Answer:
1.75L
Explanation:
Reaction of decomposition is:
2KClO₃(s) → 2KCl(s) + 3O₂(g)
We determine moles of salt:
4.35 g . 1 mol /122.55 g = 0.0355 moles
Ratio is 2:3. 2 moles of salt can produce 3 moles of oxygen
Then, our 0.0355 moles of chlorate may produce (0.0355 . 3)/ 2 = 0.0532 moles.
We have determined, moles of gas and we have data of pressure and temperature. To find out the volume, we apply the Ideal Gases Law:
We convert T° from °C to K → 27°C + 273 = 300K
P . V = n . R . T
0.75 atm . V = 0.0532 mol . 0.0821 L.atm/mol.K . 300K
V = (0.0532 mol . 0.0821 L.atm/mol.K . 300K) / 0.75 atm
V = 1.75 Liters
Dugongs are animals that live in the ocean and eat underwater grasses. The sun is shining on the shallow ocean water where the grasses and dugongs live.
What is happening to the carbon in the water around the grasses and the dugongs? Is carbon moving into the water, moving out of the water, or both?
Answer:
please mark as brainliest
Explanation:
The sun is shining on the shallow ocean water where the grasses and dugongs live. What is happening to the carbon in the water around the grasses and the dugongs? Is carbon moving into the water, moving out of the water, or both? Carbon is not moving into the water; it is only moving out of the water.
How are elements with similar properties grouped in the periodic table?
A. In the same half
B. In the same column
C. In the same row
D. In the same box
AAnswer:A
Explanation:
Liquid nitrogen becomes a gas when it is poured out of its container. The nitrogen is
Answer:
aasjajiakjka
Explanation:
g Phosphorus -32 is a commonly used radioactive nuclide in biochemical research, particularly in studies of nucleic acids. The half-life of phosphorus-32 is 14.3 days. What mass of phosphorus-32 is left from an original sample of 175 mg of Na332PO4 after 35.0 days
Answer:
6.88 mg
Explanation:
Step 1: Calculate the mass of ³²P in 175 mg of Na₃³²PO₄
The mass ratio of Na₃³²PO₄ to ³²P is 148.91:31.97.
175 mg g Na₃³²PO₄ × 31.97 g ³²P/148.91 g Na₃³²PO₄ = 37.6 mg ³²P
Step 2: Calculate the rate constant for the decay of ³²P
The half-life (t1/2) is 14.3 days. We can calculate k using the following expression.
k = ln2/ t1/2 = ln2 / 14.3 d = 0.0485 d⁻¹
Step 3: Calculate the amount of P, given the initial amount (P₀) is 37.6 mg and the time elapsed (t) is 35.0 days
For first-order kinetics, we will use the following expression.
ln P = ln P₀ - k × t
ln P = ln 37.6 mg - 0.0485 d⁻¹ × 35.0 d
P = 6.88 mg
if 7.90 mol of C5H12 reacts with excess O2, how many moles of CO2 will be produced by the following combustion reaction?
Answer:
If 7.9 moles of C₅H₁₂ reacts with excess O₂, 39.5 moles of CO₂ will be produced.
Explanation:
The balanced reaction is:
C₅H₁₂ + 8 O₂ → 5 CO₂ + 6 H₂O
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
C₅H₁₂: 1 moles O₂: 8 molesCO₂: 5 moles H₂O: 6 molesThen you can apply the following rule of three: if by stoichiometry 1 mole of C₅H₁₂ produces 5 moles of CO₂, then 7.9 moles of C₅H₁₂ will produce how many moles of CO₂?
[tex]amount of moles of CO_{2} =\frac{7.9 moles of C_{5}H_{12}*5 moles of CO_{2} }{1 mole of C_{5}H_{12} }[/tex]
amount of moles of CO₂= 39.5 moles
If 7.9 moles of C₅H₁₂ reacts with excess O₂, 39.5 moles of CO₂ will be produced.
A student named a particular compound 2-ethyl-3-methyl-2-butene. Assuming that the student's choice actually corresponded to the correct distribution of the double bond and the substituents, what is the correct IUPAC name for this compound
Answer:
2-ethyl-3-methylbut-2-ene
Explanation:
The whole idea of IUPAC nomenclature is to devise a universally accepted system of writing the name of a compound from its structure.
