Answer:
Sodium metaborate: sodium+boron+oxygen= NaBO 2.
The formula can be written also as Na 2O·B
Explanation:
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."
The graph below shows how the temperature and volume of a gas vary when
the number of moles and the pressure of the gas are held constant. What
happens to the
temperature of a gas as its volume increases?
A. The temperature decreases
B. The temperature increases
C. The temperature remains the same.
D. The temperature doubles.
Answer:
C
Explanation:
because it remains the same
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]
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
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.
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
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
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
Bond length is the distance between the centers of two bonded atoms. On the potential energy curve, the bond length is the internuclear distance between the two atoms when the potential energy of the system reaches its lowest value. Given that the atomic radii of H and Br are 37.0 pm and 115 pm , respectively, predict the upper limit of the bond length of the HBr molecule. Express your answer to three significant figures and include the appropriate units. View Available Hint(s)for Part C
Answer:
The answer is "152 pm".
Explanation:
The bond length from the values inside the atomic radii is calculated according to the query. This would be the upper limit of a molecule's binding length.
The atomic radius of [tex]H= 37.0 \ pm[/tex]
The atomic radius of [tex]Br = 115.0 \ pm[/tex]
[tex]\text{Bond length = Atomic radius of H + Atomic radius of Br}[/tex]
[tex]= 37.0\ pm + 115.0 \ pm\\\\= 152\ pm[/tex]
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-
Classify these bonds as ionic, polar covalent, or nonpolar covalent. You are currently in a sorting module.
Ionic Polar Covalent Nonpolar covalent
C-O , Mg-F , Cl-Cl
Answer:
C-O: polar covalent
Mg-F: ionic
Cl-Cl: nonpolar covalent
Explanation:
Ionic bonds are formed between an atom of a metallic element and another atom of a non-metallic element. Thus, Mg-F is an ionic bond, in which Mg is the metal and F is the nonmetal.
Covalent bonds are formed between two non-metallic elements. So, C-O and Cl-Cl are covalent bonds, because C, O, and Cl are nonmetals.
In C-O, the atom of oxygen (O) has more electronegativity than the atom of carbon (C). Thus, O will attract the electrons with more strength and a difference in charge will be established between the two bonded atoms. So, this covalent bond is polar.
In Cl-Cl, both atoms have the same electronegativity because they are from the same chemical element (Cl). Thus, this bond is nonpolar.
Rank the compounds NH3, CH4, and PH3 in order of decreasing boiling point. Choices: A) NH3 > CH4 > PH3 B) CH4 > NH3 > PH3 C) NH3 > PH3 > CH4 D) CH4 > PH3 > NH3 E) PH3 > NH3 > CH4
Answer:
C) NH3 > PH3 > CH4
Explanation:
The boiling point of a substance depends on the nature of intermolecular interaction between the molecules of the substance. The greater the magnitude of intermolecular interaction between the molecules of the substance, the higher the boiling point of the substance.
Both NH3 and PH3 have intermolecular hydrogen bonding between their molecules. However, since nitrogen is more electronegative than phosphorus, the magnitude of intermolecular hydrogen bonding in NH3 is greater than in PH3 hence NH3 has a higher boiling point than PH3.
CH4 molecules only have weak dispersion forces between them hence they exhibit the lowest boiling point.
What trends were seen in medeleevs periodic table
Answer:
groups are based on how many electrons to become stable
Explanation:
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
The enthalpy of formation for CO2 (s) and CO2 (g) is: -427.4 KJ/mole and -393.5 KJ/mole, respectively. The sublimation of dry ice is described by CO2 (s) → CO2 (g).
The enthalpy needed to sublime 986 grams of CO2 is:
(a) 181.5 Kcal
(b) 611.7 Kcal
(c) 248.3 Kcal
(d) 146.2 Kcal
Answer:
a. 181.5 kcal
Explanation:
Step 1: Calculate the enthalpy of the process (ΔH°).
Let's consider the following process.
CO₂(s) → CO₂(g)
We can calculate the enthalpy of the process using the following expression.
ΔH° = ∑ np × ΔH°f(p) - ∑ nr × ΔH°f(r)
ΔH° = 1 mol × ΔH°f(CO₂(g)) - 1 mol × ΔH°f(CO₂(s))
ΔH° = 1 mol × (-393.5 kJ/mol) - 1 mol × (-427.4 kJ/mol) = 33.9 kJ
According to the balanced equation, 33.9 kJ are required to sublime 1 mole of CO₂.
Step 2: Convert 986 g of CO₂ to moles
The molar mass of CO₂ is 44.01 g/mol.
986 g × 1 mol/44.01 g = 22.4 mol
Step 3: Calculate the enthalpy needed to sublime 22.4 moles of CO₂
22.4 mol × 33.9 kJ/1 mol = 759 kJ
We can convert it to Kcal using the conversion factor 1 kcal = 4.184 kJ.
759 kJ × 1 kcal/4.184 kJ = 181.5 kcal
Enough of a monoprotic weak acid is dissolved in water to produce a 0.0118 M solution. The pH of the resulting solution is 2.32 . Calculate the Ka for the acid.
Answer:
1.94 × 10⁻³
Explanation:
Step 1: Calculate the concentration of H⁺ ions
We will use the definition of pH.
pH = -log [H⁺]
[H⁺] = antilog -pH = antilog -2.32 = 4.79 × 10⁻³ M
Step 2: Calculate the acid dissociation constant (Ka) of the acid
For a monoprotic weak acid, whose concentration (Ca) is 0.0118 M, we can use the following expression.
Ka = [H⁺]²/Ca
Ka = (4.79 × 10⁻³)²/0.0118 = 1.94 × 10⁻³
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:
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.
All of the following statements concerning crystal field theory are true EXCEPT Group of answer choices in an isolated atom or ion, the five d orbitals have identical energy. low-spin complexes contain the maximum number of unpaired electrons. in low-spin complexes, electrons are concentrated in the dxy, dyz, and dxz orbitals. the energy difference between d orbitals often corresponds to an energy of visible light. the crystal field splitting is larger in low-spin complexes than high-spin complexes.
Answer:
low-spin complexes contain the maximum number of unpaired electrons.
Explanation:
In the crystal field theory, the magnitude of crystal field splitting and the pairing energy determines whether a complex will be low spin or high spin.
Low spin complexes often have greater magnitude of crystal field splitting energy than low spin complexes.
High spin complexes have maximum number of unpaired electrons(most of the electrons are unpaired) while low spin complexes have a minimum number of unpaired electrons in a complex(most of the electrons are paired).
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!
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.
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
How are radioactive isotopes used to diagnose thyroid problems?
Answers
A.
Production of radioactive iodine by the thyroid is measured.
B.
Uptake of radioactive iodine by the thyroid gland is measured.
C.
Destruction of damaged cells by radioisotopes is measured.
D.
Absorption of gamma rays from ingested radioisotopes is measured.
Question
Answer:
A
Explanation:
uptakw of raduoactive iofine by the thyroid gland js measurd
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]
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.
what is the charge on the Mn ions in Mn2o3? 1+, 2+, 3+,3-,4+?
Write the symbol for every chemical element that has atomic number greater than 55 and less than 140.8 u
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
https://brainly.com/question/2998270
What is the oxidation state of rubidium (Rb)?
A. +1
B. -2
C. +2
D. -1
Answer:
The answer is A. +1
Explanation:
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.
