Answer:
17
Explanation:
Step 1: Calculate the needed concentrations
[A]i = 1.00 mol/5.00 L = 0.200 M
[B]i = 1.80 mol/5.00 L = 0.360 M
[B]e = 1.00 mol/5.00 L = 0.200 M
Step 2: Make an ICE chart
A(aq) + 2 B(aq) ⇄ C(aq)
I 0.200 0.360 0
C -x -2x +x
E 0.200-x 0.360-2x x
Then,
[B]e = 0.360-2x = 0.200
x = 0.0800
The concentrations at equilibrium are:
[A]e = 0.200-0.0800 = 0.120 M
[B]e = 0.200 M
[C]e = 0.0800 M
Step 3: Calculate the concentration equilibrium constant (K)
K = [C] / [A] × [B]²
K = 0.0800 / 0.120 × 0.200² = 16.6 ≈ 17
which of the following measurements is equivalent to 5.461x10^-7m?
Answer:
B. 0.0000005461m
I used the method of moving the decimal.
Kevin's supervisor, Jill, has asked for an update on today's sales, Jill is pretty busy moving back and forth between different store locations. How can Kevin most effectively deliver an update to her ? a) Call with a quick update Ob ) Send a detailed text message c ) Book a one-hour meeting for tomorrow morning d) Send a detailed email
Answer:
d
Explanation:
since it is much convenient since the email will not get lost and it's contents will not be forgotten
how many moles of lithium atoms are contained in 5.2 g of lithium
Answer:
[tex]\boxed {\boxed {\sf 0.75 \ mol \ Li}}[/tex]
Explanation:
We are asked to convert 5.2 grams of lithium to moles of lithium.
1. Molar MassTo convert from grams to moles, we need the molar mass. This is the measurement of the mass in 1 mole of a substance. It can be found on the Periodic Table because it is the same value as the atomic mass, but the units are grams per mole instead of atomic mass units.
Look up the molar mass of lithium.
Li: 6.94 g/mol 2. Convert Grams to MolesCreate a ratio using the molar mass of lithium.
[tex]\frac { 6.94 \ g \ Li}{ 1 \ mol \ Li}[/tex]
Multiply by the value we are converting: 5.2 grams of lithium.
[tex]5.2 \ g \ Li *\frac { 6.94 \ g \ Li}{ 1 \ mol \ Li}[/tex]
Flip the ratio so the units of grams of lithium cancel.
[tex]5.2 \ g \ Li *\frac{ 1 \ mol \ Li} { 6.94 \ g \ Li}[/tex]
[tex]5.2 *\frac{ 1 \ mol \ Li} { 6.94 }[/tex]
[tex]\frac{5.2} { 6.94 } \ mol \ LI[/tex]
[tex]0.749279538905 \ mol \ Li[/tex]
3. RoundThe original measurement of grams (5.2) has 2 significant figures, so our answer must have the same. For the number we calculated, that is the hundredth place. The 9 in the thousandths place to the right tells us to round the 4 up to a 5.
[tex]0.75 \ mol \ Li[/tex]
5.2 grams of lithium is equal to 0.75 moles of lithium atoms.
A sample of aluminum, which has a specific heat capacity of , is put into a calorimeter (see sketch at right) that contains of water. The temperature of the water starts off at . When the temperature of the water stops changing it's . The pressure remains constant at .Calculate the initial temperature of the aluminum sample. Be sure your answer is rounded to significant digits.
Complete Question
A sample of aluminum, which has a specific heat capacity of 0.897 JB loc ! is put into a calorimeter (see sketch at right) that contains 200.0 g of water. The aluminum sample starts off at 85.6 °C and the temperature of the water starts off at 16.0 °C. When the temperature of the water stops changing it's 20.1 °C. The pressure remains constant at 1 atm. Calculate the mass of the aluminum sample.
Answer:
[tex]M=58g[/tex]
Explanation:
From the question we are told that:
Heat Capacity [tex]H=0.897[/tex]
Mass of water [tex]M=200g[/tex]
Initial Temperature of Aluminium [tex]T_a=85.6[/tex]
Initial Temperature of Water [tex]T_{w1}=16.0[/tex]
Final Temperature of Water [tex]T_{w2}=16.0[/tex]
Generally
Heat loss=Heat Gain
Therefore
[tex]M*0.897*(85.6-20.1) =200*4.184*(20.1-16)[/tex]
[tex]M=58g[/tex]
If we slowly add a solution of mercury(II) ions to a solution of aqueous halide ions with roughly equal concentrations, a precipitate will form. Explain what the precipitate will consist of initially. g
Water, mercury chloride and nitrogen oxide.
