For solving this exercise we need the molar mass of CuNO3 which is 125.5 g/mol
Step 1) We calculate the moles of this compound:
8.54 g CuNO3 / 125.5 g/mol = 0.0680 moles CuNO3
Step 2) We calculate the volume of the solution in L:
(Remember M is equal to mol/L)
0.0680 moles CuNO3 / 0.520 mol/L = 0.131 L
Step 3) We change L to mL:
(Remember 1L=1000mL)
0.131 L x (1000mL/1L) = 131 mL
Answer: Volume = 131 mL
A student stops a titration BEFORE theequivalence point is reached. If NaOH isbeing added to an unknown acid, HA, whations are present in the clear solution in theflask?A. H+, A-, Na+, OH-B. A-, Na+, OH-C. A-, Na+ and H+
Answer
ions are present in the clear solution in the flask, if NaOH is being added to an unknown acid, HA, before equivalence are:
. A⁻, Na⁺ and H⁺
What is the pH of a 10^-4 M HCl solution?Answers to choose:•1•2•3•4
Hydrochloric acid is a strong acid. This means that in a solution all the H+ ions will be released. Also, one mole of HCl has one hydrogen atom, so we can assume that the concentration of HCl is equal to the concentration of H+ ions.
On the other hand, the definition of pH tells us:
[tex]pH=-log\lbrack H^+\rbrack[/tex]Where [H+] is the ions H+ concentration.
H+ concentration is 10^-4M. If we replace in the equation we will have:
[tex]\begin{gathered} pH=-log10^{-4}M=4 \\ pH=4 \end{gathered}[/tex]answer: The pH of the solution is 4
Question in image can you please answer it
Answer:
C
Explanation:
The answer is c because the formula for calcium sulfate is CaSO4, and there is only one equation with this in.
Hope this helps!
What mass of iron is contained in 86.6 grams of chalcopyrite, CuFeS2?
Answer:
26.428 gm
Explanation:
molecular mass of CuFeS2 = 63.5+(2*32)+56 = 183.5 gm
Now we apply unitary method,
183.5 gm Chalcopyrite contains 56 gm Fe
∴ 86.6 gm Chalcopyrite contains [tex]\frac{56}{183.5} X 86.6[/tex]
= 26.428 gm Fe
ObjectiveYou will write a report on the states ofmatter, answering the question:How do water molecules change as theirstate of matter changes from solid, to liquid,to gas?MaterialsParagraph graphic organizerRubricThe Changing Water MoleculeReportHow do water molecules change as theirstate of matter changes? You will explorethese changes as you write your report onthe changing molecules.Hydrogen and oxygen bond together to formwater molecules. Water's state of matter willchange depending on how these moleculesare arranged. You have been learning aboutthe three states of matter - solid, liquid, andgas. Use this knowledge to write a report inresponse to this question:How do water molecules change as theirstate of matter changes from solid, toliquid, to gas?Your report should include an introduction, aparagraph for each state of matter..
Topic sentence:
All matter is composed of atoms that are in constant motion, three types of motion can be identified translation, rotation, and vibration. These particles are so small that they cannot be observed even with the smallest microscope, but they are so important that depending on their motion and interaction with other particles will be their state and how we see them at the macroscopic level. The greater the movement of the molecules the greater the disorder, so we can distinguish three states of matter presented from greater to lesser disorder: gas, liquid and solid.
Gas:
The motion in gases presents the three types of movement: translation, rotation, and vibration, and is the most intense. In the case of gases, the particles are quite far apart compared to their size. The approximate average distance between particles, under normal conditions, is ten times the particle size. In gases the particles do not occupy fixed positions, they are disordered and move randomly vibrating, rotating, and moving in all directions.
Liquid:
Liquids also exhibit all three types of motion (translation, rotation, and vibration) but with less intensity than in gases. In liquids, the particles do not occupy fixed They are disordered, although less so than in gases, and move randomly, as in gases, but with less intensity.
Solid:
The cohesive forces in solids are strong, much greater than in liquids, which are weak, and practically non-existent in gases. In solids, the particles occupy fixed positions, are ordered, and have no translational or rotational motion, only vibration.
Concluding sentence:
For all states of matter, the pressure and temperature involved will affect the order of the molecules making it possible to pass from one state to another by changing these variables.
