Answer:
16.5g of CO₂ could be produced
Explanation:
The combustion of hexane occurs as follows:
C₆H₁₄(l) + 19/2O₂ → 6CO₂ + 7H₂O
Where 1 mole of hexane reacts with 19/2 moles of O₂.
To solve this question we need to find the moles of each reactant in order to find limiting reactant. The moles of the limiting reactant will determine the moles of CO₂ produced:
Moles C₆H₁₄ -Molar mass: 86.18g/mol-:
17.2g hexane * (1mol / 86.18g) = 0.200 moles hexane
Moles O₂ -Molar mass: 32g/mol-:
19g O₂ * (1mol / 32g) = 0.594 moles oxygen.
For a complete reaction of 0.594 moles of oxygen are required:
0.594 moles O₂ * (1mol C₆H₁₄ / 19/2 moles O₂) = 0.0625 moles C₆H₁₄.
As there are 0.200 moles of hexane, hexane is the excess reactant and oxygen the limiting reactant.
The moles of CO₂ produced assuming a yield of 100% -All moles of oxygen react producing carbon dioxide.:
0.594 moles O₂ * (6mol CO₂ / 19/2 moles O₂) = 0.375 moles of CO₂ could be produced. The mass is:
0.375 moles of CO₂ * (44.01g / mol) =
16.5g of CO₂ could be producedPut these elements in order of decreasing electronegativity, with the highest electronegative element as number 1.
a. tin (Sn, Group 14, Period 5)
b. rubidium (Rb, Group 1, Period 5)
c. bromine (Br, Group 17, Period 4)
d. lithium (Li, Group 1, Period 2)
e. cadmium (Cd, Group 12, Period 5)
Answer:the answer is a c and e
Explanation:
2NH3+2O2- N2O+3H2O
If 80.0 grams of O2 reacted in above reaction ,how many grams of N2O will be produced?
Answer:
55.0125 grams NO2
Explanation:
So we have 80 grams of O2, first convert to moles
80 g O2 * 1 mol/32 g O2 = 2.5 mol O2
Next use the mole ratio of O2 to NO2
2.5 mol O2* 1 mol NO2/2 mol NO2=1.25 mol NO2
Since the question is asking how many grams, convert NO2 to grams
1.25 mol NO2 * 44.01 NO2/1 mol NO2= 55.0125 grams
A compound is 22.5% nitrogen and 77.5% oxygen, What is the empirical formula of
this compound?
Answer:
[tex]NO_3[/tex]
Explanation:
Hello!
In this case, since percent compositions are used to set up empirical formulas when assuming those percentages are masses, we can fist compute the moles of nitrogen and oxygen in the compound as shown below:
[tex]n_N=22.5g/14.01g/mol=1.61mol\\\\n_O=77.5g/16.00g/mol=4.84mol[/tex]
Now, we divide by the fewest moles to compute the subscripts:
[tex]N=\frac{1.61}{1.61} =1\\\\O=\frac{4.84}{1.61} =3[/tex]
Thus, the empirical formula turns out:
[tex]NO_3[/tex]
Best regards!
what are three limitations of current cloaking technology.
Get this correct I'll give you 40 points.
Explanation:
1.undetectable to electromagnetic waves
2.hiding an object from an illumination containing diffre t wave lengths become difficult as the object sizes grow.
3. reduce the scattering by two orders.
Electromagnetic waves are unable to detect it. With increasing item sizes, it becomes more challenging to conceal an object from an illumination with various wave lengths. two orders down on the scattering.
What is current cloaking technology?
Current clocking technology is defined as a hypothetical or made-up stealth technology that makes things like people or spaceships completely or partially invisible to certain electromagnetic (EM) spectrum wavelengths. In a specific frequency band, this gadget renders an object "invisible" to electromagnetic radiation. Obviously, cloaks that operate in the visible spectrum have the most fascinating potential uses.
These metamaterials, according to the researchers, can shield a medium-sized antenna from radio waves over a wide range of bandwidths, resulting in clearer communications. However, as visible light wavelengths are far shorter than radio waves, it is practically difficult to conceal huge things from them, such as the human body. The issue with present designs is that they can only handle a certain amount of bandwidth. Even this "perfect" 3D cloak that was previously exhibited could only conceal items from microwaves.