Water, mercury chloride and nitrogen oxide will present in the precipitate when we slowly add a solution of mercury(II) nitrate to a solution of aqueous hydrochloric acid having halide ions both in equal concentrations. The equation of this reaction is Hg2(NO3)2 + 4 HCl ----> 2 HgCl2 + 2 H2O + 2 NO so it is concluded that from this reaction we get precipitate of water, mercury chloride and nitrogen oxide.
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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:
Do you think that the human being is the center of the universe?
A 1 liter solution contains 0.370 M hypochlorous acid and 0.493 M sodium hypochlorite. Addition of 0.092 moles of barium hydroxide will: (Assume that the volume does not change upon the addition of barium hydroxide.)
In the original solution you have the mixture of a weak acid (Hypochlorous acid) and its conjugate base (Sodium hypochlorite). That is a buffer.
The barium hydroxide will react with hypochlorous acid. If this reaction cause the complete reaction of hypochlorous acid, the buffer break its capacity and the pH change in several units. In this case:
The addition of barium hydroxide will raise the pH slightly because the buffer still working.
The initial moles of those species are:
Hypochlorous acid:
[tex]1L * \frac{0.370mol}{1L} = 0.370 moles[/tex]
Sodium hypochlorite:
[tex]1L * \frac{0.493mol}{1L} = 0.493 moles[/tex]
Now, a strong acid as barium hydroxide (Ba(OH)₂) reacts with a weak acid as hypochlorous acid (HClO) as follows:
Ba(OH)₂ + 2HClO → Ba(ClO)₂ + 2H₂O
For a complete reaction of 0.092 moles of barium hydroxide are required:
[tex]0.092 moles Ba(OH)_2*\frac{2mol HClO}{1molBa(OH)_2} = 0.184 moles HClO[/tex]
As there are 0.370 moles, the moles of HClO after the reaction are:
0.370 moles - 0.184 moles = 0.186 moles of HClO will remain
As you still have hypochlorite and hypochlorous acid you still have a buffer.
Thus, the pH will raise slightly because the amount of acid is decreasing and slightly because the buffer can keep the pH.
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Equimolar solutions of A and B are mixed and the reaction is allowed to reach equilibrium. Write down the reactio that correctly describes the relative concentrations at equilibrium?
Complete Question
Complete Question is attached below
Answer:
Option A
[tex]D=A[/tex] And [tex]C>A[/tex]
Explanation:
From the question, we are told that:
The Chemical Reaction
[tex]2A_{aq}+B_{aq} \leftrightarrow 3C_{aq}+2D_{aq}[/tex]
Generally, the equation for Equilibrium constant is mathematically given by
[tex]K=\frac{C^c*D^d}{A^a*B^b}[/tex]
Therefore
[tex]K=\frac{C^3*D^d}{A^2*B^b}[/tex]
Hence we conculde
[tex]D=A[/tex] And [tex]C>A[/tex]
Therefore Option A
Determine the number of atoms of O in 60.1 moles of Fe₂(SO₃)₃.
Answer:
3.310308*10^26
Explanation:
nO=9nFe2(SO3)3=9*60.1=540.9 moles
number of atoms: 540.9*6.02*10^23
The mass of an empty flask plus stopper is 64.232g. When the flask is completely filled with water the new mass is 153.617 g. The flask is emptied and dried, and a piece of metal is added. The mass of the flask, stopper and metal is 143.557. Next, water is added to the flask containing the metal and the mass is found to be 226.196. What is the density of the metal
Answer:
11.76 g/cm^3
Explanation:
Mass of empty flask and stopper = 64.232g
Mass of flask filled with water = 153.617 g
Mass of water = 153.617 g - 64.232g = 89.385 g
Mass of flask, stopper and metal = 143.557 g
Mass of metal = 143.557 g - 64.232g = 79.325 g
Mass of water, flask, stopper and metal = 226.196 g
Mass of water = 226.196 g - 143.557 g = 82.639 g
Since mass of water =volume of water
Volume occupied by metal = 89.385 cm^3 - 82.639 cm^3 = 6.746 cm^3
Density of metal = mass/volume = 79.325 g/6.746 cm^3
= 11.76 g/cm^3
Which technique would be best for separating sand and water?
A. filtration
B. distillation
C. chromatography
D. evaporation
Answer:
A. filtration
Hope it helps
What is a litmus solution? How is it obtained?
