Answer:
1.1×10²¹ photon
Explanation:
Given data:
Wavelength of radiation = 5.10×10⁻⁴ cm
How many radiations required to liquid inorder to absorbed 42.75 J = ?
Solution:
First of all we will calculate the energy of radiation having 5.10×10⁻⁴ cm.
E = h.c/ λ
E = 6.63 ×10⁻³⁴ js × 3 ×10⁸ m/s / 0.051×10⁻⁴ m
E = 19.89 ×10⁻²⁶ j.m/ 0.051×10⁻⁴ m
E = 390 ×10⁻²² j
The energy of 1 photon with 5.10×10⁻⁴ cm. wavelength having 390 ×10⁻²² j energy.
1 photon × 42.75 J/390 ×10⁻²² j
0.11 ×10²²photon
1.1×10²¹ photon
The radioactive element carbon-14 has a half-life of 5750 years. A scientist determined that the bones from a mastodon had lost 58.9% of their carbon-14. How old were the bones at the time they were discovered?
The bones were about years old.
A scientist determined the bones from a mastodon , The bones were 9377 years old when they were first found.
Find the decay constant (λ) first by using
λ= 0.693÷t₁/₂
λ=0.693÷5750
λ= 1.205×10⁻⁴
Now,
N=N₀e⁻λt
Where, originally, N₀= Mass of radioactive element
N= mass of radioactive element after time t
Apply natural logarithm on both the sides
In N=N₀₋λt
Let N₀ be 100 then N will be 100-58.8
= 41.1
In (41.1)= In (100)₋(1.205×10⁻⁴t)
In [41.1÷100]= ₋1.205×10⁻⁴t
₋1.136= ₋1.205×10⁻⁴t
t= ₋1.13÷(₋1.205×10⁻⁴)
t= 9377 years
Radioactive : Radioactivity is the term used to describe how an atomic nucleus splits or decays. Radiation is emitted when a radioactive material decays. A few types of decay include spontaneous fission, beta decay, gamma decay, neutron emission, and alpha decay.
Bones: The majority of vertebrate animals contain bones, which are rigid organs and a part of their skeleton. The bones, which also produce red and white blood cells, store minerals, provide the body structure and support, and allow motion, protect the body's other organs.
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7.0 x 10 -3 mol of I2 in 100.00ml of solution
Determine the mass in each of the following
1) 54 molecules of CO2 (MM = 44.0 g/mol)
2) 3.65 x 1024 molecules of water (MM = 18.0 g/mol)
The mass of a substance with given number of molecules can be calculated by first finding the number of moles in the substance as follows:
No. of moles = no of molecules ÷ Avogadro's number
Moles of carbon dioxide = 54 ÷ 6.02 × 10²³ = 8.97 × 10²³ molesMoles of water = 3.65 x 10²⁴ ÷ 6.02 × 10²³ = 0.61 molesMass of these substances can be calculated by multiplying the number of moles by their molar mass as follows:
Mass of carbon dioxide = 8.97 × 10²³ moles × 44 g/mol = 3.95 × 10²⁵ grams. Mass of water = 0.61 moles × 18 g/mol = 10.98 gramsLearn more about mass at: https://brainly.com/question/21042927
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Thank me and get brainliest
Answer:
thanks
Explanation:
thanks
Calculate the proper number of significant digits, the density of a 23.23g box occupying 26.5 mL.
Answer:
0.877 mL
Explanation:
The box's density would be the ratio of the mass of the box and its volume
which is, (23.23/26.5) mL
or, 0.8766 mL
We must round this down to 3 significant figures,
which will be 0.877 mL
Why would it not be practical to count atoms or molecules in dozen, like we do donuts?
Answer: We can point out two main reasons: atoms and molecules are too small to bee seen without the help of complex equipments and there are too many particles (atoms or molecules) in a small amount of substance (6.022 x 10^23 particles per mol of substance).
Explanation:
The question requires us to explain why atoms and/or molecules can't be count in dozen.