According to IUPAC nomenclature, the root of the compound is the longest carbon chain. The substituents are named in alphabetical order and in such a way as to give each one the lowest number. The position of the functional group is indicated accordingly.
For the compound in question, its correct IUPAC name is 2-ethyl-3-methylbut-2-ene.
An atom that ______ electrons is called a positive ion. A. has 0 B. has 8 C. loses D. gains
Answer:
Gains
Explanation:
It gets more electrons
HELLP I NEEED HELP NOWW NO GUESSING FRFR
What will happen when a piece of magnesium metal is dropped into a beaker
containing a 1 M solution of copper(II) chloride?
Mg2+ + 2e → Mg(s) ° =- 2.37 V
Cu2+ + 2e → Cu(s) £° = + 0.34 V
there are tiny plants growing on rock fence after the several year . you observe something happened on the rock which of the following describes your observations ? the plants the rock
it is moss
Explanation:
What is the source of coal?
A. Compressed ancient plants
B. Crystallized magma
O C. Compressed ancient sea life
D. Modern forests
Coal and Energy obtained from coal are formed as a result of decomposed life matter such as plants
A. Compressed ancient plants
Fossil FuelsIt should be noted that coal is also a type of fossil fuel formed by prolong compaction of fossil material e.g dead plants over millions of years, forest lives etc. and as a result of high pressure and heat acting on these fossil matter they are transformed to fuel.
In summary, fossil fuels are formed as a result of age long compressed matter.
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greater than 6
Less than or equal to 3
Odd
Not 0
3or9
I think it's 9 ................'
An OH group attached to a hydrocarbon is called a _________ group whereas ______________ is a polyatomic ion with a charge of _______.
An OH group attached to a hydrocarbon is called an alkyl group whereas hydroxide is a polyatomic ion with a charge of -1.
What is OH group?OH group is also called hydroxyl group. Alcohol is a type of organic compound that is characterized by one or more hydroxyl (―OH) groups attached to a carbon atom of an hydrocarbon chain so we can conclude that an OH group attached to a hydrocarbon is called an alkyl group whereas hydroxide is a polyatomic ion with a charge of -1.
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Regions of compressed air caused by the sound of an explosion correspond to the ___?_____ of the sound waves
Explanation:
Compressions and Rarefactions
A vibrating tuning fork is capable of creating such a longitudinal wave. As the tines of the fork vibrate back and forth, they push on neighboring air particles. The forward motion of a tine pushes air molecules horizontally to the right and the backward retraction of the tine creates a low-pressure area allowing the air particles to move back to the left.

Because of the longitudinal motion of the air particles, there are regions in the air where the air particles are compressed together and other regions where the air particles are spread apart
Regions of compressed air caused by the sound of an explosion correspond to the compressions and rarefactions of the sound waves.
What is a sound wave?A sound wave is any distortion in the movement of energy transported by a suitable environment (e.g., air).
Sound waves can be transmitted by different environments such as air (atmosphere) and aquatic media.In humans, sound waves are captured by the ear and then processed by the nervous system.In conclusion, regions of compressed air caused by the sound of an explosion correspond to the compressions and rarefactions of the sound waves.
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Help me please , I got 0.003 for a I need help with b and c
Answer:
(a) The moles of CuSO₄ is 3.125 × 10⁻³ moles.
(b) The moles of Cu is 3.125 × 10⁻³ moles.
(c) The mass of Cu is 0.2 g.
Explanation:
Given:
Mass of CuSO₄ = 0.5 g
Molar mass of CuSO₄ = 160 g/mol
The given balanced chemical equation is:
[tex]2Al+3CuSO_4\rightarrow 3Cu+Al_2(SO_4)_3[/tex]
Part (a):
Calculating the moles of CuSO₄.
[tex]\text{Moles of } CuSO_4=\frac{\text{Mass of }CuSO_4}{\text{Molar mass of }CuSO_4}\\\\\text{Moles of } CuSO_4=\frac{0.5g}{160g/mol}\\\\\text{Moles of } CuSO_4=3.125\times 10^{-3}mol[/tex]
Thus, the moles of CuSO₄ is 3.125 × 10⁻³ moles.