In an experiment, hydrochloric acid reacted with different volumes of sodium thiosulfate in water. A yellow precipitate was formed during the reaction. A cross drawn at the base of each flask became gradually invisible due the formation of this yellow precipitate. The time taken for the cross to become invisible was recorded. A partial record of the experiment is shown.Experimental RecordFlaskVolume ofHClVolume ofSodium ThiosulfateVolume ofWaterTime110 mL10 mL40 mL14 seconds210 mL20 mL30 mL310 mL30 mL20 mL410 mL40 mL10 mLBased on your knowledge of factors that affect the rates of chemical reactions, predict the trend in the last column of the experimental record. Use complete sentences to explain the trend you predicted. You do not have to determine exact values for time; just describe the trend you would expect (increase or decrease) and why it occurs.
Answer
Decrease
Explanation
In the last column, the time it will take for the reaction to complete with decrease, meaning the rate of reaction will increase, this is because the reactant concentration was increased. Increasing the concentration of one or more reactants will often increase the rate of reaction. Because a higher concentration will lead to more collisions of that reactant at a specific time.
Describe two properties that the elements in the last two columns of the Periodic Table share with one another.
Two properties that the elements in the last two columns of the Periodic Table share with one another are;
they are all non-metalsthey show gradation in their physical and chemical properties.What are the elements in the last two columns of the periodic table?The elements in the last two columns of the periodic table are the group 17 and group 18 elements of the periodic table.
The elements in group 17 are known as halogens while the elements in group 18 are known as noble gases.
The halogens includes the following elements - Fluorine, Chlorine, Bromine, Iodine, Astatine, and Ununsepetium.
The noble gases include the following elements - Helium, Neon, Argon, Krypton, Xenon, and Unonoctium.
A common feature of these elements is that they are all non-metals. The elements also show gradation in their physical and chemical properties.
Learn more about halogens and noble gases at: https://brainly.com/question/10242176
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With the following information: Volume of vinegar 25 mL Mass of sodium bicarbonate 1 g Initial temperature of vinegar 17 °C Final temperature of vinegar 14 °C Change in temperature °C (∆T) ?? (answer needed) Calculate the approximate enthalpy of the reaction in joules. Estimate that 1.0 mL of vinegar has the same thermal mass as 1.0 mL of water. Ignore the thermal mass of the sodium bicarbonate. Note: It takes about 4.2 joules (J) to change 1.0 gram (1.0 mL) of water 1.0 °C.
First, let's calculate the change in temperature, which results in making the next subtraction:
[tex]\Delta T=\text{ T}_{fin}\text{ - T}_{in}=\text{ 14\degree C-17\degree C= -3\degree C}[/tex]Now, let's calculate the enthalpy of the reaction with the next assumptions:
[tex]\begin{gathered} -\Delta H_r=\text{ Q}_{sen}^{vinegar}+\text{ Q}_{sen}^{NaHCO_3} \\ \\ As\text{ we are told in the exercise, we ignore the second term. Then:} \\ \\ -\Delta H_r=\text{ Q}_{sen}^{vinegar} \end{gathered}[/tex]We also have to make the following assumptions: the vinegar's thermal mass and density are the same as the water's. Then, we can make the calculation:
[tex]\begin{gathered} Q_{sen}^{vinegar}=\text{ m*Cp*}\Delta T \\ \\ Where\text{ m is the mass. } \\ Cp\text{ is the specific thermal capacity. } \\ \Delta T\text{ is the change in temperature. } \end{gathered}[/tex][tex]Q_{sen}^{vinegar}=\text{ 25 mL * }\frac{1\text{ g}}{1\text{ mL}}*\text{ 4.2 }\frac{J}{g*^oC}*(-3^oC)=\text{ -315 J = -}\Delta H_r[/tex]So, the reaction enthalpy equals 315 J approx.
If the density of an ether was 0.727g/mL and the mass 17.7 grams, what is the volume?Report the answer in (mL) to three significant figures and also include the correct unit
Remembering the formula of density, we can clear the volume and replace the given data in the new formula:
[tex]\begin{gathered} \text{density}=\frac{mass}{volume}, \\ \text{volume}=\frac{mass}{density}. \end{gathered}[/tex]Our data is: for mass is 17.7 grams (g) and for density is 0.727 g/mL:
[tex]\text{volume}=\frac{17.7\text{ g}}{0.727\text{ }\frac{g}{mL}}=24.347mL=24.3mL.[/tex]The volume of the ether is 24.3 mL of 17.7g and 0.727 g/mL.
Calculate the moles of HCI needed to react completely with 8.25 moles of zinc. I picked C but I’m not sure if I’m correct
Answer:
D. 16.5 moles
Explanation:
From the balanced reaction we know that 1 mole of Zn reacts with 2 moles of HCl.
