Answer:
1. RbOH(s) ⇒ Rb⁺(aq) + OH⁻(aq)
2. Na₂CO₃(s) ⇒ 2 Na⁺(aq) + CO₃²⁻(aq)
3. (NH₄)₂SeO₃(s) ⇒2 NH₄⁺(aq) + SeO₃²⁻(aq)
Explanation:
Let's consider the dissolving equations for the following compounds.
1. Rubidium hydroxide
RbOH(s) ⇒ Rb⁺(aq) + OH⁻(aq)
2. Sodium carbonate
Na₂CO₃(s) ⇒ 2 Na⁺(aq) + CO₃²⁻(aq)
3. Ammonium selenite
(NH₄)₂SeO₃(s) ⇒2 NH₄⁺(aq) + SeO₃²⁻(aq)
The Nernst equation at 20oC is:
Eion= 58 millvolts/z. [log10 (ion)out/(ion)in]
Calculate the equilibrium potential for Cl- if the concentration of Cl- outside of the cell is 100 and the concentration inside of the cell is 10 mmol/liter.
a. 58 millivolts
b. +58 millivolts
c. -116 millivolts
d. 0
Answer:
a. -58 millivolts
Explanation:
The given Nernst equation is:
[tex]E_{ion} = 58 millivolts /z \Big[ log_{10} \Big( \dfrac{[ion]_{out}}{[ion]_{in}}\Big) \Big]}[/tex]
The equilibrium potential given by the Nernst equation can be determined by using the formula:
[tex]E_{Cl^-} = \dfrac{2.303*R*T}{ZF} \times log \dfrac{[Cl^-]_{out}} {[Cl^-]_{in}}[/tex]
where:
gas constant(R) = 8.314 J/K/mol
Temperature (T) = (20+273)K
= 298K
Faraday constant F = 96485 C/mol
Number of electron on Cl = -1
[tex]E_{Cl^-} = \dfrac{2.303*8.314*298} {(-1)*(96845)} \times log \dfrac{100} {10}[/tex]
[tex]E_{Cl^-} = - 0.05814 \ volts[/tex]
[tex]\mathsf{E_{Cl^-} = - 0.05814 \times 1000 \ milli volts}[/tex]
[tex]\mathsf{E_{Cl^-} \simeq - 58\ milli volts}[/tex]
The shape of a molecule is determined by:
A. All of these
B. The number of electron clouds around the atom.
C. The number of bonds.
D. Mutual repulsion between electrons.
The sample concentration was measured at 50mg/ml. The loading concentration needs to be 10mg/ml. The final volume needs to be 25ul. What is the volume of sample needed and the amount of buffer needed to reach 25ul
Answer:
a) [tex]V_1=5ul[/tex]
b) [tex]v=20ul[/tex]
Explanation:
From the question we are told that:
initial Concentration [tex]C_1=50mg/ml[/tex]
Final Concentration [tex]C_2=10mg/ml[/tex]
Final volume needs [tex]V_2 =25ul[/tex]
Generally the equation for Volume is mathematically given by
[tex]C_1V_1=C_2V_2[/tex]
[tex]V_1=\frac{C_1V_1}{C_2}[/tex]
[tex]V_1=\frac{10*25}{50}[/tex]
[tex]V_1=5ul[/tex]
Therefore
The volume of buffer needed is
[tex]v=V_2-V_1\\\\v=25-5[/tex]
[tex]v=20ul[/tex]
To prepare a standard (calibration) curve for a spectroscopy experiment, start by preparing ___________ with ______________ Then, measure the ______________ of each solution at _____________ and create a plot of ____________ for the measured values. Finally, find the best-fit line of the data set.
Answer: See explanation
Explanation:
The calibration curve is the method used for the determination of the concentration of a substance such that the unknown sample will be compared to some standard samples of the known concentration.
To prepare a standard (calibration) curve for a spectroscopy experiment, start by preparing (multiple solutions) with (different known concentrations). Then, measure the (absorbance) of each solution at (thesame wavelength) and create a plot of (absorbance vs. concentration) for the measured values. Finally, find the best-fit line of the data set.
How many neutrons does Carbon- 14 and Carbon -15 have? *
Answer: 8 for both
Explanation:
what characterizes a homogeneous mixture?
Answer:
a mixture that doesn't really show the ingredients or things put into the material or food.
