Answer:
What is the name of the substances that are used at the beginning of a chemical reaction?
The substances that go into a chemical reaction are called the reactants, and the substances produced at the end of the reaction are known as the products.
Explanation: :)
What is the ph of a 0. 10 m solution of NaCN at 25°C?(ka= 4. 9 x 10^-10for hcn at 25°C. )
The pH of a a 0. 10 m solution of NaCN at 25°C is 11.15
Calculation
The reaction in the solution is given below
[tex]CN^{-} + H_{2} O[/tex] → HCN + [tex]OH^{-}[/tex]
initial 0.1
change ( -x) (+x)
equilibrium ( 0.1 - x ) x
Kb = [HCN] [[tex]CN^{-}[/tex] ]/[[tex]CN^{-}[/tex] ]
Kb × Ka = Kw = 1.0 × [tex]10^{-14}[/tex]
Kb = 1.0 × [tex]10^{-14}[/tex] / 4.9 × [tex]10^{-10}[/tex] = [HCN] [[tex]CN^{-}[/tex] ]/[[tex]CN^{-}[/tex] ] = [tex]x^{2}[/tex]/ ( 0.1 - x )
Kb = 2.04× [tex]10^{-5}[/tex] = [tex]x^{2}[/tex]/ ( 0.1 - x )
Since , [NaCN] /Kb > 100 , we can simplify the above equation to
= 2.04× [tex]10^{-5}[/tex] = [tex]x^{2}[/tex]/ ( 0.1 )
x = 1.43 × [tex]10^{-3}[/tex] M = [HCN] = [ [tex]OH^{-}[/tex]]
Then pOH = 2.84
pH + pOH = 14
pH = 14- pOH = 14 - 2.84 = 11.15
Therefore , the pH is 11.15
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Calculate the ph of a solution formed by mixing 250. 0 ml of 0. 15 m nh4cl with 200. 0 ml of 0. 12 m nh3. the kb for nh3 is 1. 8 × 10-5.
i. 9. 45
ii. 4. 74
iii. 9. 06
iv. 04. 55
v. 9. 26
The potential of the hydrogen in water gives the pH of the acidity and basicity of the solution. The pH of the solution is 9.06. Thus, option iii is correct.
What is pH?The pH of a substance is given by the subtraction of the pOH from 14 which is the range of the pH scale.
The dissociation reaction for ammonia is given as,
NH₃ (aq) + H₂O (l) ⇄ NH₄⁺ (aq) + OH⁻ (aq)
Here, the concentration of ammonia is [NH₃] - x, ammonium ion is [NH₄⁺] + x, and hydroxide ion is x.
The molar concentration of ammonia is,
M = (0.12 M × 0.2 L) ÷ 0.45 L = 0.053 M
The molar concentration of ammonium ion is,
M = (0.15 M × 0.25 L) ÷ 0.45 L = 0.083 M
From the base dissociation constant and previous concentration from the reaction, the value of x or hydroxide ion is calculated as,
Kb = [NH₄⁺][OH⁻] ÷ [NH₃]
1.8 × 10⁻⁵ (0.053 - x) - (0.083 + x) × x = 0
x = [OH⁻] = 1.15 x 10⁻⁵
The pH from the hydroxide ion is calculated as,
pOH = - log [OH⁻]
= - log (1.15 x 10⁻⁵)
= 4.94
Further,
pH = 14 - pOH
= 14 - 4.94
= 9.06
Therefore, option iii. 9.06 is the pH of the solution.
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Determine the name for aqueous H2CO3. carbonous acid dihydrogen carbonate carbonic acid hydrocarbonic acid hydrocarbide acid
The name for aqueous [tex]H_{2} CO_{3}[/tex] is carbonic acid.
So, option C is correct one.
Binary hydrogen compounds with non metals may form H+ (proton) and an anion dissolved in water. The acidic solutions are named as if they were molecular acids , using the usual name for the compound itself , replacing hydrogen with hydro- and suffix -ide with ic. The word acid is then use.
The inorganic acid is an acid drive from one or more inorganic compounds. All inorganic acids form hydrogen ions and the conjugate base ions when dissolved in water.
Example: carbonic acid ( [tex]H_{2} CO_{3}[/tex] )
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Perform the following operationand express the answer inscientific notation.7.00x10^5 - 5.00x10^4
Taking into account the scientific notation, the result of the subtraction is 6.5×10⁵.
Scientific notationFirst, remember that scientific notation is a quick way to represent a number using powers of base ten.
The numbers are written as a product:
a×10ⁿ
where:
a is a real number greater than or equal to 1 and less than 10, to which a decimal point is added after the first digit if it is a non-integer number.n is an integer, which is called an exponent or an order of magnitude. Represents the number of times the comma is shifted. It is always an integer, positive if it is shifted to the left, negative if it is shifted to the right.Subtraction in scientific notationYou want to subtract two numbers in scientific notation. It should be noted that when the numbers to be added do not have the same base 10 exponent, the base 10 power with the highest exponent must be found. In this case, the highest exponent is 5.
Then all the values are expressed as a function of the base 10 exponent with the highest exponent. In this case: 5.00×10⁴=0.500×10⁵
Taking the quantities to the same exponent, all you have to do is subtract what was previously called the number "a". In this case:
7.00×10⁵ - 0.500×10⁵= (7.00- 0.500)×10⁵= 6.5×10⁵
Finally, the result of the subtraction is 6.5×10⁵.
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How many peaks are in the proton spin decoupled 13c nmr spectrum of 1,3,5-trinitrobenzene?
There are 2 peaks are in the proton spin decoupled 13C NMR spectrum of 1,3,5-trinitrobenzene .
The 13C NMR spectrum give the peak which is directly about the carbon skeleton not just the proton attached to it . The number of signals tell us how many different carbons or set of equivalent carbons . The splitting of a signal tells us how many hydrogens are attached to each carbon.
In 1,3,5-trinitrobenzene molecule , there are 2 peaks are in the proton spin decoupled 13C NMR spectrum three carbon give one signal and another three carbon give another one signal .
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if there are 12 moles of hydrogen how many moles of nitrogen are needed to react completely with the hydrogen? (N2(g)+3H2(g)->2NH3(g))
Answer:
Explanation:
Take a look at the balanced chemical equation for this synthesis reaction
N
2(g]
+
3
H
2(g]
→
2
NH
3(g]
Notice that you have a
1
:
3
mole ratio between nitrogen gas and hydrogen gas. This means that, regardless of how many moles of nitrogen gas you have, the reaction will always consume twice as many moles of hydrogen gas.
So, if you have
2
moles of nitrogen taking part in the reaction, you will need
2
moles N
2
⋅
3
moles H
2
1
mole N
2
=
6 moles H
2
number of oxygen atoms in 16 g of So2??
Now , we know that one mole of SO² Contains 2 Moles ( two ) of atoms of oxygen . Hence , 16 Grams of SO² Contains 0.5 Mole(s) of Oxygen
0.5
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Draw the major product formed in the reaction involving an acid anhydride dissolved in alcohol.
Ester is the major product formed in the reaction involving an acid anhydride dissolved in alcohol.
What is the chemistry of acid anhydrides?Acid chlorides and anhydrides both go through a number of similar reactions when subjected to nucleophilic action. Anhydrides react with water to generate carboxylic acids, with alcohols to form esters, and with amines to form amides, though they do so more slowly than acid chlorides. By using hydride reduction, anhydrides can also be converted to primary alcohols. As many anhydrides are created by coupling two carboxylic acids, one equivalent of the carboxylic acid is lost in anyhydride reactions as a leaving group. As a result, reactions are often only carried out using affordable, widely accessible anhydrides, such as acetic or benzoic. The advantage of anydrides over acid chlorides is that they are simpler to work with.Learn more about acid anhydride here:
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What mass of silver can be plated onto an object in 33. 5 minutes at 8. 70 a of current? ag (aq) e- → ag(s) 19. 6 g 9. 78 g 0. 326 g 3. 07 g 0. 102 g
Mass of silver can be plated onto an object in 33. 5 minutes at 8. 70 a of current is 19.55 g
Given, I=8.70A, t=33.5min=33.5*60=2010 sec, m of Ag = 19600g M=107.87 g/mol
From faradays law we can write q=I*t= 8.70*2010=17487 C.
moles of Ag =17487/ 19600=0.1812 moles
Ag deposited = 0.1812*107.87 = 19.55 g
SilverHaving the atomic number 47 and the Latin word argentum, which means "bright" or "white," silver is a chemical element. Its symbol is Ag. The highest electrical, thermal, and reflectivity of any metal are displayed by this transition metal, which is soft, white, and lustrous. The metal can be found in the Earth's crust in three different forms: as an alloy with other metals like gold and in minerals like argentite and chlorargyrite. It can also be found in its pure, free elemental form, also known as "native silver." During the refining of copper, gold, lead, and zinc, the majority of silver is created as a byproduct. As a valuable metal, silver has long been treasured.
What mass of silver can be plated onto an object in 33. 5 minutes at 8. 70 a of current? ag (aq) e- → ag(s) 19. 6 g 9. 78 g 0. 326 g 3. 07 g 0. 102 g
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Find the percent ionization of a 0. 337 m hf solution. The ka for hf is 3. 5 × 10^-4.
Percent ionization = 3.17%
We utilize the supplied acid equilibrium constant (Ka) to calculate the percentage of the acid that is ionized. It is the ratio of the acid and dissociated ion equilibrium concentrations. The HF acid would dissociate in the manner described below:
HF = H+ + F-
The following is how the acid equilibrium constant might be written:
Ka = [H+][F-] / [HF] = 3.5 x 10-4
We utilize the ICE table to determine the equilibrium concentrations.
HF H+ F-
I 0.337 0 0
C -x +x +x
---------------------------------------------
E 0.337-x x x
3.5 x 10-4 = [H+][F-] / [HF]
3.5 x 10-4 = [x][x] / [0.337-x]
Solving for x,
x = 0.01069 = [H+] = [F-]
percent ionization = 0.01069 / 0.337 x 100 = 3.17%
percent ionization = 3.17%
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A metal object is to be gold-plated by an electrolytic procedure using aqueous AuCl3 electrolyte. How much gold may be deposited in 3.0 min by a constant current of 10. A
Answer:
This is the answer
Explanation:
charges passed = current x time = 10 x 3 x 60
= 1800 C
mole of charge = 1800 / 96500
= .01865 moles
Au+3 contains 3 positive charges
3 mole of charge will deposit 1 mole of Au .01865 moles will deposit .01865 / 3 mole
= 6.2167 x 10-3 moles.
0. 34 ll of hno3hno3 is titrated to equivalence using 0. 14 ll of 0. 1 mnaohmnaoh. what is the concentration of the hno3hno3 ?
When 0.34 of HNO₃ is titrated to equivalence using 0.14 l of 0.1 m NaOH then the concentration of HNO₃ is 0.041 M
The reaction of neutralization of HNO₃ with NaOH is
HNO₃ + NaOH → H₂O + NaNo₃
When 1 mole of HNO₃ react with 1 mole of NaOH, based on chemical rection the moles of NaOH at equivalence point are equal to moles of HNO₃ present in solution: -
With the mole and volumes, we can find molarity as follows:
Moles of NaOH = moles HNO₃
⁼ 0.14 L X (0.1 mol NaOH/L) = 0.014 mole NaOH
=0.014 mol HNO₃
Molarity: -
[tex]\frac{(Mole of HNO₃)}{(volume of HNO₃)}= \frac{0.014}{0.34}[/tex]
= 0.041 M
Thus, from above solution we concluded that the concentration of HNO₃ solution is 0.041 M.
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Calculate the ph of a 0.377 m solution of ethylenediamine ( h2nch2ch2nh2 ). the pa values for the acidic form of ethylenediamine ( h 3nch2ch2nh 3 ) are 6.848 ( pa1 ) and 9.928 ( pa2 ).
The final answer is 11.71
The concentration of a given solution (H2NCH2CH2NH2) is 0.314 M
pa1 = 6.848
a1 = 1.42 X [tex]10^{-7}[/tex]
Pa2 = 9.928
a2 = 1.18 X [tex]10^{-10}[/tex]
The equation is given as;
NH2CH2CH2NH2 + H2O ---> NH2CH2CH2NH3+ + OH-
Initial 0.314 0 0
Change -x +x +x
Equilibrium 0.314 -x x x
b1 = Kw / Ka2 = 10^-14 / 1.18 X 10^-10 = 8.47 X 10^-5 = [NH2CH2CH2NH3+][OH-] / [NH2CH2CH2NH2]
8.47 X 10^-5 = x^2 / 0.314 -x
8.47 X 10^-5 X 0.314 - 8.47 X 10^-5x = x^2
x^2 - 2.66 X 10^-5 + 8.47 X 10^-5x = 0
on solving
x = 0.0051 M = [NH2CH2CH2NH3+] = [OH-]
[NH2CH2CH2NH2] = 0.314 - 0.0051 = 0.3089 M
For the second dissociation,
NH2CH2CH2NH3+ + H2O ---> +NH3CH2CH2NH3+ + OH-
Initial 0.0051 0 0.0051
Change -x +x +x
Equilibrium 0.0051 -x x 0.0051 +x
b2 = Kw / Ka1 = 10^-14 / 1.42 X 10^-7 = 7.04 X 10^-8 = [+NH3CH2CH2NH3+] [OH-] / [+NH3CH2CH2NH2]
7.04 X 10^-8 = x (0.0051+x) / (0.0051-x)
We may ignore x in the denominator as b2 is very low
7.04 X 10^-8 = x (0.0051+x) / (0.0051)
3.59 X 10^-10 = 0.0051x + x^2
x = 7.039 X 10^-8 M
[OH-] = 0.0051 + x = 0.0051 approx
[+NH3CH2CH2NH3+] = x = 7.039 X 10^-8 M
NH2CH2CH2NH3+ = 0.0051 +x = 0.0051 M approx
pOH = --log [OH-] = -log 0.0051 = 2.29
pH = 14 - pOH = 11.71
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Consider the reaction below.
H O single bonded to C double bonded above to O and to the right to the bottom left of a benzene ring. The ring is bonded from the upper right to a C double bonded above to O and to the right to O H. H O bonded to C H Subscript 2 Baseline C H Subscript 2 Baseline bonded to O H. Arrow pointing with bond H Subscript 2 Baseline O beside it pointing down to H O bonded to C double bonded above to O and to the right to the bottom left of a benzene ring. The ring is bonded from the upper right to a C double bonded above to O and to the right to O bonded to C H Subscript 2 Baseline C H Subscript 2 Baseline bonded to O H.
This reaction eventually forms this product.
A bond from left to O single bonded to C double bonded above to O and to the right to the bottom left of a benzene ring. The ring is bonded from the upper right to a C double bonded above to O and to the right to O bonded to C H Subscript 2 Baseline C H Subscript 2 Baseline bonded to O bonded to C double bonded above to O and right to the upper left corner of a benzene ring. The ring is bonded from the lower left to C double bonded above to O and single bonded right to O bonded to CH Subscript 2 Baseline C H Subscript 2 Baseline bonded right to O bonded to C double bonded above to O and right to the lower left of a benzene ring, which is bonded from upper right to C double bonded above to O and single bonded right to O bonded to C H Subscript 2 Baseline C H Subscript 2 Baseline bonded to O.
Which type of reaction is represented by these diagrams?
elimination
substitution
addition polymerization
condensation polymerization
The type of reaction that is represented by these diagrams (attached) are given as: "condensation polymerization" (Option D)
What is condensation polymerization?Any type of polymer in polymer science that undergoes a condensation reaction during the polymerization process is referred to as a condensation polymer (i.e. a little molecules, methanol or water is derived as a metabolite).
The provided reaction is an example of condensation polymerization because it involves the combination of two monomers, which produces a big polymer and water as a byproduct.
Condensation polymers make significant contributions to the packaging, insulation, and textile sectors as far as real-world application of same is concerned.
Also examples of times that are condensation polymers and that have silicon rather than carbon as part of their molecular or structural make up are;
Silicone Oils and Rubbers.In conclusion, it is to be noted that The type of reaction that is represented by these diagrams (attached) are given as: "condensation polymerization"
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Answer:
condensation polymerization
Explanation:
Calculate the ph after 10. 0 ml of 0. 40 m naoh is added to 20. 0 ml of 0. 50 m hc.
The pH is 13.31.
The smaller the pH value the more acidic, the larger the pH value the more basic.
The range is 0 to 14, with 7 representing neutrality. Acidity is indicated by pH values below 7, whereas baseness is shown by pH values above 7.
Molarity (M) is the amount of a substance in a certain volume of solution. Molarity is defined as the moles of a solute per liters of a solution.
Molarity of NaOH = 0.40 M
Volume of NaOH = 10.0 mL
Molarity of HCL = 0.50 M
Volume of HCL = 20.0 ml
[tex][OH^-] = (M_2V_2 -M_1V_1)/(V_1+V_2)\\[/tex]
[tex]= (0.50*20.0 - 0.40*10.0) / (20+10)[/tex]
[tex]= (10 -4)/ 30\\[/tex]
[tex]= 6/30[/tex]
[tex]=0.2[/tex]
[tex]pOH = - log (OH ^-)[/tex]
[tex]pOH = - log (0.2)\\[/tex]
[tex]pOH = - ( - 0.698)[/tex]
[tex]pOH = 0.69[/tex]
[tex]pH + pOH = 14[/tex]
[tex]pH = 14 -0.69[/tex]
[tex]pH = 13.31[/tex]
Therefore, the pH after 10.0 mL of 0.40 M NaOH is 13.31.
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A container holds 265 mL of chlorine gas (Cl2). If the gas sample is at standard temperature and pressure (STP), what is its mass in grams?
The mass, in grams, of the chlorine gas will be 0.84 grams
Mass of gases at STPAt standard temperature and pressure, 22.4 L of gas has 1 mole of the gas in it.
This time, what we have is 265 mL of the gas at STP. 265 ml is equivalent to: 265/1000 = 0.265 L
If.
22.4 L = 1 mole
Then,
0.265 L = 0.265 x 1/22.4 = 0.01183 moles
This means that 0.01183 moles of chlorine is present in the gas at STP.
Mass of 0.01183 moles of chlorine = moles x molar mass.
The molar mass of chlorine gas is 71
Mass of the chlorine gas = 0.01183 x 71 = 0.84 grams
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What point defects are possible for al2o3 as an impurity in mgo? how many al3 ions must be added to form each of these defects?
Point defects are possible for al2o3 as an impurity in mgo. then 2 Al3+ must be added to form 1 Mg2+ must be added to form each of these defects
Point defects are zero-dimensional lattice defects, meaning they lack lattice structure in any dimension. Impurity atoms in a pure metal, vacancies, and self-interstitials are examples of common point defects.
Point defects are zero-dimensional lattice defects, which means they lack any kind of lattice structure. Point defects include, but are not limited to, atoms of impurity in a pure metal, vacancies, and self-interstitials.
The term "point defect" refers to the flaw or imperfection that results from the placement of a point-like constituent particle, such as an atom, an ion, or a molecule, in a crystalline structure.
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A current of 5. 68 a is passed through a Fe(NO3)2 solution. How long, in hours, would this current have to be applied to plate out 7. 20 g of iron?
There are 1.2 hr would this current have to be applied to plate out 7. 20 g of iron .
Calculation ,
Given ; Current ( I ) = 5. 68 A
In [tex]Fe(NO_{3} )_{2}[/tex] , the valancy of Fe is +2 .