Part (b):
Calculating the moles of Cu.
From the balanced chemical equation, we conclude that:
As, 3 moles of CuSO₄ reacts to give 3 moles of Cu
So, 3.125 × 10⁻³ moles of CuSO₄ reacts to give 3.125 × 10⁻³ moles of Cu
Thus, the moles of Cu is 3.125 × 10⁻³ moles.
Part (c):
Calculating the mass of Cu.
Mass of Cu = Moles of Cu × Molar mass of Cu
Molar mass of Cu = 64 g/mol
Mass of Cu = (3.125 × 10⁻³ mole) × (64 g/mol)
Mass of Cu = 0.2 g
Thus, the mass of Cu is 0.2 g.
Water (H2O) is a polar solvent, and carbon tetrachloride (CCl4) is a nonpolar solvent. In which solvent is each of the following substances, found or used in the body, more likely to be soluble?
a. NaNO3, ionic
b. I2, nonpolar
c. sucrose (table sugar), polar
d. gasoline, nonpolar
e. vegetable oil, nonpolar
f. benzene, nonpolar
g. LiCl, ionic
h. Na2SO4, ionic
Answer:
a. NaNO3, ionic - water
b. I2, nonpolar - CCl4
c. sucrose (table sugar), polar - water
d. gasoline, nonpolar - CCl4
e. vegetable oil, nonpolar - CCl4
f. benzene, nonpolar -CCl4
g. LiCl, ionic - water
h. Na2SO4, ionic - water
Explanation:
Water is a polar substance. This means that it has the ability to dissolve other polar substances. Furthermore, water, even being made by covalent bonds, manages to dissolve ionic substances, because it is a molecule with a partial positive charge on one side, due to hydrogen, and a partial negative charge on the other side, due to the two molecules of oxygen. In this case, any polar or ionic substance has the ability to be dissolved in water, while any non-polar substance needs a non-polar liquid to be able to be dissolved, such as CCI4.
During a chemical reaction, an iron atom became the ion Fe2+. What happened to the iron atom?
Explanation:
Iron atom is been oxidised as it losses 2 electron to form 2 + ion.
Suppose an element has two naturally occurring isotopes
42
X
20
44
X
20
with atomic masses 42.00 amu and 45.00 amu, respectively. Determine the percent abundance of the
45
X
20
isotope given an average atomic mass of 43.22 amu.
Answer:
Chlorine
Chlorine has two naturally occurring isotopes. In a sample of chlorine, 75.77% of the atoms are Cl-35, with a mass of 34.97amu.
Explanation:
hope it helps thanks❤
Chlorine has two naturally occurring isotopes. In a sample of chlorine, 75.77% of the atoms are Cl-35, with a mass of 34.97amu.
How do you calculate the atomic mass in percentage by isotopes?
List the known and unknown values and edit the problem. Convert each percentage value into a decimal by dividing it by 100. Multiply this number by the atomic mass of that spot. Mix together each isotope to find the median atomic mass.
How do you get the maximum amount of two isotopes?To calculate the maximum percentage of each isotope in a sample of something, chemists usually divide the atomic number of a particular isotope by the total number of atoms of all that isotope and multiply the result by 100.
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The carbon-carbon bonds in benzene are:
O a) Easily broken in chemical reactions
b) A hybrid between double bonds and single bonds
c)
Identical to the carbon-carbon bonds in cyclohexene
d) Identical to the carbon-carbon bonds in cyclohexane
please hurry
Answer:
a
Explanation:
Easily broken in chemical reactions
2.The rate constant of a first order reaction is 66 s-1. What is the rate constant in units of minutes
Answer:
66s^-1 will be 1/66
then to convert to minute you multiply by 69
1/66 x 60 = 3960 mins
The product of an organic reaction is analyzed by column chromatography using silica as the stationary phase and toluene as the mobile phase.
a. True
b. False
Answer:
The product of an organic reaction is analyzed by column chromatography using silica as the stationary phase and toluene as the mobile phase.
