Answer:
5. 56.5217
Explanation:
Calculate percent yield by (actual yield / theoretical yield) times 100
(13/23) x 100 = 56.52173913
If a piece of silver specific heat .2165 j/g °C with a mass of 14.16 g and a temperature of 133.5°C is dropped into 250.0 g of fat 17.20°C what will be the final temperature of the system
Answer:
[tex]T_F=17.56\°C[/tex]
Explanation:
Hello there!
In this case, for this calorimetry problem, it is possible for us to realize that the heat lost by the hot silver is gained by the cold far whose specific heat is 3.94 J/g°c, so we can write:
[tex]-Q_{Ag}=Q_{fat}[/tex]
Which can be written in terms of mass, specific heat and temperature as shown below:
[tex]-m_{Ag}C_{Ag}(T_F-T_{Ag})=m_{fat}C_{fat}(T_F-T_{fat})[/tex]
In such a way, solving for the final temperature, we obtain:
[tex]T_F=\frac{m_{Ag}C_{Ag}T_{Ag}+m_{fat}C_{fat}T_{fat}}{m_{Ag}C_{Ag}+m_{fat}C_{fat}}}[/tex]
Then, we plug in the given data to obtain:
[tex]T_F=\frac{14.16g*0.2165J/g\°C*133.5\°C+250g*3.94J/g\°C*17.20\°C}{14.16g*0.2165J/g\°C+250g*3.94J/g\°C} \\\\T_F=17.56\°C[/tex]
Best regards!
How many molecules of N204 are in 85.0 g of N2O4?
Answer:
5.56 x 10^23
Explanation:
Just convert and cancel out.
85 g N2O4 x 1 mol/92.01 g x 6.02 x 10^23 molecules /1 mol
A moving 10 kilogram object has 20 Joules of kinetic energy. What is its velocity?
4 m/s
2 m/s
10 m/s
200 m/s
A 10.0 mL sample of HNO3 was diluted to a
volume of 100.00 mL. Then 25 mL of that
diluted solution was needed to neutralize 50.0
mL of 0.60 M KOH. What was the
concentration of the original nitric acid?
1.2 M
12 M
none of these
O 0.12 M
0.0012M
Answer:
12 M
Explanation:
The reaction between HNO₃ and KOH is:
HNO₃ + KOH → KNO₃ + H₂OFirst we calculate how many KOH moles reacted with the diluted HNO₃ sample, using the given volume and concentration:
50.0 mL * 0.60 M = 30 mmol KOHAs 1 KOH mol reacts with 1 HNO₃ mol, in 25 mL of the diluted HNO₃ solution there are 30 HNO₃ mmoles.
With that information in mind we can calculate the HNO₃ concentration in the diluted solution:
30 mmol HNO₃ / 25 mL = 1.2 MFinally we can use the C₁V₁=C₂V₂ formula to calculate the concentration of the original solution:
C₁ * 10.0 mL = 1.2 M * 100.00 mLC₁ = 12 MThe precision of a method is being established, and the
following
data are obtained 22.23, 22.18 22.25, 22.09
and 22.15%, is 22.09% a valid measurement?
Answer:
No, 22.09% is not a valid measurement
Explanation:
Precision has to do with how close a given set of measured values are to each other. It is quite different from accuracy. Accuracy refers to how close a given set of values is to the true value. A given set of values may be precise but not accurate and vice versa.
If we look at the values obtained; 22.09%, 22.15%, 22.18%, 22.23%, 22.25%, the value 22.09% is too far off the other values. This implies that it does not represent a valid measurement since it is not close to all the other values obtained.
For each of these pairs of half-reactions, write the balanced equation for the overall cell reaction and calculate the standard cell potential. Express the reaction using cell notation. You may wish to refer to Chapter 20 to review writing and balancing redox equations.
1.
Pt2+(aq)+2e-Pt(s)
Sn2+(aq)+2e-Sn(s)
2.
Co2+(aq)+2e-Co(s)
Cr3+(aq)+3e-Cr (s)
3.
