Solubility of LaF3: "S"
First, we need to write the reaction of solubility:
LaF3 (s) <=> La+3 + 3 F-
Initial - 0 +0.011 (F-, from KF)
Change +S +3.S
Equilibrium S 0.011 + 3.S
Now, we know:
Ksp = [La³⁺]x[F⁻]³ = 2×10^−19 = S x (0.011+3.S)³
We work with: 2×10^−19 = S x (0.011+3.S)³, and we clear the "S"
=> S = 1.5x10^-13 mol/L x (195.9 g/mol) = 2.93x10^-11 g/L
The molar mass of LaF3 = 195.9 g/mol
Answer: S = 2.93x10^-11 g/L
If a block has the following dimensions and a density of 7,8 g/cm what is the masst?
L = 5.0 cm
W = 6.0 cm
H = 2.0 cm
The question requires us to calculate the mass of a block, given its density (7.8 g/cm~3) and dimensions (length = 5.0 cm; width = 6.0 cm and height = 2.0 cm).
Considering that the volume of the block can be calculated following the basic formula for volume (volume = length x width x height), we can obtain the volume of the object from the dimensions given and then use the definition of density to calculate the mass from the density and volume of the block.
First, we'll calculate the volume of the block:
[tex]\begin{gathered} \text{Volume = length}\times\text{ width }\times\text{ height} \\ \text{Volume = }(5.0cm)\times(6.0cm)\times(2.0cm) \\ \text{Volume = 60 }cm^3 \end{gathered}[/tex]Next, we must consider the definition of density (mass per volume of an object) to calculate the mass of the block:
[tex]\begin{gathered} \text{density = }\frac{mass}{\text{volume}}\to\text{mass = density}\times volume \\ \end{gathered}[/tex][tex]\begin{gathered} \text{mass = (7.8g/}cm^3)\times(60cm^3) \\ \text{mass = }468g \end{gathered}[/tex]Therefore, the mass of the block is 468g.
Hellllp pleaseeeee!!!
Answer:
A
Explanation:
The molecule is an amino acid which is used to make proteins, so it is an organic molecule.
It has a double bond between c=o
and single bonds throughout the molecule
Hope this helps!
Find the formula mass of the compound, then divide the individual element total by the total mass-move the decimal over two to change it to percentage Al2O3
1) Find the formula mass of Al2O3.
Aluminum mass: 26.982 u.
Oxygen mass: 15.999 u.
Al2O3 mass = 2 * (26.982 u) + 3 * (15.999 u)
Al2O3 mass = 101.961 u.
2) Aluminum percentage.
Aluminum mass: 26.982 u.
Al2O3 mass = 101.961 u.
[tex]\frac{(2*26.982\text{ }u)}{101.961\text{ }u}=0.5293[/tex]Moving the decimal point.
52.93%
The percentage of aluminum is 52.93%
3) Oxygen percentage.
Oxygen mass: 15.999 u.
Al2O3 mass = 101.961 u.
[tex]\frac{(3*15.999\text{ }u)}{101.961\text{ }u}=0.4707[/tex]Moving the decimal point.
47.07%
The percentage of oxygen is 47.07%.
.
Which one is not an organic coumpounds
Answer:
A
Explanation:
There are many definitions, and all of them are organic under some definition. The answer is probably A because it is the only one without hydrogen, and sometimes molecules without hydrogen are counted as being inorganic. A is CCl2F2, which is comprised of two chlorine atoms, two fluorine atoms, and a carbon atom.
how many significant figures do the following numbers have?1) 5.9 x 104
Significant figures correspond to the number of digits that a number contains. Zeros at the beginning and end of the number are not counted, only zeros are counted if they are in an intermediate position.
For this case, the number is written in scientific notation, the corresponding 10 of the scientific notation is not taken into account during the digit count, therefore the significant figures of this number will be:
Answer: In 5.9 x10^4 there are 2 significant figures.
