Answer:
a. the system is at equilibrium.
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
[tex]H_2+Br_2\rightleftharpoons 2HBr[/tex]
Thus, the law of mass action is given by:
[tex]Keq=\frac{[HBr]^2}{[H_2][Br_2]} =57.6[/tex]
Nonetheless, for the given point of 4.67 × 10^-3M bromine gas, 2.14 × 10^−3 hydrogen gas, and 2.40 × 10^−2M hydrogen bromide gas we should compute the reaction quotient in order to know whether the direction of the reaction is to left or to right, thus:
[tex]Q=\frac{[HBr]^2}{[H_2][Br_2]} =\frac{(2.40x10^{-2})^2}{(4.67x10^{-3})(2.14x10^{-3})} \\\\Q=57.6[/tex]
Therefore, since Keq=Q, we say that the system is at equilibrium, for that reason, the answer is a.
Best regards.
most reactions give off energy in the form of heat and are called what
Answer:
Exothermic reaction.
Explanation:
state the importance of uric acid biomarker
Answer:
u
uric acid is a useful diagnostic tool as screening for most of purine metabolic disorders. The importance of uric acid measurement in plasma and urine with respect of metabolic disorders is highlighted. Not only gout and renal stones are indications to send blood to the laboratory for uric acid examination
The molar enthalpy of
mole of a liquid.
is the heat required to vaporize one
thermochemical equation
combustion
released
vaporization
fusion
absorbed
heat
Answer:
vaporization
Explanation:
The molar enthapy of _vaporization______ is the heat required to vaporize one mole of a liquid”
What is a good title for this chart?
Answer:
pH of the acid
Explanation:
When 106 g of water at a temperature of 21.4 °C is mixed with 64.3 g of water at an unknown temperature, the final temperature of the resulting mixture is 46.8 °C. What was the initial temperature of the second sample of water? (The specific heat capacity of liquid water is 4.184 J/g ⋅ K.)
Answer:
THE INITIAL TEMPERATURE OF THE SECOND SAMPLE IS 4.93 C OR 277.93 K
Explanation:
Mass of first sample of water = 106 g
Initial temp of first sample = 21.4 °C = 21.4 + 273 K = 294.4 K
Mass of second sample = 64.3 g
Final temp of theresulting mixture = 46.8 °C = 46.8 + 273 K = 319.8 K
Specific heat capacity of water = 4.184 J/g K
It is worthy to note that;
Heat gained by the first sample = Heat lost by the second sample
Since heat = mass * specific heat capacity * change in temperature, we have
Mass * specific heat * change in temp of the first sample = Mass * specific heat * change in temp. of the second sample
MC (T2 - T1) = MC (T2-T1)
106 * 4.184 * ( 319.8 - 294.4) = 64.3 * 4.184 * ( 319.8 - T1)
106 * 4.184 * 25.4 = 269.0312 ( 319.8 - T1)
11 265.0016 = 269.0312 (319.8 - T1)
Since the change in temperature = 319.8 -T1
Change in temperature =11265.0016 / 269.0312
Change in temperature = 41.87
Change in temperature = 319.8 -T1
41.87 = 319.8 - T1
T1 = 319.8 - 41.87
T1 = 277.93 K
T1 = 4.93 °C
So therefore, the initial temperature of the sacond sample is 4.73 °C or 277.93 K
A gaseous hydride of Nitrogen
contains its own volume of Nitrogen
and twice its volume of Hydrogen
and has vapour density 16. The
formula of the hydride is.
Select one:
a. NH2
b. NH3
c. N3H
• d. N2H4
Answer:
N2H4
Explanation:
A hydride is a binary compound of hydrogen and another element. Binary compounds contain only two atoms. We have to x-ray the hydrides of nitrogen given in the question in order to make our choice. Remember that we were told that that the hydride contains its own volume of nitrogen and twice its volume of hydrogen.
Now consider the hydride N2H4.
N2H4(g) -----> N2(g) + 2H2(g)
The volume ration of nitrogen gas to hydrogen gas in N2H4 is 1:2.
The molecular mass of the compound is;
N2H4= 2(14) + 4(1)= 28+4= 32
Since
molecular mass= 2 vapour density
Vapour density= molecular mass/2
Vapour density= 32/2
Vapour density = 16
Therefore the hydride of nitrogen referred to in the question is N2H4
The wolf gets enegry from____
The rabbit gets energy from____
The plant gets energy from___
The mushoom gets energy from___
Answer:
The wolf gets energy from other Animals through Cellular respiration. it's a carnivore
The rabbit gets energy from Carbohydrates,Fats.... obtained through different sources. A common example is the grass. It's an herbivore
The plant gets energy from the sun during photosynthesis. It's Autotrophic.