Thus, electromagnetic waves are unable to detect it. With increasing item sizes, it becomes more challenging to conceal an object from an illumination with various wave lengths. two orders down on the scattering.
To learn more about current cloaking technology, refer to the link below:
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An analytical chemist is titrating of a solution of ethylamine with a solution of . The of ethylamine is . Calculate the pH of the base solution after the chemist has added of the solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of solution added. g
Answer:
pH=11.
Explanation:
Hello!
In this case, since the data is not given, it is possible to use a similar problem like:
"An analytical chemist is titrating 185.0 mL of a 0.7500 M solution of ethylamine(C2HNH2) with a 0.4800 M solution of HNO3.ThepK,of ethylamine is 3.19. Calculate the pH of the base solution after the chemist has added 114.4 mL of the HNO3 solution to it"
Thus, for the reaction:
[tex]C_2H_5NH_2+H^+\rightleftharpoons C_2H_5NH_4^+[/tex]
Tt is possible to compute the remaining moles of ethylamine via the following subtraction:
[tex]n_{ethylamine}=0.1850L*0.7500mol/L=0.1365mol\\\\n_{acid}=0.1144L*0.4800mol/L=0.0549mol\\\\n_{ethylamine}^{remaining}=0.1365mol-0.0549mol=0.0816mol[/tex]
Thus, the concentration of ethylamine in solution is:
[tex][ethylamine]=\frac{0.0816mol}{0.1850L+0.1144L}=0.2725M[/tex]
Now, we can also infer that some salt is formed, and has the following concentration:
[tex][salt]=\frac{0.0549mol}{0.1850L+0.1144L}=0.1834M[/tex]
Therefore, we can use the Henderson-Hasselbach equation to compute the resulting pOH first:
[tex]pOH=pKb+log(\frac{[salt]}{[base]} )\\\\pOH=3.19+log(\frac{0.1834M}{0.2725M})\\\\pOH=3.0[/tex]
Finally, the pH turns out to be:
[tex]pH=14-pOH=14-3\\\\pH=11[/tex]
NOTE: keep in mind that if you have different values, you can just change them and follow the very same process here.
Best regards!
What is the percent composition of dinotrogen pentoxide? With work shown please
Answer:
%N = 25.94%
%O = 74.06%
Explanation:
Step 1: Calculate the mass of nitrogen in 1 mole of N₂O₅
We will multiply the molar mass of N by the number of N atoms in the formula of N₂O₅.
m(N): 2 × 14.01 g = 28.02 g
Step 2: Calculate the mass of oxygen in 1 mole of N₂O₅
We will multiply the molar mass of O by the number of O atoms in the formula of N₂O₅.
m(O): 5 × 16.00 g = 80.00 g
Step 3: Calculate the mass of 1 mole of N₂O₅
We will sum the masses of N and O.
m(N₂O₅) = m(N) + m(O) = 28.02 g + 80.00 g = 108.02 g
Step 4: Calculate the percent composition of N₂O₅
We will use the following expression.
%Element = m(Element)/m(Compound) × 100%
%N = m(N)/m(N₂O₅) × 100% = 28.02 g/108.02 g × 100% = 25.94%
%O = m(O)/m(N₂O₅) × 100% = 80.00 g/108.02 g × 100% = 74.06%
Decide whether a chemical reaction happens in either of the following situations. If a reaction does happen, write the chemical equation for it. Be sure your chemical equation is balanced and has physical state symbols. situation chemical reaction.
A strip of solid tin metal is put into a beaker of 0.074M MgBr2 solution .
Answer:
Sn(s) + MgBr2(aq) --> SnBr2(aq) + Mg(s)
Explanation:
Decide whether a chemical reaction happens in either of the following situations.
- A reaction happens. The type of this reaction is the single replacement reaction.
If a reaction does happen, write the chemical equation for it.
Sn(s) + MgBr2(aq) --> SnBr2(aq) + Mg(s)
Question 10 of 35
A red blood cell can be used to model a typical animal cell.
What would happen to a red blood cell placed in very salty water?
O A. It would stay the same size because salt cannot cross the cell
membrane.
B. It would swell and burst as water moves into the cell, where there
is a higher concentration of water.
C. It would swell and burst because salt would be transported across
the cell membrane.