Hg2+(aq)+2e-Hg (I)
Cr2+(aq)+2e-Cr (s)
1. The standard cell potential for this reaction is 0.14 V.
2. The standard cell potential for this reaction is 0.46 V.
3. The reduction potential for Hg2+(aq) + 2e^- → Hg(l) is 0.79 V.
1. The half-reactions are:
Oxidation: Sn2+(aq) → Sn(s) + 2e^-
Reduction: Pt2+(aq) + 2e^- → Pt(s)
To balance the charges, we multiply the oxidation half-reaction by 2:
2Sn2+(aq) → 2Sn(s) + 4e^-
Now, we can combine the half-reactions to form the overall cell reaction:
2Sn2+(aq) + Pt2+(aq) → 2Sn(s) + Pt(s)
The cell notation for this reaction is:
Sn(s) | Sn2+(aq) || Pt2+(aq) | Pt(s)
To calculate the standard cell potential (E°), we can look up the reduction potentials for each half-reaction. The reduction potential for Pt2+(aq) + 2e^- → Pt(s) is typically listed as 0.00 V. The reduction potential for Sn2+(aq) + 2e^- → Sn(s) is -0.14 V. The standard cell potential is the sum of the reduction potentials:
E° = E°(reduction) - E°(oxidation)
E° = 0.00 V - (-0.14 V) = 0.14 V
2. The half-reactions are:
Oxidation: Co2+(aq) → Co(s) + 2e^-
Reduction: Cr3+(aq) + 3e^- → Cr(s)
To balance the charges, we multiply the reduction half-reaction by 2:
2Cr3+(aq) + 6e^- → 2Cr(s)
Now, we can combine the half-reactions to form the overall cell reaction:
Co2+(aq) + 2Cr3+(aq) + 6e^- → Co(s) + 2Cr(s)
The cell notation for this reaction is:
Co(s) | Co2+(aq) || Cr3+(aq) | Cr(s)
To calculate the standard cell potential (E°), we look up the reduction potentials for each half-reaction. The reduction potential for Co2+(aq) + 2e^- → Co(s) is typically listed as -0.28 V. The reduction potential for Cr3+(aq) + 3e^- → Cr(s) is -0.74 V. The standard cell potential is the sum of the reduction potentials:
E° = E°(reduction) - E°(oxidation)
E° = -0.28 V - (-0.74 V) = 0.46 V
3. The half-reactions are:
Oxidation: Cr2+(aq) → Cr(s) + 2e^-
Reduction: Hg2+(aq) + 2e^- → Hg(l)
The balanced overall cell reaction is:
Cr2+(aq) + 2Hg2+(aq) + 4e^- → Cr(s) + 2Hg(l)
The cell notation for this reaction is:
Hg(l) | Hg2+(aq) || Cr2+(aq) | Cr(s)
To calculate the standard cell potential (E°), we look up the reduction potentials for each half-reaction. The reduction potential for Cr2+(aq) + 2e^- → Cr(s) is typically listed as -0.91 V.
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a Technical
How to write
report
what type of reaction occur in cyclohexene
Cyclohexene is a cyclic, six-membered hydrocarbon that contains one double bond. The types of reactions that can occur in cyclohexene would be those that are typical with alkenes generally.
The pi-bonded electrons in the double bond are nucleophilic. So, electrophilic addition reactions could occur with cyclohexene. For example,
cyclohexene + HBr → bromocyclohexane
cyclohexene + H2O/H+ → cyclohexanol
cyclohexene + Br2 → trans-1,2-dibromocyclohexane (racemic)
The latter is a common test for alkenes where one adds bromine to a sample to see if there is decolorization, which would indicate the presence of nucleophilic pi bonds. Bromine, which is dark reddish-brown, will become clear as it reacts with an alkene to form a colorless haloalkane.
Cyclohexene can also be converted to the fully saturated cyclohexane by hydrogenation: cyclohexene + H2/Pd → cyclohexane.
How many moles of NaOH are contained in 56.0 mL of a 2.40 M solution of 1 point
NaOH in water? (**Use only numerical answers with 3 significant figures.
The units are given in the question.)
Your answer
Answer:
1.34 mol
Explanation:
Molarity, which is the molar concentration of a solution, can be calculated by dividing the number of moles (n) by the volume (V).
That is;
Molarity (M) = n/V
According to the information provided in this question;
M = 2.40M
V = 56.0 mL = 56/1000 = 0.056 L
Since molarity = n/V
number of moles = M × V
n = 0.056 × 24
n = 1.34 mol
Plants are divided into three groups based on the
Answer:Scientists have identified more than 260,000 kinds of plants. They classify plants according to whether they have body parts such as seeds, tubes, roots, stems, and leaves. The three main groups of plants are seed plants, ferns, and mosses.
Explanation:
hope i help
In an experiment, 135.2 grams of a mystery metal is heated to 100.0*C. The metal is then dumped into a calorimeter with 59.0 grams of water at 26.0*C. The temperature of the water in the calorimeter increases to 35.0*C after the metal is dumped into it. What is the specific heat capacity of the metal? Express your answer to 2 past the decimal. Do NOT include units
How many moles of Cs are contained in 3 moles of Cs3PO4?
