Answer:
Answer:
2.77 × 10²³ Br Atoms
Solution:
Data Given:
Mass of CH₂Br₂ = 39.9 g
M.Mass of CH₂Br₂ = 173.83 g.mol⁻¹
Step 1: Calculate Moles of CH₂Br₂ as,
Moles = Mass ÷ M.Mass
Putting values,
Moles = 39.9 g ÷ 173.83 g.mol⁻¹
Moles = 0.23 mol
Step 2: Calculate number of CH₂Br₂ Molecules,
As 1 mole of any substance contains 6.022 × 10²³ particles (Avogadro's Number) then the relation for Moles and Number of CH₂Br₂ Molecules can be written as,
Moles = Number of CH₂Br₂ Molecules ÷ 6.022 × 10²³ Molecules.mol⁻¹
Solving for Number of CH₂Br₂ Molecules,
Number of CH₂Br₂ Molecules = Moles × 6.022 × 10²³ Molecules.mol⁻¹
Putting value of moles,
Number of CH₂Br₂ Molecules = 0.23 mol × 6.022 × 10²³ Atoms.mol⁻¹
Number of CH₂Br₂ Molecules = 1.38 × 10²³ CH₂Br₂ Molecules
Step 3: Calculate Number of Br Atoms:
As,
1 Molecule of CH₂Br₂ contains = 2 Atoms of Br
So,
1.38 × 10²³ Molecules of CH₂Br₂ will contain = X Atoms of Br
Solving for X,
X = (1.38 × 10²³ CH₂Br₂ × 2 Br) ÷ 1 CH₂Br₂
X = 2.77 × 10²³ Br Atoms
Explanation:
HELP FAST 100 PTS Calculate the amount of heat needed to raise the temperature of 96 g of water vapor from 124 °C to 158 °C. must provide explanation
Answer:
[tex]\huge\boxed{\sf Q = 13.7\ Joules}[/tex]
Explanation:
Given Data:
Mass = m = 96 g = 0.096 kg
[tex]T_1[/tex] = 124 °C
[tex]T_2[/tex] = 158 °C
Change in Temp. = ΔT = 158 - 124 = 34 °C
Specific Heat Constant = c = 4.186 J/g °C
Required:
Specific Heat Capacity = Q = ?
Formula:
Q = mcΔT
Solution:
Q = (0.096)(4.186)(34)
Q = 13.7 Joules
[tex]\rule[225]{225}{2}[/tex]
Hope this helped!
~AH1807Answer:
[tex]\Large \boxed{\sf 13600 \ J}[/tex]
Explanation:
Use formula
[tex]\displaystyle \sf Heat \ (J)=mass \ (g) \times specific \ heat \ capacity \ (Jg^{-1}\°C^{-1}) \times change \ in \ temperature \ (\°C)[/tex]
Specific heat capacity of water is 4.18 J/(g °C)
Substitute the values in formula and evaluate
[tex]\displaystyle \sf Heat \ (J)=96 \ g \times 4.18 \ Jg^{-1}\°C^{-1} \times (158\°C-124 \°C)[/tex]
[tex]\displaystyle Q=96 \times 4.18 \times (158-124 )=13643.52[/tex]
Consider the following equilibrium
N204(9) - 2NO2(9) Keq = 5.85 x 10-3
Which statement about this system is true?
If the equilibrium concentration of NO2 is 1.78 x 10-2 M, the equilibrium concentration of N2O4 is
The equilibrium concentration of N₂O₄, given that the concentration of NO₂ is 1.78×10⁻² is 5.42×10⁻²
Data obtained from the question N₂O₄ <=> 2NO₂Equilibrium constant (Keq) = 5.85×10⁻³ Equilibrium concentration of NO₂ = 1.78×10⁻²Equilibrium concentration of N₂O₄ =?How to determine the equilibrium concentration of N₂O₄Keq = [Product] / [Reactant]
Keq = [NO₂]² / [N₂O₄]
5.85×10⁻³ = [1.78×10⁻²]² / [N₂O₄]
Cross multiply
5.85×10⁻³ × [N₂O₄] = [1.78×10⁻²]²
Divide both sides by 5.85×10⁻³
[N₂O₄] = [1.78×10⁻²]² / 5.85×10⁻³
[N₂O₄] = 5.42×10⁻²
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Answer:
First answer is "The equilibrium lies to the left", and the second one is 5.42 x 10^-2 M.