We can point out two main reasons for why it is not practical to count atoms and molecules the way we do with donuts and eggs, for example:
- The first one is the size of this particles. We need powerful and complex equipments to be able to see atoms and molecules. The size of atoms, for example, is measured in Angstroms, which corresponds to 10^-10 meters. Although molecules are slightly bigger, they are still too small to be seen without complex microscopes.
- Considering that we were able to see atoms and molecules in order to count them, there would be another issue: the amount of atoms and/or molecules contained in small amount of substances is too big. For example: let's consider 18 g of water, which corresponds to approximately 1 mol of this substance; in 18 g of water there are 6.022 x 10^23 molecules of water - which is way more complex than counting a dozen (12) units of donuts, for example.
How many valence electrons does group 18 have
Answer:
group 18 has 8 valence electrons, that's why they are stable.
8!
Explanation:
You're looking at the very last column, right? Noble gases are known for their stability because their valence shells are full! Lucky for you, all you have to do to know the number of valence electrons is look at the column number and take ignore the 1! 16 becomes 6, 18 becomes 8.
Ocean water contains 3.3 % NaCl by mass.
How much salt can be obtained from 234g of seawater?
Answer:
Ans: 8.9 NaCl
Explanation:
Ocean water contains 3.5 nacl by mass how much salt can be obtained from 254 g of seawater
Question: Ocean water contains 3.5% NaCl by mass. How much salt can be obtained from 254g of seawater?
How is the shape of a solid object different from the shape of a liquid?
Answer:
Because the particles cannot move around, a solid has a fixed shape. Liquids do not have a fixed shape but they do have a fixed volume. The particles are very close together. ... A liquid can flow and take the shape of its container.
Explanation
PAY ATTENTION IN CLASS!!!
Glucose, C6H12O6, is used to prepare intravenous feeding solutions. What volume of 5.0 % W/V glucose solution can be prepared using 125 g of glucose? Show your working.
Please if the answer is correct, ill give brainliest
250 L of 5.0% w/v glucose solution can be prepared using 125 g of glucose.
We use the below formula to solve our problem,w/v = [ mass of solute (g) / volume of solution (mL) ] × 100
Substitute the values from our problem,5.0 % w/v = [ 125 g / volume of solution (mL) ] × 100
Rearranging the formula, we havevolume of solution (mL) = [ 125 g / 5.0 % w/v ] x 100
Substitute further for w/v,volume of solution (mL) = [ 125 g / (5.0 / 100) ] x 100
Simplify the expression,volume of solution (mL) = [ 125 g / 0.05 ] x 100
Hence, the volume of solution (mL) = 250,000 mL or 250 L1. Write the IUPAC names for the following 1.1 1.2 N 1.3 O NO2 x Y ·0 OH 5
1. The IUPAC name of N is nitrogen.
2. Nitrogen dioxide
3.The IUPAC name of O is oxygen
4.The IUPAC name of OH is hydroxyl.
The IUPAC name of ·0 is a radical. It is commonly found in organic chemistry and plays an important role in many reactions.
IUPAC names for the given compounds are:1.1. N: Nitrogen
The IUPAC name of N is nitrogen.
It is a non-metal and belongs to group 15 in the periodic table. It has an electronic configuration of 1s2 2s2 2p3.1.2. NO2: Nitrogen dioxide
Explanation: NO2 is a chemical compound that is formed by the combination of nitrogen and oxygen. It is a reddish-brown gas that has a pungent odor.
The IUPAC name of NO2 is nitrogen dioxide.1.3. O: Oxygen
Explanation: The IUPAC name of O is oxygen.
It is a non-metal and belongs to group 16 in the periodic table. It has an electronic configuration of 1s2 2s2 2p4.
X: UnknownExplanation: No IUPAC name can be given to an unknown compound as the structure and composition are not known.
Y: Hydroxyl Explanation: The IUPAC name of OH is hydroxyl.
