Answer:
The activation energy is the amount of energy needed for a chemical reaction to occur.
Answer:
Activation energy, in chemistry, the minimum amount of energy that is required to activate atoms or molecules to a condition in which they can undergo chemical transformation or physical transport.
2. The number of molecules in 1.0 mole of So2 is the same as the number of molecules in:
(1) 1.0 mole of N2
(2) 2.0 moles of Ne
(3) 0.25 mole of NO2
(4) 0.50 mole of NH3
Please show work and explain!
Answer:
1.0 moles of N2
Explanation:
since
1.0 × avogadro's no# = same answer for SO2 and N2
avogadro's no#= 6.02× 10²³
The number of molecules in 1.0 mole of SO₂ is the same as the number of molecules in (3) 0.25 mole of NO₂.
Option (3) is correct.
Avogadro's number, approximately 6.022 x 10²³, represents the number of molecules or atoms in one mole of any substance. In this case, 1.0 mole of SO₂ contains the same number of molecules as any other substance with 1.0 mole, such as 1.0 mole of N₂ or 2.0 moles of Ne.
However, when comparing different moles of substances, we can use the concept of stoichiometry and molar ratios to determine the number of molecules. Given that 1 mole of SO₂ is equivalent to 1 mole of NO₂, 0.25 mole of NO₂ would also contain the same number of molecules as 1 mole of SO₂. Therefore, the number of molecules in 1.0 mole of SO₂ is the same as the number of molecules in 0.25 mole of NO₂.
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Based on the molar masses. how can you tell that an equation is balanced
where do you think water vapor in air comes from?
Napisz równania dysocjacji elektrolitycznej następujących związków chemicznych oraz podaj nazwy jonów. Załóż, ze wszystkie substancje rozpuszczają się w wodzie:
a)H3PO4
b)FeSO3
c)Ba (OH)2
Answer is in a pho[tex]^{}[/tex]to. I can only uplo[tex]^{}[/tex]ad it to a file host[tex]^{}[/tex]ing service. link below!
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How many Liters are in 98.2 moles of neon?
Answer:
2200 L
Explanation:
There are 22.4 L in 1 mole of neon so if you have 98.2 L, then you have 98.2 mole x (22.4 L/mole) = 2200 L of neon.
Determine the number of moles of CH3Br in 47.5 grams of CH3Br?
Show Work please - NO LINKS
Answer:
0.500 moles
Explanation:
In order to convert grams of any given substance into moles, we need the substance's molar mass:
Molar Mass of CH₃Br = Molar Mass of C + (Molar Mass of H)*3 + Molar Mass of BrWe can find the molar masses of each element in the periodic table:
Molar Mass of CH₃Br = 94.94 g/molNow we can divide the given mass by the molar mass in order to calculate the number of moles:
47.5 g ÷ 94.94 g/mol = 0.500 molesgive same examples of ways that people destroy the plant animals relationship?
What effect would a decrease or increase in barometric pressure have on the boiling point
Answer:
Pressure Affects the Boiling Point
Atmospheric pressure influences the boiling point of water. When atmospheric pressure increases, the boiling point becomes higher, and when atmospheric pressure decreases (as it does when elevation increases), the boiling point becomes lower.
Explanation:
i think it will help you
One way to measure ionization energies is ultraviolet photoelectron spectroscopy (UPS, or just PES), a technique based on the photoelectric effect. In PES, monochromatic light is directed onto a sample, causing electrons to be emitted. The kinetic energy of the emitted electrons is measured. The difference between the energy of the photons and the kinetic energy of the electrons corresponds to the energy needed to remove the electrons (that is, the ionization energy). Suppose that a PES experiment is performed in which mercury vapor is irradiated with ultraviolet light of wavelength 58.4 nm.
(a) What is the energy of a photon of this light in eV?
(b) Write an equation that shows the process corresponding to the first ionization energy of Hg.
(c) The kinetic energy of the emitted electrons is measured to be 10.75 eV. What is the first ionization energy of Hg in kJ/mol?