We can calculate the moles of HCl needed to react completely with 8.25 moles of zinc, using the stoichiometry of the reaction and a mathematical rule of three:
[tex]\begin{gathered} 1moleZn-2molesHCl \\ 8.25molesZn-x=\frac{8.25molesZn*2molesHCl}{1moleZn} \\ x=16.5molesHCl \end{gathered}[/tex]So, 16.5 moles of HCl are needed.
b) Assuming a reaction starts with 1 mole of octane, which burns at 220°C, and is at a pressure of 1 atm, what volume will the products occupy upon reaction completion?
Answer
40.48 liters
Explanation
Given: that:
n = mole = 1 mol
P = Pressure = 1 atm
T = Temperature = 220 °C = (220 + 273K) = 493 K
What to find:
The volume of the product.
Step-by-step solution:
The volume, V of the product can be calculated using the ideal gas equation:
PV = nRT
Putting the values of the parameters and molar gas constant, R = 0.0821 liter·atm/mol·K into the formula, we have:
[tex]\begin{gathered} V=\frac{nRT}{P}=\frac{1\text{ }mol\times0.0821\text{ }L.atm\text{/}mol.K\times493\text{ }K}{1\text{ }atm}=40.48\text{ }L \\ \end{gathered}[/tex]Thus, the volume of the product, upon reaction completion, will occupy 40.48 liters.
(07.05 HC)In an experiment, sulfuric acid reacted with different volumes of sodium thiosulfate in water. A yellow precipitate was formed during thereaction. A cross drawn at the base of each flask became gradually invisible due the formation of this yellow precipitate. The time taken for thecross to become invisible was recorded. A partial record of the experiment is shown.Experimental RecordFlask Volume of Volume of Volume of TimeH2SO4 Sodium Thiosulfate Water5 mL 50 mL10 mL 119 seconds5 mL 40 ml10 mL3 5 ml 30 ml120 mL4 15 ml 120 mL30 mLBased on your knowledge of factors that affect the rates of chemical reactions, predict the trend in the last column of the experimental record.Use complete sentences to explain the trend you predicted. You do not have to determine exact values for time; just describe the trend youwould expect (increase or decrease) and why it occurs.
The rate of the reaction usually depends on the concentration of the reactants. We have to reactants in this reaction, sulfuric acid and sodium thiosulfate. Let's look at the values that were recorded.
The volume of sulfuric acid that is added is constant, but the volume of sodium thiosulfate is not. It decreases and the volume of water increases, so we are diluting the sodium thiosulfate. The concentration of one reactant is decreasing. So the rate of the reaction is also decreasing. Then the time that it takes for the cross to become invisible should increase.
Answer: the time should increase.
How much energy must be removed from 13.9 grams of gaseous oxygen in order to decrease the temperature from 38.8°C to 23.8°C? The specific heat of oxygen is 0.219 cal/g °C. in cal
So,
To solve this question, we could use the equation:
What we're going to do now is to replace the values that we know:
The negative sign of the result indicates that this heat will be lost.
Therefore, the amount of energy that must be removed, equals 45.6615 cal.
What must the final volume be for the pressure of the gas to be 1.56 atm at a temperature of 336 K?Express your answer in milliliters to three significant figures.
According to the Combined gas Law :
[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]The problem gives P1 = 1.32 atm, V1 = 212 ml, T1 = 292 K, P2 = 1.56 atm, T2 = 336 K.Then you just have to subtitute and solve:
[tex]V_2=\frac{P_1V_1}{T_1}\frac{T_2}{P_2}=\frac{1.32\text{ atm x 212 ml}}{292\text{ K}}\frac{336\text{ K}}{1.56\text{ atm}}=206\text{ ml}[/tex]
Calculate the number of moles of carbon dioxide gas in this fire extinguisher. Gas type CO2 temperature: 212 K pressure: 4.55 atm Volume: 3.5L
Answer
0.9154 mol CO₂
Procedure
To solve this question you will need to use the ideal gas law
PV=nRT
Data
P=4.55 atm
V= 3.5 L
T= 212 K
R=0.082057 L⋅atm⋅K⁻¹⋅mol⁻¹
Solve for n
[tex]n=\frac{PV}{RT}=\frac{4.55\text{ atm }3.5\text{L mol }\degree\text{K}}{0.082057\text{ L.atm }212\degree\text{K}}=0.9154\text{ mol }[/tex]suppose you were given three cookies containing radioactive samples, one with an alpha-emitter, the second has a beta-emitter, and the third contains that emits gamma-ray. if you were allowed to handle or do something with the cookies: (and explain the reason)1. Will you eat it2. Will you just place it in your pocket3 will you just hold it with your hand
I will eat the GAMMA-RAY because it can easily go outside my body without much harm. Those rays are just high-energy photons.