It is a functional group that is composed of an oxygen atom and a hydrogen atom (-OH). It is commonly found in alcohols and phenols. ·0: RadicalExplanation: A radical is a molecule or an ion that contains an unpaired electron.
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Note: The complete question is given below
Provide the IUPAC names for the following compounds:
\(CH_3CH_2CH(CH_3)CH_2CH_2CH_2CH_3\)
C6H5CH(CH3)2
H2NCH2CH2CH2CH2CH2NH2
CH3CH2CH2CH2CH2OH
CH3CH2CH2CHOHCH3
Element A has two isotopes. The first isotope is present 18.18% of the time and has a mass of
147.99. The second isotope has a mass of 127.76. Calculate the atomic mass of element A. (To two
decimals places)
The atomic mass of element A, given that the first isotope has abundance of 18.18% and a mass of 147.99, is 131.43 amu
How do i determine the atomic mass of element A?From the question given above, the following data were obtained:
Abundance of 1st isotope (1st%) = 18.18%Mass of 1st isotope = 147.99Mass of 2nd isotope = 127.76 Abundance of 2nd isotope (2nd%) = 100 - 18.18 = 81.82%Atomic mass of element A=?The atomic mass of the element A can be obtain as illustrated below:
Atomic mass = [(Mass of 1st × 1st%) / 100] + [(Mass of 2nd × 2nd%) / 100]
Inputting the given parameters, we have:
Atomic mass = [(147.99 × 18.18) / 100] + [(127.76 × 81.82) / 100]
Atomic mass = 26.90 + 104.53
Atomic mass = 131.43 amu
Thus, the atomic mass of element A obtained from the above calaculation is 131.43 amu
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PLEASE HELP!!
Solutions Pre-Lab Questions:
In this lab, you will make fruit drinks with powdered drink mix. Complete the pre-lab questions to get the values you need for your drink solutions.
1. Calculate the molar mass of powered fruit drink mix, made from sucrose (C12H22O11).
2. Using stoichiometry, determine the mass of powdered drink mix needed to make a 1.0 M solution of 100 mL.
(Hint: Use molarity = to find the moles of drink mix, then convert moles to grams using a mole conversion.)
3. What mass of powdered drink mix is needed to make a 0.5 M solution of 100 mL?
Answer:
Explanation:
C₁₂H₂₂O₁₁
1 )
Molar mass = 12 x 12 + 22 x 1 + 11 x 16
= 144 + 22 + 176
= 342 g
2 )
100 mL of 1.0 M will contain 1.0 x0.100 = .1 mole of sucrose
0.1 mole of sucrose = 0.1 x 342 g = 34.2 g of sucrose.
So , mass of sucrose required is 34.2 g .
3 )
100 mL of .5 M sucrose = .100 x .5 mole of sucrose
= .05 mole of sucrose
.05 mole of sucrose = .05 x 342 g = 17.1 g of sucrose .
So , mass of sucrose required is 17.1 g .
which compoud is propanic acid
Answer:
Explanation:
Propionic acid has chemical formula of C₃H₆O₂ , and smells somewhat unpleasant (like body odour).
What a mass from phenylaamine s produced from 378g of aspartame?
The molar mass of phenyl alanine is 165 g/mol and the mas of the phenyl alanine is 211.2 g /mol
What is the mass of the phenylamine?We know that the statement in the question has told that from the reaction equation. It is one mole of the aspartame that is required to be able to obtain one mole of the phenylamine.
Now we know that the number of moles of the aspartame that is required can be obtained by the use of;
Number of moles = mass/molar mass
= 378g/294.3 g/mol
= 1.28 moles
Molar mass of phenylalanine = 9(12) + 11(1) + 14 + 2(16)
=108 + 11 + 14 + 32
= 165 g/mol
Hence mass of the phenylamine = 1.28 moles * 165 g/mol
= 211.2g
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Answer this correctly please
Answer:a. 2Hg(NO3)2(s) → 2Hg(l) + 2NO2(g) + O2(g)
b. Ca3(PO4)2(aq) + 2H3PO4(aq) → 3Ca(H2PO4)2(aq)
c. 3NaOH(aq) + FeCl3(aq) → Fe(OH)3(aq) + 3NaCl(aq)
Explanation:
How many moles of nitrogen gas are produced when 36. 0 g of NH4NO3 reacts?
