All light waves can be described in terms of their speed, frequency, and___
Answer:
all light waves can be described in terms of their speed, frequency and wavelength
Explanation:
Hope it helps u.....
The total entropy of a system and its surroundings always increases for a spontaneous process. This is a statement of
Answer:
second law of thermodynamics
Explanation:
According to the second law of thermodynamics, the total entropy of a system and its surroundings always increases for a spontaneous process.
Entropy is defined as the degree of disorderliness of a system. The entropy of a system never remains constant. It either increases or decreases in a process. The total entropy is the sum of the entropy of the system and its surrounding. The total entropy must increase in a spontaneous process.
Thus, the implication of this law is that even, if the entropy of a system decreases, this must be compensated for by increase in entropy of the surroundings in order for the process to be spontaneous.
Which equation was used by Albert Einstein to explain the photoelectric effect? [E = energy, h= planck's constan, and v = frequency]
Answer:
E = hv
Explanation:
Energy = planck constant × frequency
Suppose an electron is transferred from a potassium atom to an unknown halogen atom. For which of the following halogen atoms would this process require the least amount of energy?
A. Cl
B. Br
C. I
Answer:
Cl
Explanation:
Electronegativity is the ability of an electron to attract electrons.
Now, due to the fact that halogens need just one more electron to become stable in their outermost shell, it means all halogens are electronegative.
However, the smaller the atomic number, the bigger the charge density and thus the more electronegative.
Thus, it is the halogen element with the highest atomic number further down the periodic table that will have the least electro negativity and thus require highest amount of energy to attract other electrons.
Thus, since chlorine (Cl) has the least atomic number of 17, then it means that it will be the one that will easily accept the electrons the most from other elements. Therefore the process of transferring electrons from potassium to chlorine will take the least amount of energy.
Calcium chloride and magnesium sulfate are common drying agents. What type of solvent should be dried with calcium chloride, and what type with magnesium sulfate
Answer: The type of solvent that should be dried with calcium chloride is esters while magnesium sulfate is diethyl ether
Explanation:
Drying agents are mainly hygroscopic substances that has the ability to absorb water on exposure to the atmosphere but not enough to form solutions. They are used in desiccators. Examples of drying agents include:
--> CALCIUM CHLORIDE: This is a compound of calcium that is found in soil water and sea water. It is prepared by the action of dilute hydrochloric acid on calcium trioxocarbonate(IV). The anhydrous salt is used in drying a wide variety of solvent including esters.
--> MAGNESIUM SULFATE: This is a slightly acidic drying agent. It works well in solvents like diethyl ether. It is a fast drying agent because it comes as a fine powder with a large surface area.
Which compound contains both sigma and pi bonds... HCCl3, H2CO, H2S, or HBr?
Answer:
H2CO
Explanation:
Becuase it has 2 sigma bonds plus one pi bond and one sigma bond that consitute the double bond between C and O.
Answer:
B. H2CO
Explanation:
Which Group is in the second column of the periodic table?
O A. Noble gases
O B. Alkaline earth metals
O C. Alkali metals
O D. Halogens
Answer:
O B. Alkaline earth metals
Explanation:
Noble gases → 8th column.
Alkali metal → first column.
Halogen → 7th
Answer:
B. Alkaline earth metals
Explanation:
Alkaline-earth metals: The alkaline-earth metals make up Group 2 of the periodic table, from beryllium (Be) through radium (Ra). Each of these elements has two electrons in its outermost energy level, which makes the alkaline earths reactive enough that they're rarely found alone in nature. But they're not as reactive as the alkali metals. Their chemical reactions typically occur more slowly and produce less heat compared to the alkali metals.