I think I would probably put the alpha-emitter in my pocket because those are easily stopped by a sheet of paper.
Which describe anions? (Mark ALL the correct answers.)
metals
nonmetals
positive charge
negative charge
more electrons than protons
more protons than electrons
Answer: nonmetals, negative charge, more electrons than protons
Explanation:
-4 AI (s) + 302(g)=2Al2O3(s)How many grams of Al2O3 can be formed from 43.91 g of Al and 54.13 g of O2?(Hint: You need to determine which one is the limiting reactant).
82.94g of Al2O3 will be produced.
1st) With the molar mass of each compound and the balanced equation, we need to find which reactant is the limiting reactant.
- Al molar mass: 27g/mol
- O2 molar mass: 32g/mol
- Al2O3 molar mass: 102g/mol
Calculation with Al:
According to the balanced equation, 108g of Al (4 x 27g) needs 96g of O2 (3 x 32g), so with a mathematical rule of three we can calculate the amount of oxygen that is needed to react with 43.91g of Al:
[tex]\begin{gathered} 108\text{gAl}-96gO_2 \\ 43.91\text{gAl}-x=\frac{43.91\text{gAl}\cdot96gO_2}{108\text{gAl}} \\ x=39.03gO_2 \end{gathered}[/tex]Now we know that, 43.91g of Al will need 39.03g of O2 to react, and we have 54.13g of O2, so there is excess of oxygen.
Calculation with O2:
According to the balanced equation, 108g of Al (4 x 27g) needs 96g of O2 (3 x 32g), so with a mathematical rule of three we can calculate the amount of aluminum that is needed to react with 54.13g of O2:
[tex]\begin{gathered} 96gO_2-108gAl \\ 54.13gO_2-x=\frac{54.13gO_2\cdot108gAl}{96gO_2} \\ x=60.89\text{gAl} \end{gathered}[/tex]Now we know that 54.13g of O2 will need 60.89g of Al to react, and we have only 43.91g of Al, so it is not sufficient, and that's why Al is the limiting reactant.
2nd) Now that we know that Al is the limiting reactant, we can calculate the grams of Al2O3 that will be produced in the reaction by using the limiting reactant (43.91g):
[tex]\begin{gathered} 108\text{gAl}-204gAl_2O_3 \\ 43.91\text{gAl}-x=\frac{43.91\text{gAl}\cdot204gAl_2O_3}{108\text{gAl}} \\ x=82.94gAl_2O_3 \end{gathered}[/tex]In this calculation we use the balanced equation, so we know that 108g of Al will produce 204g of Al2O3 (2 x 102g).
Finally, 82.94g of Al2O3 will be produced from 43.91g of Al and 54.13g of O2.
Oxygen and oxygen-containing compounds are involved in many different reactions. which balanced equation represents a reaction that involves 14 atoms of oxygen
Answer: 2C2H6+7O2→4CO2+6H2O
Explanation:
Answer:2C2H6+7O2→4CO2+6H2O
Explanation:
Please help me because I am unable to find help and I have to turn in this assignment or else I will fail this class for the year.
Explanation:
To solve this question, we need to use the following equation:
M = n/V
where:
M = molarity
n = number in moles
V = volume
For trial 1 we have:
M of NaOH = 0.112 mol/L
n = ??
V = 7.18 mL (we need to transform it to L, just divide by 1000)
V = 0.718 L
So:
M = n/V
n = M*V
n = 0.112*0.718
n = 0.080 moles of NaOH
The equation of dissociation of NaOH is:
NaOH --> Na+ + OH-
0.080 0.080 0.080
Answer for trial one: 0.080 moles of OH- or 8.0x10^-2 moles
For trial 2 we have:
For trial 1 we have:
M of NaOH = 0.112 mol/L
n = ??
V = 7.20 mL (we need to transform it to L, just divide by 1000)
V = 0.720 L
So:
M = n/V
n = M*V
n = 0.112*0.720
n = 0.081 moles of NaOH
The equation of dissociation of NaOH is:
NaOH --> Na+ + OH-
0.081 0.081 0.081
Answer for trial two: 0.081 moles of OH- or 8.1x10^-2 moles
At equilibrium, 1.0 mol of a 4.5 x 10-4 M solution of substance A reacts with 1 mol of a solid to form 2.0 mol of a 1.2 x 10-2 M solution of substance C and 1.0 mol of a solution of substance D. Given that K = 2.0 x 10-6, what is the equilibrium concentration of substance D?