O D. It would shrink as water moves out of the cell to where there is a
lower concentration of water.
A helium balloon has a volume of 36 L. The pressure of the helium gas in the balloon is 132 kPa and its temperature is 27 degrees Celsius. How many moles of helium are in the balloon?
Answer:
[tex]n =1.905mol[/tex]
Explanation:
Hello!
In this case, by considering helium gas as an ideal one, we can use the following equation:
[tex]PV=nRT[/tex]
Whereas P should go in atmospheres and T in Kelvins; thus we proceed as follows:
[tex]n=\frac{PV}{RT}=\frac{132kPa*\frac{1atm}{101.325kPa}*36L}{0.08206\frac{atm*L}{mol*K}(27+273)K} =1.905mol[/tex]
Best regards!
How many grams of silver chloride can be produced by reacting excess silver nitrate with 2.4 moles of zinc chloride? _____AgNO3 + ____ZnCl2 ____AgCl + _____Zn(NO3)2
690 g AgCl
General Formulas and Concepts:Math
Pre-Algebra
Order of Operations: BPEMDAS
Brackets Parenthesis Exponents Multiplication Division Addition Subtraction Left to RightChemistry
Atomic Structure
Reading a Periodic TableWriting CompoundsStoichiometry
Using Dimensional AnalysisLimiting Reactant/Excess ReactantExplanation:Step 1: Define
[RxN - Unbalanced] AgNO₃ + ZnCl₂ → AgCl + Zn(NO₃)₂
↓
[RxN - Balanced] 2AgNO₃ + ZnCl₂ → 2AgCl + Zn(NO₃)₂
[Given] 2.4 mol ZnCl₂
[Solve] x g AgCl
Step 2: Identify Conversions
[RxN] 1 mol ZnCl₂ → 2 mol AgCl
[PT] Molar Mass of Ag - 107.87 g/mol
[PT] Molar Mass of Cl - 35.45 g/mol
Molar Mass of AgCl - 107.87 + 35.45 = 143.32 g/mol
Step 3: Stoich
[DA] Set up: [tex]\displaystyle 2.4 \ mol \ ZnCl_2(\frac{2 \ mol \ AgCl}{1 \ mol \ ZnCl_2})(\frac{143.32 \ g \ AgCl}{1 \ mol \ AgCl})[/tex][DA] Multiply/Divide [Cancel out units]: [tex]\displaystyle 687.936 \ g \ AgCl[/tex]Step 4: Check
Follow sig fig rules and round. We are given 2 sig figs.
687.936 g AgCl ≈ 690 g AgCl
Which process breaks down sugars to release energy that powers bodily functions?
Answer:
Cellular respiration
Explanation:
Order the sequence of steps that occur when you add heat to a chemical reaction. Place these steps in a logical order keeping in mind how adding temperature affects the rate of reaction.
The reaction rate of the experiment is recorded and found to have occurred at a faster rate than it did without the addition of heat.
Temperature increases due to the higher ketic energy of particles.
Two chemicals are mixed together and slowly some bubbles appeanr
Heat is added to the experiment with a Bunsen burner.
More collisions of particles occurs due to the increased kinetic energy.
Answer:
C,A,B,D
Explanation:
That's the correct order, hope this helps
73. Wind blows off of the ocean cooling the land near the ocean. This is an example of what type of heat transfer?
convection
induction
conduction
radiation
Help plz
Answer:
Convection
Explanation:
Convection is the circular motion that happens when warmer air or liquid — which has faster moving molecules, making it less dense — rises, while the cooler air or liquid drops down.
Same as in the case of sea breeze where the the lighter wind in surface of land rises up and the cooler wind from the sea comes to fill up that place. This is an example of convection.
In the periodic table what is the 65th element? (pls help me)
Answer:
Terbium
Explanation:
All elements are put in squares on the periodic table. They represent the number of protons in the nucleus. They are the largest number in the square. The elements go up in order. Number 65 is in the row following Calcium. It is about in the middle of that row.Burning plants contributes to global warming because:
a) carbon dioxide is released during combustion
b) there are no trees to provide shade
c) much oxygen is used up during combustion
d) the sun beats on the bare soil.