The mushroom gets energy from the decomposition of other organic matter. It's heterotrophic.
Explanation:
In a food chain; The Wolf eats the rabbit, when the Wolf dies, decomposers such as mushrooms breaks down its body returning it to the soil, where it provides nutrients for plants
Give two examples (i.e. list 2 elements that are examples) of: a. an atom with a half-filled subshell b. an atom with a completely filled outer shell c. an atom with its outer electrons occupying a half-filled subshell and a filled subshell
Answer:
an atom with a half-filled subshell - hydrogen
an atom with a completely filled outer shell - argon
an atom with its outer electrons occupying a half-filled subshell and a filled subshell- copper
Explanation:
The outermost shell or the valence shell of the atom is the last shell in the atom. Chemical reactions occur at this outer most shell. The number of electrons on the outermost shell of an atom determines the group to which it belongs in the periodic table as well as its chemical properties.
Hydrogen has a half filled 1s sublevel. Only one electron is present in this sublevel.
Let us consider argon
1s2 2s2 2p6 3s2 3p6
The outermost ns and np levels are completely filled. Thus the outermost shell is completely filled.
In the last case; let us look at the electronic configuration of nitrogen;
1s2 2s2 2p3
The outermost 2p subshell is exactly half filled while the 2s sublevel is fully filled. The outermost shell of nitrogen is made up of 2s2 and 2p3 sublevels.
1. There are how many mol of oxygen in 3.5 mol of caffeine.
Answer:
7 mol
Explanation:
Caffeine molecular formula C8H10N4O2. It has 2 atoms of oxygen.
C8H10N4O2 - 2O
1 mol 2 mol
3.5 mol x mol
x = 3.5*2/1 = 7 mol
4 molecules of glucose has how many carbon hydrogen and oxygen
Answer:
24 carbons
48 hydrogens
24 oxygen
Explanation:
Choose the INCORRECT statement. A. Temperatures of two bodies are equal when the average kinetic energies of the two bodies become the same. B. The heat capacity is the quantity of heat required to change the temperature of the system by one degree. C. The specific heat is the heat capacity for one mole of substance. D. Most metals have low specific heats, as metals can be heated quickly. E. The law of conservation of energy can be written: qsystem qsurroundings
Answer:
Option C
The specific heat is the heat capacity for one mole of a substance.
Explanation:
The incorrect statement is The specific heat is the heat capacity for one mole of a substance.
This is because the specific heat capacity is the amount of heat required to raise the temperature of 1 gramme of a substance by 1 degree Celcius.
Note that the unit in question here for the specific heat capacity of the substance is in grammes.
The definition given in the options is actually for the molar heat capacity of the substance, not the specific heat capacity.
Calculate the pH of a 0.020 M H2CO3 solution. At 25 °C, Ka1 = 4.3 × 10-7. H2CO3(aq) + H2O(l) ↔ H3O+(aq) + HCO3-(aq)
Answer:
Explanation:
H₂CO₃(aq) + H₂O(l) ↔ H₃O⁺(aq) + HCO₃⁻(aq)
Let d be the degree of dissociation
.02( 1-d ) .02d .02d
Dissociation constant Ka₁ is given
4.3 x 10⁻⁷ = .02d x .02d / .02( 1-d )
= .004 d² / .02 ( neglecting d in denominator )
= .02 d²
d² = 215 x 10⁻⁷
d = 4.636 x 10⁻³
= .004636
concentration of H₃O⁺
= d x .02
= .004636 x .02
= 9.272 x 10⁻⁵
pH = - log [ H₃O⁺ ]
= - log ( 9.272 x 10⁻⁵ )
5 - log 9.272
= 5 - .967
= 4.033 .
Write the limiting forms (or Canonical forms) of the following ions:
i. H3O+
, ii. CO3
2-
, iii. NO3-
Answer:
Canonical structures of a chemical specie explain its observed properties from a valence bond theory perspective.
Explanation:
Resonance is a valence bond concept introduced by Linus Pauling to explain the observed properties of certain chemical species such as bond lengths, bond angles, bond order , etc.
There are certain chemical species for which a single chemical structure does not suffice in explaining its observed properties. For instance, the bond order in CO3^2- is about 1.33. Its bond length, shows that the C-O bond present in CO3^2- is neither a pure C-O single bond nor a pure C-O double bond. Hence the structure of CO3^2- is 'somewhere in between' three contributing canonical structures as shown in the image attached to this answer. The resonance structures of NO3^- are also shown.
