Answer:
O2 is the limiting reactant.
Explanation:
Hello there!
In this case, according to the given information about this reaction, one could identify the limiting reactant by performing a mole ratio of H2S to SO2 and O2 to SO2:
[tex]14.3molH_2S*\frac{2molSO_2}{2molH_2S}=14.3molSO_2 \\\\17.1molO_2*\frac{2molSO_2}{3molO_2}=11.4molSO_2[/tex]
Thus, since 17.1 moles of O2 yields fewer moles than 14.3 moles of H2S, we infer the former is the limiting reactant.
Regards!
What happens to an electroscope when a negatively charged rod is brought close to the metal sphere at the top?
Answer:
When the negatively-charged rod is brought close to the electroscope, positive charges are attracted to it and negative charges are repelled away from it. The electroscope has a net neutral charge and the rubber rod has a net negative charge. If they are brought into contact, they will both take a net negative charge.
Explanation:
I looked it up
A certain liquid has a normal freezing point of and a freezing point depression constant . A solution is prepared by dissolving some glycine () in of . This solution freezes at . Calculate the mass of that was dissolved.
The question is incomplete, the complete question is:
A certain substance X has a normal freezing point of [tex]-6.4^oC[/tex] and a molal freezing point depression constant [tex]K_f= 3.96^oC.kg/mol[/tex]. A solution is prepared by dissolving some glycine in 950. g of X. This solution freezes at [tex]-13.6^oC[/tex]. Calculate the mass of urea that was dissolved. Round your answer to 2 significant digits.
Answer: 129.66 g of glycine will be dissolved.
Explanation:
Depression in the freezing point is the difference between the freezing point of the pure solvent and the freezing point of the solution.
The expression for the calculation of depression in freezing point is:
[tex]\text{Freezing point of pure solvent}-\text{freezing point of solution}=i\times K_f\times m[/tex]
OR
[tex]\text{Freezing point of pure solvent}-\text{freezing point of solution}=i\times K_f\times \frac{\text{Given mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent (g)}}[/tex] ....(1)
where,
i = Van't Hoff factor = 1 (for non-electrolytes)
Freezing point of pure solvent = [tex]-6.4^oC[/tex]
Freezing point of solution = [tex]-13.6^oC[/tex]
[tex]K_f[/tex] = freezing point depression constant = [tex]3.96^oC/m[/tex]
[tex]M_{solute}[/tex] = Molar mass of solute (glycine) = 75.07 g/mol
[tex]w_{solvent}[/tex] = Mass of solvent = 950 g
Plugging values in equation 1:
[tex]-6.4-(-13.6)=1\times 3.96\times \frac{\text{Given mass of glycine}\times 1000}{75.07\times 950}\\\\\text{Given mass of glycine}=\frac{7.2\times 75.07\times 950}{1\times 3.96\times 1000}\\\\\text{Given mass of glycine}=129.66g[/tex]
Hence, 129.66 g of glycine will be dissolved.
What the correct answer
Answer:
[Ar] 4s²3d³
Explanation:
Vanadium has atomic number of 23. The electronic configuration of vanadium can be written as:
V (23) => 1s² 2s²2p⁶ 3s²3p⁶ 4s²3d³
NOTE: After the 18th electron, 4s will be filled before 3d.
We can also write the electronic configuration of an element in its condensed form by writing the symbol of the noble before the desired element in a squared bracket followed by the remaining electrons to complete the electronic configuration of the element.
The electronic configuration of vanadium in its condensed form is given below:
The noble gas before vanadium is Argon (Ar) with atomic number of 18. Thus, the electronic configuration of vanadium becomes:
V (23) => [Ar] 4s²3d³
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Scientists use different types of microscopes to observe objects that are not visible to the naked eye. A scientist is
reviewing various samples of pond water to determine what species of microorganisms live in the pond. The scientist
wishes to make drawings of the structure of each microorganism and study each one's method of movement. Which of
the following microscopes would be best for the scientist to use?
transmission electron microscope
b. scanning electron microscope
c. compound light microscope
d. dissecting microscope
a.
Please select the best answer from the choices provided
ОА
ОВ
D
Nox
Submit
Save and Exit
Mark this and retum
Sono
Answer:
compound light microscope
A 66.4 gram sample of Ba(ClO4)2 3 H2O was heated thoroughly in a porcelain crucible, until its weight remained constant. After heating, how many grams of the anhydrous compound remained?