Silver nitrate reacts with potassium chloride according as indicated by the balanced equation:AgNO₃ + KCl -----> AgCl + KNO₃How many grams of KCl would be required to react with 380 mL of 0.71 M AgNO₃ solution?.Answer in units of grams
Silver nitrate reacts with potassium chloride according as indicated by the balanced equation:
AgNO₃ + KCl -----> AgCl + KNO₃
How many grams of KCl would be required to react with 380 mL of 0.71 M AgNO₃ solution?.
First we have to find the number of moles in the solution of AgNO₃, then we can find the number of moles of KCl that will react with those moles of AgNO₃, and finally we can convert the moles of KCl into grams using the molar mass of KCl.
So, molarity is defined as the number of moles of solute divided the volume of solution in L.
Molarity = moles of AgNO₃/volume of solution in L
moles of AgNO₃ = molarity * volume of solution in L
We have 1000 mL in 1 L. We can use that conversion to find the volume of solution in L.
1000 mL = 1 L
Volume of solution in L = 380 mL * 1 L /1000 mL = 0.380 L
Volume of solution in L = 0.380 L
Now we can find the number of moles contained in 380 ml (or 0.380 L) of 0.71 M AgNO₃ solution.
moles of AgNO₃ = molarity * volume of solution in L
moles of AgNO₃ = 0.71 M * 0.380 L
moles of AgNO₃ = 0.270 moles
AgNO₃ + KCl -----> AgCl + KNO₃
We can compare the equation of a chemical reaction with a recipe. The coefficients are the quantities of that recipe. Since all the coefficients are 1, we can read our reaction like this: "1 mol of AgNO₃ will react with 1 mol of KCl to produce 1 mol of AgCl and 1 mol of KNO₃".
The reaction between AgNO₃ and KCl is 1 to 1. We will use that relationship to find the number of moles of KCl that will react with 0.270 moles of AgNO₃.
number of moles of KCl = 0.270 moles of AgNO₃ * 1 mol of KCl/(1 mol of AgNO₃)
number of moles of KCl = 0.270 moles of KCl
Since the reaction is 1 to 1, 0.270 moles of KCl will react with 0.270 moles of AgNO₃.
Finally we can find the mass of KCl. We need the molar mass of KCl.
atomic mass of K = 39.10 amu
atomic mass of Cl = 35.45 amu
molar mass of KCl = 74.55 g/mol
mass of KCl = number of molses of KCl * molar mass of KCl
mass of KCl = 0.270 moles * 74.55 g/mol
mass of KCl = 20.1 g
Answer: 20. g of KCl would be required.
How to draw (E)‑3‑methyl‑2‑pentene
In (E)‑3‑methyl‑2‑pentene, methyl group is attached in opposite side of the carbon that why it will be name as (E)‑3‑methyl‑2‑pentene.
The name of the given compound is (E)‑3‑methyl‑2‑pentene.
There are 5 carbon atom present in the compound. Methyl group is attached with C-2 and C-3 carbon atom in opposite direction
The structure of (E)‑3‑methyl‑2‑pentene is given as;
In the structure, it can be seen that methyl group is attached in opposite direction.
Trans molecule is named as E -compound.
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Hello , can I have help with number 43 , please? Can you give explanation for all the choices ? Tank you
Number 43:
Please look at the next drawing:
Answer: b. upper right
Accessibility ModePrintFindPart II: Limiting Reactant1. Consider the reaction: 5C +2SO₂ → CS₂ + 4COa.) If you had 10 mol of Carbon, how many moles of carbon monoxide would be produced?b.) If you had 10 mol of sulfur dioxide, how many moles of carbon monoxide would beproduced?c.) If you had 10 mol of C and SO2 which reactant would be limiting?d.) What is the theoretical yield of CO, in moles if you react 10 moles of each reactant?SAMEFaPmenSavarہےEP Immersive Reader
Answer:
a.) 8 moles of carbon monoxide (CO).
b.) 20 moles of carbon monoxide (CO)
c.) C is the limiting reactant.
d.) The theoretical yield of CO would be 8 moles.