Tech A says that hydrocarbons are a result of complete combustion. Tech B says that a catalytic converter creates a chemical reaction, changing carbon monoxide and hydrocarbons to water and carbon dioxide. Who is correct
Answer:
Neither Tech A nor B is correct
Explanation:
Combustion is a chemical reaction that occurs when a chemical molecule(s) interacts quickly with oxygen and produces heat.
When hydrocarbon undergoes a complete combustion reaction, they produce water and CO2.
Tech B is also incorrect because the main purpose of a catalytic converter is to accelerate and speed up the chemical reaction rates, Hence, they are not involved in chemical reaction formation. Catalytic converters are utilized as a control device in exhaust emission to lessen the effect of toxic gas fumes.
The freezing point of a substance is -20°C. Its boiling point is 120°C.
a. At 80°C the substance is in the state
b. At -50°C the substance is in the state.
C. At 140°C the substance is in the state.
Answer:
a. liquid
b. solid
c. gas, (should be at it's boiling point)
Explanation: If the normal melting point of a substance is below room temperature, the substance is a liquid at room temperature. Benzene melts at 6°C and boils at 80°C; it is a liquid at room temperature. If both the normal melting point and the normal boiling point are above room temperature, the substance is a solid.
if you need an explanation to each lmk
If 50.0 g of sulfuric acid and 40.0 grams of barium chloride are mixed, how many grams of sulfuric acid and how many grams of barium chloride remain after the double replacement reaction is complete?
After the double replacement reaction is complete 0 grams of BaCl₂ and 31.16 grams of H₂SO₄ will remain.
First, we will write the balanced equation for the reaction
H₂SO₄ + BaCl₂ → BaSO₄ + 2HCl
This means 1 mole of BaCl₂ is needed to react completely with 1 mole of H₂SO₄ to give 1 mole of BaSO₄ and 2 moles of HCl
From the question, 50.0g of sulfuric acid is mixed with 40.0 grams of barium chloride. To determine the quantity of each substance remaining after the complete reaction, we will first determine the number of moles present in each of the reactant.
For H₂SO₄
mass = 50.0g
Molar mass = 98.079 g/mol
From the formula
Number of moles = Mass / Molar mass
∴ Number of moles of H₂SO₄ = 50.0g / 98.079 g/mol
Number of moles of H₂SO₄ = 0.5098 mol
For BaCl₂
mass = 40.0 g
Molar mass = 208.23 g/mol
∴ Number of moles of BaCl₂ = 40.0g / 208.23 g/mol
Number of moles of BaCl₂ = 0.1921 mol
Since the number of moles of H₂SO₄ is more than that of BaCl₂, then H₂SO₄ is the excess reagent and BaCl₂ is the limiting reagent (that is, it will be used up completely during the reaction)
From the equation, 1 mole of H₂SO₄ is needed to completely react with 1 mole of BaCl₂
∴ 0.1921 mol of H₂SO₄ will be needed to completely react with 0.1921 mol of BaCl₂.
Therefore, after the reaction is complete, 0 mole (i.e 0 grams) of BaCl₂ will remain and (0.5098 mole - 0.1921 mole) of H₂SO₄ will remain.
Number of moles H₂SO₄ that will remain = 0.5098 mole - 0.1921 mole = 0.3177 moles
Now, we will convert this to grams
From the formula
Mass = Number of moles × Molar mass
Mass of H₂SO₄ that will remain = 0.3177 moles × 98.079 g/mol
Mass of H₂SO₄ that will remain = 31.1597 g
Mass of H₂SO₄ that will remain ≅ 31.16 g
Hence, after the double replacement reaction is complete 0 grams of BaCl₂ and 31.16 grams of H₂SO₄ will remain.
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HELP ASAS 15 POINTS
When using the process of evaporation to separate a mixture, what is left behind in the evaporating dish?
A. None of these.
B. The liquid evaporates and the solid is left in the dish.
C. The mixture does not separate, and the entire mixture evaporates.
D. The mixture does not separate, and the entire mixture remains in the dish.
Answer:
liquid will be evaporated while solid remains
At what velocity (m/s) must a 20.0g object be moving in order to possess a kinetic energy of 1.00J
Answer:
10 ms-1
Explanation:
Kinetic energy = 1/2 × m × v^2
1 = 1/2× 20 ×10^ -3 × v^2
v ^ 2 = 100
v = 10 ms-1
note : convert grams in to kg before substitution as above
Given:
Kinetic energy,
K.E = 1.00 JMass,
m = 20.0 gWe know the formula,
→ [tex]K.E = \frac{1}{2} mv^2[/tex]
By putting the values, we get
[tex]1 = \frac{1}{2}\times 20\times 10^{-3}\times (v)^2[/tex]
[tex]v^2 = 100[/tex]
[tex]v = \sqrt{100}[/tex]
[tex]v = 10 \ m/s[/tex]
Thus the above response is correct.