Answer:
57.2 g
Explanation:
First we convert 66.4 grams of Ba(ClO₄)₂·3H₂O into moles, using its molar mass:
Molar mass of Ba(ClO₄)₂·3H₂O = Molar mass of Ba(ClO₄)₂ + (Molar Mass of H₂O)*3Molar mass of Ba(ClO₄)₂·3H₂O = 390.23 g/mol66.4 g ÷ 390.23 g/mol = 0.170 mol Ba(ClO₄)₂·3H₂O0.170 moles of Ba(ClO₄)₂·3H₂O would produce 0.170 moles of 0.170 moles of Ba(ClO₄)₂. Meaning we now convert 0.170 moles of Ba(ClO₄)₂ into grams, using the molar mass of Ba(ClO₄)₂:
0.170 mol * 336.23 g/mol = 57.2 gA typical dollar bill is 15.50 cm by 6.50 cm.
Calculate the surface area in square meters, square centimeters and square nanometers
Answer:
0.010075 m²100.75 cm²1.0075x10¹⁶ nm²Explanation:
As the measurements are given to us in centimeters, let's start by calculating the surface area in square centimeters:
Area = 15.50 cm * 6.50 cm = 100.75 cm²Now we convert 100.75 cm² to m², as follows:
100.75 cm² * [tex](\frac{1m}{100cm}) ^2[/tex] = 0.010075 m²Finally we convert 0.010075 m² to nm², as follows:
0.010075 m² * [tex](\frac{1nm}{1x10^{-9}m}) ^2[/tex] = 1.0075x10¹⁶ nm²HELP PLS THIS IS SO HARD AHHHHHHH
1 pc
8. Some chemicals, most often inorganic salts, in the laboratory have the nasty tendency to absorb water from the
atmosphere. This property is called hygroscopicity. Anhydrous (water-free) Cupric Perchlorate [Cu(C104)2] weighs
262.447g/mol, but after sitting on out on the benchtop absorbs water from the air and now weighs 370.540g/mol. How
many water molecules did our Cupric Perchlorate absorb?
If you had used toluene instead of methyl benzoate in this reaction, what nitration product(s) would have formed? Write a stepwise mechanism for the nitration reaction of toluene.
Answer:
See explanation
Explanation:
The electrophilic substitution of aromatic compounds occurs faster in substituted aromatic compounds due to the fact that the ring becomes more or less susceptible to electrophilic attack depending on the nature of the substituent in the ring.
Electron pushing substituents such as alkyl groups stabilize the positive charge developed during electrophilic substitution hence they activate the ring towards electrophilic substitution.
The methyl group is an ortho - para directing substituent hence the product obtained by nitration of toluene is o-nitrotoluene and p-nitrotoluene.
The stepwise mechanism for obtaining these products is shown in the image attached to this answer.
4-A major textile dye manufacturer developed a new yellow dye. The dye has a percent composition of 75.95% X, 17.72% N, and 6.33% H by mass with a molar mass of about 240 g/mol. Determine the molecular formula of the dye.
Answer:
C₁₅N₃H₅
Explanation:
Let's assume we have 240 g of the dye (1 mol), in that case we'd have:
240 g * 75.95/100 = 182.28 g of C240 g * 17.72/100 = 42.53 g of N240 g * 6.33/100 = 15.19 g of HNow we convert the masses of each element into moles, using their respective molar masses:
182.28 g C ÷ 12 g/mol = 15.19 mol C ≅ 1542.53 g N ÷ 14 g/mol = 3.04 mol N ≅ 3 15.19 g C ÷ 1 g/mol = 15.19 mol H ≅ 15Thus the molecular formula is C₁₅N₃H₅.
1. Express in conventional notation (no exponents) in the space provided within the
parentheses, state how many significant figures are in the number (apis, cach)
a) 3.2 X 102
b) 2.366 X 104
C) 7.30 x 101
d) 5.325 x 102
Answer:
a) 320: two significant figures.
b) 2,366: four significant figures.
c) 73.0: three significant figures.
d. 532.5: four significant figures.