Answer:
[tex]\boxed {\boxed {\sf 9 \ moles \ Cs}}[/tex]
Explanation:
We are given the compound: Cs₃PO₄
According to the formula, 1 mole of cesium phosphate contains 3 moles of cesium, 1 mole of phosphate, and 4 moles of oxygen.
Therefore, there are 3 moles of cesium for 1 mole of cesium phosphate.
[tex]\frac {3 \ mol \ Cs}{1 \ mol \ Cs_3PO_4}[/tex]
We want to calculate the moles of cesium in 3 moles of cesium phosphate, so we multiply the ratio by 3.
[tex]3 \ mol \ Cs_3PO_4 *\frac {3 \ mol \ Cs}{1 \ mol \ Cs_3PO_4}[/tex]
[tex]3 *\frac {3 \ mol \ Cs}{1 }= 9 \ mol \ Cs[/tex]
3 moles of cesium phosphate contains 9 moles of cesium.
if 23.1g of O2 occupies a volume of 5.0L at a particular temperature and pressure, what volume will 2.50g of Oxygen gas occupy under the same conditions?
Answer:
0.54 L
Explanation:
First we convert the given masses of O₂ into moles, using its molar mass:
23.1 g ÷ 32 g/mol = 0.722 mol2.50 g ÷ 32 g/mol = 0.078 molThen we can solve the problem using Avogadro's law, which states:
V₁n₂=V₂n₁Where:
V₁ = 5.0 Ln₂ = 0.078 molV₂ = ?n₁ = 0.722 molWe input the data:
5.0 L * 0.078 mol = V₂ * 0.722 molAnd solve for V₂:
V₂ = 0.54 LWhat are the products for
Answer:
[tex]ZnCO_3(s)+2KNO_3(aq)[/tex]
Explanation:
Hello there!
In this case, for the referred reactants side, we can evidence that this is a double displacement reaction whereby Zn gets bonded with CO3 and therefore K gets bonded with NO3 to yield:
[tex]Zn(NO_3)_2+K_2CO_3\rightarrow ZnCO_3+2KNO_3[/tex]
Whereas the KNO3 product is aqueous and the ZnCO3 is solid due to the corresponding solubility rules. Thereby, the answer is the third one.
Regards!
Use the following Balanced Equation to complete the question: 2 Al + 6 HBr → 2 AlBr3 + 3 H2
If you have 10 moles of Al how many moles of H2 can be produced?
*Will give Brainly!*
What would you do during a zombie apocalypse
A. run and hide
B. fight back
C. save some people
D. raid survivor's homes
E. keep your family alive
Answer:
C
Explanation:
because then you'd be able to make a group to raid, run with you, and save more people.
Someone help thank you :)
Show the equation you will use to calculate the volume of 1 M Cu(NO3)2 (aq) needed to prepare a set of solutions that have concentrations in the range of 1 M to 1x10-4 M in a 10-mL volumetric flask. Write the reduction half-cell reaction for the copper(II) ion. What is the standard potential for an electrochemical cell that is prepared from a copper half-cell and a zinc half-cell
Answer:
Explanation:
The equation we use to calculate the volume needed to prepare other [tex](C_1,V_1)[/tex] the solution that has a concentration [tex]C_2[/tex] and volume [tex]V_2[/tex] is:
[tex]C_1V_1 =C_2V_2[/tex]
[tex]V_1=\dfrac{C_2V_2}{C_1}[/tex]
where;
[tex]C_1[/tex]= concentration of the first solution
[tex]V_1[/tex] = volume of the first solution
[tex]C_2[/tex] = concentration of the second solution
[tex]V_2[/tex] = volume of the second solution
2) Reduction half cell reaction for the copper (II) ion is:
[tex]Cu^{2+} + 2e^- \to Cu[/tex]
3) [tex]Cu^{+2} + 2e^- \to Cu \text{ \ \ \ E = 0.3370}[/tex]
[tex]Zn^{+2} + 2 e^- \to Zn \ \ \ \ \ \ E = -0.763[/tex]
[tex]Zn \to Zn^{+2} + 2 e^- \ \ \ \ \ \ E = +0.763[/tex]
Since the reduction potential of Cu is more; it means copper will go into reduction and zinc will undergo oxidation.