Explanation:
Let's write the chemical equation:
[tex]5C+2SO_2\rightarrow CS_2+4CO.[/tex]a.) You can see that 5 moles of carbon (C) reacted produces 4 moles of carbon monoxide (CO), so if we want to know how many moles of carbon monoxide can be produced, we can multiply each coefficient in the chemical equation by 2 (because we will have 10 moles of carbon):
[tex]10C+4SO_2\operatorname{\rightarrow}2CS_2+8CO.[/tex]The answer is that we will produce 8 moles of carbon monoxide (CO) by 10 moles of carbon (C).
b.) We can apply the same logic to this case. In the chemical equation, we have 2 moles of sulfur dioxide (SO2) reacted that produces 4 moles of carbon monoxide (CO), so if we want to know how many moles of carbon monoxide can be produced, we can multiply each coefficient by 5 so we will have 10 moles of sulfur dioxide:
[tex]25C+10SO_2\operatorname{\rightarrow}5CS_2+20CO.[/tex]The answer is that we will produce 20 moles of carbon monoxide (CO) by 10 moles of sulfur dioxide (SO2).
c.) We have already known that 10 moles of C will produce 8 moles of CO, and 10 moles of SO2 will produce 20 moles of CO. If we react 10 moles of C and 10 moles of SO2, you can note that we're going to have an excess of SO2 because we don't have enough amount of C to produce 20 moles of CO as SO2 do. So based on this logic, C is being consumed first and C would be the limiting reactant.
d.) Remember that the theoretical yield indicates the amount of a product obtained in a chemical reaction.
As we saw before, C is the limiting reactant if we react 10 moles of each reactant, so we have already known that 10 moles of C reacted produces 8 moles of CO, so the theoretical yield of CO would be 8 moles.
Determine the pH of a 0.536002 M C6H5CO2H M solution if the Ka of C6H5CO2H is 6.5 x 10-5. Record your answer to two decimal places
To solve this problem, we could use the ICE table as follows:
First of all, the reaction that we're working with is:
[tex]C_6H_5CO_2H\leftrightarrow C_6H_5CO^-_2+H^+_{}[/tex]Now, let's make the table:
As you can notice, the initial concentration of C6H5CO2H is 0.536002M, and as the products haven't been formed yet, their concentration is 0.
When the change happen, the concentration of the reactant will decay "x" and the concentration of the products will increase "x". (That's the reason of the signs).
And finally, the equilibrium.
We also know that:
[tex]K_a=6.5\cdot10^{-5}[/tex]And, this equilibrium constant comes from the following:
[tex]K_a=6.5\cdot10^{-5}=\frac{\lbrack H^+\rbrack\lbrack C_6H_5CO^-_{2^{}}\rbrack}{\lbrack C_6H_5CO_2H\rbrack}[/tex]Where all concentrations are the equilibrium concentrations.
If we replace, we got that:
[tex]\begin{gathered} 6.5\cdot10^{-5}=\frac{\lbrack H^+\rbrack\lbrack C_6H_5CO^-_{2^{}}\rbrack}{\lbrack C_6H_5CO_2H\rbrack} \\ \\ 6.5\cdot10^{-5}=\frac{x\cdot x}{0.536002-x} \\ \\ 6.5\cdot10^{-5}=\frac{x^2}{0.536002-x} \end{gathered}[/tex]We're going to solve this equation for x:
[tex]\begin{gathered} 6.5\cdot10^{-5}(0.536002-x)=x^2 \\ =3.484013\cdot10^{-5}-6.5\cdot10^5x-x^2=0 \end{gathered}[/tex]And, using the quadratic formula we obtain that the value of x is 0.000587014 approximately.