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to write each number by knowing we move the decimal places to the right as much as the exponent is, and also, we count every figure, even zeros, because they are to the right of the first nonzero digit:
a) 320: two significant figures because the rightmost zero is not preceded o followed by a decimal place.
b) 2,366: four significant figures.
c) 73.0: three significant figures, because the zero is followed by the decimal place.
d. 532.5: four significant figures.
Regards!
You combine 0.75 moles formate and 0.85 moles formic acid to make a buffer solution. The Ka of formic acid is 1.8x10-4 what is the pH of the solution
Answer:
pH = 3.68
Explanation:
We can solve this problem by using Henderson-Hasselbach's equation:
pH = pKa + log[tex]\frac{[Formate]}{[Formic Acid]}[/tex]Where pKa = -log(Ka)pKa = -log(1.8x10⁻⁴) = 3.74Assuming we have 1 L of the buffer solution then the molar concentrations of formate and formic acid would be:
[Formate] = 0.75 mol / 1 L = 0.75 M[Formic Acid] = 0.85 mol / 1 L = 0.85 MWe now have all required data to calculate the pH:
pH = 3.74 + log[tex]\frac{0.75}{0.85}[/tex]pH = 3.68Determine the number of significant digits in each number and list the specific significant digits. 1. 306,000 2. 0.0073 3. 39.9999 4. 25.00 5. 40,000,021 6. 45,250.0 7. 0.00011 8. 420.030700
Answer:
1. 306,000: three significant figures because the last three zeros are not preceded by a decimal point.
2. 0.0073: two significant figures because the the leftmost zeros are not significant.
3. 39.9999: six significant figures because all these numbers are nonzero digits.
4. 25.00: four significant figures because right-handed zeros, after a decimal point, are significant.
5. 40,000,021: eight significant figures because intermediate zeros are significant.
6. 45,250.0: six significant figures because al the zeros are to the right of the nonzero digits.
7. 0.00011: two significant figures as well as #2.
8. 420.030700: nine significant figures because all the zeros are to the right of the first nonzero digits and after the decimal point.
Explanation:
Hello there!
In this case, by considering the given numbers, we can proceed as follows, by keeping in mind the rules for assigning significant figures:
1. 306,000: three significant figures because the last three zeros are not preceded by a decimal point.
2. 0.0073: two significant figures because the the leftmost zeros are not significant.
3. 39.9999: six significant figures because all these numbers are nonzero digits.
4. 25.00: four significant figures because right-handed zeros, after a decimal point, are significant.
5. 40,000,021: eight significant figures because intermediate zeros are significant.
6. 45,250.0: six significant figures because al the zeros are to the right of the nonzero digits.
7. 0.00011: two significant figures as well as #2.
8. 420.030700: nine significant figures because all the zeros are to the right of the first nonzero digits and after the decimal point.
Regards!
PLS HELP The average atomic mass of carbon is 12.01 amu. Based on the atomic
masses of the two isotopes of carbon, how do the relative abundances of the
isotopes compare?
Answer:
B. There is a very large percentage of C-12.
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to realize that, since the average atomic mass is 12.01 amu, then the C-12, with an atomic mass of 12.000 am prevails over C-13 with an atomic mass of 13.003 amu as long as the average is nearer to the former.
In such a way, the answer will be B. There is a very large percentage of C-12.
Regards!
Iodide ion catalyzes the decomposition of hydrogen peroxide. The reaction is first-order in H2O2. What is the value of the rate constant, k, if the initial rate is 0.00842 mol/(L·s) and the initial concentration of H2O2 is 0.500 mol/L.
Answer:
0.01684 s⁻¹
Explanation:
In a first-order reaction, the rate is proportional to the concentration of only one reactant (raised by 1). In this case, the reactant is H₂O₂. Thus, the rate law is the following:
rate = k [H₂O₂]
We have the following data for the initial rate:
rate = 0.00842 mol/(L·s)
[H₂O₂] = 0.500 mol/L
So, we introduce the data in the expression for the rate law to calculate k:
k = rate/[H₂O₂] = (0.00842 mol/L·s)/0.500 mol/L = 0.01684 s⁻¹
What is the Name of molecule and smiles strings ?
Explanation:
Aromatic nitrogen bonded to hydrogen, as found in pyrrole must be represented as [nH] ; thus imidazole is written in SMILES notation as n1c[nH]cc1 . When aromatic atoms are singly bonded to each other, such as in biphenyl, a single bond must be shown explicitly: c1ccccc1-c2ccccc2 .