A scientists compares two samples of white powder
Answer:
answer is in exaplation
Explanation:
Answer. Chemical reaction had occurred and both the powders are different substances.
Explanation:
As density is an intensive property of the substance.Which means that different substance have different densities.
Density = \frac{mass}{volume}
volume
mass
Density of powder 1, d_1=\frac{0.5g}{45cm^3}=0.11g/cm^3d
1
=
45cm
3
0.5g
=0.11g/cm
3
Density of powder 2, d_2=\frac{1.3g}{65cm^3}=0.02g/cm^3d
2
=
65cm
3
1.3g
=0.02g/cm
3
On comparing both the densities of the powders we can say that both the substances are different. So we can conclude that the chemical reaction had occurred.
2. Points
Which of the following is not a characteristic of a transverse mechanical
wave?
A. It travels at less than the speed of light.
B. It involves displacing the medium perpendicular to the motion of
the wave
C. It looks a little bit like a snake.
D. It is also known as a compression wave.
Answer:
D
Explanation:
Logitudinal waves also known as compression waves.
It involves displacing the medium perpendicular to the motion of the wave is not a characteristic of a transverse mechanical wave. Option B is correct.
What are transverse mechanical waves?A transverse mechanical wave is a disturbance created by it to transfer energy from one point to another. while the proposition happens the particle present within the medium get vibrates.
in a transverse wave, the particle present will vibrate up and down and are perpendicular to the wave's propagation direction. The particles shake in a directional wave in the longitudinal wave propagation.
Therefore, is not a characteristic of a transverse mechanical wave. Option B is correct. It involves displacing the medium perpendicular to the motion of the wave.
Learn more about transverse mechanical waves, here:
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The mass of an object with 500 J of kinetic energy moving with a velocity of 5 m/s is kg.
Answer:
[tex]m=20kg[/tex]
Explanation:
Hello,
In this case, we define the kinnetic energy as:
[tex]K=\frac{1}{2} m*v^2[/tex]
Thus, for finding the mass we simply solve for it on the previous equation given the kinetic energy and the velocity:
[tex]m=\frac{2*K}{v^2}=\frac{500kg*\frac{m^2}{s^2} }{(5\frac{m}{s})^2} =\frac{500kg*\frac{m^2}{s^2} }{25\frac{m^2}{s^2}}\\\\m=20kg[/tex]
Best regards.
Answer:
The answer is 40 kg
Explanation:
You will this formula below:
m=[tex]\frac{2*\\KE}{v^{2} }[/tex]
Now we know our formula, now we plug in the given numbers:
m=[tex]\frac{2(500J)}{(5m/s)^2}[/tex]
Simplify and we get:
m=40 kg
I hope this was helpful.
Describe, in detail, to a freshman undergraduate how to make 1 liter of LB + Kan (100 μg/ml final concentration) + Amp (50 μg /ml final concentration) liquid media. [Include things like how many grams of each component that you use, how much antibiotic (in ml) to add (stock solutions – 100 mg/ml ampicillin, 25 mg/ml kanamycin), and in what type of container you perform the sterilization step.] Show your calculations.
Answer:
Explanation:
The objective here is to prepare 1 liter of LB + Kan (100 μg/ml final concentration) + Amp (50 μg /ml final concentration) liquid media.
Given that :
the Stock concentration of Amp: 100 mg/ml (since 1 mg/ml = 1000 μg/ml)
it is required that we convert stock concentration in μg/ml since 100 mg/ml = 100000 μg/ml
However, using formula C₁V₁=C₂V₂ (Ampicilin),
where:
C₁ = 100000 μg/ml,
V₁=?,
C₂= 50 μg/ml,
V₂=1000 ml
100000 μg/ml × V₁ = 50 μg/ml × 1000 ml
V₁ = 50 μg/ml × 1000 ml/100000 μg/ml
V₁ = 0.5 ml
Given that:
the Stock concentration of Kan: 25 mg/ml (since 1 mg/ml = 1000 μg/ml)
it is required that we convert stock concentration in μg/ml , 25 mg/ml = 25000 μg/ml
Now by using formula C₁V₁=C₂V₂ (Kanamycin),
C₁ = 25000 μg/ml,
V₁=?,
C₂= 100 μg/ml,
V₂=1000 ml
25000 μg/ml × V₁ = 100 μg/ml × 1000 ml
V₁ = 100 μg/ml × 1000 ml/25000 μg/ml
V₁ = 4 ml
Thus; in 1 lite of Lb+ Kan+Amp preparation;
0.5 ml of Amp & 4 ml of kanamycin is used for their stock preparation.