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ort
Which is a primary alcohol?
0 3-pentanol
2-propanol
1-ethanol
4-octanol
urvey
Lig A Moving to another question will save this response.
Answer:
1 ethanol is right answer
Explanation:
CH3- CH2-OH
In practice, the second law of thermodynamics means that:
a. Systems move from ordered behavior to more random behavior.
b. Systems move from random behavior to more ordered behavior.
c. Systems move between ordered and random behavior patterns based on temperature.
d. Systems are constantly striving to reach equilibrium.
Answer:
Systems move from ordered behavior to more random behavior.
Explanation:
Entropy refers to the degree of disorderliness in a system. The second law of thermodynamics can be restated in terms of entropy as follows; “any spontaneous process in any isolated system always results in an increase in the entropy of that system.''(science direct)
According to this law, systems tend towards a more disorderly behaviour (increase in entropy) hence the answer given above.
does anyone know how to solve this and what the answer would be?
Dynamic equilibrium is showed at the point at which solid liquid and gas intersect.
At the point at which solid liquid and gas intersect represents a system that shows dynamic equilibrium. There is equal amount of reactants and products at the point of dynamic equilibrium because the transition of substances occur between the reactants and products at equal rates, means that there is no net change. Reactants and products are formed at the rate that no change occur in their concentration.
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which of the following is is a chemical property of pure water
Answer:
Pure water has an acidity of about 7 on the pH scale. -is a chemical property of pure water. Pure water has an acidity of about 7 on the pH scale
Answer: không màu , không mùi không vị
Explanation:
What is Bose Einstein state of matter and their examples
Answer:
A BEC ( Bose - Einstein condensate ) is a state of matter of a dilute gas of bosons cooled to temperatures very close to absolute zero is called BEC.
Examples - Superconductors and superfluids are the two examples of BEC.
Explanation:
How many grams of H₂SO₄ are contained in 2.00 L of 6.0 M H₂SO₄?
Please explain and show work.
Answer:
1176 grams
Explanation:
nH2SO4 =2*6=12 mol
mH2SO4=12*98=1176 grams
Answer:
solution given:
molarity of H₂SO₄=6 M
volume=2L
no of mole =6M*2=12mole
we have
mass =mole* actual mass=12*98=1176g
the mass is 1176g.
Determine which choice is an example of an endothermic process.
O A. Lighting a match
B. Respiration
C. Running a gas engine
D. Baking bread
Answer:
D. Baking bread
Explanation:
In this process, energy is absorbed and in an endothermic process energy is absorbed too.
Baking bread is an example of an endothermic process, therefore option (d) is correct .
What do you mean by endothermic process ?Endothermic reactions are chemical processes in which the reactants absorb heat from the environment to produce products.
An endothermic reaction is accompanied by an absorption of heat.
Endothermic reactions cause a cooling effect by lowering the temperature of the surrounding environment.
A decrease in temperature can be observed with the progression of the reaction. The reaction is non-spontaneous in endothermic reactions .
Baking bread is an example of an endothermic process, hence option (d) is correct .
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Hãy cho biết giá trị và ý nghĩa của số lượng tử n, l, m, ms khi mô tả trạng thái của electron trong nguyên tử?
Which of the following best describes physical science?
0...
OA.
the study of motion
OB.
the study of matter and energy
O C.
the study of Earth's structure and processes
OD.
the study of reactions
O E.
the study of living things
Reset
Ne
B. The study of matter and energy.
Because physical science is everything that doesn't include organic things.
The study of matter and energy among the following best describes physical science.
What is matter?
Matter in chemistry, is defined as any kind of substance that has mass and occupies space that means it possess volume .Matter is composed up of atoms which may or not be of same type.
Atoms are further made up of sub atomic particles which are the protons ,neutrons and the electrons .The matter can exist in various states of substances such as solids, liquids and gases depending on the conditions of the temperature and the pressure.
The states of matter are inter convertible into each other by changing the parameters of temperature and pressure.Matter is always conserved by law of conservation of matter.The law was proposed by Antoine Lavoisier.