A 200. gram sample of a salt solution contains 0.050 grams of NaCl. What is the concentration of the
solution in parts per million (ppm)?
Answer:
2.5 × 10² ppm
Explanation:
Step 1: Given data
Mass of NaCl: 0.050 gMass of the sample: 200. gStep 2: Convert 0.050 g to μg
We will use the conversion factor 1 g = 10⁶ μg.
0.050 g × 10⁶ μg/1 g = 5.0 × 10⁴ μg
Step 3: Calculate the concentration of NaCl in ppm
The concentration of NaCl in ppm is equal to the micrograms of NaCl per gram of the sample.
5.0 × 10⁴ μg NaCl/200. g = 2.5 × 10² ppm
Answer:250 ppm
Explanation:
A 1.0 g sample of propane, C3H8, was burned in calorimeter. The temperature rose from 28.5 0C to 32.0 0C and heat of combustion 10.5 kJ/g. Calculate the heat capacity of the calorimeter apparatus in kJ/0C
Answer:
A 1.0 g sample of propane, C3H8, was burned in the calorimeter.
The temperature rose from 28.5 0C to 32.0 0C and the heat of combustion 10.5 kJ/g.
Calculate the heat capacity of the calorimeter apparatus in kJ/0C
Explanation:
[tex]Heat of combustion = heat capacity of calorimeter * deltaT\\[/tex]
Given,
The heat of combustion = 10.5kJ/g.
[tex]deltaT = (32.0-28.5)^oC\\deltaT = 3.5^oC[/tex]
Substitute these values in the above formula to get the value of heat capacity of the calorimeter.
[tex]deltaT =heat capacity of calorimeter * (change in temperature)\\10.5kJ/g = heat capacity of calorimeter * (3.5^oC)\\\\=>heat capacity of calorimeter = \frac{10.5kJ/g}{3.5^oC} \\=>heat capacity of calorimeter = 3.0 kJ/g.^oC[/tex]
Answer:
The heat capacity of the calorimeter is [tex]3.0kJ/g.^oC.[/tex]
1 gallon =3.785 L how many liters of gasoline will fill a 10.00 tank
Answer:
37.85 L
Explanation:
3.785 x 10.00 = 37.85 L
it would take 37.85 L to fill a 10.00 tank
(sorry if im wrong pls dont report)
(hope this helps can i plz have brainlist :D hehe)
Zinc sulfate is a 2-ion electrolyte,
dissociating 40% in a
certain concentration. Calculate its
dissociation (i) factor.
On the basis of 40% dissociation, 100
particles of zinc sulfate
will yield:
40zinc ions
40 sulfate ions
60undissociated particles
Jo 11:03
Answer: The value of i is 1.4 and 40% dissociation of 100 particles of zinc sulfate will yield 60 undissociated particles.
Explanation:
The equation used to calculate the Vant' Hoff factor in dissociation follows:
[tex]\alpha =\frac{i-1}{n-1}[/tex]
where,
[tex]\alpha [/tex] = degree of dissociation = 40% = 0.40
i = Vant' Hoff factor
n = number of ions dissociated = 2
Putting values in above equation, we get:
[tex]0.40=\frac{i-1}{2-1}\\\\0.40=i-1\\\\i=1.4[/tex]
The equation used to calculate the degee of dissociation follows:
[tex]\alpha =\frac{\text{Number of particles dissociated}}{\text{Total number of particles taken}}[/tex]
Total number of particles taken = 100
Degree of dissociation = 40% = 0.40
Putting values in above equation, we get:
[tex]0.40=\frac{\text{Number of particles dissociated}}{100}\\\\\text{Number of particles dissociated}=(0.40\times 100)=40[/tex]
This means that 40 particles are dissociated and 60 particles remain undissociated in the solution.
Hence, 40% dissociation of 100 particles of zinc sulfate will yield 60 undissociated particles.