Explanation:
The given statement is true.
In chromatography silica gel is used as the predominant stationary phase.
Since silica gel is a good adsorbent.
It is a polar adsorbent.
In order to remove polar components, silica gel is used as the stationary phase.
Answer is a.true.
3 what is the hybridization of each
of the Carbon atoms in the compound
а) +
CH₃ CH = CH CH₂=CH
HC=CH
Answer:
what is the hybridization of each of the carbon atoms in the compounds
а) CH₃ CH = CH2
b) CH₂=CH2
c) Acetylene. (CH triple bond CH).
Explanation:
The hybridization of carbon atom in the organic compounds can be determined by counting the number of surrounding atoms that are in bond with it.
If carbon is directly bonded with two atoms, then it has sp hybridization.
If carbon is directly bonded with three atoms, then it has [tex]sp^2[/tex] hybridization.
If carbon is directly bonded with four atoms, then it has [tex]sp^3[/tex] hybridization.
For the given molecules, the hybridization of each carbon atom is shown below:
The bond order denotes:_____.1. the order of filling of bonding vs. antibonding orbitals. 2. the pattern in which Lewis dot structures of atoms combine to form bonds. 3. the order that orbitals are filled in electron configurations. 4. an estimate of the polarity or dipole moment of a bond. 5. an estimate of the strength of a bond.
Answer:
an estimate of the strength of a bond.
Explanation:
The bond order is given as;
1/2(number of bonding electrons - number of anti bonding electrons)
The bond order tells us about the strength of bond. As the bond order increases, so does the strength of the bond because atoms involved in bonding come closer to each other as the bond length decreases.
Hence, bond order is an index of bond strength. Triple bonds are stronger than double bonds which are stronger than single bonds.
If you keep adding sugar to water and there comes a point that you cannot dissolve any more sugar into it then this is called?
A) solvent solution
B) solute solution
C) saturated solution
D) Dilute solution
Answer:
the answer is letter D, Dilute solution
Someone help me fill this out TY
Please fill it out in each blank :)
Answer and Explanation:
We have to identify the anion (negatively charged ion) and the positive ion to form each compound. The sum of the positive and negative charges will be equal to 0 for a neutral compound.
Chloride: the anion is Cl⁻ (1 negative charge).
Magnesium (Mg²⁺) + Chloride (Cl⁻) : MgCl₂
Sodium (Na⁺) + Chloride (Cl-): NaCl
Zinc (Zn²⁺) + Chloride (Cl-): ZnCl₂
Lithium (Li⁺) + Chloride (Cl-) : LiCl
Lead(II) (Pb²⁺) + Chloride (Cl⁻): PbCl₂
Calcium (Ca²⁺) + Chloride (Cl⁻): CaCl₂
Iron(II) (Fe²⁺) + Chloride (Cl⁻): FeCl₂
Iron(III) (Fe³⁺) + Chloride (Cl⁻): FeCl₃
Potassium (K⁺) + Chloride (Cl): KCl
Nitrate: the anion is NO₃⁻ (1 negative charge).