Answer:
Explanation:
From the given information:
The energy of photons can be determined by using the formula:
[tex]E = \dfrac{hc}{\lambda}[/tex]
where;
planck's constant (h) = [tex]6.63 \times 10^ {-34}[/tex]
speed oflight (c) = [tex]3.0 \times 10^8 \ m/s[/tex]
wavelength λ = 58.4 nm
[tex]E = \dfrac{6.63 \times 10^{-34} \ J.s \times 3.0 \times 10^8 \ m/s}{58.4 \times 10^{-9 } \ m}[/tex]
[tex]E =0.34 \times 10^{-17} \ J[/tex]
[tex]E = 3.40 \times 10^{-18 } \ J[/tex]
To convert the energy of photon to (eV), we have:
[tex]1 eV = 1.602 \times 10^{-19} \ J[/tex]
Hence
[tex]3.40 \times 10^{-18 } \ J = \dfrac{1 eV}{1.602 \times 10^{-19 } \ J }\times 3.40 \times 10^{-18 } \ J[/tex]
[tex]E = 2.12 \times 10 \ eV[/tex]
E = 21.2 eV
b)
The equation that illustrates the process relating to the first ionization is:
[tex]Hg_{(g)} \to Hg^+ _{(g)} + e^-[/tex]
c)
The 1st ionization energy (I.E) of Hg can be calculated as follows:
Recall that:
I.E = Initial energy - Kinetic Energy
I₁ (eV) = 21.2 eV - 10.75 eV
I₁ (eV) = 10.45 eV
Since ;
[tex]1 eV = 1.602 \times 10^{-19} \ J[/tex]
∴
[tex]10.45 \ eV = \dfrac{1.602 \times 10^{-19 } \ J }{ 1 \ eV}\times 10.45 \ eV[/tex]
Hence; the 1st ionization energy of Hg atom = [tex]1.67 \times 10^{-18} \ J[/tex]
[tex]1.67 \times 10^{-21} \ kJ[/tex]
Finally;
[tex]I_1 \ of \ the \ Hg (kJ/mol) = \dfrac{1.67 \times 10^{-21 \ kJ} \times 6.02 \times 10^{23} \ Hg \ atom }{1 \ Kg \ atom }[/tex]
[tex]\mathbf{= 1.005 \times 10^3 \ kJ/mol}[/tex]
The city of Annandale has been directed to upgrade its primary wastewater treatment plant to a secondary treatment plant with sludge recycle that can meet an effluent standard of 11 mg/l BOD5. The following data are available: Flow = 0.15 m3/s, MLSS = 2,000 mg/L. Kinetic parameters: K, = 50 mg/L, Hmax = 3.0 d-, kų = 0.06 d-1, Y = 0.6 Existing plant effluent BOD5 = 84 mg/L. a. Calculate the SRT (Oc) and HRT (0) for the aeration tank. b. Calculate the required volume of the aeration tank. c. Calculate the food to microorganism ratio in the aeration tank. d. Calculate the volumetric loading rate in kg BOD3/m3-d for the aeration tank. e. Calculate the mass and volume of solids wasted each day, when the underflow solids concentration is 12,000 mg/L. 10 A
Determine the mass/mass % of 4.0 g of KOH in 50.0 g of solution
Answer: 8%
Explanation:
Can't you just do 4/50= 8%? If not, just remember that for empirical formula, do percent to grams, and grams to moles, divide by smallest and multiply to whole!
The more energy that particles have, the ___ they move.
The more energy that particles have, the more they move.
1 point
Marlene sits in a roller coaster that is at the top of a 72 m hill. Marlene and
the rollercoaster have a combined mass of 120 kg. Calculate the energy.
829,785 J
311,080J
84,672 J
O 42,336 J
Answer:
84672 J
Explanation:
From the question given above, the following data were obtained:
Height (h) = 72 m
Combined mass (m) = 120 Kg
Acceleration due to gravity (g) = 9.8 m/s²
Energy (E) =?
We can obtain the energy by using the following formula:
E = mgh
Where
E => is the energy.
g => is the acceleration due to gravity
m => is the mass.
h => is the height.