Suppose a 0.034 M aqueous solution of sulfuric acid (H2SO4) is prepared. Calculate the equilibrium molarity of SO4 You'll find information on the properties of sulfuric acid in the ALEKS Data resource. Round your answer to 2 significant digits.
Answer:
[SO4²⁻] = 0.015M
Explanation:
When H2SO4 is dissolved in water, HSO4- is produced in a direct reaction as follows:
H2SO4 → HSO4- + H+
As 1 mole of H2SO4 produce 1 mole of HSO4-, the molarity of HSO4- in this first reaction is 0.034M
Now, the HSO4- is in equilibrium with SO42- and H+ as follows:
HSO4⁻ ⇄ SO4²⁻ + H⁺
Where the equilibrium constant, K, is defined as:
K = 1.2x10⁻² = [SO4²⁻] [H⁺] / [HSO4⁻]
Where [] are the equilibrium concentrations of each species in the reaction.
The equilibrium concentrations are:
[SO4²⁻] = X
[H⁺] = X
[HSO4⁻] = 0.034M - X
Where X is reaction coordinate
Replacing:
1.2x10⁻² = [X] [X] / [0.034-X]
4.08x10⁻⁴ - 1.2x10⁻²X = X²
4.08x10⁻⁴ - 1.2x10⁻²X - X² = 0
Solving for X:
X = -0.027M. False solution, there are no negative concentrations.
X = 0.015M. Right solution.
That means the equilibrium molarity of SO4²⁻,
[SO4²⁻] = X
[SO4²⁻] = 0.015M3 attempts left Be sure to answer all parts. Which indicators that would be suitable for each of the following titrations: (a) CH3NH2 with HBr thymol blue bromophenol blue methyl orange methyl red chlorophenol blue bromothymol blue cresol red phenolphthalein (b) HNO3 with NaOH thymol blue bromophenol blue methyl orange methyl red chlorophenol blue bromothymol blue cresol red phenolphthalein (c) HNO2 with KOH thymol blue bromophenol blue methyl orange methyl red chlorophenol blue bromothymol blue cresol red phenolphthalein
An indicator usually signals the endpoint of a neutralization reaction by undergoing a color change. They aid in discovering the point of equivalence of a titration.
The kind of indicator used depends on the nature of acid/base reacted.
In the case of CH3NH2 with HBr which strong acid and weak base titration, suitable indicators include; bromophenol blue, methyl orange, methyl red, and chlorophenol blue.
In the case of HNO3 with NaOH, this is a strong acid, strong base titration hence phenolphthalein, methyl red, chlorophenol, and bromothymol blue cresol red blue are suitable indicators.
In the case of HNO2 with KOH, this a weak acid, strong base titration and the suitable indicators are cresol red and phenolphthalein.
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The idea that the behavior of the states of matter is determined by the kinetic energy and movement of their particles is called _____…
A. Sublimation Theory
B. Kinetic Movement Theory
C. Kinetic Molecular Theory
D. Van der Waals Theory
Answer:
C . Kinetic Molecular Theory
Net ionic reaction of H2SO4 with Ba(OH)2
Answer:
This is an acid-base reaction (neutralization): Ba(OH) 2 is a base, H 2SO 4 is an acid. This is a precipitation reaction: BaSO 4 is the formed precipitate.
What volume of a 1.5 M KOH solution is needed to provide 3.0 moles of KOH?
3.0 L
2.0 L
4.5 L
0.50 L
0.22 L
Please explain!
Explanation:
here's the answer to your question
2.0 L
Answer:
Solution given:
no. of mole(n)=3.0mole
molarity(M)=1.5M
we have
Volume of a substance is equal to the ratio of its no of mole to its molarity.
By this we get a relation:
Volume=no.of mole/molarity
substituting value
Volume=3.0/1.5=2
The required volume is 2 litre.
It is necessary to make 225 mL of 0.222 M solution of nitric acid. Looking on the shelf, you see only 16 M nitric acid. How much concentrated nitric acid is required to make the desired solution?
Explanation:
The required concentration of [tex]HNO_3[/tex] M1 =0.222 M.
The required volume of [tex]HNO_3[/tex] is V1 =225 mL.
The standard solution of [tex]HNO_3[/tex] is M2 =16 M.
The volume of standard solution required can be calculated as shown below:
Since the number of moles of solute does not change on dilution.
The number of moles [tex]n=molarity * volume[/tex]
[tex]M_1.V_1=M_2.V_2[/tex]
[tex]V2=\frac{M_1.V_1}{M_2} \\=0.222M x 225 mL / 16 M\\=3.12 mL[/tex]
Hence, 3.12 mL of 16 m nitric acid is required to prepare 0.222 M and 225 mL of nitric acid.
What is the difference between elimination and substitution reaction
Identify the key factors that will determine if a reaction undergoes elimination or substitution mechanism.
Use the following reagents to determine the type of reaction pathway expected and determine the products in each reaction.
a. Tert BuO- in tertbutanol and chlorobutane
b. KOH in water and bromobutane
c. NaI in acetone and bromobutane
Write a conclusion of no more than two paragraphs to summarize your results
Answer:
a) E2
b) SN2
c) SN2
Explanation:
A substitution reaction involves replacement of an atom or group in a molecule by another atom or group. An elimination reaction is the loss of two atoms from the same molecule leading to the formation of a multiple bond in the molecule.
We must note that primary alkyl halides never undergo SN1/E1 reactions. However, the presence of a strong bulky base such as tert BuO- , E2 reactions predominate. In the presence of strong bases such as OH^- and good nucleophiles such as I^-, SN2 mechanism predominates.
The Ka for acetic acid (HC2H3O2) is 1.80 x 10-5 . Determine the pH of a 0.0500mol/L acetic acid solution.
I have no idea how to approach this, so If you have the answer for it, please respond as soon as you can
Answer:
pH = 3.02
Explanation:
Acetic Acid is a weak acid (HOAc) that ionizes only ~1.5% as follows:
HOAc ⇄ H⁺ + OAc⁻.
In pure water the hydronium ion concentration [H⁺] equals the acetate ion concentration [OAc⁻] and can be determined* using the formula [H⁺] = [OAc⁻] = SqrRt(Ka·[acid]) = SqrRt(1.8x10⁻⁵ x 0.0500)M = 9.5x10⁻⁴M.
By definition, pH = -log[H⁺] = -log(9.5x10⁻⁴) = 3.02
______________________________________________________
*This formula can be used to determine the [H⁺] & [Anion⁻] concentrations for any weak acid in pure water given its Ka-value and the molar concentration of acid in solution.
How many protons does Tin have?
A. 50
B. 68
C. 118
Hello There!
Tin has 50 protons.Hope that helps you!
~Just a felicitous girlie
#HaveASplendidDay
[tex]SilentNature[/tex]
GM 2 all ,What is an atom define it .Good Day
Answer:
An atom is the smallest particle of an element that can take part in chemical reaction.
Explanation:
hope it will help u Amri
Complete and balance the nuclear equations for the following fission reactions.
a. 23592U+10n→16062Sm+7230Zn+?10n. Express your answer as a nuclear equation.
b. 23994Pu+10n→14458Ce+?+210n
Answer:
23592U+10n→14454Xe+9038Sr+210n
Explanation:
a nuclear reaction for the neutron-induced fission of U−235 to form Xe−144 and Sr−90.
The half-life of radon-222 is 3.8 days. How many grams of radon-222 remain
after 15.2 days if the original amount was 6.00 g?
A. 0.750 g
B. 0.375 g
C. 1.20 g
D. 3.00 g
The mass of radon-222 that will remain after 15.2 days given that it was originally 6 g is 0.375 g (Option B)
What is half life?This is the time taken for half a substance to decay.
How to determine the number of half-lives that has elapsedWe'll begin our calculation by calculating the number of half-lives that has elapsed after 15.2 days. This is illustrated below:
Half-life (t½) = 3.8 daysTime (t) = 15.2 day Number of half-lives (n) =?n = t / t½
n = 15.2 / 3.8
n = 4
Thus, 4 half-lives has elapsed.
How to determine the amount remainingOriginal amount (N₀) = 6 gNumber of half-lives (n) = 4Amount remaining (N) = ?The amount of radon-222 remaining can be obtained as illustrated below:
N = N₀ / 2ⁿ
N = 6 / 2⁴
N = 6 / 16
N = 0.375 g
Thus, the amount of radon-222 remaining after 15.2 days is 0.375 g
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A student was given a solid containing a mixture of nitrate salts. The sample completely dissolved in water, and upon addition of dilute HCl , no precipitate formed. The pH was lowered to about 1 and H2S was bubbled through the solution. No precipitate formed. The pH was adjusted to 8 and H2S was again bubbled in. This time, a precipitate formed. Which compounds might have been present in the unknown?
a. Ca(NO3)2
b. AgNO3
c. Fe(NO3)3
d. Cr(NO3)3
e. Cu(NO3)2
f. KNO3
g. Bi(NO3)2
Answer:
Fe(NO3)3, Cr(NO3)3, Co(NO3)3
Explanation:
According to the question, no precipitate is observed when HCl was added. This means that we must rule out AgNO3.
