Answer: 5L O2 x 1 mol O2/ 22.4L O2 x 2 mol H20/ 1 mol O2 x 22.4L H20/ 1 mol H20 = 10 L H2O
Explanation:
For future reference though, since its at STP that means that the coefficient are in proportion. Since oxygen has a coefficient of 1 and water has a coefficient of 2 for every 1 liter of oxygen there is 2 liters of water. Hence you started with 5 liters and ended with 10
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
What dosage in grams per kilogram of body weight does a 134 lb woman receive if she takes two 275 mg tablets of penicillin?
How many 150. mg tablets should a 31 lb child take to receive the same dosage?
Answer:
The dosage is 0.0906 g/kgThe child should take 8.5 tablets to receive the same dosageExplanation:
First we convert 134 lb into kg:
1 lb = 0.453 kg134 lb * [tex]\frac{0.453kg}{1lb}[/tex] = 60.702 kgThen we convert 275 mg into g:
1000 mg = 1 g275 mg * [tex]\frac{1g}{1000mg}[/tex] = 0.275 gNow we can calculate the dosage in grams per kilogram of body weight, keeping in mind that two tablets are consumed:
(0.275 g) * 2 / 60.702 kg = 0.0906 g/kgAs for the second part, first we convert 31 lb into kg:
31 lb * [tex]\frac{0.453 kg}{1lb}[/tex] = 14.043 kgNow we calculate how many penicillin grams should be consumed:
14.043 kg * 0.0906 g/kg = 1.27 gWe convert 1.27 g of penicilin into mg:
1.27 g * 1000 = 1270 mgFinally we calculate how many 150 mg tables should be taken:
1270 mg / 150 mg = 8.5 tabletsHow will the temperature change if you increase the mass of the gas molecules?
The temperature of a gas sample rises, the molecules quicken, and as a result, the root mean square molecular speed rises.
What is meant by temperature?Degree of warmth or coolness determined by a specific scale
The temperature of a gas sample rises, the molecules quicken, and as a result, the root mean square molecular speed rises.
In a gas, the speed of the molecules is inversely related to the molar mass of the gas and inversely proportional to temperature.
The rate of diffusion is inversely related to the square root of the gas's molar mass, according Graham's law of Diffusion.
To learn more about temperature, refer to the below link:
https://brainly.com/question/13122916
# SPJ2
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.
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!
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.
how real gases differ from ideal gases?
An ideal gas is one that follows the gas laws at all conditions of temperature and pressure. To do so, the gas would need to completely abide by the kinetic-molecular theory. On the other hand, a real gas is a gas that does not behave according to the assumptions of the kinetic-molecular theory.
Furthermore, the particles of an ideal gas are extremely small and have a mass equivalent to practically zero. Ideal gas particles also have no volume.
An example of a real gas is helium, oxygen, and nitrogen.
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
What is the compound of c4?
Answer:
c4 is an explosive..
contains RDX, DOS, DOA, and PIB.
Explanation:
Palmitic acid, derived from palm oil, is one of the most common fatty acids found in butter, cheese, milk, and meat.
a. True
b. False
Answer:
a. True
Explanation:
Palmitic acid is the acid found in animals and plants. It is the saturated acid that contains fats. It is extracted from palm oil and from fat sources like butter, cheese, milk and meat. It is added in milk with low fat content to add Vitamin A in it. The fat content of palmitic acid is very high. excessive consumption of palmitic acid leads to heart diseases and life risks.
Based on the molar masses. how can you tell that an equation is balanced
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 molesHelp please asap will mark brainliest!
Answer:
28.466256
Explanation:
Consider the oxidation of sodium metal to sodium oxide described by the balanced equation:
4 Na + O2 → 2 Na2O. What is the theoretical yield of Na2O in grams from 9.0 mol of O2?
show steps plz
Answer:
1116g
Explanation:
We'll convert moles O2 -> moles of Na2O -> grams of Na2O.
Based on our balanced equation, we have 1 mole of O2 for every 2 moles of Na2O. This is our mole to mole ratio.
9 mol O2 x [tex]\frac{2 mol Na2O}{1 mol O2}[/tex] = 18 mol Na2O
We can convert mols -> grams using the molar mass of Na2O- 62g.
18 mol Na2O x [tex]\frac{62g Na2O}{1 mol}[/tex] = 1116g
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]
How many liters are there in 415.4 grams of carbon trioxide?
Answer:
155.1 L
Explanation:
Step 1: Given data
Mass of CO₃: 415.4 g
Step 2: Calculate the moles corresponding to 415.4 g of CO₃
The molar mass of CO₃ is 60.01 g/mol.
