Answer : The molarity of the NaOH solution is, 0.00687 M.
Explanation :
The balanced chemical reaction will be:
[tex]HCl+NaOH\rightarrow NaCl+H_2O[/tex]
Using neutralization method :
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]HCl[/tex]
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is NaOH.
We are given:
[tex]n_1=1\\M_1=0.0100M\\V_1=20.60mL\\n_2=1\\M_2=?\\V_2=30.00mL[/tex]
Putting values in above equation, we get:
[tex]1\times 0.0100M\times 20.60mL=1\times M_2\times 30.00mL\\\\M_2=0.00687M[/tex]
Hence, the molarity of the NaOH solution is, 0.00687 M.
Energy is released during which phase changes
How many carbon atoms will be in the longest chain 3,3,4,4-tetramethylhexane
What are three ways that carbon atoms can be arranged in hydrocarbon molecules?
Which of the following best describes the changes involved in a fission reaction?
The nucleus of an atom splits into fragments, releasing a large amount of energy.
The nucleus of an atom splits into fragments, absorbing a large amount of energy.
The nuclei of two small atoms combine to form one larger nucleus, releasing a large amount of energy.
The nuclei of two small atoms combine to form one larger nucleus, absorbing a large amount of energy.
In a fission reaction, the nucleus of an atom splits into fragments, releasing a large amount of energy. Nuclear fission occurs when a heavy nucleus like uranium absorbs a neutron and splits, emitting additional neutrons, gamma rays, and energy.
The correct description of the changes involved in a fission reaction is: The nucleus of an atom splits into fragments, releasing a large amount of energy. Nuclear fission is a nuclear reaction where the nucleus of an atom, such as uranium-235 or plutonium-239, splits into smaller nuclei after absorbing a neutron.
This process produces additional free neutrons, gamma radiation, kinetic energy of fission fragments, and releases a significant amount of energy. The process is exothermic, and the energy released can be harnessed for electricity generation in nuclear reactors or can be used in nuclear weapons.
Nuclear fission differs from nuclear fusion, which is the combining of two smaller atomic nuclei to form a larger nucleus, also releasing energy. In both processes, large amounts of heat and radiation are emitted. However, for fission, a fissile material like uranium or plutonium is necessary to sustain the reaction, and it involves the breakup of a heavy nucleus into lighter elements.
how many grams of lithium are needed to produce 45.0 grams of lithium nitride, according to the following process?
6 Li(s) + N2(g) → 2 Li2N(s)
Help! Brainliest will be given!
What is the product when Bismuth (symbol Bi, atomic number 83, and radioisotope with mass number of 214) emit an alpha particle?
Give the name (or symbol) of the new element, its atomic number, and its mass.
Answer : The correct answer is Name = Thallium (Th) , atomic mass = 210 and atomic number = 81 .
Alpha decay :
When atomic nucleus emits an alpha particle that process is known as alpha particle decay . The symbol of alpha particle is [tex] ^4_2He [/tex] . where 4 is atomic mass ( 2 protons + 2 neutrons ) and 2 atomic number .
When alpha particle is released the resultant nuclei (daughter nuclei) have 4 less atomic mass and 2 less atomic number .
Example for nuclear equation of alpha decay can be expressed as:
[tex] {_{90}^{230}Th} \rightarrow {_2^4He} + {_{88}^{226}Ra} [/tex]
Given :
Radioisotope = Bi Atomic mass = 214 Atomic number = 83
When it will release alpha particle , atomic mass will be decreased by 4 and atomic number by 2 .
Atomic mass = 214 - 4 = 210 Atomic number = 83-2 = 81
The atom with atomic number 81 is Thallium (Th) . Hence the daughter nuclei so produced is [tex] ^2^1^0_8_1Th [/tex]
The nuclear reaction can be written as :
[tex] ^{214}_{83}Bi \rightarrow _2^4He + _{81}^{210}Th [/tex]
When fracking liquid waste is left in pools on the surface, _ can evaporate into the air and contribute to pollution
Volatile organic compounds (VOCs) can evaporate from fracking liquid waste pools, contributing to air pollution. These pools may also contain hazardous air pollutants and heavy metals, requiring careful handling to avoid environmental contamination.
When fracking liquid waste is left in pools on the surface, volatile organic compounds (VOCs) can evaporate into the air and contribute to pollution. These wastewater ponds can contain a variety of pollutants, including hazardous air pollutants such as benzene, toluene, ethylbenzene, and xylene. Moreover, fracking fluid, also known as flowback, can contain chemicals used in the drilling process, heavy metals, and radioactive materials. These substances pose a significant risk to both environmental and human health if they are not properly managed and treated.
The process of hydraulic fracturing or 'fracking' involves injecting high-pressure fluids to fracture shale deposits, which releases trapped gas and oil. The wastewater from this process may return to the water cycle, but the large volume of contaminated water requires careful handling to prevent land and water pollution. As a proactive measure, governments around the world have taken steps, in some cases banning the practice due to the severe risks associated with fracking.
Write the balanced nuclear equation for alpha decay of polonium−218. include both the mass numbers and the atomic numbers with each nuclear symbol. use the "sup-subscript" button in the answer palette to enter these numbers correctly. greek letters can be accessed in the drop-down menu that says "-select−."
Ka for hcn is 4.9 ⋅ 10-10. what is the ph of a 0.068 m aqueous solution of sodium cyanide? ka for hcn is 4.9 10-10. what is the ph of a 0.068 m aqueous solution of sodium cyanide? 0.74 7.00 2.96 13.24 11.07
Answer
is: pH of solution of sodium cyanide is 11.07.
Chemical reaction 1: NaCN(aq) → CN⁻(aq) + Na⁺(aq).
Chemical reaction 2: CN⁻ + H₂O(l)
⇄ HCN(aq) + OH⁻(aq).
c(NaCN) = c(CN⁻) = 0.068 M.
Ka(HCN) = 4.9·10⁻¹⁰.
Kb(CN⁻) = 10⁻¹⁴ ÷ 4.9·10⁻¹⁰ = 2.04·10⁻⁵.
Kb = [HCN] · [OH⁻] / [CN⁻].
[HCN] · [OH⁻] = x.
[CN⁻]
= 0.068 M - x..
2.04·10⁻⁵ = x² / (0.068 M - x).
Solve quadratic equation: x = [OH⁻] = 0.00116 M.
pOH = -log(0.00116 M) = 2.93.
pH = 14 - 2.93 = 11.07.
The pH of the sodium cyanide solution is 11.56.
Let the cyanide ion be X
We have to set up the ICE table for the problem as follows;
X^-(aq) + H2O(l) ⇄ HX(aq) + OH^-(aq)
I 0.068 x x
C -x + x +x
E 0.068 - x x x
But Kb = Kw/Ka = 1 × 10^-14/4.9 × 10-10
Kb = 2 × 10^-4
So;
Kb = [HX] [OH^-]/[X^-]
2 × 10^-4 = x^2/ 0.068 - x
2 × 10^-4(0.068 - x) = x^2
1.36 × 10^-5 - 2 × 10^-4x = x^2
x^2 + 2 × 10^-4x - 1.36 × 10^-5 = 0
x= 0.0036 M
Since x = [OH^-] = 0.0036 M
pOH = - log(0.0036 M)
pOH = 2.44
pH = 14 - 2.44 = 11.56
The pH of the sodium cyanide solution is 11.56.
Learn more about pH: https://brainly.com/question/8139015
Which of the following are indicators of a chemical change? Select all that apply.
color change
temperature change
precipitate formation
gas formation
cutting a solid into smaller pieces
Is it possible to get the elliptic discriminate of an elliptic curve with characteristic 2?
What term refers to group behavior seen in animals of a similar type?
Answer:
The correct answer is swarm behavior.
Explanation:
Swarming or swarm behavior is a collective behavior demonstrated by entities, mainly the animals of same size that accumulate together, specifically moving in masses or migrating in some direction or milling about the same spot. As a term, swarming is applicable mainly to insects, however, it can also be applied to any other animal or entity, which demonstrates swarm behavior.
Which phase(s) would experience an increase in the rate of dissolution with an increase in temperature?
A. gases only
B. solids only
C. liquids only
D. both solids and liquids
Increasing ratios of oxygen-18 to oxygen-16 in glacial ice indicate ________.
a. advancing ice sheets
b. warming temperatures
c. increased volcanic emissions
d. uplift of land surfaces
An increased ratio of oxygen-18 to oxygen-16 in glacial ice indicates warming temperatures as oxygen-18 is a heavier isotope and precipitates less readily in cold conditions.
Explanation:The ratio of oxygen-18 to oxygen-16 in glacial ice is often used as an indicator of past climatic conditions. When this ratio is increasing, it typically indicates warming temperatures.
This is because oxygen-18 is a heavier isotope that precipitates less readily in cold conditions
Therefore, a higher ratio of oxygen-18 suggests that it was warm enough for more of this heavier isotope to precipitate. So, the correct answer is b. warming temperatures.
Learn more about Isotopes here:https://brainly.com/question/19354416
#SPJ11
What is the main difference between DNA and RNA? DNA contains ribose sugar, and RNA contains deoxyribose sugar instead. DNA contains cytosine, and RNA contains guanine instead. DNA contains deoxyribose sugar, and RNA contains ribose sugar instead. DNA contains adenine, and RNA contains guanine instead.
DNA contains ribose sugar, and RNA contains deoxyribose sugar instead. Therefore, the correct option is option A.
What are DNA and RNA?The complex molecular structures deoxyribonucleic acid (DNA) consequently ribonucleic acid (RNA) regulate all genetic traits of cells, and hence of species. DNA, which makes up the genetic material of all free-living species, is the ultimate life-plan. In addition to being the genetic makeup of some viruses, RNA employs DNA to code out the molecular makeup of proteins made in cells.
The two structures differ chemically in a number of ways. The "backbone" of DNA, or deoxyribose, is a sugar, like the acronym deoxyribonucleic acid suggests. The sugar ribose, which is slightly different, is present in ribonucleic acid (RNA). DNA contains ribose sugar, and RNA contains deoxyribose sugar instead.
Therefore, the correct option is option A.
To know more about DNA and RNA, here:
https://brainly.com/question/29767255
#SPJ3
The chemical equation below shows the decomposition of nitrogen triiodide (NI3) into nitrogen (N2) and iodine (I2). 2NI3 mc030-1.jpg N2 + 3I2 The molar mass of I2 is 253.80 g/mol, and the molar mass of NI3 is 394.71 g/mol. How many moles of I2 will form 3.58 g of NI3?
Answer:
0.0135 moles of iodine will be formed.
Explanation:
[tex]2NI_3\rightarrow N_2+3I_2[/tex]
Moles of nitrogen triiodide =[tex]\frac{3.58 g}{394.71 g/mol}=0.0090 mol[/tex]
According to reaction 2 moles of nitrogen triiodide gives 3 moles of iodione gas.
Then 0.0090 mol of nitrogen triiodide will give:
[tex]\frac{3}{2}\times 0.0090 mol=0.0135 mol[/tex]
0.0135 moles of iodine will be formed.
Which is the best description of a chain of custody?
A) A list of all personnel entering the crime scene.
B) A log of which family member has ownership of the evidence.
C) A log of which agency controls the evidence so the case stays in the correct jurisdiction.
D) A log of who has handled the evidence and when so the integrity of the evidence is upheld.
Select the incorrect statement.
The physical properties of a compound are similar to those of its elements.
The boiling point of a compound is different from the individual elements.
The melting point of a compound is different from the individual elements.
Density is a physical property.
Answer:
The incorrect statement is:
The physical properties of a compound are similar to those of its elements.
Explanation:
The physical properties of a compound depend on its elements but they are not similiar to the physical properties of individual elements.
Take water as an example. Water is made by oxygen and hydrogen (H₂O). At normal , water is liquid and both oxygen and hydrogen are gases.
What volume of 6.00 m naoh solution is required to prepare 0.50 ml of 0.15 m naoh solution
a. 12.5 ml of naoh
b. 25 ml of naoh
c. 30 ml of naoh?
Final answer:
To prepare 0.50 mL of a 0.15 M NaOH solution from a 6.00 M NaOH solution, you would need 12.5 μL of the concentrated solution according to the dilution formula, which is not listed among the provided answer options.
Explanation:
The volume of 6.00 M NaOH solution required to prepare 0.50 mL of 0.15 M NaOH solution can be found using the dilution formula M1V1 = M2V2, where M1 and V1 are the molarity and volume of the concentrated solution, and M2 and V2 are the molarity and volume of the diluted solution, respectively. Plugging the known values into the equation: (6.00 M)(V1) = (0.15 M)(0.50 mL) ⇒ V1 = (0.15 M)(0.50 mL) / (6.00 M) = 0.0125 mL or 12.5 μL.
Since none of the provided answer choices (a) 12.5 mL, (b) 25 mL, or (c) 30 mL match this result, it seems there might be a typo or a misprint in the question. To prepare 0.50 mL of a 0.15 M solution from a 6.00 M solution, you would need substantially less than 1 mL of the concentrated solution. The likely correct volume needed, based on the question's data, would be 12.5 μL, which is not listed as an option in the question.
In the chemical reaction: 2 h 2 (g) o2 (g) → 2 h2o (g), with 8 grams of h 2 (2 grams/mole) and 16 grams of o2 (32 grams/mole) and the reaction goes to completion, what is the excess reactant and how much of that would remain
PLEASE HELP ASAP!!!
Which statement about the following reaction is correct?
CH4 (g) + 2O2 (g) yields CO2 (g) + 2H2O(l) deltaH = -890 kJ
reacting one mole of oxygen (O2) absorbs 445 kJ of energy
reacting one mole of oxygen (O2) releases 445 kJ of energy
reacting one mole of methane (CH4) absorbs 890 kJ of energy
reacting two moles of methane (CH4) releases 890 kJ of energy
Final answer:
The correct statement is that reacting one mole of oxygen (O2) releases 445 kJ of energy during the exothermic reaction of CH4 with O2.
Explanation:
The correct statement about the reaction CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O(l) ΔH = -890 kJ is that reacting one mole of oxygen (O2) releases 445 kJ of energy. The reaction is exothermic, meaning it releases energy. The ΔH value of -890 kJ means that for every mole of methane (CH4) that reacts, 890 kJ of energy is released. Considering the stoichiometry of the reaction, where 2 moles of oxygen react per mole of methane, half of the total energy release (which is 890 kJ) would be associated with 1 mole of oxygen, thereby releasing 445 kJ.
What is the total number of electrons in all s orbitals of a neutral atom of phosphorus?
Calculate the mass of water produced when 4.86 g of butane reacts with excess oxygen.
When 4.86g of butane reacts with excess oxygen, 7.53g of water is produced. This calculation is based on stoichiometry principles and the balanced chemical equation of the reaction.
Explanation:The subject of this question is Chemistry and it pertains to the reaction of butane (C4H10) with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O). The balanced chemical equation for this reaction is 2C4H10 + 13O2 -> 8CO2 + 10H2O. The mass of water produced can be calculated through stoichiometry, which is the method of calculating the quantitative/weight relations of reactants and products in a chemical reaction.
Given the weight of butane is 4.86g, we first need to convert this to moles. The molar mass of butane is calculated by adding the molar masses of Carbon (C=12.01g/mol) x 4 and Hydrogen (H=1.01g/mol) x 10. This equals approximately 58.12g/mol. Therefore, the number of moles of butane we have is 4.86g / 58.12g/mol = 0.0836 moles.
From the balanced equation, we can see that two moles of butane produce ten moles of water. Therefore, 0.0836 moles of butane would produce 0.0836 x 10/2 = 0.418 moles of water. To convert this to mass, we multiply by the molar mass of water, which is 18.01g/mol. Hence, 0.418 moles x 18.01g/mol = 7.53g of water.
Learn more about Stoichiometry here:https://brainly.com/question/34828728
#SPJ12
The reaction of 4.86 g of butane with excess oxygen produces approximately 7.53 g of water. This is determined using the molar mass of butane, the balanced chemical equation, and stoichiometry.
Explanation:The chemical reaction of butane (C₄H₁₀) with oxygen (O₂) to produce carbon dioxide (CO₂) and water (H₂O) is given by the balanced equation 2C₄H₁₀ + 13O₂ -> 8CO₂ + 10H₂O. In this reaction, 2 moles of butane produce 10 moles of water. The molar mass of butane is 58.12g/mole. Thus, 4.86 g of butane is 4.86 g / 58.12 g/mole = 0.0836 moles.
By stoichiometry, 0.0836 moles of butane would produce 0.0836 moles x (10 moles of water / 2 moles of butane) = 0.418 moles of water. As the molar mass of water is approximately 18.02 g/mole, the mass-produced would be 0.418 moles x 18.02 g/mole = 7.53 g. Thus, 4.86 g of butane reacts with excess oxygen to produce approximately 7.53 g of water.
Learn more about Stoichiometry here:https://brainly.com/question/30218216
#SPJ11
The ___ is the part of the solution that is dissolved and the ___ is the part of the solution that does the dissolving.
This is for a study guide for my semester final, and I have no clue how to answer this, please help, and please at least try to help!
Besides the major types of radioactive decay, there are two others: positron emission and electron capture.
• Compare and contrast positrons with electrons.
• Explain how positron emission works and how it causes transmutations.
• Explain how electron capture works and how it causes transmutations.
• Compare the transmutations caused by positron emissions and electron capture.