Finally;
Sterilization step should be carried out in flask (Clean dry glass wares) for media in an autoclave, the container size should be twice the volume of media which is prepared.
An aqueous KNO3 solution is made using 72.5 g of KNO3 diluted to a total solution volume of 2.00 L. Calculate the molarity, molality, and mass percent of the solution. (Assume the density of 1.05 g/mL for the solution.)
Answer:
The molarity is 0.359[tex]\frac{moles}{L}[/tex]
The molality is 0.354 [tex]\frac{moles}{kg}[/tex]
The mass percent of the solution es 3.45%
Explanation:
Molarity is a unit of concentration that indicates the amount of moles of solute that appear dissolved in each liter of the mixture. It is determined by:
[tex]Molarity (M)=\frac{number of moles of solute}{volume}[/tex]
Being:
K: 39 g/moleN: 14 g/moleO: 16 g/moleThe molar mass of KNO₃ is:
KNO₃= 39 g/mole + 14 g/mole + 3*16 g/mole= 101 g/mole
You can apply the following rule of three: if 101 grams of KNO₃ are present in 1 mole, 72.5 grams in how many moles are present?
[tex]moles of KNO_{3}=\frac{72.5 grams*1 mole}{101 grams}[/tex]
moles of KNO₃= 0.718
So you have:
moles of KNO₃= 0.718volume= 2 LApplying this quantity in the definition of molarity:
[tex]molarity=\frac{0.718 moles}{2 L}[/tex]
Molarity= 0.359[tex]\frac{moles}{L}[/tex]
The molarity is 0.359[tex]\frac{moles}{L}[/tex]
Molality is a way of measuring the concentration of solute in solvent and indicates the amount of moles of solute in each kilogram of solvent.
Then the molality is calculated by:
[tex]Molality=\frac{moles of solute}{mass of solvent in kilograms}[/tex]
Density is defined as the property that matter, whether solid, liquid or gas, has to compress in a given space. That is, it is the amount of mass per unit volume. So, if the density of 1.05 g / mL for the solution indicates that in 1 mL of solution there are 1.05 grams of solution, in 2000 mL (where 2L = 2000 mL, because 1 L = 1000mL) how much mass is there?
[tex]mass=\frac{2000 mL*1.05 grams}{1 mL}[/tex]
mass= 2100 grams
Since mass solution = mass water + mass KNO₃
then mass water = mass solution - mass KNO₃
Being mass solution 2100 grams and mass KNO₃ 72.5 grams, and replacing you get: mass water= 2100 grams - 72.5 grams
mass water= 2,027.5 grams
Then, being:
moles of KNO₃= 0.718mass of solvent in kilograms= 2.0275 kg (being 2,027.5 grams= 2.0275 kilograms because 1,000 grams= 1 kilogram)Replacing in the definition of molality:
[tex]molality=\frac{0.718 moles}{2.0275 kg}[/tex]
molality= 0.354 [tex]\frac{moles}{kg}[/tex]
The molality is 0.354 [tex]\frac{moles}{kg}[/tex]
The mass percent of a solution is the number of grams of solute per 100 grams of solution. Then the mass percent is the mass of the element or solute divided by the mass of the compound or solute and the result of which is multiplied by 100 to give a percentage.
[tex]mass percent=\frac{mass of solute}{mass of solution} *100[/tex]
So, in this case:
[tex]mass percent=\frac{72.5 grams}{2100 grams} *100[/tex]
mass percent= 3.45 % KNO₃ by mass
The mass percent of the solution es 3.45%
The molarity of the solution is 0.36 mol/L. The molality of the solution is 0.34 m. The mass percent of the solution is 3.33%.
Number of moles of KNO3 = mass/molar mass = 72.5 g/101 g/mol = 0.72 moles
Molarity = Number of moles / volume = 0.72 moles/ 2.00 L = 0.36 mol/L
The molality = Number of moles of solute/Mass of solution in kilograms
mass of solution = 1.05 g/mL × 2000 mL = 21000 g or 2.1Kg
Molality of solution = 0.72 moles/2.1 Kg = 0.34 m
Mass percent of solution = mass of solute/mass of solution × 100/1
Mass percent of solution = 72.5 g/ (72.5 g + 2100 g) × 100/1
= 3.33%
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A 1.00 liter solution contains 0.42 moles nitrous acid and 0.32 moles sodium nitrite .
If 0.16 moles of nitric acid are added to this system, indicate whether the following statements are true or false.