415.4 g × 1 mol/60.01 g = 6.922 mol
Step 3: Calculate the volume occupied by 6.922 moles of CO₃
The volume of a gas depends on conditions such as Temperature and Pressure. Since the conditions are not specified, we may assume that the gas is at Standard Pressure and Temperature (1 atm and 273.15 K). At STP, 1 mole of a gas occupies 22.41 L.
6.922 mol × 22.41 L/1 mol = 155.1 L
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.
The more energy that particles have, the ___ they move.
The more energy that particles have, the more they move.
A substance in which light can travel through such as air,glass, or water
Answer:
The correct answer is - transparent medium.
Explanation:
A transparent substance or medium is the substance that allows light to pass through it. Light moves through these substances as they do not absorb the light and do not reflect too.
The example of such substances is glass, air or water. These substances allow light to pass through them.
Thus, The correct answer is - transparent medium.
The reaction between dihydrogen sulfide and sulfur dioxide is outlined below. 2 H2S(g) S02(g) -Y 3 S(s) 2 H20(g) a. Identify the limiting reactant when 3.89 g of dihydrogen sulfide react with 4.11 g of sulfur dioxide. Justify your answer. d. Based on your answer from part (a), determine the maximum mass of sulfur that can be produced in this reaction. c. Ifthe actual yield of sulfur is found to be 4.89 g, find the percent yield in this reaction.
Answer:
Explanation:
2 H₂S(g) +S0₂(g) = 3 S(s) + 2H₂0(g)
2 x 34 g 64 g 3 x 32 g
68 g of H₂S reacts with 64 g of S0₂
3.89 g of H₂S reacts with 64 x 3 .89 / 68 g of S0₂
3.89 g of H₂S reacts with 3.66 g of S0₂
S0₂ given is 4.11 g , so it is in excess .
Hence H₂S is limiting reagent .
68 g of H₂S reacts with S0₂ to give 96 g of Sulphur
3.89 g of H₂S reacts with S0₂ to give 96 x 3.89 / 68 g of Sulphur
3.89 g of H₂S reacts with S0₂ to give 96 x 3.89 / 68 g of Sulphur
5.49 g of Sulphur is produced .
Actual yield is 4.89
percentage yield = 4.89 x 100 / 5.49
= 89 % .
Considering the reaction stoichiometry and the definition of percent yield:
a. H₂S will be the limiting reagent.
b. the maximum mass of sulfur that can be produced in the reaction is 5.49 grams.
c. the percent yield of the reaction is 89%.
The balanced reaction is:
2 H₂S(g) + SO₂(g) → 3 S(s) + 2 H₂O(g)
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:
H₂S: 2 molesSO₂: 1 moleS: 3 moles H₂O: 2 molesThe molar mass of each compound is:
H₂S: 34 g/moleSO₂: 64 g/moleS: 32 g/mole H₂O: 18 g/moleSo, by reaction stoichiometry, the following amounts of mass of each compound participate in the reaction:
H₂S: 2 moles× 34 g/mole= 68 gramsSO₂: 1 moles× 64 g/mole= 64 gramsS: 3 moles× 32 g/mole= 96 grams H₂O: 2 moles× 18 g/mole= 36 gramsa. Limiting reagentThe limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.
To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 64 grams of SO₂ reacts with 68 grams of H₂S, 4.11 grams of SO₂ react with how much mass of H₂S?
[tex]mass of H_{2} S=\frac{4.11 grams of SO_{2}x 68 grams of H_{2} S }{64 grams of SO_{2}}[/tex]
mass of H₂S= 4.37 grams
But 4.37 moles of H₂S are not available, 3.89 grams are available. Since you have less mass than you need to react with 4.11 grams of SO₂, H₂S will be the limiting reagent.
b. Maximum mass of sulfur
Using the limiting reagent, you can apply the following rule of three: if by stoichiometry 68 grams of H₂S produce 96 grams of S, 3.89 grams of H₂S will produce how much mass of S?
[tex]mass of S=\frac{96 grams of Sx 3.89 grams of H_{2} S }{68 grams of H_{2}S}[/tex]
mass of S= 5.49 grams
So, the maximum mass of sulfur that can be produced in the reaction is 5.49 grams.
c. Percent yieldThe amount of product that is obtained when all the limiting reagent reacts. This is called the theoretical yield of the reaction. That is, the theoretical yield is the maximum amount of product that can be produced in a reaction.
On the other hand, the actual yield is the amount of product actually obtained from a reaction.
The percent yield determines the efficiency of the reaction, and describes the ratio of the actual yield to the theoretical yield:
[tex]percent yield=\frac{actual yield}{theoretical yield}x100[/tex]
In this case, you know:
actual yield= 4.89 gtheoretical yield= 5.49 gSo, the percent yield can be calculated as:
[tex]percent yield=\frac{4.89 grams}{5.49 grams}x100[/tex]
Solving:
percent yield= 89%
Finally, the percent yield of the reaction is 89%.