Answer:carbon dioxide is released during combustion
write the chemical equation of the reaction with a change in colour
Answer:
(i) Change in colour (ii) Change in temperature (iii) Formation of precipitate
Explanation:
(i) Change in colour: Reaction between lead nitrate solution and potassium iodide solution. Pb(NO3)2(aq)+2KI → PbI2(s)+2KNO3(aq) In this reaction, colour changes from colourless to yellow. (ii)Change in temperature: Action of dilute sulphuric acid on zinc. Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2 In this reaction, heat is evolved (iii) Formation of precipitate: Action of barium chloride on sodium sulphate. BaCl2(aq) +Na2SO4(aq) → BaSO4(s) +2NaCl(aq) BaSO4(s)
A sample of 523.1 mg of impure KBr is treated with an excess of AgNO3 and 814.5 mg of AgBr is obtained. What is the purity of KBr?
Answer:
98.68%
Explanation:
Step 1: Write the balanced equation
KBr + AgNO₃ ⇒ AgBr + KNO₃
Step 2: Calculate the moles corresponding to 814.5 mg (0.8145 g) of AgBr
The molar mass of AgBr is 187.77 g/mol.
0.8145 g × 1 mol/187.77 g = 4.338 × 10⁻³ mol
Step 3: Calculate the moles of KBr needed to produce 4.338 × 10⁻³ moles of AgBr
The molar ratio of KBr to AgBr is 1:1. The moles of KBr needed are 1/1 × 4.338 × 10⁻³ mol = 4.338 × 10⁻³ mol.
Step 4: Calculate the pure mass corresponding to 4.338 × 10⁻³ moles of KBr
The molar mass of KBr is 119.00 g/mol.
4.338 × 10⁻³ mol × 119.00 g/mol = 0.5162 g
Step 5: Calculate the purity of KBr
0.5162 g of KBr are in a 0.5231 g-sample. The purity of KBr is:
P = 0.5162 g/0.5231 g × 100% = 98.68%
When comparing the two elements K and Ge , the more metallic element is__________ based on periodic trends alone.
When comparing the two elements Sb and Pb , the more metallic element is___________ based on periodic trends alone.
a. Ge
b. Pb
c. Sb
d. K
e. Impossible to determine
Answer:
Option D and Option B
When comparing the two elements K and Ge , the more metallic element is_____K_____ based on periodic trends alone.
When comparing the two elements Sb and Pb , the more metallic element is_____Pb______ based on periodic
Explanation:
The metallic characteristic increases when we move down a column in a periodic table or when we move left in the row.
Potassium and Germanium are located on the same row, but germanium lies on the right side of potassium. Thus, potassium (K) is more metallic than Germanium (Ge)
While Lead (Pb) lies to left of Sb in the adjacent column and is also lies below Sb. Hence Pb is more metallic than Sb
1. Fill in the blanks. Example: A nitrogen atom takes on a 3- charge when it forms an anion and becomes nitride. a. A magnesium atom takes on a ___ charge when it forms a(an) ____ion and becomes . b. A chlorine atom takes on a ___ charge when it forms a(an) ____ion and becomes . c. An oxygen atom takes on a ___ charge when it forms a(an) ____ion and becomes . d. A potassium atom takes on a ___ charge when it forms a(an) ____ion and becomes .
Answer: a. +2, cation and magnesium ion .
b. -1, anion, chloride
c. -2, anion, oxide
d. +1. cation , potassium ion
Explanation:
When an atom accepts an electron negative charge is created on atom and is called as anion.
When atom loses an electron positive charge is created on atom and is called as cation.
Magnesium (Mg) with atomic number of 12 has electronic configuration of 2,8,2 and thus it can lose 2 electrons to form [tex]Mg^{2+}[/tex] cation and becomes magnesium ion.
Chlorine (Cl) with atomic number of 17 has electronic configuration of 2,8,7 and thus it can gain 1 electron to form [tex]Cl^{-}[/tex] anion and becomes chloride.
Oxygen (O) with atomic number of 8 has electronic configuration of 2,6 and thus it can gain 2 electrons to form [tex]O^{2-}[/tex] anion and becomes oxide.