(Assume that the volume does not change upon the addition of nitric acid.)
A. The number of moles of HNO2 will decrease.
B. The number of moles of NO2- will remain the same.
C. The equilibrium concentration of H3O+ will increase.
D. The pH will decrease.
E. The ratio of [HNO2] / [NO2-] will increase
Answer:
E. The ratio of [HNO2] / [NO2-] will increase
D. The pH will decrease.
Explanation:
Nitrous acid ( HNO₂ ) is a weak acid and NaNO₂ is its salt . The mixture makes a buffer solution .
pH = pka + log [ salt] / [ Acid ]
= 3.4 + log .32 / .42
= 3.4 - .118
= 3.282 .
Now .16 moles of nitric acid is added which will react with salt to form acid
HNO₃ + NaNO₂ = HNO₂ + NaNO₃
concentration of nitrous acid will be increased and concentration of sodium nitrite ( salt will decrease )
concentration of nitrous acid = .42 + .16 = .58 M
concentration of salt = .32 - .16 = .16 M
ratio of [HNO₂ ] / NO₂⁻]
= .42 / .32 = 1.3125
ratio of [HNO₂ ] / NO₂⁻] after reaction
= .42 + .16 / .32 - .16
= 58 / 16
= 3.625 .
ratio will increase.
Option E is the answer .
pH after reaction
= 3.4 + log .16 / .58
= 2.84
pH will decrease.
reasons for good care on computer
answer
1)maximise your software efficiency
2)Prevention against viruses and malware
3)Early detection of problematic issues
4)prevent data loss
5)Speed up your computer
.Draw the born-Haber lattice energy cycle for sodium chloride. Explain the concept of resonance using the nitrate ion structure.
Answer:
Born-Haber cycle is consist on four to five steps. 1: ionization energy 2: electron affinity 3: dissociation energy 4: sublimation energy and last is Hess law.Nitrate ion have 3 localized sigma bonds and 1 delocalized pie bond according to the resonance structure.Explanation:
Step 1: NaCl(s) → Na(s) + 1/2 Cl2(g) ΔHf (ionization energy) in this step energy is required to change the phase of the compound
Step 2: Na(s) + 1/2 Cl2(g) → Na(g) + 1/2 Cl2(g) ΔHa (elements needed to be in gaseous state for born-haber cycle so metal changes from solid to gas state by changing the enthalpy.
Step 3: Na(g) + 1/2 Cl2(g) → Na(g) + Cl (g) 1/2ΔHd
Step 4: Na(g) + Cl(g) → Na⁺(g) + Cl⁻(g) IE+EA ( in this step both ionization energy and electron affinity was involved because in metal (Na) electron is added which needs the energy and this energy draw from the step 3 and Chlorine require releasing electron to be in ionic state so when electron leaves the orbit energy releases.
Step 5: final step is Hess Law which is the combination of all the steps which step 4 again go back to step 5 and this cycle continues by repeating same steps Na⁺(g) + Cl⁻(g)→NaCl(s)
at this step heat of formation is calculated
Heat of formation= atomization energy+ dissociation energy+ sum of ionization energies + sum of electron affinity + lattice energy.
2: if we look at the electron configuration of the nitrogen it has 5 electrons in its outermost shell which indicates it can make 5 bonds 4 bonds and 1 lone pair usually and Oxygen has 6 electrons in its outermost shell. So nitrate ion have the total number of 24 electrons including the 1 electron which shows on the compound.
So when they make nitrate ion NO₃⁻¹ it shows that nitrate has 3 resonance structures. Nitrogen's three sigma bonds are attached to oxygen and fourth one make 1 pie bond which can rotate, delocalized and change its position anytime from one Oxygen atom to other oxygen atom.
The boiling of water is a:_______.
a. chemical change because a gas (steam) is given off.
b. chemical change because heat is needed for the process to occur.
c. physical change because the water merely disappears chemical and physical damage.
d. physical change because the gaseous water is chemically the same as the liquid.
Answer:
D
Explanation:
trust me its correct i think
A 75 gram solid cube of mercury (II) oxide has a density of 2.4 x 103 kg/m3 .
What is the length of one side of the cube in cm?
The mercury (II) oxide completely dissociates and forms liquid mercury and oxygen gas. Write a balanced chemical equation and indicate if this process is a chemical or physical change?
The oxygen gas escapes and now you are left with liquid grey substance. Is this grey substance a compound, element, homogeneous mixture or heterogeneous mixture?