Magnesium (Mg²⁺) + Nitrate (NO₃⁻) : Mg(NO₃)₂
Sodium (Na⁺) + Nitrate (NO₃⁻): NaNO₃
Zinc (Zn²⁺) + Nitrate (NO₃⁻): Zn(NO₃)₂
Lithium (Li⁺) + Nitrate (NO₃⁻) : LiNO₃
Lead(II) (Pb²⁺) + Nitrate (NO₃⁻): Pb(NO₃)₂
Calcium (Ca²⁺) + Nitrate (NO₃⁻): Ca(NO₃)₂
Iron(II) (Fe²⁺) + Nitrate (NO₃⁻): Fe(NO₃)₂
Iron(III) (Fe³⁺) + Nitrate (NO₃⁻): Fe(NO₃)₃
Potassium (K⁺) + Nitrate (NO₃⁻): KNO₃
Sulphate: SO₄²⁻ (2 negative charges)
Magnesium (Mg²⁺) + Sulphate (SO₄²⁻) : MgSO₄
Sodium (Na⁺) + Sulphate (SO₄²⁻): Na₂SO₄
Zinc (Zn²⁺) + Sulphate (SO₄²⁻): ZnSO₄
Lithium (Li⁺) + Sulphate (SO₄²⁻) : Li₂SO₄
Lead(II) (Pb²⁺) + Sulphate (SO₄²⁻): PbSO₄
Calcium (Ca²⁺) + Sulphate (SO₄²⁻): CaSO₄
Iron(II) (Fe²⁺) + Sulphate (SO₄²⁻): FeSO₄
Iron(III) (Fe³⁺) + Sulphate (SO₄²⁻): Fe₂(SO₄)₃
Potassium (K⁺) + Sulphate (SO₄²⁻): K₂SO₄
Carbonate: CO₃²⁻ (2 negative charges)
Magnesium (Mg²⁺) + Carbonate (CO₃²⁻) : MgCO₃
Sodium (Na⁺) + Carbonate (CO₃²⁻): Na₂CO₃
Zinc (Zn²⁺) + Carbonate (CO₃²⁻): ZnCO₃
Lithium (Li⁺) + Carbonate (CO₃²⁻): Li₂CO₃
Lead(II) (Pb²⁺) + Carbonate (CO₃²⁻): PbCO₃
Calcium (Ca²⁺) + Carbonate (CO₃²⁻): CaCO₃
Iron(II) (Fe²⁺) + Carbonate (CO₃²⁻): FeCO₃
Iron(III) (Fe³⁺) + Carbonate (CO₃²⁻): Fe₂(CO₃)₃
Potassium (K⁺) + Carbonate (CO₃²⁻): K₂CO₃
Hydroxide: OH⁻ (1 negative charge)
Magnesium (Mg²⁺) + Hydroxide (OH⁻): Mg(OH)₂
Sodium (Na⁺) + Hydroxide (OH⁻): NaOH
Zinc (Zn²⁺) + Hydroxide (OH⁻): Zn(OH)₂
Lithium (Li⁺) + Hydroxide (OH⁻): LiOH
Lead(II) (Pb²⁺) + Hydroxide (OH⁻): Pb(OH)₂
Calcium (Ca²⁺) + Hydroxide (OH⁻): Ca(OH)₂
Iron(II) (Fe²⁺) + Hydroxide (OH⁻): Fe(OH)₂
Iron(III) (Fe³⁺) + Hydroxide (OH⁻): Fe(OH)₃
Potassium (K⁺) + Hydroxide (OH⁻): KOH
Phosphate: PO₄³⁻ (3 negative charges)
Magnesium (Mg²⁺) + Phosphate (PO₄³⁻): Mg₃(PO₄)₂
Sodium (Na⁺) + Phosphate (PO₄³⁻): Na₃PO₄
Zinc (Zn²⁺) + Phosphate (PO₄³⁻): Zn₃(PO₄)₂
Lithium (Li⁺) + Phosphate (PO₄³⁻): Li₃PO₄
Lead(II) (Pb²⁺) + Phosphate (PO₄³⁻): Pb₃(PO₄)₂
Calcium (Ca²⁺) + Phosphate (PO₄³⁻): Ca₃(PO₄)₂
Iron(II) (Fe²⁺) + Phosphate (PO₄³⁻): Fe₃(PO₄)₂
Iron(III) (Fe³⁺) + Phosphate (PO₄³⁻): FePO₄
Potassium (K⁺) + Phosphate (PO₄³⁻): K₃PO₄
A metal X from two oxide A and B .3.oogm of A and B contain 0.72 and 1.16g of oxygen respectively.calculate the maases of metal in gm which combine with one gram of oxygen in each case
Answer:
Explanation:
Firstly, we have to determine the mass of metal X. We can do that by interpreting the first and second statement mathematically.
Metal X can form 2 oxides (A and B).
A + B = 3g
The mass of oxygen in A is 0.72g and the mass of oxygen in B is 1.16g.
The mass of metal X in the two oxides will be the same because it's the same metal.
Thus, we represent the mass of the metal in the two oxides as 2X.