E = 120 × 9.8 × 72
E = 84672 J
Thus, the energy is 84672 J
A body of mass 0.5kg is thrown vertically upwards
from the ground with an Initial velocity of 80.m/s
Calculate
" the potential energy at B 80.0cm above the ground
Here's the solution,
we know,
potential energy = mgh
where,
m = mass = 0.5 kgg = acc due to gravity = 9.8 m/s²h = height = 0.8 m (80 cm)So,
=》
[tex]p.e = 0.5 \times 9.8 \times 0.8[/tex]
=》
[tex]p.e = 3.92 \: \: joules[/tex]
If 20.00 mL of a 0.0090 M solution of (NH4)2S is mixed with 120.00 mL of a
0.0082 M solution of Al(NO3)3, does a precipitate form? The Ksp of Al2S3 is
2.00*10^-7. Included calculated ion product in answer.
Answer:
No, no precipitate is formed.
Explanation:
Hello there!
In this case, since the reaction between ammonium sulfide and aluminum nitrate is:
[tex]3(NH_4)_2S(aq)+2Al(NO_3)_3(aq)\rightarrow Al_2S_3(s)+6NH_4NO_3(aq)[/tex]
In such a way, we can calculate the concentration of aluminum and sulfide ions in the solution as shown below, and considering that the final total volume is 140.00 mL:
[tex][Al^3^+]=\frac{120.00mL*0.0082M}{140.00mL}=0.00703M[/tex]
[tex][S^2^-]=\frac{20.00mL*0.0090M}{140.00mL}=0.00129M[/tex]
In such a way, we can calculate the precipitation quotient by:
[tex]Q=[Al^3^+]^2[S^2^-]^3=(0.00703)^2(0.00129)^3=1.05x10^{-13}[/tex]
Which is smaller than Ksp and meaning that the precipitation does not occur.
Regards!
6. Calculate the mass of each product when 100.0 g of CuCl react according to the reaction
CuCl(aq) → CuCl2(aq) + Cu(s)
What do you notice about the sum of the masses of the products? What concept is being
illustrated here?
Answer:
67.91 g of CuCl2; 32.09 g of Cu.
Explanation:
The two masses add to 100.0 g, the initial amount of starting material, demonstrating the law of conservation of matter.
Hydrogen gas is collected over water in an inverted buret. If the atmospheric pressure is 745 mm Hg, the vapor pressure of water is 18 mm Hg, and a 15.0 cm-high column of water remains in the buret, the pressure of the hydrogen gas is Hydrogen gas is collected over water in an inverted buret. If the atmospheric pressure is 745 mm Hg, the vapor pressure of water is 18 mm Hg, and a 15.0 cm-high column of water remains in the buret, the pressure of the hydrogen gas is:________
a. 745 mm Hg.
b. less than 727 mm Hg.
c. 763 mm.
d. 727 mm Hg.
Answer:
[tex]P_{H_2}=727mmHg[/tex]
Explanation:
Hello there!
In this case, according to the given data, it is possible to infer that the gas mixture lies on the 15.0 cm-high column of water, so that the total pressure or atmospheric pressure is given by:
[tex]P_{atm}=P_{water}+P_{H_2}[/tex]
Thus, since the atmospheric pressure is 745 mmHg and the vapor pressure of water is 18 mmHg, the pressure of hydrogen turns out to be:
[tex]P_{H_2}=P_{atm}-P_{water}\\\\P_{H_2}=745mmHg-18mmHg\\\\P_{H_2}=727mmHg[/tex]
Best regards!
Balance this reaction:
___Naz3(PO4)+ ___K(OH) ---> ___Na(OH) +___K3(PO4)
A local orchard sells bags of red apples by the dozen. The packaging
department of the orchard determines the mass of each dozen batch of
red apples before bagging them. The bag is then labeled with the mass of
the apples. Observe the mass of the dozen red apples shown on the scale.
Based upon this mass, what would the mass of 7 red apples be in
kilograms? Assume that each of the dozen apples on the scale has the
same mass. Answer is rounded to one place after the decimal. 0.5 kg
2.00 kg
Answer:
2.0kg
Explanation:
The mass of 7 red apples in kilograms is to be considered as the 1.16 kilograms.