\(0.225 moles\) of nitrogen gas are produced when\(36.0 g\) of NH4NO3 reacts.
The reaction for the decomposition of ammonia nitrate (NH4NO3) is given by:
NH4NO3 → N2 + 2H2O
The number of moles of nitrogen gas (N2) produced when 36.0 g of NH4NO3 reacts can be determined using the following steps:
1. Calculate the molar mass of NH4NO3.
\(Molar mass of NH4NO3 = (1 *14.007 g/mol) + (4 *1.008 g/mol) + (3 * 14.007 g/mol) \\ = 80.042 g/mol\)
2. Calculate the number of moles of NH4NO3 present in the given mass.
Number of moles =\(\farc{Mass}{Molar mass}\\ = \frac{36.0 g}{80.042 g/mol}\\ = 0.45 moles\)
3. Calculate the number of moles of nitrogen gas produced.
Number of moles of N2 produced = \(\frac{0.45 moles}{2 }= 0.225 moles\)
Therefore,36.0 g of NH4NO3 react with oxygen to generate g of nitrogen gas \(0.225 moles\).
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The movement of tectonic plates in two different locations is shown below:
Two blocks labeled Location A and Location B are shown. The block labeled A has a vertical line in the middle. On the left of the line there is an arrow pointing down. On the right of the vertical line there is an arrow pointing up. At Location B the top layer of the block shows two horizontal arrows pointing towards each other. The portion of this block on the right is shown moving down.
Which statement is most likely true?
Subduction occurs in both locations.
Seafloor spreading occurs in both locations.
The magnetic orientation of rocks may change in Location A and subduction occurs in Location B.
Subduction occurs in Location A and the magnetic orientation of rocks may change in Location B.
Answer: The magnetic orientation of rocks may change in Location A and subduction occurs in Location B. I tried my best
Happy To Help ;)
Explanation:
Answer:
It is C!! Hope this helps!!
Explanation:
I got it correct on the test
Which equation shows an increase in entropy?
Hint: Look at the states of matter, g s l, of the chemicals in each equation. A C2H4(g) + H2(g) + C2H6(g) в Caco3(9) + Cao(s) - CO2(g) c Fe(s) + S (s) -+ FeS (s)
The equation C2H4(g) + H2(g) + C2H6(g) → Caco3(s) + Cao(s) + CO2(g) shows an increase in entropy due to the formation of a gas as a product. Option A
In this equation, the reactants on the left-hand side consist of gases (C2H4 and H2), while the products on the right-hand side include a solid (Caco3) and a gas (CO2).
When a reaction involves a change from gaseous to solid or liquid states, there is typically a decrease in entropy because the particles become more ordered and constrained in the solid or liquid phase.
Conversely, when a reaction involves the formation of gases, there is generally an increase in entropy because gases have higher degrees of molecular motion and greater freedom of movement compared to solids or liquids.
In the given equation, the reactants include three gaseous compounds (C2H4, H2, and C2H6), and one of the products is a gas (CO2). Therefore, the overall entropy of the system increases during this reaction.
The equation Fe(s) + S(s) → FeS(s) does not show an increase in entropy. Both the reactants (Fe and S) and the product (FeS) are solids. Since solids have lower entropy compared to gases or liquids, the entropy of the system does not increase in this reaction. Option A
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. If you apply heat to the copper bar, its density will
Answer:
Decrease
Explanation:
Density is defined as the mass per unit volume. Density is highly influenced by temperature.
