Answer:
1.59%
Explanation:
First, we convert 0.1606 g of AgCl into moles, using its molar mass:
0.1606 g ÷ 143.32 g/mol = 1.12x10⁻³ mol AgClThen we convert the calculated moles of Cl (equal to AgCl moles) into moles of DDT:
1.12x10⁻³ mol Cl * [tex]\frac{1molDDT}{5molCl}[/tex] = 2.24x10⁻⁴ mol DDTNow we convert 2.24x10⁻⁴ moles of DDT into grams, using its molar mass:
2.24x10⁻⁴ mol DDT * 354.48 g/mol = 0.0794 gFinally we calculate the percentage of DDT in the sample:
0.0794 g / 5.0 * 100% = 1.59%what is an example of an electrolyte solution?
Answer:
nacl with water
they are capable of conducting electricity
A 0.10M NH3 solution is 1.3% ionized, calculate the hydrogen ion concentration
Answer:NH3 + H2O <==> NH4+ + OH-
0.10 x 0.013 = 0.0013 M (this is the hydroxide concentration)
Kw = [H+] [H-]
1.00 x 10^-14 = (x) (0.0013)
x = 7.7 x 10^-12 M
Explanation:
A truck accelerates at a rate of 14 m/s2 . The truck weighs 17000 kg. What is the amount of force
necessary for the truck to accelerate at this rate?
Answer:
238000 N
Explanation:
Use F=ma
Mass= 17000 kg
Accel.= 14 m/s²
Pleaseee helpppp!!!!!!!!
Answer:
a covalent would be the two that are nonmetals
Wind energy is a renewable natural resource
True
False
this is true, wind energy is a renewable natural resource c:
I am doing a exam in science need help.
What type of energy comes from the motion of tiny particles of matter?
Answer:
here you are
Explanation:
atomic energy
What process is represented by this redox equation?
C6H12O6 + 602 -> 6H20 + 6CO2
A. Rusting
B. Photosynthesis
C. Cellular respiration
D. Combustion
The redox equation C6H12O6 + 6O2 -> 6H2O + 6CO2 represents the process of cellular respiration. Option C
Cellular respiration is a biochemical process that occurs in living organisms, including plants and animals, to convert organic compounds, such as glucose (C6H12O6), into usable energy in the form of adenosine triphosphate (ATP). It is the primary process by which cells derive energy to carry out their functions.
In the given equation, glucose (C6H12O6) is being oxidized, losing electrons, and releasing carbon dioxide (CO2) as a byproduct. This oxidation process results in the production of energy-rich molecules, such as ATP.
Additionally, oxygen (O2) is being reduced, accepting the electrons from glucose and combining with hydrogen (H) to form water (H2O). This reduction process allows for the transfer of electrons and the generation of energy.
The process of cellular respiration is essential for the survival and functioning of organisms, as it provides the necessary energy for various metabolic activities, growth, and maintenance of cellular processes.
It is a fundamental metabolic pathway found in both plants and animals, enabling them to extract energy from organic molecules through the oxidation of glucose or other fuel sources.
Therefore, option C, cellular respiration, is the correct answer that represents the process described by the given redox equation.
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Step 2: Now that you have counted the atoms on each side, add coefficients to balance the equation. Check your work by updating the number of atoms on each side in the table below (Remember to multiply coefficients by subscripts): NaF + NaBr + CaF2 CaBr2 Reactants Products Na- Na- Br- Br- Ca- Ca- IF-
Answer:
The answers to your questions are given below.
Explanation:
__ NaBr + CaF₂ —> __ NaF + CaBr₂
The above equation can be balance as follow:
NaBr + CaF₂ —> NaF + CaBr₂
There are 2 atoms of F on the left side and 1 atom on the right side. It can be balance by writing 2 before NaF as shown below:
NaBr + CaF₂ —> 2NaF + CaBr₂
There are 2 atoms of Na on the right side and 1 atom on the left side. It can be balance by writing 2 before NaBr as shown below:
2NaBr + CaF₂ —> 2NaF + CaBr₂
Now, the equation is balanced.