Answer:
0.031 m
HgO(s) ⇒ Hg(l) + 1/2 O₂(g)
Chemical change
Element
Explanation:
A 75 gram solid cube of mercury (II) oxide has a density of 2.4 × 10³ kg/m³. What is the length of one side of the cube in cm?
Step 1: Convert the mass to kilograms
We will use the relationship 1 kg = 1,000 g.
[tex]75g \times \frac{1kg}{1,000g} = 0.075kg[/tex]
Step 2: Calculate the volume (V) of the cube
[tex]0.075kg \times \frac{1m^{3} }{2.4 \times 10^{3} kg} = 3.1 \times 10^{-5} m^{3}[/tex]
Step 3: Calculate the length (l) of one side of the cube
We will use the following expression.
[tex]V = l^{3} \\l = \sqrt[3]{V} = \sqrt[3]{3.1 \times 10^{-5} m^{3} }=0.031m[/tex]
The mercury (II) oxide completely dissociates and forms liquid mercury and oxygen gas. Write a balanced chemical equation and indicate if this process is a chemical or physical change?
The balanced chemical equation is:
HgO(s) ⇒ Hg(l) + 1/2 O₂(g)
This is a chemical change because new substances are formed.
The oxygen gas escapes and now you are left with liquid grey substance. Is this grey substance a compound, element, homogeneous mixture or heterogeneous mixture?
The liquid gray substance is Hg(l), which is an element because it is formed by just one kind of atoms.
ilicon has three naturally occurring isotopes (Si-28, Si-29, and Si-30). Masses and natural abundances for two isotopes are listed here. Isotope Mass (amu) Abundance (%) Si-28 27.9769 92.2 Si-29 28.9765 4.67 Si-30 ? ? Part A Find the natural abundance of Si-30 . Express your answer using two significant figures.
Answer:
Part A
The natural abundance of Si-30 = 3.1
Part B
The mass of Si-30 = 29.9551 amu
Explanation:
Part A
The sum of the natural abundances = 100
Si-28 has a mass of 27.9769 amu and a natural abundance of 92.2%.
Si-29 has a mass of 28.9765 amu and a natural abundance of 4.67%
Si-30's mass and natural abundance are unknown.
Natural abundance for Si-30 = 100% - 92.2% - 4.67% = 3.13% = 3.1% to 2 s.f.
Part B
The total atomic mass of an element is an addition combination of the mass and natural abundances of all the isotopes of that element.
Molar mass of Silicon normally = 28.0855 amu
Let the mass of Si-30 be m
28.0855 = (27.9769×0.922) + (28.9765×0.0467) + (m×0.0313)
28.0855 = 27.14790435 + 0.0313m
0.0313m = 28.0855 - 27.14790435 = 0.93759565
m = (0.93759565/0.0313) = 29.9551325879 amu = 29.9551 amu
Hope this Helps!!
Part A: The natural abundance of Si-30 = 3.1
Lets solve the question:
The sum of the natural abundances = 100 Si-28 has a mass of 27.9769 amu and a natural abundance of 92.2%. Si-29 has a mass of 28.9765 amu and a natural abundance of 4.67% Si-30's mass and natural abundance are unknown.Natural abundance (NA) refers to the abundance of isotopes of a chemical element as naturally found on a planet.
Natural abundance for Si-30 = [tex]100\% - 92.2\% - 4.67\% = 3.13\% = 3.1\%[/tex] in significant figures.
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81. Find the pH of each mixture of acids. a. 0.115 M in HBr and 0.125 M in HCHO2 b. 0.150 M in HNO2 and 0.085 M in HNO3 c. 0.185 M in HCHO2 and 0.225 M in HC2H3O2 d. 0.050 M in acetic acid and 0.050 M in hydrocyanic acid
Answer:
See explanation below
Explanation:
This problem is a little long so I'm gonna be as clear as possible.
a) In this case we have two acids, HBr and HCHO2. Between these two acids, the HBr is the strongest, and does not have a Ka value to dissociate, while HCHO2 do.