Calculation of the mass:Since the mass of a dozen apples is 2 kg.
we know that
1 dozen is 12 units
So the mass of 12 apples = 2kg
So mass of 1 apple = 2/12 = 1/6 kg
Now the mass of 7 apple is to be
= 7/6 kg
= 1.16 kg
hence, The mass of 7 red apples in kilograms is to be considered as the 1.16 kilograms.
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7. Which object would be the least dense in a tub of
water?
A. Golf ball
B. Glass marble
C. Rubber ball
D. Beach ball
When 16.35 moles of SI reacts with 11.26 moles of N2, how many moles of SI3N4 are formed
Answer:
5.450 mol Si₃N₄
Explanation:
Step 1: Write the balanced equation
3 Si + 2 N₂ ⇒ Si₃N₄
Step 2: Establish the theoretical molar ratio between the reactants
The theoretical molar ratio of Si to N₂ is 3:2 = 1.5:1.
Step 3: Establish the experimental molar ratio between the reactants
The experimental molar ratio of Si to N₂ is 16.35:11.26 = 1.45:1. Comparing both molar ratios, we can see that Si is the limiting reactant.
Step 4: Calculate the moles of Si₃N₄ produced from 16.35 moles of Si
The molar ratio of Si to Si₃N₄ is 3:1.
16.35 mol Si × 1 mol Si₃N₄/3 mol Si = 5.450 mol Si₃N₄
Can someone please help me with this
Explanation:
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Explain the oxygen cycle. I’m in the 6th grade I need a answer please
Answer:
Oxygen cycle, along with the carbon cycle and nitrogen cycle plays an essential role in the existence of life on the earth. The oxygen cycle is a biological process which helps in maintaining the oxygen level by moving through three main spheres of the earth which are:
Atmosphere Lithosphere BiosphereExplanation:
The atmosphere is the layer of gases presents above the earth’s surface. The sum of Earth’s ecosystems makes a biosphere. Lithosphere is the solid outer section along with the earth’s crust and it is the largest reservoir of oxygen.
Answer:
Oxygen cycle, circulation of oxygen in various forms through nature. ... Free in the air and dissolved in water, oxygen is second only to nitrogen in abundance among uncombined elements in the atmosphere. Plants and animals use oxygen to respire and return it to the air and water as carbon dioxide (CO2).
Explanation:
Write the balanced equation for the equilibrium reaction for the dissociation ofsilver chloride in water, and write the K expression for this reaction. Then create an ICE chart. Since we know the equilibrium concentration of the silver ion, we can solve for Ksp.Does it agree with the literature value
Answer:
See explanation
Explanation:
Hello there!
In this case, since the the concentrations are not given, and not even the Ksp, we can solve this problem by setting up the chemical equation, the equilibrium constant expression and the ICE table only:
[tex]AgCl(s)\rightleftharpoons Ag^+(aq)+Cl^-(aq)[/tex]
Next, the equilibrium expression according to the produced aqueous species as the solid silver chloride is not involved in there:
[tex]Ksp=[Ag^+][Cl^-][/tex]
And therefore, the ICE table, in which x stands for the molar solubility of the silver chloride:
[tex]\ \ \ \ \ \ \ \ \ \ \ \ \ \ AgCl(s)\rightleftharpoons Ag^+(aq)+Cl^-(aq)[/tex]
I - 0 0
C - +x +x
E - x x
Which leads to the following modified equilibrium expression:
[tex]Ksp=x^2[/tex]
Unfortunately, values were not given, and they cannot be arbitrarily assigned or assumed.
Regards!
You have a cold gas of atoms, and you observe that if you shine light consisting of photons with energy 10 eV through the gas, some free electrons are observed, implying that a photon of this energy is able to ionize an atom in the gas. (a) If you find that the emitted electrons from the gas have a kinetic energy of 1 eV, what is the ionization energy of the cold atom
Answer:
Ionization Energy = 9 eV
Explanation:
If we apply the law of conservation of energy to the given situation, we will get the following equation:
[tex]Energy\ of\ Photon = Ionization\ Energy + Kinetic\ Energy\ of\ Electron\\[/tex]
where,
Energy of Photon = 10 eV
Ionization Energy = ?