Atom >>>> Reactant >>>> Product
Na >>>>>> 2 >>>>>>>>>>> 2
Br >>>>>>> 2 >>>>>>>>>>> 2
Ca >>>>>> 1 >>>>>>>>>>>> 1
F >>>>>>> 2 >>>>>>>>>>>> 2
Bacteria can be heterotrophs or autotrophs
true or false?
Answer:
True
Explanation:
I learned about it a few years ago
arrange the following group of atoms in order of increasing atomic size:B,Al,Ga
Answer:
Al,Ga,B
Explanation:
Now since i helped you can you help me with this plz
Matteo took 5 math quizzes. The mean of the 5 quizzes was 8.2. Here are four of his quiz scores 7, 7, 8, 10. What is the 5th quiz score? Show work.
Boron ( B )
Aluminium ( Al )
Gallium ( Ga )
Aqueous solutions containing approximately 3 percent(w/w) H2O2 are sold in drug stores as a disinfectant. Determination of the actual concentration of H2O2 in a bottle of peroxide solution was done by diluting 1.50 mL to 100 mL with water, acidifying with dilute H2SO4 and titrating with a 0.01411 M KMnO4 solution. 32.17 mL of the permangate solution was needed to reach the end point of the titration. What is the actual molar concentration of the H2O2 solution]
Answer:
0.01804 M KMnO4 Solution.
Explanation:
3 Percent(w/w) H2O2 Are Sold In Drug Stores As A Disinfectant. ... H2O2 In A Bottle Of Peroxide Solution Was Done By Diluting 1.50 ML To 100 ML With Water, Acidifying With Dilute H2SO4 And Titrating With A 0.01804 M KMnO4 Solution.
how many molecules of sodium chloride are in 2.5 moles??
Answer:
1.5x1024
Explanation:
calculate the value of (Ag+) in a saturated solution of AgCl in distilled water
Answer:
[tex][Ag^+]=1.3x10^{-5}M[/tex]
Explanation:
Hello there!
In this case, according to the dissociation of silver chloride in aqueous solution, we can write:
[tex]AgCl(s)\rightarrow Ag^+(aq)+Cl^-(aq)[/tex]
Whereas the equilibrium expression is:
[tex]Ksp=[Ag^+][Cl^-][/tex]
And the solubility product constant is 1.7 x10⁻¹⁰; thus, by inserting x as the concentration of both silver and chloride ions as they are the same at equilibrium, we obtain:
[tex]1.7x10^{-10}=x^2\\\\x=\sqrt{1.7x10^{-10}} \\x=1.3x10^{-5}M[/tex]
Which is also equal to the concentration silver ions at equilibrium in a saturated solution because we considered the Ksp.
[tex][Ag^+]=1.3x10^{-5}M[/tex]
Best regards!
8. If 134 L of Nitrogen gas at STP is combined with 99 L of Chlorine gas at STP,
Which is the limiting reactant?
N2 + Cl2 → NCI3
.
Answer:
Cl₂ is the limiting reactant.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
N₂ + 3Cl₂ —> 2NCl₃
From the balanced equation above,
1 L of N₂ reacted with 3 L of Cl₂.
Finally, we shall determine the limiting reactant. This can be obtained as follow:
From the balanced equation above,
1 L of N₂ reacted with 3 L of Cl₂.
Therefore, 134 L of N₂ will react with = 134 × 3 = 402 L of Cl₂.
From the calculation made above, we can see that a higher volume (i.e 402 L) of Cl₂ than what was given (i.e 99 L) is needed to react completely with 134 L of N₂.
Therefore, Cl₂ is the limiting reactant and N₂ is the excess reactant.
True/False –Similar to liquids, gases cannot be compressed.
Answer:
False
Explanation:
A liquid has a definite shape and takes on the volume of its container. A gas has both the shape and the volume of its container. The particles in a gas cannot be compressed into a smaller volume. Liquids tend to contract when heated.
Hope this helps
2. Calculate the concentration (in molarity) of a NaOH solution if 25.0 mL of the solution are needed to neutralize 17.4 mL of a 0.312 M HCl solution.
Answer:
0.217 M NaOH
Explanation:
M1V1 = M2V2
M1 = 0.312 M HCl
V1 = 17.4 mL HCl
M2 = ?