Explanation:
Hope i helped yall :D
A sample of gas has an initial volume of 30.8L AND an initial temperature of -67 degree Celcius. What will be the temperature of the gas if the volume is 21.0L?
Answer:
–132.55 °C
Explanation:
From the question given above, the following data were obtained:
Initial volume (V₁) = 30.8 L
Initial temperature (T₁) = –67 °C
Final volume (V₂) = 21.0 L
Final temperature (T₂) =?
Next, we shall convert –67 °C to Kelvin temperature. This can be obtained as follow:
T(K) = T(°C) + 273
Initial temperature (T₁) = –67 °C
Initial temperature (T₁) = –67 °C + 273
Initial temperature (T₁) = 206 K
Next, we shall determine the final temperature of the gas. This can be obtained as follow:
Initial volume (V₁) = 30.8 L
Initial temperature (T₁) = 206 K
Final volume (V₂) = 21.0 L
Final temperature (T₂) =?
V₁/T₁ = V₂/T₂
30.8 / 206 = 21 / T₂
Cross multiply
30.8 × T₂ = 206 × 21
30.8 × T₂ = 4326
Divide both side by 30.8
T₂ = 4326 / 30.8
T₂ = 140.45 K
Finally, we shall convert 140.45 K to celsius temperature. This can be obtained as follow:
T(°C) = T(K) – 273
T₂ = 140.45 K
T₂ = 140.45 K – 273
T₂ = –132.55 °C
Thus, the new temperature of the gas is –132.55 °C
A sample of calcium fluoride was decomposed into the constituent elements. If the sample produced 294 mg of calcium, how many g of fluorine were formed?
Answer:
Approximately [tex]279\; \rm mg[/tex] of fluorine would be produced.
Explanation:
Look up the relative atomic mass of these calcium and fluorine on a modern periodic table:
[tex]\rm Ca[/tex]: [tex]40.078[/tex].[tex]\rm F[/tex]: [tex]18.998[/tex].In other words, the mass of each mole of calcium atoms is (approximately) [tex]40.078\; \rm g[/tex]. Similarly, the mass of each mole of fluorine atoms is (approximately) [tex]18.998\; \rm g[/tex].
Calculate the number of moles of calcium atom that were produced:
[tex]\begin{aligned} &n({\rm Ca}) \\ &= \frac{m({\rm Ca})}{M({\rm Ca})} \\ &= \frac{294\; \rm mg}{40.078\; \rm g \cdot mol^{-1}} \\ &= \frac{0.294\; \rm mg}{40.08 \; \rm g \cdot mol^{-1}}\\ & \approx 7.3357 \times 10^{-3}\; \rm mol\end{aligned}[/tex].
Every formula unit of calcium fluoride, [tex]\rm CaF_2[/tex], contains twice as many fluorine atoms as calcium atoms.
Hence, if the sample contained approximately [tex]7.33570 \times 10^{-3}\; \rm mol[/tex] calcium atoms, it would contain twice as many fluorine atoms:
[tex]\begin{aligned}n({\rm F}) &= 2\, n({\rm Ca}) \\ &\approx 2 \times 7.33570\times 10^{-3}\; \rm mol \\ &\approx 1.4671 \times 10^{-2}\; \rm mol\end{aligned}[/tex].
Calculate the mass of that many fluorine atoms:
[tex]\begin{aligned}& m({\rm F}) \\ &= n({\rm F}) \cdot M({\rm F}) \\ &\approx 1.4671 \times 10^{-2}\; \rm mol \times 18.998\; \rm g \cdot mol^{-1} \\ &\approx 278.72 \times 10^{-3}\; \rm g \\&\approx 279\; \rm mg\end{aligned}[/tex].