Kinetic Energy of Electrons = 1 eV
Therefore,
[tex]10\ eV = Ionization\ Energy + 1\ eV\\Ionization\ Energy = 10\ eV - 1\ eV[/tex]
Ionization Energy = 9 eV
is scandium a transition metal?
Answer:no
Explanation:
Answer:
Scandium is a transition metal
Explanation:
What is the compound of c4?
Answer:
c4 is an explosive..
contains RDX, DOS, DOA, and PIB.
Explanation:
You are managing a city that needs to upgrade its disinfection basin at your 40 MGD surface water drinking water treatment plant. You would like to use chlorine (Cl2) as your disinfectant and you need to achieve a 4-log removal of E. coli. You are deciding between a traditional 750,000 gallon PFR contact basin (serpentine flow) and a newer system which contains three 150,000 gallon CSTRs in series, each receiving an equal injection of Cl2. Your final decision is going to be based on which system requires the least amount of Cl2 to achieve a 4-log removal of E. coli.
Required:
What is the amount of Cl2 required to operate each system (please answer in units of kg Cl/day)?
Solution :
According to Chick's law
[tex]$\frac{N_t}{N_0}=e^{-k'C^n t}$[/tex]
where, t = contact time
c = concentration of disinfectant
k' = lethality coefficient = 4.71
n = dilution coefficient = 1
4 log removal = % removal = 99.99
[tex]$\frac{N_t}{N_0}=\frac{\text{bacteria remaining}}{\text{bacteria initailly present}}$[/tex]
= 1 - R
= 1 - 0.9999
Now for plug flow reactor contact time,
[tex]$\tau =\frac{V}{Q} =\frac{75000}{40 \times 10^6}$[/tex]
= 0.01875 days
= 27 minutes
For CSTR, [tex]$\tau =\frac{V}{Q} =\frac{150000}{40 \times 10^6}$[/tex]
[tex]$=3.75 \times 10^{-3}$[/tex] days
= 5.4 minute
There are 3 reactors, hence total contact time = 3 x 5.4
= 16.2 minute
Or [tex]$\frac{N_t}{N_0}=e^{-k'C^n t}$[/tex]
or [tex]$(1-0.9999)=e^{-4.71 \times C \times t}$[/tex]
∴ C x t = 1.955
For PFR, [tex]$t_1 = 27 $[/tex] min
∴ C [tex]$=\frac{1.955}{27}$[/tex] = 0.072 mg/L
For CSIR, [tex]$t_2=16.2$[/tex] min
[tex]$C=\frac{1.955}{16.2} = 0.1206$[/tex] mg/L
∴ Chlorine required for PFR in kg/day
[tex]$=\frac{0.072 \times 40 \times 10^6 \times 3.785}{10^6}$[/tex] (1 gallon = 3.785 L)
= 18.25 kg/day
Therefore we should go for PFR system.
Rubbing alcohol and vegetable oil soluble on insoluble?
Use the following balanced equation to answer the question: 2 ZnS + 3 O2 →2 ZnO + 2 SO2 When 56.25 grams of ZnS react, how many moles of O2 are produced?
Answer:
[tex]n_{O_2}=0.866molSO_2\\\\n_{SO_2}=0.577molSO_2[/tex]
Explanation:
Hello there!
In this case, according to the given balanced chemical reaction by which ZnS reacts with O2, it is possible to calculate the moles of the latter that are consumed, not produced, according the 2:3 mole ratio between them and the following stoichiometric set up:
[tex]n_{O_2}=56.25gZnS*\frac{1molZnS}{97.47gZnS}*\frac{3molO_2}{2molZnS} \\\\n_{O_2}=0.866molO_2[/tex]
But also, we can compute the moles of SO2 that are produced via the the 2:2 mole ratio of ZnS to SO2:
[tex]n_{SO_2}=56.25gZnS*\frac{1molZnS}{97.47gZnS}*\frac{2molSO_2}{2molZnS} \\\\n_{SO_2}=0.577molSO_2[/tex]
Regards!