Potassium (K) with atomic number of 19 has electronic configuration of 2,8,8,1 and thus it can lose 1 electron to form [tex]K^{+}[/tex] cation and becomes potassium ion.
can you help me with my science
For the following reaction, 29.9 grams of sulfur dioxide are allowed to react with 6.26 grams of oxygen gas . sulfur dioxide(g) + oxygen(g) sulfur trioxide(g) What is the maximum mass of sulfur trioxide that can be formed? grams What is the FORMULA for the limiting reagent? What mass of the excess reagent remains after the reaction is complete? grams
Answer: a) The maximum mass of sulfur trioxide that can be formed is 31.4 grams
b) The FORMULA for the limiting reagent is [tex]O_2[/tex]
c) Mass of excess reagent remains is 4.8 grams
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} SO_2=\frac{29.9g}{64g/mol}=0.467moles[/tex]
[tex]\text{Moles of} O_2=\frac{6.26g}{32g/mol}=0.196moles[/tex]
[tex]2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)[/tex]
According to stoichiometry :
1 mole of [tex]O_2[/tex] require = 2 moles of [tex]SO_2[/tex]
Thus 0.196 moles of [tex]O_2[/tex] will require=[tex]\frac{2}{1}\times 0.196=0.392moles[/tex] of [tex]SO_2[/tex]
Thus [tex]O_2[/tex] is the limiting reagent as it limits the formation of product and [tex]SO_2[/tex] is the excess reagent as (0.467-0.392) = 0.075 moles or [tex]0.075mol\times 64g/mol=4.8g[/tex] are left.
As 1 mole of [tex]O_2[/tex] give = 2 moles of [tex]SO_3[/tex]
Thus 0.196 moles of [tex]O_2[/tex] give =[tex]\frac{2}{1}\times 0.196=0.392moles[/tex] of [tex]SO_3[/tex]
Mass of [tex]SO_3=moles\times {\text {Molar mass}}=0.392moles\times 80g/mol=31.4g[/tex]
Thus 31.4 g of [tex]SO_3[/tex] will be produced from the given masses of both reactants.
Specific Gravity is also known as ______________
It is often possible to change a hydrate into an anhydrous compound by heating it to drive off the water (dehydration). Write an equation that shows the dehydration of manganese(II) sulfate pentahydrate . Use an asterisk to enter the dot.
Answer:
MnSO4.5H2O(s) ---------> MnSO4(s) + 5H2O(g)
Explanation:
The dehydration of a hydrate implies that the water of crystallization is lost. Water of crystallization is included in the crystal structure as it is formed and is incorporated into the structure of the compound.
Now the equation for the dehydration of manganese(II) sulfate pentahydrate is;
MnSO4.5H2O(s) ---------> MnSO4(s) + 5H2O(g)
The MnSO4 is now said to be anhydrous.
PLS ANSWER QUICK!!! ILL GIVE WAY MORE PIONTS!!!!! and ill give brainliest!, JUST GIVE RIGHT ANSWER!!
The following diagram shows a stage of a cell during mitosis.
Which of the following happens to the cell during the stage shown in the diagram? The chromosomes are replicated.
The chromosomes start to condense.
The chromosomes line up in the middle.
The chromosomes begin to uncoil and form chromatin.
Answer:
I promise it is the 2nd one
Explanation:
The following diagram shows a stage of a cell during mitosis. Two identical circles are joined in the middle and they have the nucleolus inside the nucleus in the center of the circles.
Answer:
The Second One
Explanation:I did the test and got this question correct
Pls comment :D
The chemical equation below shows the reaction of magnesium and oxygen which combine to form magnesium oxide.
Mg + O = MgO
In this equation, magnesium and oxygen are the _____
•reactants
•products
Help plz
To run a spectrophotometry experiment, begin by _________ the spectrophotometer and preparing the samples. Be sure to select the correct ________, then run a measurement on the _________ solution. Follow up by running measurements on _______solutions. Once data is collected, turn off the instrument, clean the area, and discard the samples.
Answer:
To run a spectrophotometry experiment, begin by warming up the spectrophotometer and preparing the samples. Be sure to select the correct wavelength, then run a measurement on the blank solution. Follow up by running measurements on sample solutions. Once data is collected, turn off the instrument, clean the area, and discard the samples.