Liquid ethyl mercaptan, C2H6S, has a density of 0.84 g/mL. Assuming that the combustion of this compound produces only CO2 , H2O, and SO2 , what masses of each of these three products would be produced in the combustion of 3.15 mL of ethyl mercaptan
Answer:
Mass CO2 = 3.75 grams
Mass H2O = 2.30 grams
Mass SO2 = 2.73 grams
Explanation:
Step 1: Data given
Density of Liquid ethyl mercaptan, C2H6S = 0.84 g/mL
Volume of ethyl mercaptan = 3.15 mL
Step 2: The reaction
2C2H6S + 9O2 → 4CO2 + 6H2O + 2SO2
Step 3: Calculate mass of ethyl mercaptan
Mass = Volume * density
Mass ethyl mercaptan = 3.15 mL * 0.84 g/mL
Mass ethyl mercaptan = 2.646 grams
Step 4: Calculate moles ethyl mercaptan
Moles = mass / molar mass
Moles ethyl mercaptan = 2.646 grams / 62.13 g/mol
Moles ethyl mercaptan = 0.04259 moles
Step 5: Calculate moles of other products
For 2 moles ethyl mercaptan we need 9 moles O2 to produce 4 moles CO2, 6 moles H2O and 2 moles SO2
For 0.04259 moles we need 0.1917 moles O2 to produce:
2*0.04259 = 0.08518 moles CO2
3*0.04259 = 0.1278 moles H2O
1*0.04259 = 0.04259 moles SO2
Step 6: Calculate mass produced
Mass = moles * molar mass
Mass CO2 = 0.08518 moles * 44.01 g/mol
Mass CO2 = 3.75 grams
Mass H2O = 0.1278 moles * 18.02 g/mol
Mass H2O = 2.30 grams
Mass SO2 = 0.04259 moles * 64.07 g/mol
Mass SO2 = 2.73 grams
c. rubidium sulfate
2. Write balanced molecular equation, complete ionic equation, and net ionic
equations for the mixing of the following solutions. Show states. If no reaction
occurs, show the ionic equation. (8 marks)
a.
NaNO3 +
Ag2SO4 → AgNO3 + Na2SO4
What Is The Name For CH3(CH2)4CH3
Answer:
hexane
I hope it's helps you
Diethyl ether (C2H5 )2O vaporizes at room temperature. If the vapor exerts a pressure of 233 mm Hg in a flask at 25 °C, what is the density of the vapor?
Answer: The density of the given vapor is 0.939 g/L.
Explanation:
Given: Pressure = 233 mm Hg (1 mm Hg = 0.00131579 atm) = 0.31 atm
Temperature = [tex]25^{o}C[/tex] = (25 + 273) K = 298 K
According to the ideal gas equation,
[tex]PV = \frac{m}{M}RT[/tex]
where,
P = pressure
V = volume
m = mass
M = molar mass
R = gas constant = 0.0821 L atm/mol K
T = temperature
This formula can be re-written as follows.
[tex]PM = \frac{m}{V}RT[/tex] (where, [tex]Density = \frac{mass (m)}{Volume (V)}[/tex] )
Hence, formula used to calculate density of diethy ether (molar mass = 74.12 g/mol) vapor is as follows.
[tex]d = \frac{PM}{RT}[/tex]
Substitute values into the above formula as follows.
[tex]d = \frac{PM}{RT}\\= \frac{0.31 atm \times 74.12 g/mol}{0.0821 L atm/mol K \times 298 K}\\= \frac{22.9772}{24.4658}\\= 0.939 g/L[/tex]
Thus, we can conclude that the density of the given vapor is 0.939 g/L.
How do you predict the geometrical shape of NH3 on VSEPR model
Answer:
NH3 Ammonia
Explanation:
Ammonia has 4 regions of electron density around the central nitrogen atom (3 bonds and one lone pair). These are arranged in a tetrahedral shape. The resulting molecular shape is trigonal pyramidal with H-N-H angles of 106.7°.
Calculate the Experimental Molar Volume in L/mol of the Hydrogen gas, H2, if the volume of H2 at STP is 52.8 mL and the mass of Magnesium metal, Mg, used in the experiment is 0.055 g.
Answer:
The Experimental Molar Volume in L/mol of the Hydrogen gas=23.36L/mol
Explanation:
We are given that
Volume of H2 at STP=52.8mL
Mass of magnesium metal ,M(Mg)=0.055g
We have to find the Experimental Molar Volume in L/mol of the Hydrogen gas.
Molar mass of Mg=24.305 g/mol
Number of moles=[tex]\frac{given\;mass}{molar\;mass}[/tex]
Using the formula
Number of moles of Mg=[tex]\frac{0.055}{24.305}[/tex]moles
Number of moles of Mg=0.00226moles
Number of moles of Mg=Number of moles of H2
Number of moles of H2=0.00226moles
Molar volume of Hydrogen gas (H2)=[tex]\frac{volume\;at\;STP}{No\;of\;moles\;H_2}[/tex]
Molar volume of Hydrogen gas (H2)=[tex]\frac{52.8}{0.00226}mL/mol[/tex]
Molar volume of Hydrogen gas (H2)=[tex]\frac{52.8}{0.00226}\times 10^{-3}L/mol[/tex]
[tex]1L=1000mL[/tex]
Molar volume of Hydrogen gas (H2)=23.36L/mol
Hence, the Experimental Molar Volume in L/mol of the Hydrogen gas=23.36L/mol
Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid, HCl(aq) , as described by the chemical equation
MnO2(s)+4HCl(aq)⟶MnCl2(aq)+2H2O(l)+Cl2(g)
How much MnO2(s) should be added to excess HCl(aq) to obtain 185 mL Cl2(g) at 25 °C and 745 Torr ?