In order to calculate pH we need the [H₃O⁺], and in this case, as HBr is stronger, the contribution of the weaker acid can be negligible, therefore, the pH of this mixture will be:
pH = -log[H₃O⁺]
pH = -log(0.115)
pH = 0.93
b) In this case it happens the same thing as part a) HNO₃ is the strongest acid, so the contribution of the HNO₂ which is a weak acid is negligible too, therefore the pH of this mixture will be:
pH = -log(0.085)
pH = 1.07
c) Now in this case, HCHO2 and HC2H3O2 are both weak acids, so to determine which is stronger, we need to see their Ka values. In the case of HCHO2 the Ka is 1.8x10⁻⁴ and for the HC2H3O2 the Ka is 1.8x10⁻⁵. Note that the difference between the two values of Ka is just 10¹ order, so, we can neglect the concentration of either the first or the second acid. We need to see the contribution of each acid, let's begin with the stronger acid first, which is the HCHO2, we will write an ICE chart to determine the value of the [H₃O⁺] and then, use this value to determine the same concentration for the second acid and finally the pH:
HCHO₂ + H₂O <-------> CHO₂⁻ + H₃O⁺ Ka = 1.8*10⁻⁴
i) 0.185 0 0
c) -x +x +x
e) 0.185-x x x
1.8*10⁻⁴ = x² / 0.185-x
As Ka is small, we can assume that "x is small" too, therefore the (0.185-x) can be rounded to just 0.185 so:
1.8*10⁻⁴ = x²/0.185
1.8*10⁻⁴ * 0.185 = x²
x² = 3.33*10⁻⁵
x = 5.77*10⁻³ M = [H₃O⁺]
Now that we have this concentration, let's write an ICE chart for the other acid, but taking account this concentration of [H₃O⁺] as innitial in the chart, and solve for the new concentration of [H₃O⁺] (In this case i will use "y" instead of "x" to make a difference from the above):
HC₂H₃O₂ + H₂O <--------> C₂H₃O₂⁻ + H₃O⁺ Ka = 1.8x10⁻⁵
i) 0.225 0 5.77x10⁻⁶
c) -y +y +y
e) 0.225-y y 5-77x10⁻³+y
1.8x10⁻⁵ = y(5.77x10⁻³+y) / 0.225-y ---> once again, y is small so:
1.8x10⁻⁵ = 5.77x10⁻³y + y² / 0.225
1.8x10⁻⁵ * 0.225 = 5.77x10⁻³y + y²
y² + 5.77x10⁻³y - 4.05x10⁻⁶ = 0
Solving for y:
y = -5.77x10⁻³ ±√(5.77x10⁻³)² + 4*4.05x10⁻⁶ / 2
y = -5.77x10⁻³ ±√4.95x10⁻⁵ / 2
y = -5.77x10⁻³ ± 7.04x10⁻³ / 2
y₁ = 6.35x10⁻⁴ M
y₂ = -6.41x10⁻³ M
We will take y₁ as the value, so the concentration of hydronium will be:
[H₃O⁺] = 5.77x10⁻³ + 6.35x10⁻⁴ = 6.41x10⁻³ M
Finally the pH for this mixture is:
pH = -log(6.41x10⁻³)
pH = 2.19
d) In this case, we have the same as part c, however the Ka values differ this time. The Ka for acetic acid is 1.8x10⁻⁵ while for HCN is 4.9x10⁻¹⁰. In this ocassion, we the difference in their ka is 10⁵ order, so we can neglect the HCN concentration and focus in the acetic acid. Let's do an ICE chart and then, with the hydronium concentration we will calculate pH:
HC₂H₃O₂ + H₂O <--------> C₂H₃O₂⁻ + H₃O⁺ Ka = 1.8x10⁻⁵
i) 0.050 0 0
c) -y +y +y
e) 0.050-y y y
1.8*10⁻⁵ = y² / 0.050-y
As Ka is small, we can assume that "y is small" too
1.8*10⁻⁵ = y²/0.050
1.8*10⁻⁵ * 0.050 = y²
y² = 9*10⁻⁷
y = 9.45*10⁻⁵ M = [H₃O⁺]
Finally the pH:
pH = -log(9.45x10⁻⁵)
pH = 3.02
Fishing trawlers in a certain bay catch a large variety of marketable fish along with a species of eel that is toxic. They normally kill the eels and throw them back into the sea. What term is used to refer to the eel? The eel species is called a(n) ______ of the fishing operation.
Answer:
1. Non-target
2. Bycatch
Explanation:
In the fishing industry, the main aim of the industry is to capture fishes that can be used or eaten and sell. A variety of fishes are captured for this purpose and since they are used therefore are known as Target catch.
But there are some species which has to be discarded because they are toxic and not useful. These non-useful species like eel which gets captured in the net while capturing other fishes are known as Non-target fish.
The eel fish which gets captured is known as bycatch fishes in the fishing operation.
Thus, Non-target and Bycatch are the correct answer.
Answer:
Its just bycatch
Explanation:
The eel species is called a(n) bycatch of the fishing operation.