Now, if x=0.000587014, that means that [H+] = 0.000587014
And finally, remember that the pH of a solution is defined as:
[tex]\begin{gathered} pH=-\log \lbrack H^+\text{\rbrack} \\ pH=-\log \lbrack0.000587014\rbrack \\ pH=3.23 \end{gathered}[/tex]Therefore, the pH is 3.23.
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To answer this question we have to find the value of ka, to do it, we can use the value of kb. The product of the kb of a base and the ka of its conjugated base is kw(1x10^-14):
[tex]\begin{gathered} k_a\cdot k_b=k_w \\ k_a=\frac{k_w}{k_b} \\ k_a=\frac{1\times10^{-14}}{4.38\times10^{-5}} \\ k_a=2.88\times10^{-10} \end{gathered}[/tex]Now, we have to use the equation of Hendersson and Hasselbach:
[tex][/tex]Problem set3. What is the [OH-] of a solution that has a pH of 2?
We have a solution that pH = 2
We can use this relation which says:
pH + pOH = 14
If we clear pOH from here:
pOH = 14 - pH = 14 - 2 = 12
So, pOH = 12 and pOH could be calculated as:
pOH = - log [OH-] => we clear [OH-] from this,
10^(-pOH) = [OH-] => 10^(-12) = [OH-] => 1x10^-11 mol/L = [OH-]
(mol/L is generally the unit for [OH-])
Answer: [OH-] = 1x10^-11 mol/L
In the following chemical wording, which atom gains electrons?(The formula is in the picture)The answers to choose are:•none•aluminum•oxygen•iron
Hierro
El número de oxidación del hierro en el lado izquierdo de la reacción es +3 mientras que en el lado derecho está como hierro metálico, es decir su número de oxidación es 0
After being ignited in a Bunsen burner flame, a piece of magnesium ribbon burns brightly, giving off heat and light. In this situation, the Bunsen burner flame providesA)ionization energyB)activation energyC)heat of reactionD)heat of vaporization
The reaction that involves the combustion of Magnesium will not start unless some amount of energy is added to it, therefore we need the Bunsen burner flame to be the source of the activation energy for the reaction properly occur, since the reaction would not occur in a spontaneous way. Therefore answer letter B
Why does the ice pack get colder when the inside barrier is broken and the chemicals mix together?
The ice pack gets colder when the inside barrier is broken and the chemicals mix together as instant cold packs employ this type of endothermic reaction.
Endothermic reactions are those that involve the absorption of heat.When the chemical ammonium nitrate is dissolved in water, the resultant solution is colder than either of the starting ingredients. This is an illustration of an endothermic process. Instant cold packs employ this type of endothermic reaction.Water is the substance that fills the cold pack. Another plastic bag or tube carrying fertilizer ammonium nitrate is submerged in the water. The tube is broken when you strike the cold pack, allowing the water and fertilizer to mix. The endothermic reaction this mixture produces results in heat absorption.
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6. What would be the concentration of a solution formed when 1.00 g of NaCl are dissolved in water to make100 mL of solution?
To find the concentration we can do it in terms of molarity. Molarity is a way of expressing the concentration of a solute in a solution, it is expressed with the term M and can be described by the following equation:
[tex]Molarity=\frac{MolesSolute}{Lsolution}[/tex]So first we must find the moles of NaCl present in 1.00 grams, the moles of NaCl will be:
[tex]\begin{gathered} molNaCl=givengNaCl\times\frac{1molNaCl}{MolarMass,gNaCl} \\ molNaCl=1.00gNaCl\times\frac{1molNaCl}{58.44gNaCl}=0.017molNaCl \end{gathered}[/tex]So, the molarity of the solution will be:
[tex]Molarity=\frac{0.017molNaCl}{0.100Lsolution}=0.17M[/tex]Answer: The concentration of the solution in molarity will be 0.17M
The Haber process for producing ammonia commercially is represented by the equation N2(g) + 3H2(g) → 2NH3(g). If 7 L of NH3 are consumed, how many liters of H2 gas are required?