mass of MnO2:
Answer:
0.605 g
Explanation:
MnO₂(s) + 4HCl(aq) ⟶ MnCl₂(aq) + 2H₂O(l) + Cl₂(g)First we calculate how many Cl₂ moles need to be produced, using the PV=nRT formula:
P = 745 Torr ⇒ 745 / 760 = 0.980 atmV = 185 mL ⇒ 185 / 1000 = 0.185 Ln = ?R = 0.082 atm·L·mol⁻¹·K⁻¹T = 25 °C ⇒ 25 + 273.16 = 298.16 KInputting the data:
0.980 atm * 0.185 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 298.16 Kn = 0.00696 molThen we convert 0.00696 moles of Cl₂ to MnO₂ moles:
0.00696 mol Cl₂ * [tex]\frac{1molMnO_2}{1molCl_2}[/tex] = 0.00696 mol MnO₂Finally we convert 0.00696 moles of MnO₂ to grams, using its molar mass:
0.00696 mol MnO₂ * 86.94 g/mol = 0.605 g2- A 0.60 sample an unknown organic acid found in muscle cells is burned in air and found to contain 0.24 grams of carbon, 0.040 grams of hydrogen, with the rest being oxygen. If the molecular weight of the substance is 90 grams/n, what is the molecular formula
Answer:
C₃H₆O₃
Explanation:
To solve this question we need to find, as first, the moles of each atom in order to find empirical formula (Simplest whole-number ratio of atoms present in a molecule).
With the molar mass of the substance and the empirical formula we can find the molecular formula as follows:
Moles C -Molar mass:12.0g/mol-
0.24g * (1mol/12.0g) = 0.020 moles C
Moles H = Mass H because molar mass = 1g/mol:
0.040 moles H
Moles O -Molar mass: 16g/mol-
Mass O: 0.60g - 0.24g - 0.040g = 0.32g O
0.32g O * (1mol/16g) = 0.020 moles O
Ratio of atoms (Dividing in moles of C: Lower number of moles):
C = 0.020 moles C / 0.020 moles C = 1
H = 0.040 moles H / 0.020 moles C = 2
O = 0.020 moles O / 0.020 moles C = 1
Empirical formula:
CH₂O.
Molar mass CH2O:
12g/mol + 2*1g/mol + 16g/mol = 30g/mol
As molecular formula has a molar mass 3 times higher than empirical formula, the molecular formula is 3 times empirical formula:
C₃H₆O₃The molecular formula of the organic acid would be C3H6O3
Molecular formulaMolecular formula = [empirical formula]n
Where n = molar mass/mass of empirical formula
Empirical formula
C = 0.24/12 = 0.02
H = 0.040/1 = 0.04
O = 0.6 - (0.24+0.04) = 0.32/16 = 0.02
Divide by the smallest
C = 1
H = 2
O = 1
Empirical formula = CH2O
Empirical formula mass = 12 + 2 + 16 = 30
n = 90/30 = 3
Molecular formula = [CH2O]3
= C3H6O3
More on molecular formula can be found here: https://brainly.com/question/1247523
A student measured the gram weight of a metal object to be 5.88g. According to the supplier the object weighs 5.97g. What is the error in the student's measurement?
A. -0.09
B. +0.09
Answer:
–0.09
Explanation:
From the question given above, the following data were obtained:
Measured value = 5.88 g
Actual value = 5.97 g
Error =?
The error in the student's measurement can be obtained as follow:
Error = Measured value – Actual value
Error = 5.88 – 5.97
Error = –0.09
Therefore, the error in the student's measurement is –0.09
Silver has two naturally occurring isotopes with the following isotopic masses: 10747Ag 10947Ag 106.90509 108.9047 The average atomic mass of silver is 107.8682 amu. The fractional abundance of the lighter of the two isotopes is ________.
