Answer:
The mole ratio of PbO2 to H2O is 1 : 2.
Explanation:
The balanced reaction equation is:
Pb + PbO2 + 2H2SO4 → 2PbSO4 + 2H2O
On the reactant side, we have 1 mole of Pb, 1 mole of PbO2, 2 moles of H2SO4.
On the product side, we have 2 moles of PbSO4 and 2 moles of H2O.
This means that for ever 1 mole of PbO2 consumed, 2 moles of water is formed as product.
Hence, the mole ratio of PbO2 to H2O is 1 : 2.
Do the following calculation and express the answer using correct scientific notation. (6.00x10^23)y(3.00)/284)
To calculate the given expression, multiply 6.00x10^23 by 3.00 and divide the result by 284 to get 6.33x10^21.
Explanation:To calculate the given expression and express the answer using scientific notation, we need to perform the following steps:
Multiply 6.00x10^23 by 3.00 to get 1.80x10^24.Divide the result by 284 to get 6.33x10^21.So, the answer is 6.33x10^21.
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What does STP mean? What are the values for an ideal gas? Are the values different for a real gas?
Answer: STP is a standard reference point of temperature and pressure, used when measuring gases. The universal value of STP is 1 atm (pressure) and 0o C. Note that this form specifically stated 0o C degree, not 273 Kelvin, even thought you will have to convert into Kelvin when plugging this value into the Ideal Gas equation or any of the simple gas equations. yes they are deal gases are those which follow ideal gas equation (which is PV=nRT).
Explanation:
Which molecular formula is also an empirical formula?
A. PCl3
B. C2H4
C. H2O2
D. C6H12O6
Answer:
The molecular formula [tex]PCl_{3}[/tex] is also an empirical formula.
Explanation:
Empirical formula of a compound represents the simplest positive integer ratio of constituting atoms.
Molecular formula may or may not be equal to empirical formula.
In general, if [tex]A_{x}B_{y}[/tex] is the empirical formula then molecular formula is [tex]A_{nx}B_{ny}[/tex] where n = 1, 2, 3......
Empirical formula Molecular formula
[tex]PCl_{3}[/tex] [tex]PCl_{3}[/tex]
[tex]CH_{2}[/tex] [tex](CH_{2})_{2}[/tex] or [tex]C_{2}H_{4}[/tex]
OH [tex](OH)_{2}[/tex] or [tex]H_{2}O_{2}[/tex]
[tex]CH_{2}O[/tex] [tex](CH_{2}O)_{6}[/tex] or [tex]C_{6}H_{12}O_{6}[/tex]
Hence the molecular formula [tex]PCl_{3}[/tex] is also an empirical formula.
A molecular formula which also represents the empirical formula has its atoms in the simplest whole number ratio. For the given options, PCl3 is an example of this as its phosphorus to chlorine ratio is 1:3.
Explanation:The empirical formula represents the simplest whole number ratio of its atoms while the molecular formula represents the actual number of atoms in a molecule. A molecular formula will also represent the empirical formula if the ratio of its atoms are in their simplest form. Considering the options, we can say that A. PCl3 is also an empirical formula because the ratio of phosphorus to chlorine atoms is in its simplest form, 1:3.
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. Solve: Turn off Show summary. Use the Choose reaction drop down menu to see other
equations, and balance them. Check your answers and then write the balanced equations.
2
6
2
Alt
HCI->
AlCl3 +
Hz
NaCl →
Na +
Cl₂
Na2S +
HCI →
NaCl +
H₂S
CHA +
O₂ →
CO2 +
НО
Practice Balance the following chemical equations. (These equations are not in the Gizmo.) help plz
Balancing chemical equations involves adjusting the coefficients of the reactants and products in order to ensure adherence to the Law of Conservation of Mass. An example of a balanced chemical equation is 2NaOH(aq) + H₂(g) + Cl₂(g) which can also be written in complete ionic form and net ionic form. Always remember to first balance individual atoms, then oxygen, and lastly hydrogen.
Explanation:The subject of your question pertains to balancing chemical equations, which is a key concept in the field of Chemistry. Balancing chemical equations is the process of adjusting the coefficients of reactants and products in a chemical reaction to ensure that the law of conservation of mass is adhered to. This law states that matter cannot be created nor destroyed, meaning every atom we start with, we must end with and in the same quantity.
For example, let's look at the equation 2NaOH(aq) + H₂(g) + Cl₂(g). In its balanced form, this equation would look like this:
An important tip to remember while balancing equations is to first check individual atom balances (aside from O and H), then balance O, and finally, H.
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Cardiac muscle causes movement in the .