Step 1
The reaction:
N2(g) + 3H2(g) → 2NH3(g) (balanced and completed)
All gases are assumed to be ideal and to be at STP conditions.
----------------
Step 2
STP conditions:
1 mole of gas = 22.4 L
----------------
Step 3
Information provided:
7 L of NH3 produced (ammonia)
Procedure:
By stoichiometry,
1 mole H2 = 22.4 L
1 mole NH3 = 22.4 L
N2(g) + 3H2(g) → 2NH3(g)
3 x 22.4 L H2 -------- 2 x 22.4 L NH3
X -------- 7 L NH3
X = 7 L NH3 x 3 x 22.4 L H2/2 x 22.4 L NH3 = 10.5 L
Answer: 10.5 L of H2 are consumed.
FeCl3 (aq) undergoes a double displacement reaction with Ba(OH)2 (aq). If this reaction produces a solid product containing iron, and the other product is in an aqueous solution, what's the balanced chemical equation for this chemical reaction? Briefly explain how to arrive at the balanced reaction equation.
The displacement reaction is given as; [tex]2FeCl_{3} + 3Ba(OH)_{2} ----- > 2Fe(OH)_{2} (s) +3 BaCl2[/tex]. We can see that a precipitate was obtained in the reaction.
What is a displacement reaction?We know that a displacement reaction is a reaction that involves the combination of two compounds in which there is a kind of exchange between the reactants. We know in this case that this is a double replacement reaction. The anions that are in the system would exchange partners in the products.
Let us now look at the reaction and try to see how to be able to obtain the reaction. One more time the reaction is going to occur between the aqueous solutions of iron III chloride and barium hydroxide. We must recall that the insoluble product is iron III hydroxide.
We know that the product that is observed is a solid and an aqueous products from the reaction.
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Determine the following elements using their quantum #s of the elements last placed e-.include the e- configuration of the last placed electron.A. n = 4 l = 1, m = +1 , s = - ½,
If the question is asking us to find the element based on the quantum numbers of the last electron, we have these informations:
n = 4
l = 1
m = +1
s = -1/2
n represents the shell or level value, this value will generally go from 1 to 7, and since our value is 4, we are talking about an element whose valence shell is 4, so this element is in the 4th period
l represents the value of the orbital, this value can range from 0 to 3,
0 is the s orbital
1 is the p orbital
2 the d orbital
3 = f orbital
In out question we have l = 1, therefore the last electron of this element is in the p orbital and in the 4th period, now we are down to 6 elements, Gallium, Germanium, Arsenic, Selenium, Bromine and Krypton, all 6 elements are in the 4th period and have the p orbital as its last one
m represents the location of the electron within the orbitals, each orbital can hold 2 electrons only, p orbitals can hold 6 electrons in 3 pairs of 2, and the m value represents in which orbital the electron is, for l = 1, m can be -1, 0, +1
and in our question we have m = +1, therefore the electron is located in the last orbital, it can only be the 3rd electron or the 6th electron now, and now we are down to Arsenic and Krypton
s is the spin value of the electron, this value will tell us if the electron is pointing upwards or downwards in the orbital, this value can only be +1/2 (upwards) and -1/2 (downwards), and in our question, we have -1/2, which is the last electron of the orbital
Based on all that information, we conclude that this element is Krypton, with 36 of atomic number and electron configuration of [Ar]3d10 4s2 4p6
What is the molality of a solution containing 10.0 g Na»SO dissolvedin 1000.0 g of water?
Given Data
mass of solute Na 10.0 g
mass of solvent 1000.0g water
to calculate:
molality of solution
molalilty is defined as the mols of a solute divided by the Kg of the solvent
therefore mols of Na = mass/ molar mass
= 10.0g/22.98g/mol
= 0.44 mols
1 Kg = 10000. grams
Therefore molality of solution
= 0.44 mols/ 1 Kg
= 0.44 mols/Kg
When 1.72 grams of copper metal react with 12.33 grams of silver nitrate how much silver is produced?