Answer: The fractional abundance of lighter isotope is 0.518
Explanation:
Average atomic weight is the sum of the masses of the individual isotopes each multiplied by its fractional abundance. The equation used is:[tex]\text{Average atomic weight}=\sum_{i=1}^{n}\text{(Atomic mass of isotope)}_i\times \text{(Fractional abundance)}_i[/tex] ......(1)
Let the fractional abundance of Ag-107 isotope be 'x'
For Ag-107 isotope:Atomic mass = 106.90509 amu
Fractional abundance = x
For Ag-109 isotope:Atomic mass = 108.9047 amu
Fractional abundance = (1 - x)
Average atomic mass of silver = 107.8682 amu
Plugging values in equation 1:
[tex]107.8682=(106.90509 \times x) + (108.9047 \times (1-x))\\\\107.8682=106.90509x+108.9047-108.9047x\\\\1.99961x=1.0365\\\\x=0.518[/tex]
Fractional abundance of Ag-107 isotope (lighter) = x = 0.518
Hence, the fractional abundance of lighter isotope is 0.518
formula of
Al³⁺ and SO₄²⁻
Answer:
The formula of Al³⁺ and SO₄²⁻ is aluminum sulfate.
Explanation:
The formula for aluminum sulfate is Al₂(SO₄)₃. If we say in terms of ions. The ions are Al³⁺. It is a positive ion or the cation. Other ion is SO₄²⁻. It is sulfate ion. It is anion.
Aluminum sulphate is used in water purification and as a mordant in dyeing and printing textiles.
Hence, the formula of Al³⁺ and SO₄²⁻ is aluminum sulfate.
a laser emits light with a frequency of 4.69 x 10 to the 14th power s - 1 calculate the wavelength of this light.
Answer:
6.40x10^-7
Explanation:
answer with work is attached.
Is it possible for number of moles to be less than one?
Answer:
yes is very possible to be
If I have 21 liters of gas held at a pressure of 78 atm and a temperature of 900. K, what will be the volume of the gas if I decrease the pressure to 52 atm and decrease the temperature to 750 K?
SHOW YOUR WORK
Answer:
txyxyc8uviciycuyc
Explanation:
gghvj ucy7cyvyfy
What is oxygen's half-equation?
answer; 1/ 20_2[2-] +2e - ->0.
The number 0.0007270 is larger than the number 5.7 × 10–3.
Answer:
Yes
Explanation:
=5.7×10-3
=5.7×7=39.9
=0.0007270 greater than 5.7× 10-3
OR
=5.7×10
=57-3
=54
So it is greater number
A 15.4 mL aliquot of 0.204 MH3PO4(aq) is to be titrated with 0.17 MNaOH(aq). What volume (mL) of base will it take to reach the equivalence point?
Answer:
55.44L of the 0.17M NaOH are required
Explanation:
Phosphoric acid, H3PO4, reacts with NaOH as follows:
H3PO4 + 3NaOH → Na3PO4 + 3H2O
Where 1 mole of H3PO4 reacts with 3 moles of NaOH
To solve this question we need to find the moles of H3PO4 in the aliquot. Using the balanced equation we can find the moles of NaOH and its volume with the concentration (0.17M) as follows:
Moles H3PO4:
15.4mL = 0.0154L * (0.204mol/L) = 0.00314 moles H3PO4
Moles NaOH:
0.00314 moles H3PO4 * (3mol NaOH / 1mol H3PO4) = 0.009425moles NaOH
Volume NaOH:
0.009425moles NaOH * (1L/0.17moles NaOH) = 0.05544L 0.17M NaOH =
55.44L of the 0.17M NaOH are requiredWhich equation obeys the law of conservation of
mass?
Answer:2C4H10+2C12+12O2 4CO2+CC14+H20
Draw 2,3-dichloro octane
Answer:
Hi friend
I hope this image will help you if not I'm sorry
if this help you please mark me as brinalist or vote me.
Thankyou
You titrate 41.27 mL of 0.108 M Ca(OH)2 into 25.00 mL of citric acid (C6H307) (triprotic). What is the balanced equation and the molarity of the acid?
Select both an equation and a molarity.
CoH2O2 (aq) + Ca(OH)2 (aq) + H20 (1) + Ca(C6H507)2 (aq)
2C6H807 (aq) + 3Ca(OH)2 (aq) + 6H20 (1) + Ca3(C6H507)2 (aq)
C6H2O7 (aq) + Ca(OH)2 (aq) + H20 (1) + CaC6H50% (aq)
3C6H307 (aq) + 2Ca(OH)2 (aq) + 6H20 (1) + Caz(C6H507)2 (aq)
0.267 M
0.178 M
0.0654 M
0.119 M
Answer:
3Ca(OH)2 + 2C6H8O7 → 6H2O + Ca3(C6H5O7)2
And 0.119M is the concentration of the citric acid.