V2 = 25.0 mL NaOH
Solve for M2 --> M2 = M1V1/V2
M2 = (0.312 M)(17.4 mL) / (25.0 mL) = 0.217 M NaOH
what are the reasons that you cannot see many stars in the city nights?
Answer:
Because of the artificial lights influencing the dark night, turning it into a grey.
how to calculate relative abundance of copper isotopes
Answer: Atomic mass (Cu) = (x)(63.0 amu) + (1 – x)(65.0 amu) = 63.5 amu 63.0x + 65 – 65.0x = 63.5 –2x = –1.5 x = 0.75 The percent abundance of each isotope is 75.0 % (Cu-63) and 25.0 % (Cu-65).
Explanation: As you know, the average atomic mass of an element is determined by taking the weighted average of the atomic masses of its naturally occurring isotopes.
Simply put, an element's naturally occurring isotopes will contribute to the average atomic mass of the element proportionally to their abundance.
If 50.0 g of silicon dioxide is heated with an excess of carbon, 27.9 g of silicon carbide is produced. What is the percent yield of the reaction
Answer: The percent yield of the reaction is 83.5 %
Explanation:
The given balanced equation is
[tex]SiO_2+3C\rightarrow SiC+2CO[/tex]
[tex]SiO_2[/tex] is the limiting reagent as it limits the formation of product and [tex]C[/tex] is the excess reagent.
According to stoichiometry :
60.08 g of [tex]SiO_2[/tex] produce = 40.11 of [tex]SiC[/tex]
Thus 50.0 of [tex]SiO_2[/tex] will produce=[tex]\frac{40.11}{60.08}\times 50.0=33.4[/tex] of [tex]SiC[/tex]
Experimental yield of SiC = 27.9 g
Percent yield = [tex]\frac{\text {Experimental yield}}{\text {theoretical yield}}\times 100=\frac{27.9g}{33.4g}\times 100=83.5\%[/tex]
Thus percent yield of the reaction is 83.5 %
The percent yield of 83.5 % of 50.0 g of silicon dioxide is heated with an excess of carbon, and 27.9 g of silicon carbide is produced in the reaction.
What is the chemical balance of the equation?
The chemical equations are balanced when the reactants react to form products. The reactants and products react in proper ratios and if they are not in ratio then we balance them by adding the required quantity in the reactants and the products.
The given balanced equation is
[tex]\rm SiO_2+3C---- > SiC+2CO[/tex]
[tex]SiO_2[/tex] is the limiting reagent as it limits the formation of product and is the excess reagent.
According to stoichiometry
60.08 g [tex]SiO_2[/tex] of produce = 40.11 of [tex]SiC[/tex]
Thus 50.0 of [tex]SiO_2[/tex] will produce= [tex]\dfrac{40.11}{60.08} \times 50=33.4\ SiC[/tex]
The experimental yield of SiC = 27.9 g
The percentage yield will be calculated as
[tex]\rm Percentage \ Yield = \frac{Experimental\ yield}{Theoretical \ yield }\times 100[/tex]
[tex]\rm Percentage \ yield =\dfrac{27.9}{33.49} \times 100=83.5[/tex]
Thus the percent yield of 83.5 % of 50.0 g of silicon dioxide is heated with an excess of carbon, and 27.9 g of silicon carbide is produced in the reaction.
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when 70.0 grams of mno2 reacted with 128.0 grams of hcl, the reaction resulted in a 62.7% yield of chlorine gas. what is the actual yield of chlorine gas in grams? Mno2 + HCI —> MnCl2 + h2o + cl2
Answer:
35.8g of Cl₂ is the yield
Explanation:
Based on the reaction:
MnO₂ + 4HCl → MnCl₂ + 2H₂O + Cl₂
1 mole of MnO₂ and 4 moles of HCl react producing 1 mole of Cl₂
To solve this question we must find limiting reactant. with limiting reactant we can find the theoretical yield of Cl₂. As the actual yield is the 62.7% we can find actual yield of Cl₂ in grams:
Moles MnO₂ -Molar mass: 86.9368g/mol-:
70.0g * (1mol / 86.9368g) = 0.805 moles
Moles HCl -Molar mass: 36.46g/mol-:
128.0g * (1mol / 36.46g) = 3.51 moles
For a complete reaction of 3.51 moles of HCl are required:
3.51 moles HCl * (1mol MnO₂ / 4mol HCl) = 0.878 moles MnO₂.