Classify an element having the following ground state electron configuration as an alkali metal, alkaline earth metal, nonmetal, halogen, transition metal, or noble gas.
a. [Ne]3s1
b. [Ne]3s23p3
c. [Ar]4s23d104p5
d. [Kr]5s24d1
e. [Kr]5s24d105p6
Explanation:
Alkali metal refers to group1 elements.
Alkali earth metal refers to group 2 elements.
Non metals refers to elements in grouos 4 to group 7.
Halogen refers to group 17 elements
Transition Metal refers to group 3 to group 12 elements
Noble gases refer to elements in group 18.
To obtain the group number from the electronic configuration, we calculate the total number of electrons in the principal quantum number (coefficient of the letters).
a. [Ne]3s1
Principal quantum number = 3
Number of electrons present = 1
This element belongs to group 1. It is an Alkali Metal.
b. [Ne]3s23p3
Principal quantum number = 3
Number of electrons present = 2 + 3 = 5
This element belongs to group 15 (5A). It is a Non metal
c. [Ar]4s23d104p5
Principal quantum number = 4
Number of electrons present = 2 + 5 = 7
This element belongs to group 17 (7A). It is an Halogen.
d. [Kr]5s24d1
This configuration belongs to the element yttrium and has an incomplete d shell. Hence it is a transition metal.
e. [Kr]5s24d105p6
Principal quantum number = 5
Number of electrons present = 2 + 6 = 8
This element belongs to group 18 (8A). It is a Noble gas.
Unscramble the following words to form a complete
sentence about the cycles of nature:
limited is through environment Matter recycled the on Earth is and
Answer:
recycled is limited through enviroment and matter on earth
Explanation:
Arrange the following compounds in order of increasing solubility in
water and explain your sequence.
C7H15OH C6H13OH C6H6 C2H5OH
Answer:
C6H6<C7H15OH<C6H13OH<C2H5OH
Explanation:
Organic substances are ordinarily nonpolar. This means that they do not dissolve in water. However, certain homologous series of organic compounds actually dissolve in water because they possess certain functional groups that effectively interact with water via hydrogen bonding.
A typical example of this is alcohol family. All members of this homologous series contain the -OH functional group. This group can effectively interact with water via hydrogen bonding, leading to the dissolution of low molecular weight alcohols in water.
Low molecular weight alcohols are miscible with water in all proportions. This implies that they are highly soluble in water. However, as the size of the alkyl moiety in the alcohol increases, the solubility of the alcohol in water decreases due to less effective interaction of the -OH group with water via hydrogen bonding. This explains the fact that C2H5OH is the most soluble alcohol in the list.
C6H6 is insoluble in water since it is purely a hydrocarbon with no -OH group capable of interaction with water via hydrogen bonding.
Acetonitrile, CH3CN, is a polar organic solvent that dissolves many solutes, including many salts. The density of a 1.80 M acetonitrile solution of LiBr is 0.826 g/mL. Calculate the concentration of the solution in units of (a) molality; (b) mole fraction of LiBr; (c) mass percentage of CH3CN.
Answer:
(a) [tex]m=2.69m[/tex]
(b) [tex]x_{LiBr}=0.099[/tex]
(c) [tex]\% LiBr=18.9\%[/tex]
Explanation:
Hello,
In this case, given the molality in mol/L, we can compute the required units of concentration assuming a 1-L solution of acetonitrile and lithium bromide that has 1.80 moles of lithium bromide:
(a) For the molality, we first compute the grams of lithium bromide in 1.80 moles by using its molar mass:
[tex]m_{LiBr}=1.80mol*\frac{86.845 g}{1mol}=156.32g[/tex]
Next, we compute the mass of the solution:
[tex]m_{solution}=1L*0.826\frac{g}{mL}*\frac{1000mL}{1L}=826g[/tex]
Then, the mass of the solvent (acetonitrile) in kg:
[tex]m_{solvent}=(826g-156.32g)*\frac{1kg}{1000g}=0.670kg[/tex]
Finally, the molality:
[tex]m=\frac{1.80mol}{0.670kg} \\\\m=2.69m[/tex]
(b) For the mole fraction, we first compute the moles of solvent (acetonitrile):
[tex]n_{solvent}=669.68g*\frac{1mol}{41.05 g} =16.31mol[/tex]
Then, the mole fraction of lithium bromide:
[tex]x_{LiBr}=\frac{1.80mol}{1.80mol+16.31mol}\\ \\x_{LiBr}=0.099[/tex]
(c) Finally, the mass percentage with the previously computed masses:
[tex]\% LiBr=\frac{156.32g}{826g}*100\%\\ \\\% LiBr=18.9\%[/tex]
Regards.