Step 1
The reaction:
1 A (aq) + 1 B (s) <=> 2 C (aq) + 1 D (aq)
A, C and D participate in the equilibrium constant K
B doesn't participate
--------------------
Step 2
Data provided:
At equilibrium:
[A] =4.5 x 10^-4 M
[C] = 1.2 x 10^-2 M
[D] = Unknown
-------------------
Step 3
[tex]K\text{ = }\frac{\lbrack D\rbrack^1\lbrack C\rbrack^2}{\lbrack A\rbrack^1}=\frac{\lbrack D\rbrack.(1.2x10^{-2}M)^2}{(4.5x10^{-4}M)}=\text{ 2.0 x 10}^{-6}[/tex]Answer:
[D] = 6.25x10^-6 M
Calculate the quantity (in grams) of sucrose (C12H22011) required to make a 1.00 M strength solution with a volume of 500. mL.
Let us first define the term molarity. Molarity is a way of expressing the concentration of a compound in solution and is defined as the number of moles per liter of solution.
We are given the molarity value, 1.00M, and the volume of the solution, 500 mL=0.5L. We can find the number of moles by clearing them from the following equation:
[tex]\begin{gathered} Molarity=\frac{MolesSolute}{Lsolution} \\ MolesSolute=Molarity\times Lsolution \end{gathered}[/tex]We replace the known data:
[tex]\begin{gathered} MolesSolute=1.00M\times0.5L \\ MolesSolute=1.00\frac{mol}{L}\times0.500L=0.500molC_{12}H_{22}O_{11} \end{gathered}[/tex]Now, the grams of sucralose are found by multiplying the moles found by the molar mass of sucralose, which is 342.3 g/mol:
[tex]gC_{12}H_{22}O_{11}=0.500molC_{12}H_{22}O_{11}\times\frac{342.3gC_{12}H_{22}O_{11}}{1molC_{12}H_{22}O_{11}}=171gC_{12}H_{22}O_{11}[/tex]The quantity of sucrose required will be 171 g of sucrose
If the following solutions are mixed, is the resulting solution acidic, basic, or neutral?60.0 mL of 0.0500 M HClO4 and 40.0 mL of 0.0750 M NaOH
Answer
The resulting solution will be neutral.
Explanation
Given:
Volume of HClO4 = 60.0 mL = 0.06 L
Molarity of HClO4 = 0.0500 M
Volume of NaOH = 40.0 mL = 0.04 L
Molarity of NaOH = 0.0750 M
What to find:
To determine if the resulting solution from (HClO4 and NaOH) acidic, basic, or neutral.
Solution;
The first step is to write a balanced chemical equation for the reaction.
HClO4 + NaOH → NaClO4 + H2O
The reaction is a neutralization reaction (also a double displacement reaction). The strong acid (HClO4) and a strong base react to produce salt (NaClO4) and water (H2O).
The next step is to determine the moles of the acid and the base using the formula below:
[tex]\begin{gathered} Moles=Molarity\times Volume\text{ }in\text{ }L \\ \\ For\text{ }HClO_4: \\ \\ Moles=0.0500M\times0.06L=0.003\text{ }mol \\ \\ For\text{ }NaOH: \\ \\ Moles=0.0750M\times0.04L=0.003\text{ }mol \end{gathered}[/tex]From the balanced chemical equation, 1 mol HClO4 neutralized 1 mol NaOH.
Therefore, 0.003 mol HClO4 will completely neutralize 0.003 mol NaOH and the resulting solution will be neutral.
If you need 300g of NH3 how many liters of N2 are required? N2+3H2=2NH3
Answer:
197.12L of N2 are required.
Explanation:
1st) From the balanced reaction we know that 1 mol of N2 reacts with 3 moles of H2 to produce 2 moles of NH3.
2nd) It is necessary to convert moles to grams with the molar mass, and use the stoichiometry of the balanced reaction to find the amount of N2 moles that are required to produce 300g of NH3:
- N2 molar mass: 28g/mol
- NH3 molar mass: 17.01g/mol
[tex]300gNH_3*\frac{1mol}{17.01g}=17.64mol[/tex]300g of NH3 is equal to 17.64 moles of NH3.