As there are just 0.805 moles of MnO₂, MnO₂ is limiting reactant.
1 mole of MnO₂ produce 1 mole of Cl₂. The theoretical moles of Cl₂ produced are 0.805 moles.
As the yield is of 62.7%, the yield of Cl₂ is:
0.805 moles * (62.7 / 100) = 0.505 moles Cl₂. In grams:
0.505 moles Cl₂ * (70.906g / mol) =
35.8g of Cl₂ is the yield3. The name of the functional group in the following compound
CH3 - CH2 - CH2 - CH-CH2 - CH3
1
CH3 – CH2
a. Methyl
b. Ethyl
C. Propyl
d. Isopropy!
Answer: a.) Methyl would be the answer.
Explanation:
1. What organ system is responsible for controlling all of the body
functions?
Answer:
The human brain is the body's control center, receiving and sending signals to other organs through the nervous system and through secreted hormones. It is responsible for our thoughts, feelings, memory storage and general perception of the world. The human heart is a responsible for pumping blood throughout our body
Can you watch the video and help me fill in the blanks ? Thank you
1. If I have 5 moles of a gas at a pressure of 7.6 atm and a volume of 12 liters, what is the temperature?
Answer:
221.22K or -51°C
Explanation:
We will be using the Ideal Gas Law to calculate the temperature of the gas. It is a mathematical relationship that describes the behavior of ideal gas ample for any combo of varying pressure, volume, temperature, and # of moles (n). It is derived by combing Boyle's Law, Charles' Law, Gay-Lussac's & Avogadro's Law.
Note: As always, remember that temperature must be in Kelvin not Celsius when using this equation.
Ideal Gas Law: [tex]PV = nRT[/tex], where P = pressure, V = volume (in Liters), n = # of moles, R = the ideal gas constant, and T = temperature (in Kelvin).
Based on the problem, we are given the pressure, volume, and # of moles. We are asked to find the temperature. What about R you ask? Well, R is a constant that is the value of 1 mole of gas at STP. R has various values depending on the pressure units. In this case, our pressure is in atm so the R value = 0.0821.
Onto the math - all that needs to be done now is to plug and chug. Plug in the given values to find the temperature:
Set up: [tex](7.6 atm)(12L) = (5 mol)(0.0821 L*atm/(mol*K))(T)[/tex]
==> [tex]T = \frac{(7.6 atm)(12L)}{(5 mol)(0.0821 L*atm/(mol*K))}[/tex]
==> T = 221.17K
The answer is 221.17K. To convert into Celsius, subtract by 273.15 to get -50.99 or -51°C.
which is the most fluorescent molecule?
How does increasing the pressure affect the reaction rate? A. The activation energy of the reaction is changed. B. The concentration of reactants is changed. C. The temperature of the reaction is changed. D. The phase the reactants are in is changed
Answer:
Explanation:
b
The rate of the reaction determines the product formation and depends on various factors. The increase in pressure will change the concentration of reactants. Thus, option B is correct.
What is the reaction rate?The rate of the reaction is the speed at which the chemical reaction moves in a direction to yield products. The reaction rate is influenced by the surface area of the substance, temperature, and concentration.
The pressure directly affects the concentration which in turn affects the reaction rate. The increased pressure raises the concentration and more collisions will occur between the particles resulting in a fast rate of reaction.
Therefore, the increase in pressure increases the concentration and reaction rate.
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How many moles of magnesium nitrate are produced when he reacts 0.34 moles of nitric acid with excess magnesium?
What are the final units when you have a mass (in grams), and then divide it by molar
mass
Answer:
The final units are mol
Explanation:
The units of molar mass are grams per mole, g/mol.
This means that when dividing mass by molar mass, the units of grams cancel each other, leaving only units of moles:
Grams ÷ Grams/molAnother way of writing that might help is as follows:
[tex]\frac{g}{\frac{g}{mol}}[/tex]Thus, the final units are moles.
