Answer:
[tex]T_2=637\°C[/tex]
Explanation:
Hello there!
In this case, considering this problem as pressure constant, since the change is exhibited in temperature and volume only, it is possible for us to use the Charles' law as shown below:
[tex]\frac{V_2}{T_2}=\frac{V_1}{T_1}[/tex]
Thus, by solving for the final temperature, we obtain:
[tex]T_2=\frac{T_1V_2}{V_1}\\\\T_2=\frac{273K*1L}{0.30L}\\\\T_2=910K-273\\\\T_2=637\°C[/tex]
Best regards!
How much of the excess reagant in Problem 1 is left over?
Answer:
4 Al + 3 O2 → 2 Al2O3
(10.0 g Al) / (26.98154 g Al/mol) = 0.37062 mol Al
(19.0 g O2) / (31.99886 g O2/mol) = 0.59377 mol O2
0.37062 mole of Al would react completely with 0.37062 x (3/4) = 0.277965 mole of O2, but there is more O2 present than that, so O2 is in excess.
((0.59377 mol O2 initially) - (0.277965 mol O2 reacted)) x (31.99886 g O2/mol) =
10.1 g O2 left over
Explanation:
i really hope this is what you were looking for <3 :)
What happens if more solute is added to a saturated solution?
Answer:
A saturated solution is a mixture in which the maximum amount of a given solute has been dissolved into the solvent. ... At this point adding more solute will not change the concentration of the solution; adding more solute will simply result in more solid at the bottom of the solution.
Answer:
will not change
Explanation:
A saturated solution is a mixture in which the maximum amount of a given solute has been dissolved into the solvent. ... At this point adding more solute will not change the concentration of the solution; adding more solute will simply result in more solid at the bottom of the solution.
Many bones get their name from the bone they are
Which energy source produces less negative
environmental impacts
A)Renewable Energy Sources
B)Fossil Fuels
Help
Answer:
A)Renewable Energy Sources
Explanation:
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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The elemental composition of propane gas (C3H8) is 81.68% C and 18.32% H by mass. What is the maximum amount of C3H8 in grams that can be formed from 160.0 g C and 160.0 g H
Molar mass of C3H8 = C 3 (12.01 g/mol) = 36.03 (g/mol)
H 8 (1.008 g/mol) = 8.064 (g/mol)
44.09 (g/mol)
74.6 g propane x 1 mole propane x 6.022 x 10
23
molecules
44.09 g propane 1 mole propane
= 1.02 x 10
24 molecules propane
How many moles are contained in 2.3 liters of a 1.2M solution?
Answer:
[tex]\boxed {\boxed {\sf 2.76 \ mol}}[/tex]
Explanation:
Molarity is found by dividing the moles of solute by liters of solution.
[tex]molarity = \frac {moles}{liters}[/tex]
We know the molarity is 1.2 M (mol\liter) and there are 2.3 liters of solution. Substitute the known values into the formula.
[tex]1.2 \ mol/liter= \frac {x}{2.3 \ liters}[/tex]
Since we are solving for x, we must isolate the variable. It is being divided by 2.3 and the inverse of division is multiplication. Multiply both sides by 2.3 liters.
[tex]2.3 \ liters *1.2 \ mol/liter= \frac {x}{2.3 \ liters}* 2.3 \ liters\\2.3 *1.2 \ mol= x\\2.76 \ mol =x[/tex]
In a solution with a molarity of 1.2 and 2.3 liters of solution, there are 2.76 moles.
What kind of waste is shown in the photograph?
A. Composted waste
B. E-waste
C. Hazardous waste
O D. Septic system waste
Two elevators carry five passengers to the fifth floor. However, the elevators do not do the same work. Choose the best factor for
determining the amount of work the elevators did.
A.the speed of the elevator
B.the weight of the passengers
C.the number of buttons pressed
Will mark brainlist pls help!
Answer:
B the weight of the passengers
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
How many grams of Co, are produced when 88 g of o, is reacted with an excess of
butane?