Explanation:
The spectrophotometer is a device used in laboratories to carry out analysis in experiments that need to measure and compare the amount of light absorbed or transmitted between samples, regardless of whether the samples are transparent or opaque solid. The spectrophotometer can be single-beam and double-beam. Regardless of the nature of the beam, to perform a spectrophotometry experiment, you must start by heating the spectrophotometer and preparing the samples. Make sure to select the correct wavelength and perform a measurement on the blank solution. Follow up by performing measurements on sample solutions. Once the data is collected, turn off the instrument, clean the area and discard the samples.
A normal adult jawbone contains 200 mg of Carbon-14 in a living person. If scientists found a jawbone that only had 50mg of Carbon-14, how old is the bone? (The half-life of C-14 is 5730 years).
Carbon-14 is a radioisotope of carbon that decays following first-order kinetics. There are four values of interest in this problem: the "normal" (or original) amount of carbon-14 for a jawbone ([tex]\mathrm{N_0}[/tex]), the actual amount of carbon-14 in a jawbone ([tex]\mathrm{N}[/tex]), the half-life of carbon-14 ([tex]\mathrm{t_{1/2}}[/tex]), and the actual time elapsed ([tex]\mathrm{t}[/tex]) from the original time. There is an equation that ties all these values in together,
[tex]N= N_0 e^{-kt}[/tex]
where k is the rate constant, which, for first-order decay, is related to the half-life by
[tex]k = \dfrac{\ln 2}{ t_{1/2} }.[/tex]
What you want to find here is the time elapsed (t). So, you can substitute the latter equation for k into the k in the former equation to get
[tex]N= N_0 e^{\frac{-\ln 2 \;t}{t_{1/2}}.[/tex]
Rearranging to solve for t, the equation becomes
[tex]t = \left(\dfrac{\ln \dfrac{N_0}{N}}{\ln 2} \right) t_{1/2}.[/tex]
You are given all three of the values necessary to solve for t: The normal amount of carbon-14 is 200 mg; the actual amount of carbon-14 in the sample is 50 mg; and the half-life of carbon-14 is 5730 years. Plugging them into the above equation, we get
[tex]t = \left(\dfrac{\ln \dfrac{200 \text{ mg}}{50 \text{ mg}}}{\ln 2} \right) \left(5730 \text{ years} \right) = 11460 \text{ years}.[/tex]
So the jawbone found is 11460 years old (or 11000 if accounting for sig figs).
A sample of potassium has an average atomic mass of 39.0983amu. There are three isotopic forms of potassium element in the sample. About 93.2581% of the potassium atoms are 38.9637amu mass (39K); 0.0117% are 39.9639amu mass (40K), and the remaining isotope 41K is of 40.9618amu mass. Calculate the percentage composition of 41K. Report your answer with five significant figures.
Answer:
Percent Composition of 41K = 6.7302%
Explanation:
The explination is in the image.
The percentage composition of 41K isotope of potassium according to the given data is 6.7302 percent.
Composition of isotopes:Given that the atomic masses of the potassium isotopes are:
mass of 39K = 38.9637amu
mass of 40K = 39.9639amu
mass of 41K = 40.9618amu
and the percentages are:
percentage of 39K = 93.2581%
percentage of 40K = 0.0117%
percentage of 41K = x
Given average mass of potassium = 39.0983amu
sum (mass of isotopes × percentage abundance) = average mass
38.9637 × 93.2581% + 39.9639 × 0.0117% + 40.9618 × x = 39.0983
x = 6.7302%
So the percentage composition of 41K is 6.7302%
Learn more about composition of isotopes:
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How many grams of water are formed if 3.5 moles of ammonia are used?
Ammonia chemical formula : NH3
Ammonia molar mass : 14 + (1×3) = 17
Mass = Mol × mm
= 3.5mol × 17 = 59.5g
See pic for formula
nvm i think your question is incomplete. pls report my answer i thought u are asking the mass of ammonia
You are boiling water to cook some noodles. You notice bubbles forming on the bottom of the pot. Your brother thinks it is a chemical change. Is he correct?
Yes. The bubbles contain hydrogen and oxygen that separated from the water.
Yes. The bubbles are now filled with air.
No. The water changes into carbon dioxide.
No. It is a physical change when water is heated and is converted into water vapor.
Answer:
i think its Yes. The bubbles contain hydrogen and oxygen that separated from the water but i could be wrong so i would just wait to see if anybody else says anything
Explanation: