Hydrogen (H-1), (H-2), and (H-3) are the three isotope of hydrogen. the values of a,b, and c in the table are: a=1 , b=1, c=2.
no. of proton no. of electron no. of neutron
H-1 1 1 0
H-2 a= 1 1 b= 1
H-2 1 1 c= 2
Isotope can be defined as the element having same no. of protons but different no. of neutrons. we can also defined as elements having same atomic number but different mass no. Hydrogen (H-1), (H-2), and (H-3) are the three isotopes of hydrogen. ¹H has only 1 proton and zero neutron having ,mass number 1. ²H has one proton and one neutron having mass no. 2. ³H has one proton and two neutron and having mass number 3.
Thus, Hydrogen (H-1), (H-2), and (H-3) are the three isotope of hydrogen. the values of a,b,and c in the table are: a=1,b=1,c=2.
no. of proton no. of electron no. of neutron
H-1 1 1 0
H-2 a = 1 1 b = 1
H-2 1 1 c = 2
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on a solid mixture, X. The inferences made are recorded in Table 2. Complete Table 2 by filling in the observations based on the inferences made. TABLE 2: TESTS ON MIXTURE X Test Observation Inferences a) Distilled water was added to a portion of X and the resulting mixture stirred and filtered. (The residue was set aside for use later.) The filtrate was divided into 3 equal portions and tests (b) to (d) done on separate portions. Cl- ions are present. b) Dilute nitric acid followed by a few drops of silver nitrate solution was added. Ammonium hydroxide solution was added to the resulting mixture. (3 marks)
Answer:
a) Upon adding silver nitrate, a white precipitate is observed
b) Upon adding ammonium hydroxide, the white precipitate dissolves to give a clear, colorless solution
Explanation:
Here, we want to state the observations when testing for chloride ions
From what we have:
a) When silver nitrate is added, a white precipitate is formed
This is as a result of the following chemical reaction:
[tex]Ag\placeholder{⬚}_{(aq)}^+\text{ + Cl}_{(aq)}^-\text{ }\rightarrow\text{ AgCl}_{(s)}[/tex]The AgCl is the white precipitate formed
b) Upon the addition of the ammonium hydroxide solution, a colorless and clear solution is observed showing that the white precipitate has dissolved
What is the volume, in mL, of a solution that is 0.421 M K2SO3 and was prepared by dissolving 16.7 g of solid potassium sulfite in water?
If carbon has undergone neutron capture what happened?
Answer:
It would generate a electro lithium nucleous
In aqueous solution, 5.2 mg of iron(III) chloride reacts with excess ammonium hydroxide.A. Write and kalance the equation, including phasesB. Is this a limiting reactant problem? Why or why not?C. Calculate the moles of each product that are formedD. Calculate the grams of each product that are formedI already answered part a and b. I need help on part c and d.
With a balanced reaction, we can determine the moles of products. The balanced equation will be:
FeCl3 + 3NH4OH → Fe(OH)3 + 3NH4Cl
They says that NH4OH is in excess, so, the limiting reactant will be FeCl3 and we will do all the calculations with this reactant.
We have to calculate the moles of FeCl3, we will use the molar mass:
[tex]\begin{gathered} molFeCl_3=5.2mg\times\frac{1g}{1000mg}\times\frac{1molFeCl_3}{MolarMass,gFeCl_3} \\ molFeCl_3=5.2mg\times\frac{1g}{1,000mg}\times\frac{1molFeCl_3}{162.2gFeCl_3}=3.2\times10^{-5}molFeCl_3 \end{gathered}[/tex]Now, to calculate the moles of the products we must take into account the product/reactive ratios, for this we are guided by the coefficients that accompany the molecules.
Ratio Fe(OH)3 to FeCl3 = 1/1
Ratio NH4Cl to FeCl3 = 3/1
Moles of each product
Moles of Fe(OH)3
[tex]molFe(OH)_3=3.2\times10^{-5}molFeCl_3\times\frac{1molFe(OH)_3}{1molFeCl_3}=3.2\times10^{-5}molFe(OH)_3[/tex]Moles of NH4Cl
[tex]molNH_4Cl=3.2\times10^{-5}molFeCl_3\times\frac{3molNH_4Cl}{1molFeCl_3}=9.6\times10^{-5}molNH_4Cl[/tex]The grams of each product we will find by multiplying the moles by the molar mass. So we have.
g of Fe(OH)3
[tex]\begin{gathered} gFe(OH)_3=3.2\times10^{-5}molFe(OH)_3\times106.87g/molFe(OH)_3 \\ gFe(OH)_3=3.4\times10^{-3}g=3.4mg \end{gathered}[/tex]g of NH4Cl
[tex]\begin{gathered} gNH_4Cl=molNH_4Cl\times MolarMassNH_4Cl \\ gNH_4Cl=9.6\times10^{-5}molNH_4Cl\times53.491g/molNH_4Cl=5.1\times10^{-3}g=5.1mg \end{gathered}[/tex]Acetone has a density of 0.7857 g/cm^3. What is the volume in mL of 5.52 g of acetone?
Let's see that the formula to find the volume using density and mass is:
[tex]V=\frac{m}{d}\begin{cases}V=\text{volume} \\ m=\text{mass} \\ d=\text{density}\end{cases},[/tex]In this case, the problem is asking about the volume in mL but remember that mL is the same that cm^3, so using the formula we're going to obtain:
[tex]V=\frac{5.52\text{ g}}{0.7857\text{ }\frac{g}{mL}}=7.02\text{ mL.}[/tex]The volume of 5.52 g of acetone is 7.02 mL.
Scientists are looking for ways to help increase the amount of calcium carbonate in oceans so that coral reefs can be healthy ecosystems again. Why is the amount of calcium carbonate in the ocean decreasing? (1 point)
Responses
Increased amounts of carbon dioxide cause reactions to happen in the water that decrease the pH and the amount of calcium carbonate.
Decreased amounts of carbon dioxide cause reactions to happen in the water that decrease the pH and the amount of calcium carbonate.
Increased amounts of carbon dioxide cause reactions to happen in the water that increase the pH and decrease the amount of calcium carbonate.
Decreased amounts of carbon dioxide cause reactions to happen in the water that increase the pH and decrease the amount of calcium carbonate.
Scientists are looking for ways to help increase the amount of calcium carbonate in oceans so that coral reefs can be healthy ecosystems again mount of calcium carbonate in the ocean decreasing because increased amounts of carbon dioxide cause reactions to happen in the water that decrease the pH and the amount of calcium carbonate
Calcium carbonate is a compound CaCO₃ found in nature as calcite and aragonite and in plant ashes, bones and shells and used especially in making lime and portland cement and as a gastric antacid and as ocean acidification increases available carbonate ion bond with excess hydrogen resulting in fewer carbonate ion available for calcifying organisms to build and maintain their shells, skeletons, and other calcium carbonate structures
Ocean acidification describe the lowering of sweater pH and carbonate saturation that result from increasing atmospheric CO₂ concentration and that's why increased amounts of carbon dioxide cause reactions to happen in the water that decrease the pH and the amount of calcium carbonate.
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Answer: Its C I did my research and that's what makes sense to me. because increased amount of carbon dioxide decreases the PH and the amount of calcium carbonate.
PLEASE HELP!
In a monomolecular reaction A->B , at t =250C, the initial concentration decrease at 25% in t =52 min. Calculate:
a) the constant rate;
b) the time after the initial concentration decrease with 75%;
c) the initial reaction rate, if the initial concentration of the reactant is 2.5 mol/L·s
1) The calculate rate constant is 9.22 * 10^-5 s-1
2) In the period of is 15033 s 75% is used up
3) From the calculation, there is an initial 9.22 * 10^-5 M.
How can we find reaction rate?In chemistry, the rate of reaction would give the idea that the reaction is proceeding quickly or slowly. If a reaction has a large rate of reaction then it tends to move on to completion.
From the question, it is clear that there is 25% in t =52 min.
Initial concentration [A]o = [A]o
Final concentration = [A]o - 0.25 [A]o = 0.75 [A]o
Time taken = 52 min or 3120 s
From the formula that can be applied to a first order reaction;
ln[A] = -kt + ln[A]0
k = -(ln[A] - ln[A]0)/t
k = - (ln0.75 [A]o/A]0)/3120
k = 9.22 * 10^-5 s-1
b)
Then we have to find the time after the initial concentration decrease with 75%
[A] = [A]o - 0.75 [A]o = 0.25 [A]o
ln[A] = -kt + ln[A]0
t = -(ln[A] - ln[A]0)/k
t = - (ln0.25 [A]o/A]0)/9.22 * 10^-5
t = 15033 s
c) Given that in this case, the initial concentration of the reactant is 2.5 mol/L·s
k = - (ln0.25 (2.5)/ln(2.5))/15033
k = 9.22 * 10^-5 M
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Calculate the number of mol corresponding to 18.8 g Na2SO4.
Answer: 0.132 moles of Na2SO4
Explanation:
Multiply the grams of Na2SO4 by the 1/molar mass of Na2SO4, which is the sum of all the elements 2(Na)+S+4(O)= 142.04.
In the following equation, how many grams of O2 are needed to react with 24.0 g of NH3?
4 NH3(g)+302(g) → 2N2(g)+6H₂O(1)
a) 45.1
b) 22.6
c) 33.9
d) 45.0
e) 135.5
33.6 g grams of O2 are needed to react with 24.0 g of NH3.
The molecular mass of NH3 = 14 + 3 = 17g
Given mass of NH3 = 24g
Firstly, we will calculate the number of moles.
Moles is defined as the ratio of given mass of substance to the molecular mass of substance.
Moles = given mass/ molecular mass
Number of moles of NH3 = 24/17
= 1.4 mole
Chemical reaction4 NH3(g)+302(g) → 2N2(g)+6H₂O(1)
4 moles of NH3 require 3 moles of O₂ to react.
1 moles of NH3 require 3/4 moles of O₂ to react.
1.4 moles of NH3 require 1.05 moles of O₂ to react.
Now we calculate the grams of O₂.
1.05 × 32 = 33.6 g.
Thus, we concluded that the 33.6 g grams of O2 are needed to react with 24.0 g of NH3.
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Explain how you would calculate the total change in bond energy for the reaction H2+Cl2->2HCI. How would you know if the reaction was endothermic or exothermic?
To calculate the total change in binding energy we must apply the following equation:
[tex]\Delta H=\Delta H_{f(\text{Reagents)}}-=\Delta H_{f(\text{Products)}}[/tex]This equation tells us that the energy change will be equal to the sum of the potential energy of the product bonds minus the sum of the potential energy of the reagent bonds.
Now, let's calculate each term separately.
[tex]\begin{gathered} \Delta H_{f(\text{Reagents)}}=n_{H2}\times\text{Energy bond H-H + }n_{Cl2}\text{Energy bond Cl-Cl} \\ \Delta H_{f(\text{Reagents)}}=1molH_2\times432\frac{kJ}{mol}+1molCl_2\times239\frac{kJ}{mol} \\ \Delta H_{f(\text{Reagents)}}=671kJ \end{gathered}[/tex][tex]\begin{gathered} \Delta H_{f(\text{Products)}}=n_{\text{HCl}}\times EnergyBond\text{ H-Cl} \\ \Delta H_{f(\text{Products)}}=2molHCl\times427\frac{kJ}{mol} \\ \Delta H_{f(\text{Products)}}=854kJ \end{gathered}[/tex]So, the change in bond energy will be:
[tex]\begin{gathered} \Delta H=\Delta H_{f(\text{Reagents)}}-=\Delta H_{f(\text{Products)}} \\ \Delta H=671kJ-854kJ=-183kJ \end{gathered}[/tex]We have a negative value in the result, when this happens it means that the reaction is exothermic, that is to say, that it releases heat and the energy of the products is greater than that of the reagents.
When we have a positive value the reaction will be endothermic, this means that it needs energy.
Phthalic acid ( H2C8H4O4) is a diprotic acid with a1=1.12×10−3 and a2=3.90×10−6. Determine the pH of a 0.294M phthalic acid ( H2C8H4O4 ) solution.
pH of the acid can be calculated from the first and second ionization constants of the acid. The pH of 0.294 M of Phthalic acid is 0.53.
What is pH?pH of a solution is the measure of its H+ ion concentration.It measures the acidity or basicity of the solution. Mathematically it is the negative logarithm of hydrogen ion concentration.
pH = -log [H+]
Give that the first ionization constant of the acid is 1.12 × 10 ⁻³ and second ionization constant is 3.90 × 10⁻⁶. The diprotic acid can be represented as H2A and after its first ionization it produce HA- and H+ On the second ionization the base A- and H+ ion is produced.
The second ionization constant a₂ is written as follows:
a₂ = [A-] [H+] / [HA - [A]] = 3.90 × 10⁻⁶
From this simply we can get [A] = 3.90 × 10⁻⁶
Thus [H+] = H₂A - [A]
The concentration of the diprotic acid H₂A is given 0.294 M.
Thus [H+] = 0.294 - 3.90 × 10⁻⁶
= 0.2933 M
Now, the pH is calculated as follows:
pH = -log (0.2933)
= 0.53.
Therefore, the pH of 0.294 M of Phthalic acid is 0.53.
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Find the volume in liters of the 0.505 molar NaOH solution needed to react with 40 milliliters of the 0.505 molar H2SO4 solution.
A//: 80 mililiters
Please, it's urgent, I need it ASAP
The volume of NaOH required is 80ml.
Sulfuric acid is a dibasic acid and NaOH is monoacidic base.
M1V1 = M2V2
where, M1 = initial concentration,
V1 = initial volume,
M2 = concentration after mixing
V2 = total final volume.
This formula is used for calculating the final volume or molarity after mixing two solutions.
Given the question,
M1 = 0.505M
V1 =?
M2 = 0.505M
V2 = 40 ml
M1V1 = 2×M2V2
0.505 × V2 = 2 × 0.505 × 40
V2 = 2×0.505×40
0.505
V2 = 80ml
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What are the strengths, weaknesses, and implications of risk/benefit analysis as a method for deciding whether a technology should be employed?
A risk-benefit analysis compares the risks and benefits of a situation and determines whether the advantages outweigh the disadvantages.
What is Risk-benefit analysis in technology?Strengths
Risk-benefit analysis calculates the amount of time will be worth it to the production of technology and whether the technology will have a healthy impact on the industry or not.
Weaknesses
Risk-benefit analysis cannon determine product implementation and the outcomes of real life experiences of individual customers. It has some drawbacks like benefits of customer is take under consideration but pollution in nature is not calculated.
Implication
Risk-benefit analysis is implied in almost all technical industry as it the decision maker of any developing team to work on a particular project or not. Some examples are automobile industry and smartphone industry.
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At equilibrium, a reaction vessel contains 4.50 atm of Br₂ and 1.10 atm of NBr₃. According to the reaction: 2 NBr₃ (g) ⇌ N₂ (g) + 3 Br₂ (g) Kp = 4.8
Determine the equilibrium partial pressure of N₂.
Equilibrium constant is the only concept that is to be used here to calculate partial pressure of nitrogen. The partial pressure of Nitrogen comes out to be 1.17atm
What is equilibrium constant?Equilibrium constant is a rate constant at equilibrium shows the values of reaction with respect to the atmospheric pressure and concentration
[tex]K_{c}[/tex] is equilibrium constant with respect to concentration.
[tex]K_{p}[/tex] is equilibrium constant with respect to atmospheric pressure.
Mathematically,
[tex]K_{p}=\frac{[N_{2}][Br_{2} ] }{[NBr_{3}] }[/tex]
Substituting values
(4.8×1.10atm)÷ 4.50atm=[tex]{[N_{2}][/tex]
1.17atm=[tex]{[N_{2}][/tex]
Thus the equilibrium partial pressure of N₂ is 1.17atm.
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2Al + 3Cl₂ 2AlCl₃molar mass of AlCl3 => 133 g/molIf 25.0 moles of chlorine is used, how many grams of aluminum chloride can be produced?
Answer
Explanation
Given:
2Al + 3Cl₂ 2AlCl₃
Molar mass of AlCl3 => 133 g/mol
Moles of Cl₂ = 25.0 mol
What to find:
The grams of aluminum chloride that can be produced.
Step-by-step solution:
Step 1: Determine the mole of AlCl₃ produced.
From the given chemical equation for the reaction,
3 mol Cl₂ produce 2 mol AlCl₃
So 25.0 mol Cl₂ will produce
[tex]\frac{25.0mol\text{ }Cl_2\times2mol\text{ }AlCl₃}{3mol\text{ }Cl_2}=16.67mol\text{ }AlCl₃[/tex]16.67 moles of AlCl₃ produced.
Step 2: Calculated the grams of AlCl₃ produced.
Using the mole formula
[tex]undefined[/tex]Uranium-238 is an unstable nuclide that emits an alpha particle, three neutrons, and
gamma particles. A = 231 and Z = 90. What daughter nuclide is formed from this
reaction?
OA) Ra
232U2a + 2x + 3 n + Y
-
B) Pu
C) Th
D) Fr
Uranium-238 is an unstable nuclide that emits an alpha particle, three neutrons, and
gamma particles and produces a daughter nuclide Thorium.
The atomic mass of alpha particle is 4g and atomic number 2. Since an alpha particle emits, so there is reduction in the atomic mass 4 and atomic number 2.
The atomic number of neutron is 0 and the atomic mass of neutron is 1. When three neutrons emit the atomic mass reduces to 3 while atomic number remains same.
Gamma particles have no mass and charge. So, atomic number and atomic mass remains same.
Total reduction in atomic mass : 238 - 4 - 3 = 231.
Total reduction in atomic number = 92- 2 = 90
Therefore, the daughter nuclide is Thorium.
Thus, we concluded that the Uranium-238 is an unstable nuclide that emits an alpha particle, three neutrons, and gamma particles and produces a daughter nuclide Thorium.
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Convert 0.05090Kg/mol to dg/mmol
Answer: .509
Explanation: Im pretty sure
How should the electrode of a pH meter be preserved? Explain your answer.
We can care for the pH meter electrode by;
1) Making sure that the electrode remains moist
2) The electrode should be stored in a 4M solution of KCl
3) Electrodes should not come in contact with deionized water.
What is pH?The term pH has to do with the negative logarithm of the hydrogen ion concentration. Now we know that the pH of a solution tells us the amount of the hydrogen ions or the hydroxyl ions that is present in the solution. The pH mete is the instrument that we could use to be able to measure the pH of the solution.
The pH scale runs between 0 - 14. The points on the scale that have been labeled from 0 - 6 tells us that the solution is an acidic solution and contains more hydrogen ions. If the solution has a pH of 7, then it is neutral and contains equal concentration of hydrogen and hydroxyl ions. A solution of pH 8 - 14 is basic and contains more hydroxyl than hydrogen ions.
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How many grams of Br are in 395 g CaBr2 ?
In order to answer this question we will need to use the molar mass of Calcium bromide (CaBr2), which is 199.89 g/mol, now we can find how much of this mass is only Bromide, so what is the percentage of Bromide that makes up this whole compound, to do so we will also use the molar mass of Bromide (Br2) which is 159.8 g/mol
199.89 g/mol = 100% of the compound
159.8 g/mol = x %
x = roughly 80% (is like 79.9%)
So, regardless of the mass of CaBr2, bromide needs to be 80% of its mass, now solving our question
395 g = 100%
x grams = 80%
x = 316 grams is the mass of Bromide in the compound
Gaseous ethane (CH3CH3) reacts with gaseous oxygen gas (O2) to produce gaseous carbon dioxide (CO2) and gaseous water (H2O). if 2.46 g of carbon dioxide is produced from the reaction of 2.71 g of ethane and 16.7 g of oxygen gas, calculate the percent yield of carbon dioxide. Round to 3 sig figs
Taking into account definition of percent yield, the percent yield of carbon dioxide is 30.90%
Reaction stoichiometryIn first place, the balanced reaction is:
2 CH₃CH₃ + 7 O₂ → 4 CO₂ + 6 H₂O
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:
CH₃CH₃: 2 molesO₂: 7 molesCO₂: 4 molesH₂O: 6 molesThe molar mass of the compounds is:
CH₃CH₃: 30 g/moleO₂: 32 g/moleCO₂: 44 g/moleH₂O: 18 g/moleThen, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
CH₃CH₃: 2 moles ×30 g/mole= 60 gramsO₂: 7 moles ×32 g/mole= 224 gramsCO₂: 4 moles ×44 g/mole= 176 gramsH₂O: 6 moles ×18 g/mole= 108 gramsLimiting 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.
