Answer:
Carbon dioxide becomes a poisonous gas when there is too much of it in the air you breathe. Besides the effects it can have on the planet and the atmosphere, carbon dioxide poisoning can lead to central nervous system damage and respiratory deterioration in humans and other breathing creatures.
Answer:
Carbon dioxide becomes a poisonous gas when there is too much of it in the air you breathe. Besides the effects it can have on the planet and the atmosphere, carbon dioxide poisoning can lead to central nervous system damage and respiratory deterioration in humans and other breathing creatures.
Explanation:
Hope This Helps! [ :
Which organelle in the table is correctly matched with its function?
which one is it??
Answer:
[tex]\huge\boxed{\sf Ribosomes}[/tex]
Explanation:
Organelles and their function:Lysosomes:Lysosomes functions in the digestion of food of the cell.It contains hydrolytic enzymes.Vacuole:Vacuole mostly functions in storage.Mitochondrion:Mitochondrion is the power house of the cell.Ribosome:Ribosome functions in protein synthesis.[tex]\rule[225]{225}{2}[/tex]
liquid octane(CH3)(CH2)6CH3) reacts with gaseous oxygen gas(O2) to produce gaseous carbon dioxide(CO2) and gaseous water(H2O). If 5.18 g of carbon dioxide is produced from the reaction of 3.43 g of octane and 19.1 g of oxygen gas, calculate the percent yield of carbon dioxide. Round your answer to 3 significant figures.
The percent yield of carbon dioxide will be 49.0 %.
Percent yieldFirst, let's look at the equation of the reaction:
[tex]2C_8H_1_8 + 25O_2 -- > 16CO_2 + 18H_2O[/tex]
The mole ratio of octane to oxygen is 2:25.
Mole of 3.43 g octane = 3.43/114.23 = 0.03 mol
Mole of 19.1 g oxygen = 19.1/32 = 0.60 mol
Thus, octane is limiting.
Mole ratio of octane to carbon dioxide = 2:16.
Equivalent mole of carbon dioxide = 0.03 x 8 = 0.24 mol
Mass of 0.24 mol carbon dioxide = 0.24 x 44.01 = 10.5624 grams
Percent yield of carbon dioxide = 5.18/10.5624 = 49.0 %
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Which of the following rules is applicable for balancing a chemical equation?
Change only the coefficients
Add the coefficients and change the subscripts
Change only the subscripts
Change the coefficients and the subscripts
Answer:
A.) Change only the coefficients
Explanation:
An equation is balanced when there is an equal quantity of each type of element on both sides of a reaction. When balancing an equation, the only way to manipulate the amounts of each element is by changing the coefficient values. The coefficients alter the amount of each molecule in the reaction.
The subscripts cannot be altered. If you were to change the subscripts, you would be altering the amount of atoms in a particular molecule.
When the soil is saturated in a gently sloping area, any additional rainfall in the area will most likely
When the soil is saturated in a gently sloping area, any additional rainfall in the area will most likely become surface run off.
What is super saturation?
Super saturation occurs with a chemical solution when the concentration of a solute exceeds the concentration specified by the value equilibrium solubility.
Additional water into the soil will cause the soil to be super saturated and eventually runs off due to the steepness of the area.
Thus, when the soil is saturated in a gently sloping area, any additional rainfall in the area will most likely become surface run off.
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The molecular orbitals depicted below are derived from 2p atomic orbitals in F2+, (a) Give the orbital designations. (b) Which is occupied by at least one electron in F2+,? (c) Which is occupied by only one electron in F2+,?
a) Orbital designation of F₂⁺ is shown below:
b) σ(2p) is occupied by at least one electron in F₂⁺
c) π*(2p) is occupied by only one electron in F₂⁺
What is Molecular orbital diagram?The molecular orbital diagram can be used to show the bonds inside a diatomic molecule. The magnetic characteristics of a molecule and their ionization-related changes can be determined using MO diagrams. The number of bonds shared between the two atoms, or the bond order of the molecule, is likewise shown by them.
The structure A represents an anti-bonding MO of side-wise overlapping of 2p-orbital. Thus designation of this orbital diagram is: π*(2p).