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A sample of gas is held at constant volume. If the number of moles of this sample of gas is doubled and the pressure of this sample of gas is halved, what happens to the absolute temperature of the gas?
Select one
a. The absolute temperature is doubled.
b. The absolute temperature is halved.
c. The absolute temperature is quadrupled.
d. The absolute temperature is quartered.
e. The absolute temperature stays the same.
Answer:
number of moles of gas increases the volume also increases.
I need to know what is the median of the data
Answer:
The median is also the number that is halfway into the set. To find the median, the data should be arranged in order from least to greatest. If there is an even number of items in the data set, then the median is found by taking the mean (average) of the two middlemost numbers.
I hope it helps
The functional groups in an organic compound can frequently be deduced from its infrared absorption spectrum.
a. True
b. False
Answer:
a. True
Explanation:
The main information that gives an infrared absorption spectrum is the type of functional groups that are present in an organic compound. The infrared (IR) spectroscopy is based on the fact that functional groups absorb light in the IR region of the electromagnetic spectrum (approximately at 2,500-16,000 nm) and induces a vibrational excitation of the covalently bonded atoms in the group. The vibration of the atoms can be of different types, such as stretching, bending, etc. Each functional group (such as the carbonyl group) in an organic compound absorbs at a specific IR frequency so they can be distinguished from an IR spectrum.
Nitric acid can be formed in two steps from the atmospheric gases nitrogen and oxygen, plus hydrogen prepared by reforming natural gas. In the first step, nitrogen and hydrogen react to form ammonia: N2 (g) + 3H2 (g) â 2NH3 (g) =ÎHâ92.kJ In the second step, ammonia and oxygen react to form nitric acid and water:
NH3 (g) + 2O2 (g) â HNO3 (g) + H2O (g) =ÎHâ330.kJ
Required:
Calculate the net change in enthalpy for the formation of one mole of nitric acid from nitrogen, hydrogen and oxygen from these reactions.
Answer:
-376 kJ
Explanation:
The first step equation:
[tex]\mathsf{N_{2(g)} + 3H_2{(g)} \to 2NH_3{(g)} \ \ \ \Delta H = -92\ kJ}[/tex] ---- (1)
The second step equation:
[tex]\mathsf{NH_{3(g)} + 2O_2{(g)} \to HNO_3{(g)} +H_2O_{(g)} \ \ \ \Delta H = -330\ kJ}[/tex] ---- (2)
To determine the enthalpy of formation for 1 mole of HNO₃ (nitric acid), we have the following.
From the above equations; let multiply equation (1) by 1 and equation (2) by 2.
[tex]\mathsf{N_{2(g)} + 3H_2{(g)} \to 2NH_3{(g)} \ \ \ \Delta H = -92\ kJ}[/tex] ---- (3)
[tex]\mathsf{2NH_{3(g)} + 4O_2{(g)} \to 2HNO_3{(g)} +2H_2O_{(g)} \ \ \ \Delta H = 2(-330)\ kJ}[/tex] ----- (4)
adding the above two equations, we have:
[tex]\mathsf{N_{2(g)} + 3H_2{(g)}+ 2NH_{3(g)} + 4O_{2(g)} \to 2HNO_{3(g)} + 2NH_3{(g)} +2H_2O_{(g)} \ \ \ \Delta H = (-660 \ kJ -92\ kJ)}[/tex][tex]\mathsf{N_{2(g)} + 3H_2{(g)} + 4O_{2(g)} \to 2HNO_{3(g)} +2H_2O_{(g)} \ \ \ \Delta H = (-752 \ kJ)}[/tex]
Now, from the recent equation, we have:
2 moles of nitric acid = -752 kJ
∴
1 mole of nitric acid will be: = (1 mole × (-752 kJ)) ÷ 2 moles
1 mole of nitric acid will be: = -376 kJ
A sample of Kr gas is observed to effuse through a pourous barrier in 8.15 minutes. Under the same conditions, the same number of moles of an unknown gas requires 4.53 minutes to effuse through the same barrier. The molar mass of the unknown gas is ____________ g/mol.
Answer:
25.88 g/mol
Explanation:
Graham's law is a famous law which states that the diffusion rate or the effusion rate of any gas varies inversely to the square root of the molecular weight the gas.