Answer:
organ
Explanation:
it causes movement in the heart and the heart is an organ
What are the possible phenotypes of the offspring?
freckles with curly hair
freckles with straight hair
no freckles with curly hair
no freckles with straight hair
SHIRTS
TUNG
W
WORD
CLE
COLE
ROUTE
G
LOBREGA
Savo
WORLD
BAL
What’s the answer
Answer:freckles with straight hair
Explanation:
Answer:
The Answer is B Freckles with Straight Hair
Explanation:
Early Merry Christmas Everyone
Use the observation in the first column to answer the question in the second column. observation question The enthalpy of vaporization of Substance C is bigger than that of Substance D. Which has the higher boiling point? Substance C Substance D Neither, C and D have the same boiling point. It's impossible to know without more information. At 1atm pressure, Substance E boils at 89.°C and Substance F boils at 128.°C. Which has a higher enthalpy of vaporization? Substance E Substance F Neither, E and F have the same enthalpy of vaporization. It's impossible to know without more information. At −3°C, Substance A has a vapor pressure of 147.torr and Substance B has a vapor pressure of 177.torr. Which has a higher boiling point? Substance A Substance B Neither, A and B have the same boiling point. It's impossible to know without more information.
Answer:
See explaination
Explanation:
1. Substance C
With Vapor pressure (VP) of C = 136 torr and VP of D = 186 torr. Since VP (C) < VP (D), this entaills that one needs to supply more energy to get C into the vapor phase. Hence, enthalpy of vaporization of C will be higher.
2. Substance A
B.pt (A) = 109 C
B,pt (B) = 125 C
Having A boils at a lower temperature than B, this automatically will make A to have a higher vapor pressure.
3. Substance E
Since E has a lower enthalpy of vaporization, it can be brought into the vapor phase with lower energy as compared to F. Hence, E will have a higher vapor pressure.
Substance C has a higher boiling point than Substance D, Substance F has a higher boiling point (and thereby a higher enthalpy of vaporization) than Substance E, and Substance A has a higher boiling point than Substance B due to its lower vapor pressure.
Explanation:The boiling point of a substance is directly related to its enthalpy of vaporization. In other words, a substance with a higher enthalpy of vaporization requires more heat to transform from a liquid to a gaseous state, making it have a higher boiling point.
So, based on the given observations:
Substance C has a higher enthalpy of vaporization than Substance D, therefore Substance C has a higher boiling point.Substance E boils at 89°C and Substance F boils at 128°C, therefore Substance F has a higher boiling point and consequently a higher enthalpy of vaporization.Substance A has a lower vapor pressure than Substance B at the same temperature (-3°C). Therefore, Substance A would require more energy to convert to the gaseous phase and so between Substance A and Substance B, Substance A has a higher boiling point.Learn more about Enthalpy of Vaporization here:https://brainly.com/question/32361849
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Complete the following sentences. Match the words in the left-hand column with the appropriate blank in the sentences in the right-hand column. View Available Hint(s) ResetHelp 1. A monosaccharide is a(n) glucose if the carbonyl group is on the end of the carbon chain. 2. If a carbohydrate, like xylulose, has five carbon atoms and a carbonyl group on the second carbon, it is called a(n) aldotriose. 3. Any carbohydrate with the carbonyl group on the second carbon is a(n) ketopentose. 4. With the carbonyl group on the end of a six-carbon chain, the carbohydrate would be classified as a(n) aldose. 5. The most common carbohydrate, ketose, has six carbon atoms. 6. Glyceraldehyde is an example of a(n) aldohexose, because it has three carbon atoms.
1. If a carbohydrate, like xylulose, has five carbon atoms and a carbonyl group on the second carbon, it is called a(n) ketopentose.
2. Any carbohydrate with the carbonyl group on the second carbon is a(n) ketose.
3. The most common carbohydrate, glucose, has six carbon atoms.
4. A monosaccharide is a(n) aldose if the carbonyl group is on the end of the carbon chain.
5. Glyceraldehyde is an example of a(n) triose, because it has three carbon atoms.
6. With the carbonyl group on the end of a six-carbon chain, the carbohydrate would be classified as a(n) aldohexose.
1. The term "ketopentose" refers to a five-carbon carbohydrate with a carbonyl group on the second carbon.
2. A "ketose" is a carbohydrate with a carbonyl group (ketone group) on an internal carbon atom.
3. "Glucose" is a common hexose sugar with six carbon atoms.
4. An "aldose" is a monosaccharide with a carbonyl group at the end of the carbon chain.
5. "Triose" denotes a three-carbon sugar, and glyceraldehyde is an example.
6. An "aldohexose" is a six-carbon sugar with the carbonyl group at the end of the carbon chain.
In summary, Carbohydrates are classified based on carbon atom count and carbonyl group placement. Xylulose is a ketopentose. Glucose is a common hexose. An aldose has the carbonyl group at the chain end, while glyceraldehyde is a triose. An aldohexose has a six-carbon chain with the carbonyl group at the end.
The question probable may be:
Complete the following sentences. Match the words in the left-hand column with the appropriate blank in the sentences in the right-hand column.
aldose, ketose, ketopentose, aldohexose, triose, glucose
1. 1. If a carbohydrate, like xylulose, has five carbon atoms and a carbonyl group on the second carbon, it is called a(n)
2. 2. Any carbohydrate with the carbonyl group on the second carbon is a(n)
3. The most common carbohydrate,_________, has six carbon atoms.
4. A monosaccharide is a(n)______ if the carbonyl group is on the end of the carbon chain.
5. Glyceraldehyde is an example of a(n) _________ , because it has three carbon atoms.
6. With the carbonyl group on the end of a six-carbon chain, the carbohydrate would be classified as a(n)____________.