCl5
Step 1: Data given
Partial pressure before adding chlorine gas:
Partial pressure of PCl5 = 217.0 torr
Partial pressureof PCl3 = 13.2 torr
Partial pressureof Cl2 = 13.2 torr
A quantity of Cl2(g) is injected into the mixture, and the total pressure jumps to 263.0 Torr at the moment of mixing
Step 2: The equation
PCl3(g)+Cl2(g) ⇔ PCl5(g)
Step 3: The expression of an equilibrium constant before adding chlorine gas
Kp = pPCl5 / (pPCl3 * pCl2)
Kp = 217.0 / (13.2 * 13.2)
Kp = 1.245
Step 4: The expression of an equilibrium constant after adding chlorine gas
Partial pressure of PCl5 = 217.0 torr
Partial pressure of PCl3 = 13.2
Partial pressure of Cl2 = TO BE DETERMINED
Step 5: The total pressure of the system
Ptotal = pPCl5 + pPCl3 + pCl2
263.0 torr = 217.0 torr + 13.2 torr + pCl2
pCl2 = 263.0 - 217.0 -13.2 = 32.8 torr
Step 6: The initial pressure
The equation: PCl3(g)+Cl2(g) ⇔ PCl5(g)
pPCl3 = 13.2 torr
pCl2 = 32.8 torr
pPCl5 = 217.0 torr
Step 7: The pressure at the equilibrium
p'PCl3 = (13.2 -x) torr
p'Cl2 = (32.8 - x) torr
p'PCl5 = (217.0 + x) torr
Step 8: The equilibrium constant
'Kp = p'PCl5 / (p'PCl3 * p'Cl2)
1.245 = (217.0+x) / ((13.2-x)(32.8-x)
x = 6.40 torr
p'PCl3 = 13.2 -6.40 = 6.8 torr
p'Cl2 = 32.8 - 6.40 =26.4 torr
p'PCl5 = 217.0 + x) 6.4 = 223.4 torr
How many electrons will one atom of element with 6 protons and 9 neutrons .
Answer:
The atomic mass of this element would be 12 amu
Explanation:
The subatomic particles provide several bits of information about a given element.
The number of protons is equal to the atomic number of the element. For this element having 6 protons makes the atomic number 6 and makes the element Carbon.
The number of protons plus the number of neutrons equals the atomic mass of the element based upon atomic mass units (amus)
For this element 6 protons and 6 neutrons combine to make an atomic mass of 12 amus.
Lastly, the values of protons and electrons tell whether the atom is an ion or neutral. When protons equal electrons the atom is neutral. When protons are greater than neutrons the atom is a positive charge or cation. When the protons are less than the electrons the atom is a negative charge or anion. For this example the Carbon atom has 6 protons and 6 electrons making it neutral.
A 13.4 mL sample of CO2 gas was collected in an experiment.
What is this volume in liters (L)? Use significant figures, do NOT include the units.
Explanation:
here's the answer to your question
how many moles of CO2 are formed from 3.0 mol of C2H2
Answer:
50.0 moles C02
Explanation:
First write down the CORRECTLY balanced equation. NOTE: The equation you provide is incorrect.
2C2H2(g) + 5O2(g) ==> 4CO2(g) + 2H2O(g) CORRECT EQUATION
Next, look at the stoichiometric ratio of C2H2 to CO2. You can see it is 2 moles C2H2 produces 4 moles CO2.
Thus, 25.0 moles C2H2 x 4 moles CO2/2 moles C2H4 = 50.0 moles CO2
importance of hematology
Answer:
Haematology is the specialty important for the diagnosis and management of a wide range of benign and malignant disorders of the red and white blood cells, platelets and the coagulation system in adults and children.
The decomposition of ammonia is: 2 NH3(g) ⇌ N2(g) + 3 H2(g). If Kp is 1.5 × 103 at 400°C, what is the partial pressure of ammonia at equilibrium when N2 is 0.20 atm and H2 is 0.15 atm?
Answer:
"[tex]6.7\times 10^{-4} \ atm[/tex]" is the right answer.
Explanation:
Given:
Partial pressure of [tex]N_2[/tex],
= 0.20 atm
Partial pressure of [tex]H_2[/tex],
= 0.15 atm
[tex]K_p = 1.5\times 10^3[/tex] at [tex]400^{\circ} C[/tex]
As we know,
⇒ [tex]K_p = \frac{pN_2\times pH_2^3}{pNH_3^2}[/tex]
By putting the values, we get
[tex]1.5\times 10^3=\frac{0.20\times (0.15)^3}{pNH_3^2}[/tex]
[tex]pNH_3^2 = \frac{0.000675}{1.5\times 10^3}[/tex]
[tex]=6.7\times 10^{-4} \ atm[/tex]
When water reaches the metastable state, the phase of water will be?
Answer:
water, when the metastable state is reached, is cooled below the zero temperature. It freezes abruptly. this is called metastable. They are not at equilibrium per se; as at negative temperatures the only equilibrium state of water is ice.
Explanation:
Sodium acetate is produced by the reaction of baking soda and vinegar. The resultant solution is then heated until it becomes saturated and allowed to cool. As a result, the solution has become supercooled. Upon addition of a small seed crystal, the solution temperature increases as sodium acetate trihydrate crystallizes. Its molar enthalpy of fusion is 35.9 kJ/mol. How much thermal energy would be released by 276.0 g of sodium acetate trihydrate (molar mass
Answer: The thermal energy that would be released by 276.0g of sodium acetate trihydrate is 71.8kJ.
Explanation:
Supercooling is the process of lowering the temperature a liquid below its freezing point, without it becoming solid. A liquid below its freezing point will crystallize in the presence of a seed crystal because it serves as a structure for formation of crystals. From the question,
The given mass of sodium acetate trihydrate
(CH3COONa.3H2O)= 276.0g
Molar mass of sodium acetate
trihydrate= 136.08g/mol
Thermal heat of fusion of sodium acetate
trihydrate = 35.9 kJ/mol
From the given mass the number of moles present= 276.0/ 136.08
= 2.0moles
Therefore the heat (thermal) energy of the given mass of sodium acetate
trihydrate = 2.0 × 35.9
= 71.8kJ
Therefore, upon addition of a small seed crystal, the solution temperature increases as sodium acetate trihydrate crystallizes.
Write a balanced half-reaction for the oxidation of manganese ion (mn2 ) to solid manganese dioxide (mno2) in acidic aqueous solution. Be sure to add physical state symbols where appropriate.
Answer:
Mn2+(aq) + 2H2O(l) ⇒ MnO2(s) + 4H+(aq) + 2e-
Explanation:
Step 1: Data given
The oxidation number of manganese ion (Mn2+ ) is +2
The oxidation number of manganese dioxide is +(MnO2)4
This means the oxidation number from Mn will go from +2 to +4, since it's increased, this is an oxidation reaction
Mn2+(aq) ⇒ MnO2(s)
We have to balance both sides. Mn is already the same. But on the right side we have O atoms. T obalance both sides we have to add O atoms to the left side. This by adding 2x H2O
Mn2+(aq) + 2H2O(l) ⇒ MnO2(s)
Now the amount of O atoms is balanced, but we have H- atoms at the left side. To balance we have to add 4 H atoms to the right side
Mn2+(aq) + 2H2O(l) ⇒ MnO2(s) + 4H+(aq)
Now the amount of atoms is balanced at both sides. We also have to check if the charge on both sides is the same.
Since the left side has a charge of +2, and right has a charge of +4, we have to add 2 electrons to balance this.
Mn2+(aq) + 2H2O(l) ⇒ MnO2(s) + 4H+(aq) + 2e-
Hydrogen chloride decomposes to form hydrogen and chlorine, like this:
2HCl(g) + H2(g) â Cl2(g)
Also, a chemist finds that at a certain temperature the equilibrium mixture of hydrogen chloride, hydrogen, and chlorine has the following composition:
compound pressure at equilibrium
HCl 84.4 atm
H2 77.9 atm
Cl2 54.4
Required:
Calculate the value of the equilibrium constant for this reaction. Round your answer to significant digits.
Solution :
Given :
Partial pressure of HCl, [tex]$P_{HCl}$[/tex] = 84.4 atm
Partial pressure of [tex]H_2[/tex], [tex]$P_{H_2}$[/tex] = 77.9 atm
Partial pressure of [tex]Cl_2[/tex], [tex]$P_{Cl_2}$[/tex] = 54.4 atm
Reaction :
[tex]$2HCl (g) \leftrightharpoons H_2(g) + Cl_2(g)$[/tex]
Using equilibrium concept,
[tex]$k_p=\frac{(P_{H_2})(P_{Cl_{2}})}{(P_{HCl})^2}$[/tex]
[tex]$k_p=\frac{77.9 \times 54.4}{(84.4)^2}$[/tex]
[tex]$k_p=0.594$[/tex]
[tex]k_p=0.59[/tex] (in 2 significant figures)
or [tex]k_p=5.9 \times 10^{-1}[/tex]
A solution is made by mixing 34.5 g of sugar with 75.0 g of water. What is the mass percent of sugar in this solution?
Please explain and show work.
the mass percent of sugar in this solution is 46%.