Explanation:
In an acid-base reaction, the proton H+ and the hydroxil ion OH- reacts producing water. The ions of the acid and base (C6H5O7³⁻ and Ca²⁺ ions produce the respective salt) as follows:
Ca(OH)2 + C6H8O7 → H2O + Ca3(C6H5O7)2
To balance the Calcium ions:
3Ca(OH)2 + C6H8O7 → H2O + Ca3(C6H5O7)2
To balance the C6H5O7³⁻ ions:
3Ca(OH)2 + 2C6H8O7 → H2O + Ca3(C6H5O7)2
And to balance the oxygens of water:
3Ca(OH)2 + 2C6H8O7 → 6H2O + Ca3(C6H5O7)2
And this is the balanced reaction.
The moles of Ca(OH)2 that reacts are:
41.27mL = 0.04127L * (0.108mol/L) = 0.004457 moles Ca(OH)2
Moles of citric acid:
0.004457 moles Ca(OH)2 * (2mol C6H8O7 / 3mol Ca(OH)2) = 0.002971 moles C6H8O7
In 25.00mL = 0.02500L:
0.002971 moles C6H8O7 / 0.0250L =
0.119M
Where do most organisms that live in water get oxygen from?Give a word, not a formula.
Answer:
Surely with water
Ok, but how?
There are many Hydrogen Bond between H2O moleculs and london bonds. When fishes take water with their gill,they are broke london bonds. And they can take their needs, Oxygen. Only this.
Good luck :D
The energy levels of hydrogenlike one-electron ions of atomic number Z differ from those of hydrogen by a factor of Z^2. Predict the wavelength of the 2s--->1s transition in He+.
Answer:
[tex]\mathbf{\lambda \simeq 3.039 \times 10^{-8} \ m}[/tex]
Explanation:
For a hydrogen-like atom, the spectral line wavelength can be computed by using the formula:
[tex]\bar v = Z^2 R_H \Big(\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2}\Big)[/tex]
where:
emitted radiation of the wavenumber [tex]\bar v[/tex] = ???
atomic no of helium Z = 2
Rydberg's constant [tex]R_H = 1.097*10^7 \ m^{-1}[/tex]
the initial energy of the principal quantum [tex]n_1[/tex] = 2
the initial energy of the principal quantum [tex]n_1[/tex] = 2
Now, the emitted radiation of the wavenumber can be computed as:
[tex]\bar v = (2)^2 (1.097*10^7 \ m^{-1} ) \Big(\dfrac{1}{1^2}-\dfrac{1}{2^2}\Big)[/tex]
[tex]\bar v = 3.291 \times 10^ 7/m[/tex]
Now, the wavelength for the transition can be computed by using the relation between the wavelength λ and the emitted radiation of the wavenumber [tex]\bar v[/tex], which is:
[tex]\bar v = \dfrac{1}{\lambda}[/tex]
[tex]\lambda = \dfrac{1}{\bar v}[/tex]
[tex]\lambda = \dfrac{1}{3.291 \times 10^{7}}\times \dfrac{m}{1}[/tex]
[tex]\mathbf{\lambda =3.03859 \times 10^{-8} \ m}[/tex]
[tex]\mathbf{\lambda \simeq 3.039 \times 10^{-8} \ m}[/tex]
A sample of calcium fluoride was decomposed into the constituent elements. Write a balanced chemical equation for the decomposition reaction. If the sample produced 294 mg of calcium, how many g of fluorine were formed
Answer:
A sample of calcium fluoride was decomposed into the constituent elements. Write a balanced chemical equation for the decomposition reaction. If the sample produced 294 mg of calcium, how many g of fluorine was formed
Explanation:
The balanced chemical equation for the decomposition of calcium fluoride is shown below:
[tex]CaF_2(s)->Ca(s)+F_2(g)[/tex]
The sample produced 294 g of calcium then, how many grams of fluorine is formed?
From the balanced chemical equation,
1 mol of CaF2 forms 1mol of calcium and 1 mol of fluorine.
That is:
40g of calcium and 38.0 g of fluorine are formed.
then,
If 294 g of calcium is formed then how many grams of fluorine is formed?
[tex]294g Ca * 38g F2 / 40g Ca\\=279.3 g F_2[/tex]
Hence, 279.3 g of fluorine will be formed.