The density of metals and copper is inversely proportional to the temperature, it means density of copper decreases with an increase in temperature or when heated. This is so because when the copper is heated, the copper atoms start vibrating faster that allows the collision of atoms with each other faster and occupying more volume than before. As the copper atoms occupy more volume, the density of copper decreases.
Hence, the correct answer is "On applying heat the density of copper decreases".
In a common reaction in living things, glucose combines with oxygen to form carbon dioxide and water vapor. In an experiment on this reaction, the masses of glucose, oxygen, and carbon dioxide are measured. This data table shows the measurements. What is the mass of the water vapor? please help
Multiply the number of moles of glucose by the molar mass of water (18.015 g/mol) to find the mass of water vapor produced.
To determine the mass of water vapor produced in the reaction, we need to analyze the data provided in the data table. The reaction mentioned is the process of cellular respiration, where glucose (C6H12O6) combines with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O). We can use the principle of conservation of mass to calculate the mass of water vapor produced.The data table should provide the initial masses of glucose and oxygen, as well as the final mass of carbon dioxide. From these measurements, we can calculate the mass of water vapor.First, determine the number of moles of glucose and oxygen by dividing their respective masses by their molar masses. Then, use the balanced equation for cellular respiration to determine the mole ratios between glucose, oxygen, carbon dioxide, and water.
The balanced equation for cellular respiration is:
C6H12O6 + 6O2 -> 6CO2 + 6H2O
Based on this equation, we know that for every mole of glucose, 6 moles of water are produced.In conclusion, by using the principles of conservation of mass and stoichiometry, you can calculate the mass of water vapor produced in the reaction based on the given measurements.
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b. After 20,000 L of ethylene oxide at 748 kPa and 525 K is cooled to 293 K it is transferred to a 110,000 L tank. what is the new pressure?
Answer:
75.9 KPa
Explanation:
From the question given above, the following data were obtained:
Initial volume (V1) 20000 L
Initial pressure (P1) = 748 kPa
Initial temperature (T1) = 525 K
New temperature (T2) = 293 K
New volume (V2) = 110000 L
New pressure (P2) =?
The new pressure pressure of the gas can be obtained by using the general gas formula as illustrated below:
P1V1 / T1 = P2V2 / T2
748 × 20000 / 525 = P2 × 110000 / 293
14960000 / 525 = P2 × 110000 / 293
Cross multiply
P2 × 110000 × 525 = 14960000 × 293
P2 × 57750000 = 4383280000
Divide both side by 57750000
P2 = 4383280000 / 57750000
P2 = 75.9 KPa
Thus, the new pressure of the gas is 75.9 KPa
How many meters is 32 kilometers?
A.
3.2 × 10-5 meters
B.
3.2 × 10-4 meters
C.
3.2 × 103 meters
D.
3.2 × 104 meters
E.
3.2 × 105 meters
Answer:
B
Explanation:
3.2 × 10⁴
= 3.2 × 10000
= 32000 m
Part 2 The student wanted to know if the value obtained from their experiment (part 1) is similar to that calculated using average bond enthalpy data.
a) Using the balanced equation and the data in the table below, calculate the theoretical enthalpy of combustion.
Note: you will need to include the enthalpy of vaporisation for the liquid components which are also given.
C₂H5OH()+302(g) → 2CO2(g) + 3H₂O(1)
Average Bond Enthalpies (kJ mol-¹)
C-H 412
C-C 348
C-O 358
O=O 496
C=O 743
O-H 463
Enthalpy of Vaporisation (kJ mol-¹)
Ethanol 42.5
Water 41
-1113.5kJ is the theoretical enthalpy of combustion.
What makes energy different from enthalpy?
The entire amount of heat energy that is either absorbed or released in a thermodynamic system is measured by enthalpy. Internal energy denotes all of the potential or moving energy present in a thermodynamic system.
Enthalpy of combustion is the term used to describe the change in a system's enthalpy that occurs when one mole of a substance fully burns in oxygen or air at a specific temperature.