So from Graham's law, we have,
[tex]$\frac{\text{time}}{M^{1/2}}=\text{constant}$[/tex]
Using the sample of Kr gas having M = 83.8
[tex]$\frac{8.15}{(83.8)^{0.5}}= \frac{4.53}{M^{0.5}}$[/tex]
[tex]$M^{0.5}= 5.088$[/tex]
M = 25.88 g/mol
4. A sample of ammonia, NH3, contains 3.3 x 1021 hydrogen atoms. How many NH; molecules are in this sample?
Answer:
1.1 × 10²¹ NH₃ molecules
Explanation:
From the given information:
We were being told that the number of the hydrogen (H) atoms present in the sample of NH3 = 3.3 × 10²¹ hydrogen.
However, it signifies that each molecule of ammonia harbors 3hydrogen (H) atoms.
Hence, the number of molecules of NH₃ present;
[tex]\mathsf{=\dfrac{3.3\times 10^{21}}{3} \ molecules \ of \ {NH_3}}[/tex]
= 1.1 × 10²¹ NH₃ molecules
P.1 is a variant of SARS-CoV2. This is the so-called "Brazil variant". It has the amino acid the substitution N501Y in the spike protein on the virus surface. In cellular fluids at ~pH 7 the wild type spike protein has a net negative charge. This substitution will make the spike protein
more negative
less hydrophobic
able to absorb move UV light
less negative
There are various variants of Cov id virus. The Brazilian variant P also known as Gamma variant is the third variant of the original SARS-CoV2.
The correct answer is more negative
This variant has raised concerns since it has ability to spread more quickly then previous variants and this is more negative variant.
It is assumed that Cov id variant Gamma and Delta has ability to absorb move UV light but this is not proved yet and research is underway.
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If the balance were not tared prior to weighing out the KHP.... how would you expect this to affect the molarity of NaOH calculated? What type of error is this?
Answer:
Following are the response to the given question:
Explanation:
In the given scenario, When the balance has never been tainted before the KHP is weighted, which can affect the molar concentration of NaOH because its molarity is directly proportional to the weight including its substance. In this question it is the mistake is systemic because it may be corrected by modifying balancing parameters.
In the reoxidation of QH2 by purified ubiquinone-cytochrome c reductase (Complex III) from heart muscle, the overall stoichiometry of the reaction requires 2 mol of cytochrome c per mole of QH2 because:
Answer: Options related to your question is missing below are the missing options
a. cytochrome c is a one-electron acceptor, whereas QH2 is a two-electron donor.
b. cytochrome c is a two-electron acceptor, whereas QH2 is a one-electron donor.
c. cytochrome c is water soluble and operates between the inner and outer mitochondrial membranes
d. heart muscle has a high rate of oxidative metabolism, and therefore requires twice as much cytochrome c as QH2 for electron transfer to proceed normally.
e. two molecules of cytochrome c must first combine physically before they are catalytically active.
answer:
cytochrome c is a one-electron acceptor, whereas QH2 is a two-electron donor. ( A )
Explanation:
The overall stoichiometry of the reaction requires 2 mol of cytochrome per mole of QH2 because a cytochrome is simply a one-electron acceptor while QH2 is not a one-electron donor ( i.e. it is a two-electron donor )
An electron donor in a reaction is considered a reducing agent because it donates its electrons to another compound thereby self oxidizing itself in the process.
At 445oC, Kc for the following reaction is 0.020. 2 HI(g) <--> H2 (g) + I2 (g) A mixture of H2, I2, and HI in a vessel at 445oC has the following concentrations: [HI] = 1.5 M, [H2] = 2.50 M and [I2] = 0.05 M. Which one of the following statements concerning the reaction quotient, Qc, is TRUE for the above system?
a. Qc = Kc; the system is at equilibrium.
b. Qc is less than Kc; more H2 and I2 will be produced.
c. Qc is less than Kc; more HI will be produced.
d. Qc is greater than Kc; more HI will be produced.
Explanation:
The given balanced chemical equation is:
[tex]2 HI(g) <--> H_2 (g) + I_2 (g)[/tex]
The value of Kc at 445oC is 0.020.
[HI]=1.5M
[H2]=2.50M
[I2]=0.05M
The value of Qc(reaction quotient ) is calculated as shown below:
Qc has the same expression as the equilibrium constant.
[tex]Qc=\frac{[H_2][I_2]}{[HI]^2} \\Qc=(2.50Mx0.05M)/(1.5M)^2\\Qc=0.055[/tex]
Qc>Kc,
Hence, the backward reaction is favored and the formation of Hi is favored.
Among the given options, the correct answer is option d. Qc is greater than Kc; more HI will be produced.