Learn more about reaction stoichiometry:
brainly.com/question/16487206?referrer=searchResults brainly.com/question/14446695?referrer=searchResults brainly.com/question/11564309?referrer=searchResults brainly.com/question/4025026?referrer=searchResults brainly.com/question/18650135?referrer=searchResultsCan someone please help me with this
Explanation:
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The Swedish chemist Karl Wilhellm was the first to produce chlorine in the lab
2NaCl + 2H2SO4 + MnO2 -----> Na2SO4 + MnSO4 + H2O + Cl2
If Dr. Wilhellm started with 50.0 g of each reactant, which reactant is the limiting reactant?
Answer:
Explanation:
Remark
Interesting que8stion. You have to figure out how many mols are present in each reactant. Since all periodic tables are different, I'm going to use rounded numbers. If it is too close, I will go further.
NaCl
Na = 23
Cl = 35.5
1 mol = 58.5 grams
given = 50.0 grams
Mols for the reaction = 50/58.5 = 0.855
H2SO4
H2 = 2*1 2
S = 1 * 32 32
O4 = 4*16 64
1 mol = 98 grams
mols present = 50/98 = 0.510
MnO2
Mn = 1 * 55 = 55
O2 = 2*16 = 32
1 mol = 87 grams
mols available = 50/87 = 0.5747
Discussion
Na Cl and H2SO4 both require 2 moles for every mol of Cl2 produces.
H2SO4 has 0.51 mols available for a reaction
NaCl has 0.855 moles available for a reaction
MnO2 has 0.575 moles available for a reaction.
Given those numbers 0.510 mols of H2SO4 will only produce 0.255 mols of chlorine and the rest will be reduced in a similar manner. H2SO4 is the limiting reagent (reactant).
In other words only 0.510 moles of NaCl will be used and 0.855 - 0.510 moles will be left over on the reactants side.
only 0.575 moles of MnO2 will be used and 0.065 moles will be left over.
The oddity in the result shows up because the balance numbers in the equation give a ratio of 2 to 1 for H2SO4 and NaCl The 2 belongs to the reactants and the 1 for the chlorine.
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!
How much energy (kJ) is required to change 0.18 mole of ice (s) at 0 C to water (l) at 0 C?
Answer:25,06 kJ of energy must be added to a 75 g block of ice.
ΔHfusion(H₂O) = 6,01 kJ/mol.
T(H₂O) = 0°C.
m(H₂O) = 75 g.
n(H₂O) = m(H₂O) ÷ M(H₂O).
n(H₂O) = 75 g ÷ 18 g/mol.
n(H₂O) = 4,17 mol.
Q = ΔHfusion(H₂O) · n(H₂O)
Q = 6,01 kJ/mol · 4,17 mol
Q = 25,06 kJ.
Explanation:
Oil is
so it will
dissolve in water.
Answer:
liquid, and of course
Explanation:
Which is an example of health technology?
A. Television
B. Vaccines
C. Light bulbs
D. Swimming pools
Answer:
B
Explanation:
Vaccines prevent illness and disease
Which of the following choices has the ecological levels listed from largest to smallest?
Answer:
the answer is something
Explanation:
What is the oxidation number of iron in FeO?
Answer:
+2
Explanation:
Answer:
2 electrons
Explanation:
In Iron (II) oxide, or Ferrous Oxide, or FeO, the Iron element (Fe) is bonded to the Oxygen, in the oxidation state of "2". This means that the Iron has accepted 2 electrons from the Oxygen.
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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Identify the balanced equation for the following reaction:
SO2(g) + O2(g) → SO3(g)
Answer: The balanced equation for the given reaction is
[tex]2SO_{2}(g) + O_{2}(g) \rightarrow 2SO_{3}(g)[/tex].
Explanation:
A chemical equation which contains same number of atoms on both reactant and product side.
For example, [tex]SO_{2}(g) + O_{2}(g) \rightarrow SO_{3}(g)[/tex]
Here, number of atoms on reactant side are as follows.
S = 1O = 4Number of atoms on product side are as follows.
S = 1O = 3To balance this equation, multiply [tex]SO_{2}[/tex] by 2 on reactant side and multiply [tex]SO_{3}[/tex] by 2. Hence, the equation will be re-written as follows.
[tex]2SO_{2}(g) + O_{2}(g) \rightarrow 2SO_{3}(g)[/tex]
Here, number of atoms on reactant side are as follows.
S = 2O = 6Number of atoms on product side are as follows.
S = 2O = 6Now, there are same number of atoms on both reactant and product side. So, this equation is balanced.
Thus, we can conclude that the balanced equation for the given reaction is [tex]2SO_{2}(g) + O_{2}(g) \rightarrow 2SO_{3}(g)[/tex].