2X + 0.72 + 1.16 = 3
2X + 1.88 = 3
2X = 3 - 1.88
2X = 1.12
X = 0.56
Thus, 0.56 g of the metal combines with 0.72g of oxygen in A and 1.16 g of oxygen in B.
Thus, mass of metal (X) in 1g of oxygen in A is
0.56g ⇒ 0.72g
X ⇒ 1
X = 1 × 0.56/0.72
X = 0.78 g
Hence, 0.78g of the metal will combine with 1g of oxygen for A
Also, mass of metal (X) in 1g of oxygen in B is
0.56g ⇒ 1.16g
X ⇒ 1g
X = 1×0.56/1.16
X = 0.48 g
Thus, 0.48g of the metal will combine with 1g of oxygen for B
Guide Questions:
1. How do you compare the distance travelled by the object
when you pushed it with a weak/gentle, strong and
strongest?
Refle
2. Which amount of force applied made the materials travelled
the farthest from the starting point. Nearest the starting
point?
3. What factors do you think affected the movement of the
materials?
Answer:
Kindly check explanation
Explanation:
The force applied is directly proportional to the distance moved by an object, the larger the applied force, the greater the distance moved.
a = f/m
a = acceleration ; f = applied force ; m = mass
From the relation, we can see that acceleration is directly proportional to force applied.
The ball will travel farthest with the greatest applied force while, nearest distance will be attained with the smallest applied force.
The distance covered is affected by both the mass of the object and the applied force
Steel is an alloy consisting of Fe with a small amount of C. Elemental Cr can be added to steel to make the steel less likely to rust
a. True
b. False
To make the steel less likely to rust, element like chromium is added to it. Therefore, the given statement is correct.
What is steel?Steel can be described as an alloy composed of iron with a few tenths of a percent of carbon to enhance its strength and fracture resistance. In steel, alloying elements inclusions within the iron act as hardening agents as they stop the movement of dislocations.
Stainless steels are corrosion- and oxidation-resistant but they need an additional 11% chromium. This type of steel is used in buildings, infrastructure, ships, trains, machines, electrical appliances, weapons, and rockets because of its high tensile strength and low cost.
It can take two crystalline forms body-centered cubic and face-centered cubic depending on the temperature. The interaction of the allotropes of iron with elements such as carbon gives the steel its range of unique properties.
Therefore, the given statement is true that chromium is used in steel formation.
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Calculate the number of molecules of carbon (IV) oxide
produced when 10g of CaCO3 is treated with 100cm of
0.2mol dm HCl. The equation of the reaction is
CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(1) + CO2(g)
ICa=40 (=12 O=161
Answer: The mass of [tex]CO_2[/tex] produced is 0.44 g
Explanation:
For calcium carbonate:The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)
Given mass of calcium carbonate = 10 g
Molar mass of calcium carbonate = 100 g/mol
Plugging values in equation 1:
[tex]\text{Moles of calcium carbonate}=\frac{10g}{100g/mol}=0.1 mol[/tex]
For HCl:The formula used to calculate molarity:
[tex]\text{Molarity of solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (mL)}}[/tex] .....(2)
Molarity of HCl = [tex]0.2mol/dm^3=0.2mol/L[/tex] (Conversion factor: [tex]1L=1dm^3[/tex]
Volume of solution = [tex]100cm^3=100mL[/tex] (Conversion factor: [tex]1mL=1cm^3[/tex]
Putting values in equation 2, we get:
[tex]0.2=\frac{\text{Moles of HCl}\times 1000}{100}\\\\\text{Moles of HCl}=\frac{0.2\times 100}{1000}=0.02mol[/tex]
For the given chemical reaction:
[tex]CaCO_3(s)+2HCl(aq)\rightarrow CaCl_2(aq)+H_2O(l)+CO_2(g)[/tex]
By stoichiometry of the reaction:
If 2 moles of HCl reacts with 1 mole of calcium carbonate
So, 0.02 moles of HCl will react with = [tex]\frac{1}{2}\times 0.02=0.01mol[/tex] of calcium carbonate
As the given amount of calcium carbonate is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.
Thus, HCl is considered a limiting reagent because it limits the formation of the product.