Positron emission = emission of a positron and a neutrino when a proton is convert into a neutron. The total number of particles in the nucleus doesn't change, -1 proton +1 neutron
It's a spontaneous reaction for some nucleus.
eg:
Positron = e+
Neutrino=ve
O-15 --> N-15 + e+ +ve
Electron capture= A nucleus absorb an electron while a proton is convert in a neutron and emit a neutrino. The total number of particles in the nucleus doesn't change, -1 proton +1 neutron
eg:
Al-26 +e- --> Mg-26 + ve
Electron capture and positron emission are two mechanisms to explain the decay of some unstable isotopes. Electron capture is usually observed when the energy difference between the initial and final state is low. Mainly because of the larger amount of kinetic energy need for the expulsion two particles with the positron emission mechanism.
Which aqueous solution has the lowest freezing point c6h12o6, c2h5oh, ch3cooh, or nacl?
Among C₆H₁₂O₆, C₂H₅OH, CH₃COOH, and NaCl, NaCl is the one that will provide the aqueous solution with the lowest freezing point.
We have 4 aqueous solutions and we want to determine which has the lowest freezing point.
C₆H₁₂O₆C₂H₅OHCH₃COOHNaClWhat is the freezing point depression?Freezing-point depression is a drop in the temperature at which a substance freezes, caused when a smaller amount of another, non-volatile substance is added.
We can calculate the freezing point depression (ΔT) using the following expression.
ΔT = Kf × b × i
where,
Kf is the cryoscopic constant for water.b is the molality of the solution.i is the Van't Hoff factor.Assuming all the solutions have the same molality, the freezing point depression will be a function of the van't Hoff factor.
What is the van't Hoff factor?The van 't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass.
For non-electrolytes, such as C₆H₁₂O₆ and C₂H₅OH, i = 1. Comparing NaCl and CH₃COOH, we can determine that i(NaCl) > i(CH₃COOH) because NaCl is a strong electrolyte and CH₃COOH a weak one.Thus, NaCl, with the highest Van't Hoff factor, will have the lowest freezing point.
Among C₆H₁₂O₆, C₂H₅OH, CH₃COOH, and NaCl, NaCl is the one that will provide the aqueous solution with the lowest freezing point.
Learn more about freezing point depression here: https://brainly.com/question/2292439
Identify the element that has a ground state electronic configuration of [ar]4s23d104p 3 .
Determine the molecular formula of a compound having the empirical formula C9H17O and a molar mass of 847.56g/mol.
Determine the hydroxide ion concentration in a solution that is 0.0033 m hno3. answer in units of m.
We wish to dilute some 18.4 m h2so4 solution to make 600 ml of a 0.1 m h2so4 solution. how much of the 18.4 m solution should we start with?
We should start with approximately [tex]3.10 ml[/tex] of the [tex]18.4 M[/tex] [tex]H\(_2\)SO\(_4\)[/tex] solution to make [tex]600 ml[/tex] of a [tex]0.1 M[/tex] [tex]H\(_2\)SO\(_4\)[/tex] solution.
To solve this problem, we can use the dilution formula:
[tex]\[ C_1V_1 = C_2V_2 \][/tex]
where:
[tex]\( C_1 \)[/tex] is the concentration of the stock solution ([tex]18.4 M[/tex] [tex]H\(_2\)SO\(_4\)[/tex]).
[tex]\( V_1 \)[/tex] is the volume of the stock solution we need to use (unknown).
[tex]\( C_2 \)[/tex] is the concentration of the final solution ([tex]0.1 M[/tex] [tex]H\(_2\)SO\(_4\)[/tex]).
[tex]\( V_2 \)[/tex] is the volume of the final solution ([tex]600 ml[/tex] or [tex]0.6 L[/tex]).
We want to find [tex]\( V_1 \)[/tex], so we rearrange the formula:
[tex]\[ V_1 = \frac{C_2V_2}{C_1} \][/tex]
Now, plug in the values:
[tex]\[ V_1 = \frac{(0.1 \text{ M})(0.6 \text{ L})}{18.4 \text{ M}} \][/tex]
[tex]\[ V_1 = \frac{0.06}{18.4} \][/tex]
[tex]\[ V_1 =0.00326 \text{ L} \][/tex]
Since [tex]1 L = 1000 ml[/tex], we convert [tex]\( V_1 \)[/tex] to milliliters
[tex]\[ V_1 = 0.00326 \text{ L} \times 1000 \text{ ml/L} \][/tex]
[tex]\[ V_1 = 3.26 \text{ ml} \][/tex]
Rounding to two decimal places, we find that we need approximately [tex]3.10 ml[/tex] of the [tex]18.4 M[/tex] [tex]H\(_2\)SO\(_4\)[/tex] solution to make [tex]600 ml[/tex] of a [tex]0.1 M[/tex][tex]H\(_2\)SO\(_4\)[/tex] solution.