279 mg of Fluorine was formed in the decomposition reaction
The equation of the reaction is;
CaF2 -----> Ca + 2F
Mass of calcium formed = 294 mg
Number of moles of Ca in 294 mg = 294 × [tex]10^-3[/tex] g/40 g/mol = 0.00735 moles
From the equation, the products are in ration 1 : 2
Since 1 mole of Ca is to 2 moles of F
0.00735 moles of Ca is to 0.00735 moles × 2 moles/1 mole
= 0.0147 moles of F
Mass of F = 0.0147 moles × 19 g/mol
= 0.279g or 279 mg of F
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how many grams of hcl would be produced if 54 grams of water were used
Answer:
i dont know
Explanation:
its simple i dont know
Answer:
Explanation:Red
According to the text, fission occurs when the nucleus
two lighter nuclei.
Answer:
Splits into
Explanation:
Because fission reaction is divided into two or more pieces.
FREE BRAINLIST NEED HELP ASAP
Answer: F. Electrons
Explanation: hope it helped .u.
Calculate the mass of calcium chloride that contains 3.20 x 1024 atoms of chlorine.
Answer:
294.87 gm CaCl_2
Explanation:
The computation of the mass of calcium chloride is shown below:
But before that following calculations need to be done
Number of moles of chlorine atom is
= 3.20 × 10^24 ÷ 6.022 × 10^23
= 5.314 moles
As we know that
1 mole CaCl_2 have the 2 moles of chlorine atoms
Now 5.341 mole chloride atoms would be
= 1 ÷ 2 × 5.314
= 2.657 moles
Now
Mass of CaCl_2 = Number of moles × molar mass of CaCl_2
= 2.657 moles × 110.98 g/mol
= 294.87 gm CaCl_2
Calculate the molarity of a solution that contains 3.00x10-2 mol NH4Cl in exactly 450 cm3 of solution.
Answer:
0.06moles/litres
Explanation:
molarity = no of moles / volume in litres = 3.00*10^-2/(450/1000)=0.06 moles/litres
Answer:
0.07mol/dm^3
Explanation:
Please mark as brainliest
what career field/job would utilize ppm units
A environmental jobs
B Medical fields
C Jobs at the brewery
D Jobs in analytical chemistry
A chemistry student is given 4.00 L of a clear aqueous solution at 22°C. He is told an unknown amount of a certain compound X is dissolved in the solution. The student allows the solution to cool to 22 C. The solution remains clear. He then evaporates all of the water under vacuum. A precipitate remains. The student washes, dries and weighs the precipitate. It weighs 0.56 kg.
Required:
Using only the information above, Calculate the solubility of X in water at 22° C. If you said yes, calculate it Be sure your answer has a unit symbol and 2 no .0 it. significant digits.
Answer:
The responses to the given points can be defined as follows:
Explanation:
In point 1:
The answer is "No".
In point 2:
The mass of solute = The mass of precipitate [tex]= 0.56\ kg =0.56\ \times 1000 = 560 \ g[/tex]
Calculating the solubility:
[tex]= \frac{\text{mass of solute in g}}{\text{volume of solution in ml}}\\\\ = \frac{560}{ 4000} \\\\ = 0.14 \ \frac{g}{ml}[/tex]
Which of the following is NOT part of STP?
O O Fahrenheit
O 1 Atmosphere
O 273 Kelvin
O 101.3 kiloPascals
Answer:
Farenheit
Explanation:
STP is used to measure standard temperature and pressure -- it uses Kelvin for temperature, not Farenheit.
Lets make this one quick.
How many grams are in 1.2 x 10^24 atoms of sodium?