Again, the sulphides of Cu^2+, Bi^3+ are soluble in acidic medium but according to the question, the sulphides do not precipitate at low pH hence Cu(NO3)2 and Bi(NO3)3 are both ruled out.
The sulphides of Fe^3+, Cr^3+ and Co^3+ all form precipitate in basic solution hence Fe(NO3)3, Cr(NO3)3, Co(NO3)3 may be present.
The presence of Ca(NO3)2 and KNO3 may be confirmed by flame tests.
How many grams of copper(II) phthalocyanine would be produced by the complete cyclotetramerization of 0.125 moles of phthalonitrile in the presence of excess copper(II) chloride?
Answer:
it is 11.55 and ik because I just had that question
18.0 grams of copper(II) phthalocyanine would be produced by the complete cyclotetramerization of 0.125 moles of phthalonitrile in the presence of excess copper(II) chloride.
Let's consider the following balanced equation.
4 C₈H₄N₂(l) + CuCl₂(s) → Cu(C₃₂H₁₆N₈)(s) + Cl₂(g)
The molar ratio of C₈H₄N₂ to Cu(C₃₂H₁₆N₈) is 4:1. The moles of Cu(C₃₂H₁₆N₈) produced from 0.125 moles of C₈H₄N₂ are:
[tex]0.125 mol C_8H_4N_2 \times \frac{1molCu(C_{32}H_{16}N_8)}{4mol C_8H_4N_2} = 0.0313 molCu(C_{32}H_{16}N_8)[/tex]
The molar mass of Cu(C₃₂H₁₆N₈) is 576.07 g/mol. The mass corresponding to 0.0313 moles of Cu(C₃₂H₁₆N₈) is:
[tex]0.0313 moles \times \frac{576.07g}{mol} = 18.0 g[/tex]
18.0 grams of copper(II) phthalocyanine would be produced by the complete cyclotetramerization of 0.125 moles of phthalonitrile in the presence of excess copper(II) chloride.
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Which substance would be the most soluble in gasoline?
Select one:
A. hexane
B. NaNO3
C. HCI
D. water
E. Nacl
I think the answer most be d
In chemistry like dissolves like hence hexane will dissolve in gasoline.
Dissolution stems from intermolecular interaction between solute and solvent molecules.
If this interaction can not occur, dissolution of one substance in another is impossible.
Hexane dissolves in gasoline because the both substances are non-polar and can interact with each other effectively.
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The element Co exists in two oxidation states, Co(II) and Co(III), and the ions form many complexes. The rate at which one of the complexes of Co(III) was reduced by Fe(II) in water was measured. Determine the activation energy of the reaction from the following data:
T(K) K(s^-1)
293 0.054
298 0.100
We measured the Fe(II) reduction of one of the Co(III) complexes by water at a rate of about 0.545 kJ/mol (to three significant figures).
How is activation energy determined?Calculating a Reaction's Activation Energy A reaction's rate is influenced by the temperature at which it is carried out. The molecules travel more quickly and clash more frequently as the temperature rises. Moreover, the molecules contain greater kinetic energy.
We can use the Arrhenius equation to calculate the reaction's activation energy:
k = A × exp(-Ea/RT)
When the activation energy Ea, the rate constant k, the gas constant R, and the temperature T in Kelvin are all present.
Finding the natural logarithm of the equation's two sides results in:
ln(k) = ln(A) - (Ea/RT)
This equation can be rearranged to take a linear form:
ln(k) = (-Ea/R) × (1/T) + ln(A)
y = mx + b, where (1/T) is x, (-Ea/R) is the slope, and ln(A) is the y-intercept, has the form of a linear equation.
We can get the slope of the line using the given data:
slope = (-Ea/R) = (ln(k2/k1)) / (1/T2 - 1/T1)
where the rate constants for temperatures T1 and T2, respectively, are k1 and k2.
substituting the specified values:
k1 = 0.054s⁻¹ at 293 K
k2 = 0.100s⁻¹ at 298 K
T1 = 293 K
T2 = 298 K
slope = (-Ea/R)
= (ln(0.100/0.054)) / (1/298 - 1/293)
= 65.5 kJ/mol
Therefore, the activation energy of the reaction is:
Ea = slope * R = 65.5 kJ/mol × 8.314 J/mol-K = 545 J/mol
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Give the IUPAC and common name
Answer:
IUPAC Name:
N-ethyl-N-methylaniline
Common Name:
Benzenamine