The silver produced is 7.77 g.
Given
Mass of Cu (copper) = 1.72 g.
Mass of AgNO3 (silver nitrate) = 12.33 g.
Molar mass of Cu = 63.5 g/mol.
Molar mass of AgNO3 = 170 g/mol.
Molar mass of Ag = 108 g/mol.
Step-by-step solution:
Cu + 2AgNO₃ --> Cu(NO₃)₂ + 2 Ag
First, let's state the chemical equation with copper (Cu) and silver nitrate (AgNO3) reacting:
1.72 g Cu . [tex]\frac{1 mol Cu}{63.5 g Cu}[/tex] = 0.027 moles of Cu
12.33 g AgNO₃ . [tex]\frac{1 mol AgNO_{3} }{170 g AgNO_{3} }[/tex] = 0.072 moles of Ag
Now, let's find the number of moles of each reactant using their molar mass:
0.027 moles Cu . [tex]\frac{2 moles Ag}{1 mol Cu}[/tex] = 0.054 moles Ag.
Now, let's see how many moles of Ag are being produced based on the number of moles that we found in each reactant. By doing this calculation, we will find the limiting reactant.
You can see in the chemical equation that 1 mol of Cu reacted, produces 2 moles of Ag, so the moles of Ag produced are:
And you can see that moles of AgNO₃ reacts, produces 2 moles of Ag too. This means that the molar ratio between them is 2, more simply sis 1:1. This is telling u that 0.072 moles of AgNO₃ will produce 0.0072 moles of Ag too.
In this case, as you can see the limiting reactant is AgNO₃ because this is being consumed first in the reaction, so the final step is to find the mass using its molar mass. The conversion from 0.0072 moles of Ag to mass in grams is:
0.072 mol Ag . [tex]\frac{108 g Ag}{1 mol Ag}[/tex] = 7.77
The silver that is produced by 0.0072 moles is 7.77 g of Ag (silver).
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15. NI3 decomposes into Nitrogen gas and Iodine. If you start with 0.02 grams of NI3,How many moles of Nitrogen are produced at STP?
0.0000253moles
Explanations:The decomposition of NI3 is given as shown below;
[tex]2NI_3(s)→N_2(g)+3I_2(g)[/tex]Given the following parameters
Mass of NI3 = 0.02 grams
Determine the moles of NI3
[tex]\begin{gathered} moles\text{ of NI}_3=\frac{mass}{molar\text{ mass}} \\ moles\text{ of NI}_3=\frac{0.02}{394.719} \\ moles\text{ of NI}_3=0.0000507moles \end{gathered}[/tex]Acoording to stochiometry, 2 moles of NI3 preoduces 1 mole of nitrogen. The mole of nitrogen produced is;
[tex]\begin{gathered} mole\text{ of nitrogen}=\frac{1}{2}\times0.0000597 \\ mole\text{ of nitrogen}=\text{0.0000253moles} \end{gathered}[/tex]Hence the moles of nitrogen produced is 0.0000253moles
Can you describe the two different kinds of intermolecular forcesthat occur between diol molecules within a liquid?