Answer:
Solution given:
mass of solute=34.5g
mass of solvent=75g
mass percent=[tex]\frac{mass\:of\:solute}{mass\:of \:solvent}*100\%[/tex]
=[tex]\bold{\frac{34.5}{75.0}*100\%=46\%}[/tex]
We know that,
[tex]mass \: solute = \frac{mass \: of \: solute}{mass \: of \: solvent} [/tex]×100
[tex] \frac{34.5}{75.0} = 0.46[/tex]
hope it helps..
For each reaction, write the chemical formulae of the oxidized reactants.
a. ZnCl2 (aq) + 2Na(s) → Zn(s) + 2NaCl(aq)
b. Al(s) + FeBrz (aq) → AlBrz (aq) + Fe(s)
c. FeSO4 (aq) + Zn (s) → Fe(s) + ZnSO4(aq)
Answer:
a. Na(s); b. Al(s); c. Zn(s)
Explanation:
Let's consider the following redox reactions.
a. ZnCl₂ (aq) + 2 Na(s) → Zn(s) + 2 NaCl(aq)
Na is oxidized because its oxidation number increases from 0 to +1 (in NaCl) whereas Zn is reduced because its oxidation number decreases from 2+ (in ZnCl₂) to 0.
b. Al(s) + FeBr₃ (aq) → AlBr₃ (aq) + Fe(s)
Al is oxidized because its oxidation number increases from 0 to +3 (in AlBr₃) whereas Fe is reduced because its oxidation number decreases from 3+ (in FeBr₃) to 0.
c. FeSO₄ (aq) + Zn(s) → Fe(s) + ZnSO₄(aq)
Zn is oxidized because its oxidation number increases from 0 to +2 (in ZnSO₄) whereas Fe is reduced because its oxidation number decreases from 2+ (in FeSO₄) to 0.
A certain watch’s luminous glow is due to zinc sulfide paint that is energized by beta particles given off by tritium, the radioactive hydrogen isotope 3 H, which has a half-life of 12.3 years. This glow has about 1/10 of its initial brightness. How many years old is the watch? g
Answer:
The watch is 40.9 years old.
Explanation:
To know how many years old is the watch we need to use the following equation:
[tex] I_{(t)} = I_{0}e^{-\lambda t} [/tex] (1)
Where:
[tex]I_{(t)}[/tex]: is the brightness in a time t = (1/10)I₀
[tex]I_{0}[/tex]: is the initial brightness
λ: is the decay constant of tritium
The decay constant is given by:
[tex] \lambda = \frac{ln(2)}{t_{1/2}} [/tex] (2)
Where:
[tex]t_{1/2}[/tex]: is the half-life of tritium = 12.3 years
By entering equation (2) into (1) we have:
[tex] I_{(t)} = I_{0}e^{-\lambda t} = I_{0}e^{-\frac{ln(2)}{t_{1/2}}t} [/tex]
[tex] \frac{I_{(t)}}{I_{0}} = e^{-\frac{ln(2)}{t_{1/2}}t} [/tex]
By solving the above equation for "t" we have:
[tex] ln(\frac{I_{(t)}}{I_{0}}) = -\frac{ln(2)}{t_{1/2}}t [/tex]
[tex] t = -\frac{ln(\frac{I_{(t)}}{I_{0}})}{\frac{ln(2)}{t_{1/2}}} = -\frac{ln(\frac{1}{10})}{\frac{ln(2)}{12.3}} = 40.9 y [/tex]
Therefore, the watch is 40.9 years old.
I hope it helps you!
Balance the following reaction:
_______ CO₂ + _______ H₂O + heat ↔ _______ C₆H₁₂O₆ + _______ O₂
Please explain!
*Note: If any of the coefficients are the number one. Please, write "1" in the space. Thanks!
Answer:
6CO2+6H2O+heat" C6H12O6+6O2
the best way to balance a chemical reaction is to start with balancing the hydrogen followed by the other elements then lastly oxygen.so in this case if you put a 6 in front of carbon dioxide,water and oxygen you will definitely balance it.cause at the first side you have 6 carbons similar to the product,12 oxygen similar to the product and 18 oxygen similar to the products.
I hope this helps
Answer:
Explanation:
I saw this after answering your other question on the same reaction.
To balance the chemical reaction, look at the reactants and products. As O is part of both products, focus on C and H instead.
On the products side, 1 C6H12O6 has 6 C and 12 H. So that requires the same numbers of C and H on the reactant side because of mass conservation.
That gives 6 CO2 and 6 H2O as the reactants. Counting the number of O in the reactants, there are 6*2 + 6 = 18 O. Subtracting the 6 O in C6H12O6, that leaves 12 O so there are 12/2 = 6 O2 in the products.
Combining the numbers above, the balanced equation is:
___6___ CO₂ + ___6___ H₂O + heat ↔ ___1___ C₆H₁₂O₆ + ___6___ O₂