Answer:
[tex]74.5gCO_2[/tex]
Explanation:
Hello there!
In this case, for the described reaction, it is possible to realize there is a 13:8 mole ratio between oxygen (O2) and carbon dioxide (CO2); moreover, since the molar mass of the former is 32.00 g/mol and that of latter is 44.01 g/mol, the produced mass of the required product turns out to be:
[tex]88gO_2*\frac{1molO_2}{32.00gO_2}*\frac{8molCO_2}{13molO_2}*\frac{44.01gCO_2}{1molCO_2} \\\\=74.5gCO_2[/tex]
Best regards!
What is orbit? A. An increase in centripetal motion and mass friction. B. Resistance of an object to avoid friction. C. Gravity causing a curved path as an object tries to go straight. D. How well an object floats.
Answer:
C. gravity causing a curved path as an object tries to go straight.
Atoms in the same GROUP have the same...
A . Size
B. Temperature
C. Number of energy levels
D. Reactivity
Answer:
reactivity
Explanation:
for example atoms in group 7 react by gaining 1 electron to become stable but they do not have the same number of energy levels
What is the molarity of the resulting solution when 300. mL of a 0.400 M solution is diluted to 800.
mL?
Answer:
0.150M
Explanation:
hope it helps!!!!!!!!!!!!
The molarity of the resulting solution when 300. mL of a 0.400 M solution is diluted to 800 mL is 0.15M.
Molarity is defined as the ratio of the number of moles of solute present in a solution, to the volume of that particular solution in litres.
It is calculated by using the formula
M₁V₁ = M₂V₂
M₁= initial molarity of the solution
V₁ = initial volume of the solution
M₂ = final molarity of the solution
V₂ = final volume of the solution
Given data,
M₁ = 0.400 M (initial molarity)
V₁ = 300 mL (initial volume)
V₂ = 800 mL (final volume)
Substituting the values we get
M₁V₁ = M₂V₂
.4× 300/800 = M₂
Therefore, molarity of the resulting solution = .15
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What is the percent yield of water from the combustion of propane (C3H8), if my theoretical yield is 8.17g of water and my actual yield was 6.13 g water?
A)Answer not shown
B)0.75%
C)75%
D)1.33%
Answer:
im not sure
Explanation:
i went on here looking nobody answered yet
T/F___ At the eutectic composition, an alloy can solidify at a constant temperature.___ For effective dispersion strengthening, the dispersed phase should be needle-like, as opposed to round___ Intermetallic compounds are usually hard and brittle.___ For effective dispersion strengthening, the dispersed phase should be continuous.___ Stoichiometric intermetallic compounds exist overa range of compositions.___ Faster solidification results in smaller interlamellar spacing
Answer:
TRUEFALSETRUEFALSEFALSETRUEExplanation:
At the eutectic composition, an alloy can solidify at a constant temperature : TRUE . this is because at eutectic composition the type of reaction that takes place there is invariant reaction in its thermal equilibrium For effective dispersion strengthening, the dispersed phase should be needle-like, as opposed to round : FALSE. because the rounded shape will not cause a crack. Intermetallic compounds are usually hard and brittle : TRUE. because Intermetallic compounds prevents dislocation movements and this makes them brittle and hardFor the effective dispersion and strengthening, the dispersed phase should be continuous : FALSE. this is because the dispersed precipitate must be small and not continuous Stoichiometric intermetallic compounds exist over a range of compositions : FALSEFaster solidification results in smaller interlamellar spacing : TRUEWhich of these benefits the grassland by allowing light, warmth, moisture, and nutrients back into the soil?
fires
ponds
trees
rainfall
Answer:
Trees
Explanation:
I think sorry if I get it wrong
14. If one half-life is the amount of time it takes for a radioactive isotope to decay by a factor of 2,
approximately which percent describes how much of the initial quantity of a substance would
remain after 2.3 half-lives?