Calculate the amount of heat associated with cooling a 350.0 g aluminum bar from 70.0 oC to 25.0 oC. The specific heat of aluminum is 0.897 J/g oC. (–14,127.75 J)
Explanation:
[tex]q = 350 \times0.897 \times (70 - 25) \\ q = 14127.75[/tex]
PLEASE TELL ME THE AWNSERS ITS A DOC FILE SO OPEN IT I WILL GIVE BRAINLIEST PLS HURRY
I NEED THIS NOW NO LINKS OR ILL REPORT
What is a substance that has multiple elements in one area but are not
chemically combined; such as air? *
atom
element
compound
mixture
The correct answer is mixture
g 1. Write a mechanism for the Grignard reaction of benzophenone with phenylmagnesium bromide. Be as complete as possible and show electron flow for all steps.
Answer:
See explanation and image attached
Explanation:
The reaction between benzophenone and phenylmagnesium bromide is a Grignard reaction.
A Grignard reagent is any alkyl magnesium halide compound. In this case, the Grignard reagent is phenylmagnesium bromide.
Reaction of Grignard reagent with a ketone yields all alcohol. Thus, the reaction of benzophenone with phenylmagnesium bromide yields triphenyl methanol.
The mechanism of the reaction and all electron movements are shown in the image attached to this answer.
electron affinity of lithium is -60 whereas of cesium is -45.this trend is due to... plz give me accurate answer
what are the answers for these I did four through 10 but I think I got four through nine wrong because I mixed up the radius and the atomic radius but I’m not sure can you please tell me the answers I can send you a picture of my work as well!
Answer:
1) B
2) D
3) A
4) Ga
5) K
6)Po
7) Atomic size increases down the group
8) B<Al<Ga<In<Tl
9)Se<C<Ga
10) ionization energy is the energy required to remove electrons from the outermost shell of an atom.
Explanation:
In the periodic table, the properties of elements reoccur ''periodically'' throughout the table, hence the name 'periodic table'.
Ionization energy increases across the period hence the noble gas He has the highest ionization energy.
Since ionization energy increases across the period, group 1 elements possess the lowest ionization energy.
Since atomic size increases down the group and decreases across the period, gallium is smaller than indium, potassium is smaller than caesium, polonium is smaller than titanium and iodine is larger than bromine.
This explanation above justifies the order of increasing atomic radius of group 13 elements shown in answer number 8 above.
Since atomic size decreases across the period, the order of increasing atomic size shown in answer number 9 above is correct.
Ionization energy is the energy required to remove electrons from the outermost shell of an atom.
28) Which of the following is true about the modern model of the atom?
A) Electrons orbit the nucleus in shells at fixed distances
B) Protons, neutrons, and electrons are located in the nucleus.
C)Atoms are hard, indivisible spheres.
D)A positive nucleus is surrounded by negative electrons.
Answer:
A) Electrons orbit the nucleus in shells at fixed distances
Answer:
the answer is A
Explanation:
g There are two substances, 1 and 2, that diffuse across identical surface areas. The substances have diffusion constants D1 and D2, and D1 > D2. The substances have identical concentration gradients. Which substance will diffuse at a faster rate
Answer:
Substance 1 will diffuse at a faster rate.
Explanation:
We can solve this problem by keeping in mind Fick's law, which states:
J = -D * (dc/dx)Where:
J is the fluxD is the diffusion constant(dc/dx) is the concentration gradientsAs (dc/dx) is equal for both substances, as stated by the problem, the substance with the higher diffusion constant will diffuse at a faster rate.
Thus the answer is substance 1.
Gold is a precious metal that is chemically unreactive. It is used mainly in jewelry, dentistry, and electronic devices. A piece of gold ingot with a mass of 301 g has a volume of 15.6 cm3. Calculate the density of gold.
Answer:
Density = 19.3 g/cm³
Explanation:
In order to answer this question we need to keep in mind the following definition of density:
Density = Mass / VolumeAs both the mass and the volume are given by the problem, we can proceed to calculate the density of gold:
Density = 301 g / 15.6 cm³Density = 19.3 g/cm³Methanol and butanol are alcohols. Alcohols have the same_______________ as alkanes and the __________ identifies the compound as an alcohol.
Answer:
Methanol and butanol are alcohols. Alcohols have the same_______________ as alkanes and the __________ identifies the compound as an alcohol.
Explanation:
Alcohols belong to a group of organic compounds which contain -OH group as the functional group.
So alcohols have the same carbon -hydrogen bonds as alkanes and the -OH functional group identifies the compound as an alcohol.