C₂H5OH()+302(g) → 2CO2(g) + 3H₂O(1)
Reactants:
5 C-H : 5*412
1 C-C : 348
1 C-O: 358
3 O=O: 3* 496
1 O-H: 463
Products:
2*2 C=O : 4*743
2*3 O-H: 6*463
Enthalpy of Vaporization (kJ mol-¹) for :
Ethanol 42.5
Water 41
Enthalpy of combustion : (5*412 + 348 + 358 + 3* 496 + 463 + 42.5) - ( 3*41 + 4*743 + 6*463)
: -1113.5kJ
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A certain ionic compound X has a solubility in water of 0.773 g/mL. Calculate the greatest mass of X that could be dissolved in 500 mL of water. You may assume the volume of the solution doesn't change as the X is dissolved.
Answer
387 g
Procedure
We need to use a dimensional analysis considering the given solubility to solve this question. Also, we will consider 3 significant digits.
\(500\text{ mL }\frac{0.773\text{ g}}{1\text{ mL}}=386.5\approx387\text{ g}\)17. Which of the following
hydrocarbon undergo addition
reaction:
С3Н6
С2Н6
ОООО
С3Н8
CH4
Answer:
С3Н6.
Explanation:
Hello!
In this case, since addition reactions imply that a radical or some radicals are added to the parent chain, we notice that only unsaturated hydrocarbons are able to undergo addition whereas saturated ones undergo substitution reactions as they already have all the carbon bonds bonded to leaving groups.
In such a way, we can rule out C2H6, C3H8 and CH4 as they are all alkanes; therefore, only С3Н6 is able to undergo an addition reaction due to the C=C which is able to lose one of those bonds and allow an incoming radical to get included into the parent chain.
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Why is HCI a strong acid and HCIO a weak acid?
Strong Acid: dissolves and dissociates 100% to produce protons (H+) 1. seven strong acids: HCl, HBr, HI, HNO3, H2SO4, HClO4, & HClO3 2. ... Weak Acid: dissolves but less than 100% dissociates to produce protons (H+) 1.
Which of the following amino acids would be easiest to distinguish from the other three in a titration?
A. F - Phenylalanine.
B. K - Lysine.
C. V - Valine.
D. G - Glycine.
Answer:
B. K - Lysine.
Explanation:
Required
The easiest amino acid to distinguish
The 4 amino acids in the question can be grouped into 2. The groups are:
(1) Group I - Nonpolar amino acids and (2) Group IV - Basic amino acids
Three of the amino acids belong to group I while the last belongs to group IV.
The acids according to their categories are:
(1) Group I - Nonpolar amino acids:- Phenylalanine, Valine and Glycine
(2) Group IV - Basic amino acids:- Lysine
Because Lysine belongs to a different group different from the other three, it will be easily distinguished because it will exhibit a different property from the other three acids.
? Question
In an ozone molecule, the three atoms must be connected, so there must at least be a single bond between them. Place
dots in pairs around the oxygen atoms until each oxygen atom has eight valence electrons, starting with the atoms on the
outside and doing the central atom last if there are enough. Do not exceed the total number of valence electrons
identified in part A. Remember that the dashes between the oxygen atoms, which represent single bonds, each indicate
the presence of two valence electrons.
A is the answer
In an ozone molecule, the three atoms must be connected, so there must at least be a single bond between them. Place
dots in pairs around the oxygen atoms until each oxygen atom has eight valence electrons, starting with the atoms on the
outside and doing the central atom last if there are enough. Do not exceed the total number of valence electrons
identified in part A. Remember that the dashes between the oxygen atoms, which represent single bonds, each indicate
the presence of two valence electrons
Answer:
Explanation: i did it
In an alkalinity test, a sample is titrated with a stand-ardized acid. If 7.5 milliliters of an acid with a knownstrength of 0.02 equivalents per liter (or 0.02 N) areneeded to complete a titration with 60 mL of sample,what is the sample's alkalinity in mg/L as CaCO3?
Answer:
Explanations: