Answer:
a measure of concentration equal to the gram equivalent weight per liter of solution.
Explanation:
Gram equivalent weight is the measure of the reactive capacity of a molecule. The solute's role in the reaction determines the solution's normality. Normality is also known as the equivalent concentration of a solution.
hope it helped
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
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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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]
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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313.9 liters of a gas has a pressure of 390.89 kPa at 76.6°C. If the pressure increases to 718.3 kPa and the temperature to 154.2°C, what would be the new volume of the gas?
A.) 210
B.) 353
C.) 470
D.) 209.92
Explanation:
P1V1/T1 = P2V2/T2
=((390.89×313.9)/76.6)×((718.3×V2)/154.2)
= (122700.371/76.7) × ((718.3×V2)/154.2)
make V2 the subject of the formula...
V2 =(122700.371×154.2)/(76.6×718.3)
V2 =18920397.21/55021.78
V2 = 343.87
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
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.
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
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[tex]SilentNature[/tex]
Calculate the mass percent of each component in the following solution.
159 g NiCl2 in 500 g water
% Nicla
% water
Answer:
% NiCl2 = 24.13%
% water = 78.57%
Explanation:
Mass percentage = mass of solute/mass of solution × 100
According to this question, a solution contains 159 g of NiCl2 in 500 g of water. Hence, mass of the solution is calculated as follows:
Mass of solution = 159g + 500g
Mass of solution = 659g
Therefore;
A) % Mass of NiCl2 in solution = mass of NiCl2/mass of solution × 100
% Mass of NiCl2 in solution = 159/659 × 100
% Mass of NiCl2 in solution = 0.2413 × 100
= 24.13%
B) % Mass of water in solution = mass of water/mass of solution × 100
% Mass of water in solution = 500/659 × 100
% Mass of water in solution = 0.7587 × 100
% Mass of water in solution = 75.87%
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
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.
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 from mixing 50.0g of sulfuric acid and 40.0 grams of barium chloride is complete, 31.16 grams of sulfuric acid and 0 grams of barium chloride remain.
The equation of the reaction between sulfuric acid and barium chloride is
BaCl₂ + H₂SO₄ → BaSO₄ + 2HCl
From this equation of reaction, it means 1 mole of barium chloride will completely react with 1 mole of sulfuric acid.
From the question, we have 50.0g of sulfuric acid and 40.0 grams of barium chloride.
First, we will determine the number of moles of the sulfuric acid and barium chloride present.
Number of moles is given by the formula
Number of moles = Mass / Molar mass
For sulfuric acid
Mass = 50.0 g
Molar mass = 98.079 g/mol
∴ Number of moles = 50.0 / 98.079
Numbers of moles of sulfuric = 0.509793 mol
For barium chloride
Mass = 40.0 grams
Molar mass of barium chloride = 208.23 g/mol
∴ Number of moles = 40.0 / 208.23
Number of moles of barium chloride = 0.192095 mol
Since the number of moles of sulfuric acid is more than that of barium chloride, then the limiting reagent is barium chloride and the excess reagent is sulfuric acid
NOTE: A limiting reagent is the reactant that is completely used up in a reaction, and it determines when the reaction stops.
Hence, barium chloride will be used up during the reaction (that is, 0 grams will remain after the reaction is complete).
For the mass of sulfuric acid that will remain,
First, we will determine the number of mole that will remain.
Since 1 mole of barium chloride completely reacts with 1 mole of sulfuric acid, then 0.192095 mol of barium chloride will react with 0.192095 mol of sulfuric acid.
∴ The remaining number moles of sulfuric acid = 0.509793 mol - 0.192095 mol
The remaining number moles of sulfuric acid = 0.317698 mol
Then, from
Mass = Number of moles × Molar mass
Mass = 0.317698 mol ×98.079 g/mol
Mass = 31. 1595 gram
Mass ≅ 31.16 grams
∴ 31.16 grams remains after the reaction is complete.
Hence, 31.16 grams of sulfuric acid and 0 grams of barium chloride remain after the double replacement reaction is complete.
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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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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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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.
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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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
Name the following compound: CH3CH2CH2CH2CH2CH2CCCH3
2-nonene
7-nonyne
2-heptyne
2-nonyne
Finding Nemo??? sorry I really need these points
Answer:
2-nonyne
explanation:
First consider the type of bonds, there are tripple bonds of carbon to carbon, in position 2 from the right.
hence it us alkyn.