A diol molecule is a compound that contains two Hydroxyl groups (OH), and these hydroxyl groups are attached to a hydrocarbon chain, which makes the main chain of the compound. In this type of molecule, we can find two intermolecular forces
Between the Carbons and Hydrogens in the hydrocarbon chain we have a Van der Waals force, which is a weak force
In the Hydroxyl groups, we will have a Hydrogen bond, which will be connecting Hydrogens and Oxygens
One of the reactions used to inflate automobile airbags involves sodium azide (NaN3): 2NaN3(s) ⇨ 2Na(s) + 3N2(g)Determine the mass of N2 produced from the decomposition of NaN3 shown at right
Step 1
The reaction provided:
2 NaN3 (s) ⇨ 2 Na (s) + 3 N2 (g) (completed and balanced)
---------------------
Step 2
Information provided:
100.0 g NaN3 decomposes
-----
Information needed:
The molar masses of:
NaN3) 65.00 g/mol
N2) 28.00
(please, the periodic table is useful here)
---------------------
Step 3
Procedure
By stoichiometry,
2 NaN3 (s) ⇨ 2 Na (s) + 3 N2 (g)
2 x 65.00 g NaN3 ----------------- 3 x 28.00 g N2
100.0 g NaN3 ----------------- X
X = 100.0 g NaN3 x 3 x 28.00 g N2/2 x 65.00 g NaN3
X = 64.62 g
Answer: 64.62 g of N2 produced
Dichlorobenzene exists in three forms. For each of the following write the name of the compound with a numerical designation indicating the location of the chlorine atoms A. Ortho - dichlorobenzene B. Para - dichlorobenzeneC. Meta - diclorobenzine
Please, look at the next drawing:
what is the water cycleA process by which water moves from cloud in the skyB A process by which water is absorbed into soil on earth surfaceC A process by which water moves from earth surface into the sky and back againD A process by which wind washes large amounts of water onto land from the oceans
Since water cycle is a multi-process cycle, the best option that describes and summarize all the cycle, will be letter C, in which we have water moving from the surface of the Earth to the sky and back again as rain, and then back to the sky again as water vapor.
Answer:
(C) A process by which water moves from Earth's surface into the sky and back again
Explanation:
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An unstable nucleusA-increases its half-lifeB - emits energy when it decaysC- increases its nuclear mass by fissionD- expels all of its protons
Explanation:
A nucleus consists o both electrons and proton. Therefore, an unstable nucleus undergoes radiation. And this causes a formation of a new atom. This is known as radioactive decay. Therefore, an unstable nucleus emits energy as it decays.
Answer:
The correct answer is B.herefore, an unstable nucleusherefore, an unstable nucleusR
Engineers increase the concentration of uranium-235 atoms in nuclear fuelduring what process?A. Uranium-235 modificationB. Half-lifeC. Nuclear fuel modificationD. Enrichment
Explanation:
The majority of nuclear reactors use the isotope uranium-235 as fuel; however, it is only a small part of the natural uranium mined. The concentration must be increased through a process called enrichment. Enrichment is the process of increasing the concentration of uranium-235 atoms in nuclear fuel.
Answer: D. Enrichment
Read the water level with the correct number of significant figures.
Reading significant figures. When you're reading signif
I believe the answer is 4 fold but I'm not sure
Its volume will be 12-fold.
From the Ideal Gas formula we know that:
[tex]\frac{P_{1\text{ }}\cdot V_1}{T_1}=\frac{P_2\cdot V_2}{T_2}[/tex]- P is the pressure os the gas.
- V is the volume of the gas.
- T is the temperature of the gas.
So, we can replace the hypothetical values to calculate the volume of the gas:
[tex]\begin{gathered} \frac{P_{1\text{ }}\cdot V_1}{T_1}=\frac{P_2\cdot V_2}{T_2} \\ \\ \frac{P_{1\text{ }}\cdot V_1}{T_1}=\frac{P_{1\cdot}\frac{1}{3}\cdot V_2}{T_1\cdot4} \\ V_1=\frac{P_{1\cdot}\frac{1}{3}\cdot V_2\cdot T_1}{P_{1\text{ }}\cdot T_1\cdot4} \\ V_1=\frac{_{}\frac{1}{3}\cdot V_2_{}}{_{}4} \\ V_1\cdot4\cdot3=V_2 \\ V_1\cdot12=V_2 \end{gathered}[/tex]Finally, its volume will be 12-fold.