The structure B represents a bonding MO of head-on overlapping of 2p-orbital. Thus designation of this orbital diagram is: σ(2p)
The structure C represents a bonding MO of side-wise overlapping of 2p-orbital. Thus designation of this orbital diagram is: π(2p)
The structure D represents an anti-bonding MO of head-on overlapping of 2p-orbital. Thus designation of this orbital diagram is: σ*(2p)
a) Thus, orbital designation of F₂⁺ is shown below:
b) σ(2p) is occupied by at least one electron in F₂⁺
c) π*(2p) is occupied by only one electron in F₂⁺
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The complete question is as follows:
The molecular orbitals depicted below are derived from 2p atomic orbitals in F₂⁺, (a) Give the orbital designations. (b) Which is occupied by at least one electron in F₂⁺? (c) Which is occupied by only one electron in F₂⁺?
Balance the following reaction: NH3 + I2 --> N2I6 + H2
The balanced equation will be [tex]2NH_3 + 3I_2 -- > N_2I_6 + 3H_2[/tex]
What are balanced equations?They are chemical equations that obey the law of conservation of atoms.
In other words, they are equations in which the number of atoms before and after reactions are the same.
Thus, the balanced equation for the reaction will be [tex]2NH_3 + 3I_2 -- > N_2I_6 + 3H_2[/tex]
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The second order reaction A → Products takes 13.5 s for the concentration of A to decrease from 0.740 M to 0.319 M. What is the value of k for this reaction?
The rate constant of the second order reaction is 0.137 M-1s-1.
What is the rate constant?For the second order reaction we can write;
1/[A] = kt + 1/[A]o
[A]o = initial concentration
[A] = final concentration
k = rate constant
t = time
Now;
1/0.319 = 13.5k + 1/ 0.740
1/0.319 - 1/0.740 = 13.5k
3.13 - 1.35 = 13k
k = 3.13 - 1.35/13
k = 0.137 M-1s-1
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An ion of an element can have the same as another element.
A. number of protons
B. None of these
OC. electron configuration
D. atomic mass
Answer:
A
Explanation:
In a simple perspective, an ion of an element is an element +- [tex]n[/tex] amount of electrons.
We know that electrons have atomic mass, so we can mark off D
We know that with more or fewer electrons, the electron configuration would change
However, the number of protons won't change in the ion of an element, this is because an ion is only the change in electrons, not protons or neutrons (the particles that make up a nucleus of an atom)
balance the redox reaction in alkaline medium & identify the oxidizing & reducing agents
I- + MnO4- ——> IO3- + MnO2
The balanced redox equation of the reaction is given below:
I⁻ + 2 MnO₄ + H₂O → IO₃⁻ + 2 MnO₂ + 2 OH⁻The oxidizing agent is MnO₄ while the reducing agent is I⁻.
What are redox equations?Redox equations are equations in which oxidation and reduction reactions occur together.
Redox reactions can take place in alkaline or acidic mediums.
The balanced redox equation of the reaction is given below:
I⁻ + 2 MnO₄ + H₂O → IO₃⁻ + 2 MnO₂ + 2 OH⁻The oxidizing agent is MnO₄ while the reducing agent is I⁻
In conclusion, a balanced redox equation is one in which the atoms and the change in oxidation state is equal on both sides of the reaction.
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3. What is the molar mass of Li2SO4? (2 points)
O 54.9 g/mol
103.0 g/mol
O109.9 g/mol
O 206.1 g/mol
Answer:
C.) 109.9 g/mol
Explanation:
The molar mass is the sum of each element's atomic weight times their quantity. There is one atom of each element unless denoted by subscripts.
Atomic Mass (Li): 6.9410 g/mol
Atomic Mass (S): 32.065 g/mol
Atomic Mass (O): 15.998 g/mol
Molar Mass (Li₂SO₄): 2(6.9410 g/mol) + 32.065 g/mol + 4(15.998 g/mol)
Molar Mass (Li₂SO₄): 109.939 g/mol
• An alkane with the formula C5H12 undergoes chlorination to give only one product with the formula C5H11Cl. What is the structure of this alkane? •
Answer:
When an alkane undergoes a halogenation reaction, the majority of the products will add the halogen to the most substituted and most stable carbon. In this case, this would likely be the third carbon (middle).