An unknown concentration of sodium thiosulfate, Na2S2O3, is used to titrate a standardized solution of KIO3 with excess KI present. Suppose 15.65 mL of the Na2S2O3 solution is required to titrate the iodine formed from 21.55 mL of 0.0131 M KIO3. What is the molarity of the Na2S2O3 solution
Answer: The molarity of [tex]Na_2S_2O_3[/tex] is 0.108 M
Explanation:
[tex]KIO_3+5KI+3H_2SO_4\rightarrow 3K_2SO_4+3H_2O+3I_2[/tex]
[tex]2Na_2S_2O_3+I_2\rightarrow Na_2S_4O_6+2NaI[/tex]
To calculate the number of moles for given molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]
[tex]\text{Moles of }KIO_3=\frac{0.0131mol/L\times 21.55}{1000}=2.8\times 10^{-4}mol[/tex]
1 mole of [tex]KIO_3[/tex] produces = 3 moles of [tex]I_2[/tex]
[tex]2.8\times 10^{-4}[/tex] moles of [tex]KIO_3[/tex] produces = [tex]\frac{3}{1}\times 2.8\times 10^{-4}=8.4\times 10^{-4}[/tex] moles of [tex]I_2[/tex]
Now 1 mole of [tex]I_2[/tex] uses = 2 moles of [tex]Na_2S_2O_3[/tex]
[tex]8.4\times 10^{-4}[/tex] moles of [tex]I_2[/tex] uses = [tex]\frac{2}{1}\times 8.4\times 10^{-4}=1.69\times 10^{-3}[/tex] moles of [tex]Na_2S_2O_3[/tex]
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution in L}}=\frac{1.69\times 10^{-3}\times 1000}{15.65}=0.108M[/tex]
The molarity of [tex]Na_2S_2O_3[/tex] is 0.108 M
A certain metal M forms a soluble nitrate salt M NO Suppose the left half cell of a galvanic cell apparatus is filled with a 3.00 mM solution of M(NO) and the right half cell with a 3.00 M solution of the same substance. Electrodes made of M are dipped into both solutions and a voltmeter is connected between them. The temperature of the apparatus is held constant at 20.0 °C.
1. Which electrode will be positive? What voltage will the voltmeter show?
a. left
b. right
2. Assume its positive lead is connected to the positive electrode. Be sure your answer has a unit symbol, if necessary, and round it to 2 significant digits.
Answer:
Explanation:
From the information given:
left half-cell = 3.00 mM M(NO)
right half-cell = 3.00 M M(NO)
Since the electrode are the same in both cells, then the concentration for the cell are also the same.
Negative electrode = Anode = lower concentration = 3.00 mM
Positive electrode = cathode = higher concentration = 3.00 M
Thus, right half cell will be postive electrode.
To determine the concentration cell:
[tex]Ecell =\Big( \dfrac{2.303\times R\times T}{nF} \Big)log\Big(\dfrac{[cathode]}{[anode]}\Big)[/tex]
SInce [Cathode] > [anode],
[tex]R = 8.314 J/K/mol, \\ \\ T = 273+20 = 293 K \\ \\ Faraday's constant (F)= 96500 C/mol[/tex]
n = 3
[tex]Ecell ={ \dfrac{2.303\times 8.314\times 293 }{(3\times 96500)}} log\dfrac{3}{0.003}[/tex]
[tex]\mathbf{E_{cell} = 0.0582 V } \\ \\ \mathbf{E_{cell} = 0.06 V}[/tex]
The half-life of 226Ra is 1.60x103 years. How long until only 12.5% of the original sample of 226Ra remains?
Answer:
4.80×10³ years
Explanation:
Let the original amount (N₀) of ²²⁶Rn = 1 g
Therefore,
12.5% of the original amount = 12.5% × 1 = 12.5/100 × 1 = 0.125 g
Next, we shall determine the number of half-lives that has elapse. This can be obtained as follow:
Original amount (N₀) = 1
Amount remaining (N) = 0.125 g
Number of half-lives (n) =?
N = 1/2ⁿ × N₀
0.125 = 1/2ⁿ × 1
0.125 = 1/2ⁿ
Cross multiply
0.125 × 2ⁿ = 1
Divide both side by 0.125
2ⁿ = 1/0.125
2ⁿ = 8
Express 8 in index form with 2 as the base
2ⁿ = 2³
n = 3
Thus, 3 half-lives has elapsed.