Several elements must be transmuted to sea the real implications of this event
a.surpassed
b.estimated
c.changed
d.summed
thanks guys :)
The iodide in a sample that also contained chloride was converted to iodate by treatment with an excess of bromine: The unused bromine was removed by boiling; an excess of barium ion was then added to precipitate the iodate: In the analysis of a 1.54-g sample, 0.0596 g of barium iodate was recovered. Express the results of this analysis as percent potassium iodide.
Answer: The percentage of potassium iodide in the sample is 2.63 %.
Explanation:
The chemical equation for the reaction of iodide ions with bromine gas follows:
[tex]I^-+3Br_2+3H_2O\rightarrow 6Br^-+IO_3^-+6H^+[/tex] (i)
The chemical equation for the reaction of iodate ions with barium ions follows:
[tex]Ba^{2+}+2IO_3^-\rightarrow Ba(IO_3)_2[/tex] ......(ii)
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass of barium iodate = 0.0596 g
Molar mass of barium iodate = 487.13 g/mol
Using equation 1:
[tex]\text{Moles of barium iodate}=\frac{0.0596 g}{487.13 g/mol}\\\\\text{Moles of barium iodate}=1.22\times 10^{-4} moles[/tex]
By Stoichiometry of the reaction (ii):
1 mole of barium iodate is produced by 2 moles of iodate ions
So, [tex]1.22\times 10^{-4} moles[/tex] of barium iodate will be produced by [tex]\frac{2}{1}\times 1.22\times 10^{-4} =2.44\times 10^{-4}moles[/tex] of iodate ions
By the stoichiometry of the reaction (i):
1 mole of iodate ions are produced by 1 moles of iodine ions
So, [tex]2.44\times 10^{-4} moles[/tex] of iodate ions will be produced by [tex]\frac{1}{1}\times 2.44\times 10^{-4} =2.44\times 10^{-4}moles[/tex] of iodine ions
Moles of potassium iodide = Moles of iodide ions = [tex]2.44\times 10^{-4}[/tex]
Since, the molar mass of potassium iodide = 166 g/mol
Using equation 1:
[tex]\text{Mass of potassium iodide}=2.44\times 10^{-4}mol\times 166 g/mol\\\\\text{Mass of potassium iodide}=0.0405 g[/tex]
To calculate the percentage by mass of a substance, the equation used is:
[tex]\text{Percent by mass}=\frac{\text{Mass of a substance}}{\text{Mass of solution}}\times 100[/tex]
Mass of a solution = 1.54 g
Mass of potassium iodide = 0.0405 g
Using above equation:
[tex]\text{Percent potassium iodide}=\frac{0.0405 g}{1.54g}\times 100\\\\\text{Percent potassium iodide}=2.63\%[/tex]
Hence, the percentage of potassium iodide in the sample is 2.63 %.
Expressing the results of potassium iodide in percentage = 2.63%
The chemical reaction of iodine ions with Bromine gas can be expressed as :
I⁻ + 3Br₂ + 3H₂O -- > 6Br⁻ + IO₃ + 6H⁺ ----- ( 1 )
Chemical reaction between iodate ions with barium ions can be expressed as : Ba²⁺ + 2IO⁻₃ ------> Ba ( IO₃ )₂ --------- ( 2 )
Step 1 : Calculate the number of Barium iodate moles
mass of Barium iodate = 0.0596 g
molar mass of Barium iodate = 487.13 g/mol
from equation ( 1 )
moles of Barium iodate = ( 0.0596 ) / ( 487.13 ) = 1.22 * 10⁻⁴ moles
also from equation ( 1 ) the moles of potassium iodide = moles of iodide ions
= 2.44 * 10⁻⁴
molar mass of potassium iodide = 166 g/mol
Next step : Determine the mass of potassium iodide
moles of potassium * molar mass
= 2.44 * 10⁻⁴ * 166 g/mol = 0.0405 g
Final step : Determine the percentage of potassium iodide in the solution
Percentage = ( mass of potassium iodide / mass of solution ) * 100
= ( 0.0405 / 1.54 ) * 100
= 2.63%
Hence we can conclude that potassium iodide in percentage = 2.63%
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