Key:
Leftmost molecule = C₅H₁₂
Rightmost molecule = C₅H₁₁Cl
*Not pictured = HCl (another product)
An ideal gas in a closed container initially has a volume V and Temperature T the final tempera is 5/4T and the final pressure is 2P what is the final gas
Answer:
[tex]V_2 = \frac{5V}{8}[/tex]
Explanation:
I am assuming you are saying what is the final volume of the gas
Known :
Initial volume (V1) = V
Initial temperature (T1) = T
Final temperature (T2) = 5/4 T
Initial pressure (P1) = P
Final pressure (P2) = 2P
Wanted: Final volume (V2)
[tex]\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\\\frac{PV}{T} = \frac{(2P)V_2}{(5/4)T}\\\frac{V}{1} = \frac{(2)V_2}{5/4}\\5/4V = 2V_2\\\\V_2 = \frac{5V}{8}[/tex]
At 1755 oC the equilibrium constant for the reaction: 2 IBr(g) I2(g) + Br2(g) is KP = 0.748. If the initial pressure of IBr is 0.00465 atm, what are the equilibrium partial pressures of IBr, I2, and Br2? p(IBr) = . p(I2) = . p(Br2) = .
From the calculation, the equilibrium partial pressure of IBr is 6.5 * 10^-4 atm while that of I2 and Br2 is 0.002 atm.
What is the equilibrium pressure?We must set up the ICE table as shown hence;
2 IBr(g) < ------> I2(g) + Br2(g)
I 0.00465 0 0
C - 2x + x + x
E 0.00465 - 2x + x + x
Kp = pI2. pBr2/pIBr^2
pI2 = pBr2 = x
0.748 = x^2/ 0.00465 - 2x
0.748 (0.00465 - 2x) = x^2
3.5 * 10^-3 - 1.496x = x^2
x^2 + 1.496x - 3.5 * 10^-3 = 0
x=0.002 atm
Hence;
For IBr = 0.00465 - 2(0.002) = 6.5 * 10^-4 atm
For I2 and Br2 = 0.002 atm
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how much is 5 moles of lithium phosphide in grams
Answer:
we are given that mass of lithium four spade that is equal to 77.7 grams
Which equation obeys the law of conservation of mass? H2(g) + O2(g) → H2O(g)
H2(g) + O2(g) → H2O(g) +4He(g)
2H2(g) + O2(g) → 2H2O(g)
H2(g) → H2O(g)
H2(g) + O2(g) → 2H2O(g)
Answer:
2H2(g) + O2 -> 2H2O
Explanation:
According to the law of conservation of mass, mass cannot be created nor destroyed in a chemical formula.
The only chemical reaction of the five options that follows this law is
2H2(g) + O2 -> 2H2O because the mass of the compounds stays the same before the reaction (arrow) and after.
Before reaction (reactants) we have:
2 (H2) = 4H
1 (O2) = 2O
After reaction takes place (products) we have:
2 (H2O) = 4H and 2O
and so mass is conserved.
Answer: its the 3rd option
Explanation: on edge hope this helps
For any spontaneous process, universe entropy intensifies is known as the __________ law of thermodynamics.
Group of answer choices
fourth
second
tenth
For any spontaneous process, universe entropy intensifies is known as the second law of thermodynamics.
What is entropy?Entropy is defined as the degree of randomness or disorderliness of a system.
The entropy of a system generally increases for any spontaneous process.
This is according to the second law of thermodynamics.
In conclusion, the entropy of a system is the a measure of randomness of the system.
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Which of these compounds is the ester formed from the reaction of acetic acid and 1-propanol? O 1
O 2
O 3
O 4
O 5
The ester formed from the reaction of acetic acid and 1-propanol is O3, which is propyl acetate.
What is ester of acetic acid ?The chemical compound ethyl acetate, also known as ethyl ethanoate or EtOAc or EA, has the formula C4H8O2 and is an organic compound. This clear liquid is used to decaffeinate tea and coffee as well as in glues and nail paint removers. It has a distinctive sweet scent akin to pear drops. The ester of ethanol and acetic acid, ethyl acetate is produced in huge quantities for use as a solvent.
Due to its affordability, low toxicity, and pleasant smell, ethanol acetate is mostly employed as a solvent and diluent. It is frequently utilized, for instance, to clean circuit boards and in some nail polish removers (acetone is also used). This solvent is used to decaffeinate tea leaves and coffee beans. Additionally, it serves as a hardener or activator in paints. Fruits, colognes, and confections all contain ethyl acetate. In perfumes, it swiftly evaporates, leaving the fragrance on the skin.