Finally, we shall determine the time taken for only 12.5% of the original sample of ²²⁶Rn to remain.
This can be obtained as follow:
Half-life (t½) = 1.60×10³ years
Number of half-lives (n) = 3
Time (t) =?
t = n × t½
t = 3 × 1.60×10³
t = 4.80×10³ years.
Thus, it will take 4.80×10³ years for 12.5% of the original sample of ²²⁶Rn to remain.
For a nucleus to go through mitosis,
A. it must start with two sets of chromosomes (diploid)
B. it must start with one set of chromosomes (haploid)
C. it can start with one set or two sets of chromosomes (haploid or diploid)
D. it can start with any number of chromosomes
Answer:
A
Explanation:
I had this in a test
how many atoms are in 0.5 mole of aluminum
According to the Avogadro's number, there are 3.011×10²³ atoms in 0.5 mole of aluminium.
What is Avogadro's number?Avogadro's number is defined as a proportionality factor which relates number of constituent particles with the amount of substance which is present in the sample.
It has a SI unit of reciprocal mole whose numeric value is expressed in reciprocal mole which is a dimensionless number and is called as Avogadro's constant.It relates the volume of a substance with it's average volume occupied by one of it's particles .
According to the definitions, Avogadro's number depend on determined value of mass of one atom of those elements.It bridges the gap between macroscopic and microscopic world by relating amount of substance with number of particles.
Number of atoms can be calculated using Avogadro's number as follows: mass/molar mass×Avogadro's number, on substituting values in formula, 0.5×6.023×10²³=3.011×10²³ atoms.
Thus, there are 3.011×10²³ atoms in 0.5 mole of aluminium.
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Do molecules move in all phases? (yes/no)
Answer: sorta
Explanation:
In fact, they do not flow at all: they simply vibrate back and forth. Because the atoms in a solid are so tightly packed, solid matter holds its shape and cannot be easily compressed
What are the types of forces involved in the production of hydro-electricity
Answer:
There are three types of hydropower facilities: impoundment, diversion, and pumped storage. Some hydropower plants use dams and some do not. The images below show both types of hydropower plants.
Explanation:
Based on the graph, which statement about human use of natural resources is most likely false?
Hey love! <3
Answer:
C. Resource use by humans has gone up and down over the years is a false statement
Explanation:
In the graph you can make an observation on how the increase of human population is directly positive.
The resource use either has a positive or negative correlation, making it impossible for the usage to fluctuate over the years.
Hope this brightens up your day a little! ❀‿❀ Sincerely, Kelsey from Brainly.
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8. Zeolite is used to remove moisture from methane. A vertical column is filled with 1000.0 kg of dry zeolite. The zeolite has the capacity to hold 0.100 kg water/kg dry zeolite. Once the zeolite becomes saturated with moisture, it must be regenerated by heating. The inlet moisture content of the methane is 7.00% (by mass) and the outlet moisture content is 0.05% (by mass). How much methane (kg) will be produced before the zeolite must be regenerated
Answer:
Follows are the solution to this question:
Explanation:
Methane inlet humidity content of [tex]=7.00\%[/tex]
Methane moisture outlet content [tex]=0.05\%[/tex]
Zeolite absorption humidity [tex]= 6.95\%[/tex]
Dry zeolite 1 kg will accommodate water[tex]= 0.1000 \ kg[/tex]
One kilogram of Dry Zeolite will carry water from [tex]=0.1000\ kg[/tex]
The water can contain 1000 kg of zeolite [tex]= 100 \ kg[/tex]
Methane which would be made [tex]=\frac{100}{6.95\%}= 1,439 \ kg[/tex]
That's why it will be producing 1439 kg of methane.
The amount of methane that can be produced before the regeneration of zeolite is 1,439 kg.
What is zeolite?
Zeolite belongs from the family of hydrated aluminosilicate minerals, in which alkali and alkaline earth metals are present.