3rd) Now, with a mathematical rule of three we can find the moles of N2 that are required to produce 17.64 moles of NH3:
[tex]\begin{gathered} 2molesNH_3-1molN_2 \\ 17.64molesNH_3-x=\frac{17.64molesNH_3*1molN_2}{2molesNH_3} \\ x=8.82molN_2 \end{gathered}[/tex]Now we know that 8.82 moles of N2 are needed.
4th) Finally, using the molar volume value (22.4L), we can calculate the liters of N2 that are required.
The molar volume indicates that 1 mole is equivalent to a volume of 22.4L under normal conditions.
[tex]8.80mol*\frac{22.4L}{1mol}=197.12L[/tex]So, 197.12L of N2 are required.
Perform the followingmathematical operation, andreport the answer to theappropriate number ofsignificant figures.7.55 + 17.002 = [ ?=
7.55 + 17.002 = ?
First, we have to determine the number of significant figures that each number has.
7.55: 3 SF and 2 SF after the decimal point.
17.002: 5 SF and 3 SF after the decial point.
Second, perform the mathematical operation as we usually do.
7.55 + 17.002 = 24.552
Finally we determine the correct number of SF. When adding or substracting the rule says: "The final answer may have no more SF after the decimal point than the least number of significant figures in the operation". Since 7.55 has 2 SF after the decimal point, we must round the result of the addition to the second place to the right of the decimal point.
Rounding 24.552 we get 24.55.
Answer: 24.55
7. Which is the best description of when water boils?a. It is when the vapor pressure of water equals the surrounding atmospheric pressure.b. It is when the first bubble appears in a sample that is being heated.c. It is when bubbles are continually forming in a sample that is being heated.d. It is when the temperature of the water reaches 100 °C.
The alternative b is not a good description because you can form bubles without the water being boiling yet.
Also, alternative c is also not a good option because it is just a description of what is being seen, it could be replicated in other ways.
Alternative d only gives a termperature, and the boiling temperature can change depending on other conditions.
The alternative the best describes is alternative A.
A block of al with a density of 2.7/ml has a mass of 549 g. what's the volume. ( show steps)
Explanation:
We are given: density of Al = 2.7g/mL
: mass of Al = 549g
[tex]\begin{gathered} \rho\text{ = }\frac{m}{V} \\ \\ \therefore\text{ V =}\frac{m}{\rho} \\ \\ \text{ =}\frac{549}{2.7} \\ \\ \text{ = 203.33mL} \end{gathered}[/tex]Answer:
The volume is 203.33mL
What is the mole fraction of a solute in solution?
The mole fraction of a solute is the ratio of the number of moles of that solute to the total number of moles of the solution.
mole fraction of A = moles of A / total number of moles.
Moles and mass and help with ratios eg 3:4 3:7
The balanced equation is:
[tex]Mg_{(s)}+H_2O_{(g)}\to MgO_{(s)}+H_{2(g)}[/tex]The first step is to convert the given mass of magnesium to moles (MW=24.3g/mol):
[tex]1g\cdot\frac{1mol}{24.3g}=0.04mol[/tex]Now, use the ratio given by the chemical equation:
[tex]0.04molMg\cdot\frac{1molMgO}{1molMg}=0.04molMgO[/tex]Now, use the molecular weight of magnesium oxide to convert it to grams (MW=40.3g/mol):
[tex]0.04mol\cdot\frac{40.3g}{1mol}=1.612g[/tex]The maximum mass of magnesium oxide produced is 1.612g.
Iodine-131 has a half-life of 8 days. If you start with a sample of 150 grams, how much of the original isotope will remain after 30 days?A. 144 gramsB. 150 gramsC. 11 gramsD. 8 grams
The first step is to remember the equation for radioactive disintegration:
[tex]M(t)=\text{ M}_{0\text{ }}*\text{ }e^{\frac{-t\text{ * ln 2}}{T}}[/tex]Where M(t) is the mass of the atom at any time t, Mo is the initial mass of the element, and T is the half-life time.
In this case, we have to calculate M(t), which is the remaining mass of the radioactive element. We have all the data from the right part of the equation (initial mass Mo, the time t at we want to calculate the remaining mass, and the half-life T of the element):
[tex]\begin{gathered} M(8)=\text{ 150 g * }e^{\frac{-30\text{ d }*\text{ }ln\text{ 2}}{8\text{ d}}\text{ }} \\ M(8)=\text{ 11.1488 g} \end{gathered}[/tex]Then, the answer is that the remaining mass of iodine-131 is C. 11 g approx