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Two isotopes of lithium are found in nature Li6 has a mass of 6. 02u and Li7 has a mass of 7.02u . Use the atomic weight of lithium found on the periodic table to determine which isotope is more abundance
The isotope that is more abundant, given the data is isotope Li7
AssumptionLet Li6 be isotope ALet Li7 be isotope BHow to determine whiche isotope is more abundantMolar mass of isotope A (Li6) = 6.02 uMolar mass of isotope B (Li7) = 7.02 uAtomic mass of lithium = 6.94 uAbundance of A = A%Abundance of B = (100 - A)%Atomic mass = [(mass of A × A%) / 100] + [(mass of B × B%) / 100]
6.94 = [(6.02 × A%) / 100] + [(7.02 × (100 - A)) / 100]
6.94 = [6.02A% / 100] + [702 - 7.02A% / 100]
6.94 = [6.02A% + 702 - 7.02A%] / 100
Cross multiply
6.02A% + 702 - 7.02A% = 6.94 × 100
6.02A% + 702 - 7.02A% = 694
Collect like terms
6.02A% - 7.02A% = 694 - 702
-A% = -8
A% = 8%
Thus,
Abundance of B = (100 - A)%
Abundance of B = (100 - 8)%
Abundance of B = 92%
SUMMARY
Abundance of A (Li6) = 8%Abundance of B (Li7) = 92%From the above, isotope Li7 is more abundant.
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Lactic acid has a pKa of 3.08. What is the approximate degree of dissociation of a 0.35 M solution of lactic acid?
A) 1.1%
B) 2.2%
C) 4.8%
D) 14%
E) none of the above
Your answer would be C, 4.8%.
15. When traveling through a smoke-filled area, a person must be able
distance of at least
in order to safely reach a fire exit.
O A. 10 feet
B. 12 feet
C. 4 feet
D. 8 feet
When traveling through a smoke-filled area, a person must be able to have distance of at least 4 feet in order to safely reach a fire exit. That is option C.
What is a smoke-filled area?Smoke-filled area is an enclosed area that is filled with smoke gas that is usually made up of carbon monoxide.
The smoke-filled area is usually caused by
Fire out breakBurning of plastics or chemical productsRelease of fuming gases.When there is smoke-filled area die to fire out break, there are some guidelines to follow to safely reach the fire exit early enough.
The importance of these guidelines include the following:
enable rescuers to navigate a scene more quickly,identify risks and hazards,locate safety points, determine the safest way in and out of a building, and map out evacuation routes.The affected individual should get down and crawl while taking short breath through the nose because cleaner air is nearest to the floor.
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Suppose a gas mixture used for anesthesia contains 4.60 mol oxygen (O₂) and 6.00 mol nitrous oxide (N₂O). The total pressure of the mixture is 2.00 atm. A mole fraction is defined as the moles of a specific component divided by the total number of moles present. What is the mole fraction of O₂ in this mixture?
Considering the definition of mole fraction, the mole fraction of O₂ in the mixture is 0.434.
Definition of mole fractionThe molar fraction is a way of measuring the concentration that expresses the proportion in which a substance is found with respect to the total moles of the solution.
In other words, the mole fraction expresses the concentration of solute in a solution as the ratio of moles of substance to total moles of solution:
[tex]mole fraction=\frac{moles of substance}{moles of solution}[/tex]
Mole fraction of O₂ in this mixtureIn this case, you know a gas mixture used for anesthesia contains 4.60 mol oxygen (O₂) and 6.00 mol nitrous oxide (N₂O).
So, the total moles of the solution can be calculated as:
Total moles = moles of oxygen (O₂) + moles of nitrous oxide (N₂O)
Then:
Total moles= 4.60 moles + 6 moles
Total moles= 10.60 moles
Finally, the more fraction of O₂ can be calculated as follow:
[tex]Mole fraction of O_{2} =\frac{moles of O_{2}}{total moles}[/tex]
[tex]Mole fraction of O_{2} =\frac{4.60 moles}{10.6o moles}[/tex]
Solving:
Mole fraction O₂ = 0.434
Finally, the mole fraction of O₂ in the mixture is 0.434.
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An analyst prepared a sucrose solution by weighing 1kg of water and add 1.5kg of sucrose. what is the concentration of the resultant solution in mass percent
The concentration of solution in mass percent is 60%.