By the stoichiometry of the reaction:
If 2 moles of [tex]HCl[/tex] produces 1 mole of [tex]CO_2[/tex]
So, 0.02 moles of [tex]HCl[/tex] will produce = [tex]\frac{1}{2}\times 0.02=0.01mol[/tex] of [tex]CO_2[/tex]
We know, molar mass of [tex]CO_2[/tex] = 44 g/mol
Putting values in equation 1, we get:
[tex]\text{Mass of }CO_2=(0.01mol\times 44g/mol)=0.44g[/tex]
Hence, the mass of [tex]CO_2[/tex] produced is 0.44 g
draw the structure of two acyclic compounds with 3 or more carbons which exhibits one singlet in the 1H-NMR spectrum
Answer:
attached below
Explanation:
Structure of two acyclic compounds with 3 or more carbons that exhibits one singlet in 1H-NMR spectrum
a) Acetone CH₃COCH₃
Attached below is the structure
b) But-2-yne (CH₃C)₂
Attached below is the structure
(A) Calculate the wavelength (in nm) of light with energy 1.89 × 10–20 J per photon, (b) For light of wavelength 410 nm, calculate the number of photons per joule, (c) Determine the binding energy (in eV) of a metal if the kinetic energy possessed by an ejected electron [using one of the photons in part (b)] is 2.93 × 10–19 J.
Answer:
For A: The wavelength of the light is [tex]1.052\times 10^4nm[/tex]
For B: The number of photons per joule is [tex]2.063\times 10^{18}[/tex]
For C: The binding energy of a metal is 1.197 eV.
Explanation:
The equation used to calculate the energy of a photon follows:
[tex]E=\frac{hc}{\lambda}[/tex] ......(1)
where,
E = energy of a photon
h = Planck's constant = [tex]6.626\times 10^{-34}J.s[/tex]
c = speed of light = [tex]3\times 10^{8}m/s[/tex]
[tex]\lambda[/tex] = wavelength
For A:Given values:
E = [tex]1.89\times 10^{-20}J[/tex]
Putting values in equation 1, we get:
[tex]\lambda=\frac{(6.626\times 10^{-34}J.s)\times (3\times 10^8m/s)}{1.89\times 10^{-20}J}\\\\\lambda=1.052\times 10^{-5}m[/tex]
Converting the wavelength into nanometers, the conversion factor used is:
[tex]1m=10^9nm[/tex]
So, [tex]\lambda=1.052\times 10^{-5}m\times \frac{10^9nm}{1m}=1.052\times 10^4nm[/tex]
Hence, the wavelength of the light is [tex]1.052\times 10^4nm[/tex]
For B:Given values:
[tex]\lambda=410nm=410\times 10^{-9}m[/tex]
Putting values in equation 1, we get:
[tex]E=\frac{(6.626\times 10^{-34}J.s)\times (3\times 10^8m/s)}{410\times 10^{-9}m}\\\\E=4.848\times 10^{-19}J[/tex]
To calculate the number of photons, we use the equation:
[tex]\text{Number of photons}=\frac{\text{Total energy}}{\text{Energy of a photon}}[/tex]
Total energy = 1 J
Energy of a photon = [tex]4.848\times 10^{-19}J[/tex]
Putting values in the above equation:
[tex]\text{Number of photons}=\frac{1J}{4.848\times 10^{-19}J}\\\\\text{Number of photons}=2.063\times 10^{18}[/tex]
Hence, the number of photons per joule is [tex]2.063\times 10^{18}[/tex]
For C:To calculate the binding energy of a metal, we use the equation:
[tex]E=K+B[/tex] .....(2)
E = Total energy
K = Kinetic energy of a photon
B = Binding energy of metal
Converting the energy from joules to eV, the conversion factor used is:
[tex]1eV=1.602\times 10^{-19}J[/tex]
Using the above conversion factor:
[tex]K=2.93\times 10^{-19}J=1.829eV\\\\E=4.848\times 10^{-19}J=3.026eV[/tex]
Putting values in equation 2:
[tex]B=(3.026-1.829)eV=1.197eV[/tex]
Hence, the binding energy of a metal is 1.197 eV.