Limiting reagent in this caseTo determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 224 grams of O₂ reacts with 60 grams of CH₃CH₃, 16.7 grams of O₂ reacts with how much mass of CH₃CH₃?
mass of CH₃CH₃= (16.7 grams of O₂× 60 grams of CH₃CH₃)÷ 224 grams of O₂
mass of CH₃CH₃= 4.47 grams
But 4.47 grams of CH₃CH₃ are not available, 2.71 grams are available. Since you have less mass than you need to react with 16.7 grams of O₂, CH₃CH₃ will be the limiting reagent.
Percent yieldThe percent yield is the ratio of the actual return to the theoretical return expressed as a percentage.
The percent yield is calculated as the experimental yield divided by the theoretical yield multiplied by 100%:
percent yield= (actual yield÷ theoretical yield)×100%
where the theoretical yield is the amount of product acquired through the complete conversion of all reagents in the final product, that is, it is the maximum amount of product that could be formed from the given amounts of reagents.
Theoretical yield of CO₂Considering the limiting reagent, the following rule of three can be applied: if by reaction stoichiometry 60 grams of CH₃CH₃ form 176 grams of CO₂, 2.71 grams of CH₃CH₃ form how much mass of CO₂?
mass of CO₂= (2.71 grams of CH₃CH₃× 176 grams of CO₂)÷ 60 grams of CH₃CH₃
mass of CO₂= 7.96 grams
Then, the theoretical yield of CO₂ is 7.96 grams.
Percent yield for the reaction in this caseIn this case, you know:
actual yield= 2.46 gramstheorical yield= 7.96 gramsReplacing in the definition of percent yields:
percent yield= (2.46 grams÷ 7.96 grams)×100%
Solving:
percent yield= 30.90%
Finally, the percent yield for the reaction is 30.90%.
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If 335g water at gains 102.3 J of heat, how much does the temperature of the water change? The specific heat of water is 4.184 J/g*C
Temperature is a measure of how hot a substance or radiation is expressed numerically.
There are three different types of temperature scales:
those that depend only on macroscopic properties and thermodynamic principles, like Kelvin's original definition;
those that depend on practical empirical properties of particles rather than theoretical principles;
and those that are defined by the average translational kinetic energy per freely moving microscopic particle, like an atom, molecule, or electron, in a body, like the SI scale.
How much heat is gained or lost by a sample can be calculated using the equation q = mcΔT, where m is the mass of the sample, c is the specific heat, and T is the temperature change (q).
Therefore,
q = m*c*ΔT
102.3 = 335 * 4.184 * ΔT
ΔT = 0.073 °c
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Which of the following changes are chemical changes?
Answer:
flammability, toxicity etc..
Explanation:
are chemical changes
You have 0.834 moles of Potassium. The NIH recommends 2,600mg of potassium as a daily intake recommendation. Convert the moles to mg.
Answer:
0.834 mole = 13293.32 mg
2,600 mg = 0.16 mole
Explanation:
1 mole of K = 39.098 g
=> 0.34 mole has (0.34)(39.098) = 13.29332g
13.29332 g x 10^3 = 13293.32 mg
so 0.834 mole of K = 13293.32 mg
mole of 2,600mg = (0.834)(2,600)/(13293.32) = 0.16311952168 or 0.16 mole
Which is an example of plasmas in nature?
Answer: nbhhvyyvub
Explanation:
What is the oxidation number or Br in NaBrO?
Give the names of following compounds (in aqueous solution):
HBr
7 HBrO
2 HCN
3 H₂CO3
4 HC₂H302
5 H₂SO4
6 H3ASO3
8 HNO2
9 HC104
10 H₂C2O4
11 H3PO4
12 H₂CrO4_
Here are all names of compounds in aqeous solution.
What is aqeous solution?
An aqeous solution is one in which the solvent is liquid water .
In this solution water act as a solvent.
Sol-
HBr- Hydrogen bromide
1-HBrO- hydrobromic acid
2-HCL-hydrochloric acid
3-H2O3- hydrogen peroxide
4- HC2H3O2- glacial acetic acid
5-H2SO4-sulfuric acid
6- H3ASO3- Arsenous acid
7- HNO2- Nitrous acid
8- HC104- Perchloric acid
9-H2C2O4- oxalic acid
10- H3pO4- phosporic acid
11-H2CrO4- chromic acid
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can atoms of different elements have the same atomic number?
No, Atoms of different element can not have same atomic number because only same type of atoms combine to form element. Atoms belonging to different element can have different atomic number.