What is mass percentage?Mass percentage refer to the percentage of solute present in solution.
The concentration of substance can be expressed in mass percent.
So, we can write,
Mass percent = mass of solute / mass of solution x 100.
Mass of solution = mass of solute + solvent.
Here, sucrose is the solute and water is solvent.
Mass of solute is 1.5Kg and mass of solvent is 1Kg.
Mass of solution = 1.5 + 1 = 2.5 Kg.
Mass percent = 1.5/2.5x100
So, Concentration of solution in mass percent = 60%.
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The diagram shows the setup of an experiment. A few observations of the experiment are listed in table below the diagram.
Experimental Observations
1. Color of solution turned blue
2. Shiny hair like crystals deposited on the copper strip
Which of the following is the correct explanation for one of the given observations?
Observation 1 is a result of silver ions moving into the solution.
Observation 2 is a result of nitrate ions moving into the solution.
Observation 1 is a result of silver ions changing their oxidation state.
Observation 2 is a result of silver ions changing their oxidation state.
Answer:
Observation 2 is a result of silver ions changing their oxidation state.
Explanation:
Silver is reduced and forms silver precipitate. Observation 1 is a result of coppe being oxidized and forming copper nitrate
Which of the following arranges the bonds in order of decreasing bond length?
The order of decreasing bond length is; C- P > C - N > C=N. Option B
What is bond length?A chemical bond is formed when two compounds combine together. We know that the strength of a bond is shown by the bond length and the bond order. Clearly, bonds that have shorter bond lengths tend to be stronger than longer bonds that have a longer bond distance.
This explains why triple bonds are shorter and stronger bonds. This stems for the high s contribution (50%) in the formation of the bond. Therefore, the order of decreasing bond lengths is single bonds > Double bonds > triple bonds.
Now applying this background to the case in point, we can see that the decreasing order of bond energy must have to do with increasing bond order and decreasing bond distance.
Hence, the order of decreasing bond length is; C- P > C - N > C=N.
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A solution is prepared by mixing 200.0 g of water, H2O, and 300.0 g of
ethanol, C2H5OH. Determine the mole fractions of each substance.
Answer:
Mole Fraction (H₂O) = 0.6303
Mole Fraction (C₂H₅OH) = 0.3697
Explanation:
(Step 1)
Calculate the mole value of each substance using their molar masses.
Molar Mass (H₂O): 2(1.008 g/mol) + 15.998 g/mol
Molar Mass (H₂O): 18.014 g/mol
200.0 g H₂O 1 mole
--------------------- x ------------------ = 11.10 moles H₂O
18.014 g
Molar Mass (C₂H₅OH): 2(12.011 g/mol) + 6(1.008 g/mol) + 15.998 g/mol
Molar Mass (C₂H₅OH): 46.068 g/mol
300.0 g C₂H₅OH 1 mole
---------------------------- x -------------------- = 6.512 moles C₂H₅OH
46.068 g
(Step 2)
Using the mole fraction ratio, calculate the mole fraction of each substance.
moles solute
Mole Fraction = ------------------------------------------------
moles solute + moles solvent
11.10 moles H₂O
Mole Fraction = -------------------------------------------------------------
11.10 moles H₂O + 6.512 moles C₂H₅OH
Mole Fraction (H₂O) = 0.6303
6.512 moles C₂H₅OH
Mole Fraction = -------------------------------------------------------------
11.10 moles H₂O + 6.512 moles C₂H₅OH
Mole Fraction (C₂H₅OH) = 0.3697
When 50 electrons drop from energy level 4 to energy level 2, how many photons of light are given off?
1 photon
100 photons
2 photons
50 photons
The number of photons given off will be 50 photons.
To find the answer, we need to know about the plank's equation.