Answer:
Since the equation for amount remaining is (1/2)^n, where n stand for the number of half lives, after 2.3 half lives the equation is (1/2)^2.3. This makes the answer about 1/4.92, which converts to about 20% of the original amount.
Explanation:
Since the equation for amount remaining is (1/2)^n, where n stand for the number of half lives, after 2.3 half lives the equation is (1/2)^2.3. This makes the answer about 1/4.92, which converts to about 20% of the original amount.
. Which law of motion relates the action of a stretched rubber band
Answer:
newtons 3rd law of motion
Explanation:
What is the entropy of this collection of training examples with respect to the positive class B. What are the information gains of A1 and A2 relative to the training dataset For A3, which is a continuous attribute, compute the information gain for every possible split. C. What is the best split (among A1,A2, and A3) according to the information gain
The data set is missing in the question. The data set is given in the attachment.
Solution :
a). In the table, there are four positive examples and give number of negative examples.
Therefore,
[tex]$P(+) = \frac{4}{9}$[/tex] and
[tex]$P(-) = \frac{5}{9}$[/tex]
The entropy of the training examples is given by :
[tex]$ -\frac{4}{9}\log_2\left(\frac{4}{9}\right)-\frac{5}{9}\log_2\left(\frac{5}{9}\right)$[/tex]
= 0.9911
b). For the attribute all the associating increments and the probability are :
[tex]$a_1$[/tex] + -
T 3 1
F 1 4
Th entropy for [tex]$a_1$[/tex] is given by :
[tex]$\frac{4}{9}[ -\frac{3}{4}\log\left(\frac{3}{4}\right)-\frac{1}{4}\log\left(\frac{1}{4}\right)]+\frac{5}{9}[ -\frac{1}{5}\log\left(\frac{1}{5}\right)-\frac{4}{5}\log\left(\frac{4}{5}\right)]$[/tex]
= 0.7616
Therefore, the information gain for [tex]$a_1$[/tex] is
0.9911 - 0.7616 = 0.2294
Similarly for the attribute [tex]$a_2$[/tex] the associating counts and the probabilities are :
[tex]$a_2$[/tex] + -
T 2 3
F 2 2
Th entropy for [tex]$a_2$[/tex] is given by :
[tex]$\frac{5}{9}[ -\frac{2}{5}\log\left(\frac{2}{5}\right)-\frac{3}{5}\log\left(\frac{3}{5}\right)]+\frac{4}{9}[ -\frac{2}{4}\log\left(\frac{2}{4}\right)-\frac{2}{4}\log\left(\frac{2}{4}\right)]$[/tex]
= 0.9839
Therefore, the information gain for [tex]$a_2$[/tex] is
0.9911 - 0.9839 = 0.0072
[tex]$a_3$[/tex] Class label split point entropy Info gain
1.0 + 2.0 0.8484 0.1427
3.0 - 3.5 0.9885 0.0026
4.0 + 4.5 0.9183 0.0728
5.0 -
5.0 - 5.5 0.9839 0.0072
6.0 + 6.5 0.9728 0.0183
7.0 +
7.0 - 7.5 0.8889 0.1022
The best split for [tex]$a_3$[/tex] observed at split point which is equal to 2.
c). From the table mention in part (b) of the information gain, we can say that [tex]$a_1$[/tex] produces the best split.
Which best describes the law of conservation of mass? 0 The coefficients in front of the chemicals in the reactants should be based on the physical state of the products. O Products in the form of gases are not considered a part of the total mass change from reactants to products. O When reactants contain both a solid and a liquid, the solid counts toward the overall mass and the liquid does not. O The mass of the reactants and products is equal and is not dependent on the physical state of the substances.
Explanation:
pdrias darme la traduccion no te entiendo
One liter of gas A at 1.0 atm pressure has a temperature of 25°C while the same volume and pressure of gas B has a temperature of 50°C. What else do you know about these two gases?
A) There are twice as many moles of gas B than gas A.