There are 9 carbons.
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 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.
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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Do you think that the human being is the center of the universe?
Write balanced equations for the reaction of each of the following carboxylic acids with NaOH. Part A formic acid Express your answer as a chemical equation. A chemical reaction does not occur for this question. Request Answer Part B 3-chloropropanoic acid Express your answer as a chemical equation. nothing A chemical reaction does not occur for this question.
Answer:
Part A
HCOOH(aq) + NaOH(aq) → HCOONa(aq) + H2O(l)
Part B
ClCH2CH2CO2H(aq) + NaOH(aq) ------> ClCH2CH2CO2Na(aq) + H2O(l)
Explanation:
The reaction between an alkanoic acid and a base is a neutralization reaction. The reaction occurs as follows;
RCOOH + NaOH ----> RCOONa + H2O
We have to note the fact that the net ionic reaction still remains;
H^+(aq) + OH^-(aq) ---> H2O(l)
In both cases, the reaction can occur and they actually do occur as written.
90
1
39
is the
In the following decay equation,
90
Sy →
38
et
90
39
-1
A. alpha particle
B. parent element
C. daughter element
D. beta particle
Answer:
D. beta particle
Explanation:
Number of protons increases from 38 to 39 indicating beta decay (only one proton up from parent isotope to daughter isotope) Also atomic mass (on top of an isotope), 90 stays the same as beta particle is very small.
oxidation number of Ni in Ni(CO)4 is
Answer:
0
Explanation:
answer from gauth math
which of the following measurements is equivalent to 5.461x10^-7m?
Answer:
B. 0.0000005461m
I used the method of moving the decimal.
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.
The reaction for photosynthesis producing glucose sugar and oxygen gas is:
__CO2(g) + __H2O(l) UV/chlorophyl−→−−−−−−−−−−−−−− __C6H12O6(s) + __O2(g)
What is the mass of glucose (180.18 g/mol) produced from 2.20 g of CO2 (44.01 g/mol)?
a. 66.1 g C6H12O6
b. 396 g C6H12O6
c. 54.0 g C6H12O6
d. 1.50 g C6H12O6
e. 9.01 g C6H12O6
The correct option is d.: 1.5 grams of glucose is produced from 2.20 g of CO₂.
To find the mass of glucose produced, first you must know the balanced reaction. For this, the Law of Conservation of Matter is followed.
The law of conservation of matter states that since no atom can be created or destroyed in a chemical reaction, the number of atoms that are present in the reagents has to be equal to the number of atoms present in the products.
So, in this case, the balanced reaction is:
6 CO₂ + 6 H₂O → C₆H₁₂O₆ + 6 O₂
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the amounts of moles of each reactant and product participate in the reaction:
CO₂: 6 moles H₂O: 6 moles C₆H₁₂O₆: 1 mole O₂: 6 molesSo, you know that 2.20 g of CO₂ react, whose molar weight is 44.01 g/mole. By definition of molar mass, 1 mole of CO₂ has 44.01 g. So, the number of moles that 2.20 grams of the compound represent is calculated as:
[tex]moles of CO_{2} =2.20 grams*\frac{1 mole}{44.01 grams}[/tex]
moles of CO₂= 0.05 moles
Now you must follow the following rule of three: if by stoichiometry of the reaction 6 moles of CO₂ produce 1 mole of C₆H₁₂O₆, 0.05 moles of CO₂ produce how many moles of C₆H₁₂O₆?
[tex]moles of C_{6} H_{12} O_{6} =\frac{0.05moles of CO_{2} *1 mole of C_{6} H_{12} O_{6}}{6moles of CO_{2}}[/tex]
moles of C₆H₁₂O₆= 8.33*10⁻³
Being the molar mass of glucose 180.18 g/mole, the mass that 8.33*10⁻³ moles of the compound represent is calculated as:
[tex]mass of glucose =8.33*10^{-3} moles*\frac{180.18 grams}{1 mole}[/tex]
mass of glucose= 1.5 grams
In summary, the correct option is d.: 1.5 grams of glucose is produced from 2.20 g of CO₂.
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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