How to find the number of photons emitted?We have the expression for energy of a single electron in eV as,[tex]E=-13.6(\frac{1}{(n_f)^2}- \frac{1}{(n_i)^2})eV\\[/tex]
We have,[tex]n_f=4\\n_i=2\\N=50[/tex]
Substituting values, we get,[tex]E=-13.6(\frac{1}{(4)^2}- \frac{1}{(2)^2})eV\\\\E=-13.6*-0.188=2.55eV[/tex]
For N electrons,[tex]E=50*2.55eV=127.5eV[/tex]
We have the plank's equation,E=nhf
From this, the number of photons emitted from 50 electrons will be,[tex]n=\frac{E}{h*f} =\frac{127.5*1.67*10^{-19}J}{(6.63*10^-34)Js} \\[/tex]
To find n, we have to find the frequency f. For that, we have the equation,[tex]\frac{1}{wave length}=R_H(\frac{1}{(n_i)^2}- \frac{1}{(n_f)^2})\\\\1/wv= 1.1*10^5(\frac{1}{(2)^2}- \frac{1}{(4)^2})=20625cm^{-1}.\\wavelength=486nm.[/tex]
Thus, frequency will be,[tex]f=\frac{c}{wavelength} =\frac{3*10^8}{486*10^{-9}} =6.172*10^{14}s{-1}[/tex]
Then, the number of photons will be,[tex]n=\frac{E}{h*f} =\frac{127.5*1.67*10^{-19}J}{(6.63*10^-34)Js*6.17*10^{14} s^{-1}} \\\\n=52.05 photons[/tex]
Thus, we can conclude that, the number of photons given off will be 50 photons.
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Stibnite, Sb2S3, is a dark gray mineral from which antimony metal is obtained. If you have 2.00 kg of an ore that contains 15.1% antimony, what mass of Sb2S3 (in grams) is in the ore?
Answer: 421.264 grams.
Explanation: This problem is essentially testing you on your knowledge of percent mass except with one extra step.
We are given that there are 2kg or 2000 grams of ore. This ore is %15.1 Antimony. We can multiply 2000 grams by .151 to find the amount of Antimony which is 302 grams. We don't want the amount of Antimony though, we want the amount of Stibnite. To find the amount of Stibite, we can multiply the amount of Antimony by 1/(the percent mass of Antimony in 1 molecule of Stibnite. To find the percent mass we divided 2(molar mass of Sb) by the molar mass of one molecule of Stibnite. This calculation is 2(121.67)/339.69. This is the percent mass of Antimony in Stibnite. We take the inverse because we want to find out how much Stibite there is per amount of Antimony. We do 1/(((2(121.67))/339.69) and multiply that by 302 to find the total amount of Sb2S3.
Hope this helps!
How many liters of fluorine gas are needed to
form 919 L of sulfur hexafluoride gas if the
following reaction takes place at 2.00 atm
and 273.15 K:
S(s) + 3F₂(g) → SF. (g)?
2756.1 liters of fluorine gas is needed to produce 919 liters of sulfur hexafluoride.
Given data:The volume of SF₆ = 919 L
Pressure = 2 atm
Temperature = 273.15 k
The volume of fluorine required =?
Balance chemical equation:
S (s) + F₂(g) → SF₆(g)
First of all, we will calculate the moles of SF₆.
PV = nRT
n = PV/RT
n = 2. atm× 919L / 0.0821 L. atm. mol⁻¹. K⁻¹ × 273.15 K
n = 1838 atm. L/ 22.43 L. atm. mol⁻¹
n = 81.9 mol
81.9 moles of SF₆ will produce.
Now we will compare the moles of SF₆ and fluorine from the balanced chemical equation.
SF₆: F
1 : 3
81.9 : 3/1 × 81.9 = 245.7 moles
Now we will calculate the volume of fluorine.
PV = nRT
V = nRT / P
V= 245.7 mol × 0.0821 L. atm. mol⁻¹. K⁻¹ × 273.15 K / 2 atm
V = 5512.2 / 2
V = 2756.1 L
2756.1 liters of fluorine gas are needed to produce 919 liters of sulfur hexafluoride.
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please help me with this question as soon as possible
Answer:
nah what questionExplanation:
If U-235 decays into Cs-135 and 4 neutrons, what other nuclide will be produced?
Fr-223
Pu-244
Ra-226
Rb-96
If U-235 decays into Cs-135 and 4 neutrons, the other nuclide that will be produced is Rb-96 (option D).
What is radioactive decay?A radioactive decay is the process by which an unstable large nuclei emit subatomic particles and disintegrate into one or more smaller nuclei.
According to this question, a radioactive material Uranium- 235 undergoes radioactive decay into Cs- 135 and 4 neutrons (1/0n).
This means that the mass of the products we have is 135 + 4 = 139.
The mass of the nuclide left must be 235 - 139 = 96, hence, the other nuclide that will be produced is Rb-96.
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