What is element?Element generally consist of atoms or we can atoms combine to form element. Atoms of an element is always same, means all the properties of all atoms of one type of element is same. Two or more than two atoms with different physical or chemical properties can not combine together to form an element.
So we can say that atomic number of all atoms that is constituting an element is same. It can never be different.
Thus Atoms of different element can not have same atomic number.
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How does temperature of land far from water compare to that of land near water?
The temperature of land far from water is hotter than land near water.
The temperature of land far from water is less stable compared to land near water.
The temperature is similar for land both near and far from water.
The temperature is either too hot or too cold near water.
The temperature of land far from water is hotter than land near water.
Why the temperature of land far away from water different from land near water?The lower heat capacity of land often allows them to cool the nearby water temperature so it takes less energy to change the temperature of land compared to water bodies. This means that land heats and cools more quickly as compared to water. This difference affects the climate of different areas on Earth. Large bodies of water like oceans, seas and large lakes can affect the climate of the nearby regions such as coastal regions. Water heats and cools more slowly than land regions. The coastal regions will stay cooler in summer season and warmer in winter season, which creates moderate climate on the coastal regions.
So we can conclude that the temperature of land far from water is different from land near water because of cooling effect of water bodies.
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13. A gas of unknown molecular mass was allowed to effuse through a small opening under
constant pressure conditions. It required 72 s for the gas to effuse. Under identical
experimental conditions, it required 28 s for O₂ gas to effuse. Determine the molar mass
of the unknown gas.
The molar mass of unknown gas is 211.59 g/mol.
What is effusion of gases?Effusion occurs when a gas pass through an opening that is smaller than the mean free path of the particles, which is the average distance traveled between collisionsGraham's law is an empirical relationship which states that ratio of the rates of diffusion or effusion of two gases is the square root of the inverse ratio of their molar masses.Higher the molar mass of a gas, slower the effusionSince both the gases are present in identical experimental condition, using Grahams law of effusion:
[tex]\frac{time req for unknown gas}{time of oxygen}[/tex] = [tex]\sqrt{\frac{Molar mass of unknown gas}{molar mass of Oxygen} }[/tex]
Given:
Time of unknown gas = 72 sec
Time of Oxygen = 28 sec
we know the molar mass of oxygen = 32 g/mol
Now substituting:
[tex]\frac{72 sec}{28 sec} = \sqrt{\frac{M.M of unknown}{32g/mol} }[/tex]
M.M of unknown = (72 / 28)² × 32 g/mol
= 211.59 g/mol
So the molar mass of unknown gas is 211.59 g/mol
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Following the Experiment 3 procedure, you combine 50.0 mL H2O, 50.0 mL of 2.2 M HCl(aq), and 4.039 g NaOH(s) (molar mass of NaOH = 40.00 g/mol) in a Styrofoam calorimeter, recording a maximum solution temperature change of 22.9 oC and a final solution mass of 104.153 g (csolution = 4.184 J/g.oC). Calculate the molar change in enthalpy of reaction
The molar change in enthalpy of reaction is 99.6 kJ/mol.
the formula for the specific heat capacity expressed as :
Q = mcΔT
where,
m , mass = 104.153 g
c, specific heat = 4.184 J/g °C
dt, change in temperature = 22.9 °C
Q = 104.153 × 4.184 × 22.9
Q = 9969.7 J = 9.96 kJ
now, the molar change in enthalpy is give by:
Q = ΔH / n
n is no. of moles
morality of HCl = 2.2 M
v = 50 mL = 0.05 L
n = 0.05 × 2.2
= 0.11 moles
no. of moles of NaOH = mass / molar mass
= 4.039 / 40
= 0.100
NaOH is limiting reactant.
using the formula we get:
Q = ΔH / n
ΔH = Q / n
ΔH = 9.96 kJ/ 0.10
ΔH, change in enthalpy = 99.6 kJ/mol
Thus, combine 50.0 mL H₂O, 50.0 mL of 2.2 M HCl(aq), and 4.039 g NaOH(s) (molar mass of NaOH = 40.00 g/mol) in a Styrofoam calorimeter, recording a maximum solution temperature change of 22.9 °C and a final solution mass of 104.153 g (c solution = 4.184 J/g°C). the molar change in enthalpy of reaction is 99.6 kJ/mol.
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