Standard Potential =[tex]E^0_{left} - E^0_{right}[/tex]
[tex]= -0.763 -0.337[/tex] ( since both are reduction potential)
[tex]\mathbf{E^0_{cell} = -1.100 volt}[/tex]
Tarnish on silver is the compound Ag2OAg2O. A tarnished silver spoon is placed in an aluminum pan of boiling water. When enough salt is added so that the solution conducts electricity, the tarnish disappears. Imagine that the two halves of this redox reaction were separated and connected with a wire and a salt bridge.
Answer:
The answer is "2.459".
Explanation:
The cell potential or emf norm = oxidation pot. of anode -oxidation pot. of cathode
When the pot oxidizer is the opposite of a red pot size it's been transformed to either the redox reaction so that anode was its higher oxide pot material that means (AL) so,
[tex]\to emf=1.66-(-0.799) \\\\ \to emf=1.66+0.799) \\\\ \to emf =2.459[/tex]
DUE IN 1 MINUTE HELP I'M DESPERATE
How do I solve stoichiometry problems? Please I need help with at least one of these stoichiometry problems with work and steps so I understand better. Thank you!
Answer:
2. m = 426.6 gr
4. m = 143 gr
Do 5. What is the total number of atoms contained in a 1.00-mole sample of
helium?
Answer: 1 mole = 6.02 x 10^ 23
Explanation:
The number of atoms in one mole of helium is equal to 6.022×10²³ He atoms.
What is Avogadro's number?Avogadro’s constant can be demonstrated as the proportionality factor that utilizes to count the number of particles such as ions, molecules, atoms, or ions in a sample of the substance.
Avogadro's number can be defined as the approximate count of nucleons in 1 gram of substance. The value of the Avogadro number is the mass of 1 mole of a compound, in grams, and is the number of nucleons in one particle.
The value of Avogadro’s number is equal to 6.022×10²³ per mole.
Given, the number of moles of the helium = 1 mol
Helium is a monatomic gas so it contains only one atom in its one molecule.
The number of atoms of Helium in one mole = 6.022 × 10²³
Therefore, the total number of atoms contained in a 1 mole sample of
helium is 6.022 × 10²³.
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Choose all the answers that apply.
The atmosphere:
is made mostly of nitrogen
can be used to transmit radio signals
traps heat from the sun
protects the earth from harmful radiation
is important in the water cycle
Answer:1 2 and 4
Explanation:
The trait that seems to fade into the background.
Dominant
Recessive
Why is SpaceX likely to succeed in a mission to Mars?
A. It only hires expert NASA employees as its employees.
B. It takes more risks than NASA, which cannot afford them.
C. It is run by Elon Musk, who is determined to get to Mars.
D. It has fewer restrictions than NASA does.
Answer:
C
Explanation:
It is run by Elon Musk, who is determined to get to Mars.
How many moles of sodium sulfate are produced along with 5.86 moles of iron(III)
hydroxide?
Fe2(SO4)3 + 6NaOH → 3Na SO4 + 2Fe(OH)3
Answer:
5.86mol 2Fe(OH)3 * [3 mols Na SO4 / 2 mols Fe(OH)3] = 8.79 mols sodium sulfate
Explanation:
To solve this question you need to look at the ratio of sodium sulfate to iron(III) hydroxide. The ratio is 3 to 2, meaning that for every 3 moles of Na SO4 there are also 2 moles of Fe(OH)3. So, you can multiply 5.86 by (3/2) to find the mols of sodium sulfate.
What kind of waste is shown in the photograph?
A. Composted waste
B. E-waste
C. Hazardous waste
O D. Septic system waste
Pls someone help me with this question pls
Answer:
So confusing but I'll try
which part of the human nervous system links the brain to the rest of the body
Digaments
spindi cord
heart
Answer: The Spinal Cord
Explanation: The spinal cord runs along the dorsal side of the body and links the brain to the rest of the body.
Suppose that you have 70 grams of calcium chloride (CaCl2) contaminated with 5% sodium chloride (NaCl) and 2% sand. You dissolve the impure solid in 54 ml of hot water, and allow it to cool and crystallize. What would be the maximum mass (grams) of purified CaCl2 that would you would be able to recover
Answer:
65.1 grams
Explanation:
The percentage of impurities in the contaminated CaCl₂ sample is:
% of NaCl + % of sand = 7%Meaning that the purity of the sample is (100 - 7) 93%.
Now we calculate the mass of pure CaCl₂ in the contaminated sample:
70 g * 93/100 = 65.1 gThus the answer is 65.1 grams, as that amount is how much CaCl₂ is in the sample and it would be impossible to obtain more than that.