B) Gas A has twice the molar mass of gas B.
C) Gas A has more kinetic energy than gas B.
D) Gas B has more kinetic energy than gas A.
Answer:
b
Explanation:
What is the speed of sound in dry air at 20°C?
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.
Question 10 of 23
Which characteristic makes a digital signal more useful than an analog signal
for storing information?
A. A digital signal includes a spectrum of colors.
B. A digital signal is more difficult to copy.
C. A digital signal cannot be heard by humans.
O D. A digital signal moves between a discrete number of values.
Answer: O D. A digital signal moves between a discrete number of values.
Explanation:
The digital signals are considered more reliable over the analog signals as they encode the information in a coded form. The bits or samples of the data are transmitted and converted into digital and numerical value. The stream of encoded data is in the form of continuous data at regular time intervals. It provides information in waveform and the data is in compact form. The data is in the form of binary bits 0 and 1 so greater the number of bits greater will be the greater will be the resolution of the information.
Using the rules that we developed in this chapter (ARIO), we might have expected these two compounds to have the same pKa. Nevertheless, they are different. Salicylic acid is apparently more acidic than its constitutional isomer. Can you offer an explanation for this observation
Answer:
The correct answer is
- after deprotonation carboxylic acid stabilizes by binding -OH group with hydrogen bonding in salicylic acid and while it's not possible with its constitutional isomer (para-hydroxy benzoic acid).
The more electronegative atoms are replaced by the H atom of the carboxylic group and form more stabilize carboxylic acid which not takes place in para-hydroxy benzoic acid.
For the reaction 2 Cr(s) + 3 Pb²⁺(aq) ⟶ 3 Pb(s) + 2 Cr³⁺(aq), what is the value of n in the Nernst equation?
Answer:
The value of n is 6
Explanation:
The half-reactions of the problem are:
Cr(s) → Cr³⁺ + 3e⁻
Pb²⁺ + 2e⁻→ Pb(s)
To balance the electrons we must multiply the half-reactions as follows:
2 * (Cr(s) → Cr³⁺ + 3e⁻)
3 * (Pb²⁺ + 2e⁻→ Pb(s))
2Cr(s) → 2Cr³⁺ + 6e⁻
3Pb²⁺ + 6e⁻→ 3Pb(s)
In Nernst equation, the value of n are the electrons used to balance the reaction, as in this problem, the electrons are 6:
The value of n is 6what is bond stability??
Answer:
Bond order is a counting method that gives an idea about numbers of electrons shared between atoms. A species with a higher bond order is more stable. A bond order equal to 2 is a double bond, and a bond order of 3 is a triple bond.
Explanation:
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.
A mole equals 6.02 x 10^23 . Answer these questions below.
1. How many ants are in 1.25 moles of ants?
2. How many moles of pencils are in 4.92 * 10^26 pencils?
3. How many molecules are in 0.26 moles of molecules?
4. How many moles of molecules are in 3.46 * 10^19 molecules?
5. 5.3 * 10^20 atoms are equal to how many moles of atoms?
6. 0.11 moles of atoms are equal to how many atoms?
Answer:
1. 1.25 mol ants x 6.02*10^23 ants/1 mol ants = 7.53*10^23 ants
2. 4.92*10^26 pencils x 1 mol pencils/6.02*10^23 pencils = 817 mol pencils
3. 0.26 mol molecules x 6.02*10^23 molecules/1 mol molecules = 1.6*10^23 molecules
4. 3.46*10^19 molecules x 1 mol molecules/6.02*10^23 molecules = 5.75*10^-5 mol molecules
5. 5.3*10^20 atoms x 1 mol atoms/6.02*10^23 atoms = 8.8 mol atoms
6. 0.11 mol atoms x 6.02*10^23 atoms/1 mol atoms = 6.6*10^22 atoms
I would suggest looking into "dimensional analysis" for help with this type of material. Dimensional analysis will stick with you all throughout chemistry, so picking it up will be extremely beneficial.