In the question, it is given that:
Capacity of 1kg of zeolite to hold water = 0.100kg
Capacity of 1000kg of zeolite to hold water = 100kg
Inlet moisture content of the methane = 7.00 % (by mass)
Outlet moisture content of the methane = 0.05% (by mass)
Capacity of zeolite to absorb methane content = 6.95 %
Amount of methane before the regeneration of zeolite = 100kg / 6.95% = 14.39% of kg = 1,439 kg
Hence, 1,439 kg methane will be produced before the zeolite must be regenerated.
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Estimate the crystal field stabilization energy for the octahedral ion hexacyanomanganate(III) , if the wavelength of maximum absorption for the ion is 600 nm. [Note: This is a high-field (low-spin) complex.]
Answer:
The crystal field stabilization energy for the octahedral ion hexacyanomanganate(III) , if the wavelength of maximum absorption for the ion is 600 nm is - 1987.59kJ/mole
Explanation:
Lets calculate -:
Crystal field stabilization energy -
[tex]E=\frac{hc}{\pi }[/tex]
where h = planks constant = [tex]6.626\times10^-^3^4 Js[/tex]
c= velocity of light = [tex]3\times10^8m/second[/tex]
[tex]\pi =wavelength=600nm=600\times10^-^9m[/tex]
[tex]E=\frac{6.626\times10^-^3^4\times3\times10^8}{600\times10^-^9}[/tex]
=[tex]0.033\times10^-^1^6 J/ion[/tex]
= [tex]0.033\times10^-^1^6\times6.023\times10^2^3[/tex]
= [tex]0.198759\times10^7 J/mole[/tex] = [tex]1987.59\times10^3J/mole[/tex]
= [tex]1987.59kJ/mole[/tex]
Thus , the crystal field stabilization energy for the octahedral ion hexacyanomanganate(III) is 1987.59kJ/mole
i
CH3-CH2-CO-CH2-CH2-CH3 + H2
Pd
Answer:
i don't know but I just want to have coins
Convert 146 calories to kilojoules need to know ASAP
An iron sulfide compound is analyzed, and found to contain 11.26 g iron and 9.70 g sulfur. Determine the molar ratio of sulfur to iron in this compound, and hence its chemical formula.
Answer:
Molar ratio = 1.5Chemical formula = Fe₂S₃Explanation:
First we convert the given masses of each element into moles, using their respective molar masses:
Fe ⇒ 11.26 g ÷ 55.845 g/mol = 0.202 mol FeS ⇒ 9.70 g ÷ 32.065 g/mol = 0.302 mol SNow we divide them in order to calculate the molar ratio of S to Fe:
0.302 / 0.202 = 1.5Meaning that for each 1 Fe mol, there's 1.5 S moles. We can write that as Fe₁S₁.₅
Finally we double those subscripts so that they become the lowest possible integers: Fe₂S₃.
The empirical formula for compound is P2O5. The molar mass of the compound is 283.89g. What is the molecular formula?
Answer:
P4O10
Explanation:
(P2O5)n=283.89
62+80=283.89
142n=283.89
n=2
P2O10
The empirical and molecular formula are related by the formula, molecular formula=2×empirical formula ,meaning double the atoms present in empirical formula,hence molecular formula is P₄O₁₀.
What is empirical formula?Empirical formula of a compound is defined as the simplest whole number ratio of atoms which are present in a compound.It does not make any mention of the arrangement of atoms or the number of atoms. The empirical formula gives information about the ratio of number of atoms which are present in a compound.
Molecular formula is determined from the empirical formula by dividing the molar mass of a compound by the empirical formula mass. The resultant which should be a whole number or very close to the whole number , then the subscripts are multiplied by the whole number to get the molecular formula.
For the compound P₂O₅ the subscripts of phosphorous and oxygen are multiplied by 2 and hence the molecular formula P₄O₁₀.
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If 3 moles of NaOH is dissolved in 4L of water, what is the
molarity of the solution?
Answer:
0.75mol/LNaOH
Explanation:
Data: Soln:
M= ? M= n/ V(l)
n= 3moles. M= 3mol/ 4L
V= 4L. M = 0.75 mol/LNaOH
What’a the answer!!?
Answer:
2,1,2
1,2,1,1
the answer