2 moles of [tex]e^{-}[/tex] are required for the decomposition of 1 mole of Fe .
7. 20 g of Fe in moles = 7. 20 g /55.845 g/mol =0.12 mole
x moles of [tex]e^{-}[/tex] are required for the decomposition of 0.128 mole of Fe .
moles of [tex]e^{-}[/tex] are required = 0.256 moles
Charge on 1 mole of [tex]e^{-}[/tex] = 96500 C
Charge on 0.256 mole of [tex]e^{-}[/tex] = 24704 C
Current ( I )= Q/t
t =Q / I = 24704 C/5. 68 A = 4349 sec = 1.2 hr
Therefore , there are 1.2 hr would this current have to be applied to plate out 7. 20 g of iron .
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While working in a pharmaceutical laboratory, you need to prepare 1.50 L of a 2.20-M NaCl solution. What mass of NaCl would be required to prepare this solution
193 grams of NaCl are needed to make 1.50 L of a 2.20-M NaCl solution.
Calculation of the mass of NaCl required:
The only thing we need to know is that a solution's molarity informs us of the exact amount of moles of solute that are contained in 1 L of a solution.
In this situation, a NaCl (sodium chloride) solution with a 2.20-M concentration will have 2.20 moles of the solute (sodium chloride) in every 1 L of the solution.
The solution's molarity of 2.20 M requires that every 1 L of this solution include 2.20 moles of sodium chloride, which implies that every 1.50 L of this solution needs to contain
[tex]1.5 L solution .\frac{2.20 moles NaCl}{1L solution} = 3.30 moles NaCl[/tex]
Utilize the compound's molar mass to translate the amount of NaCl (sodium chloride) in moles to grams.
[tex]3.30 moles NaCl . \frac{58.4 gm}{1 mole NaCl} = 192.7 gm[/tex] ≈ 193 gm
Therefore it is concluded that the final answer is 193 gm.
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Draw the alkane formed when 4,5,5‑trimethyl‑1‑hexyne or 4,5,5‑trimethylhex‑1‑yne is treated with two equivalents of hbr.
Answer:
2,2 - dibromo - 4,5,5 - trimethylhexane
Explanation:
When 2 equivalents of HBr are added to 4,5,5‑trimethyl‑1‑hexyne, the alkane (at the top right of the image) is formed. The bromine branches are added to the second carbon because the intermediate carbocations are the most stable on the secondary carbon. Below I have drawn the mechanism.
Which best explains why the trend in noble gas boiling points increases down the group?.
The atoms become larger, the intermolecular forces between the atoms become stronger and more energy is needed to overcome these forces. Hence, the boiling point of noble gases increases down the group.
What are noble gases?
Group 18 of the modern periodic table consists of noble gases. Helium, neon, argon, krypton, xenon, and radon are noble gases. These gases are monoatomic and chemically inert under normal conditions and because of this they are also called inert gases. These gases are present in very small quantities in the atmosphere, and so they are also called rare gases.Trends in Physical Properties of Noble Gases:
Atomic Radii (Atomic size): As we move down the group from helium to radon the atomic size of Noble Gasses keeps on increasing. This is because when going down the group, the number of occupied shells with valence electrons increases.Boiling and Melting Points: As we move down the group, the boiling and melting point of the Noble Gasses increases due to the following reason:When moving down in the group, the atomic radius increases which results in the formation of a strong Van Der Waals force of attraction between the atoms.As to overcome the interatomic force of attraction which becomes stronger, it also requires more energy during melting and boiling.Density: All elements of group 18 have low densities. When going down the group, density increases as the atomic mass keeps on increasing.Ionization Energy: In the periodic table, As we move down the group the first ionization energy keeps on decreasing.Ionization Potential: When going down the group, the atomic radii increases which increases the attractive force and ultimately results in the increase of polarity and decrease in the ionization potential. This is because the larger atom of the group in the valence electrons are held together less tightly by the atom as they are situated far away from the nucleus.To learn more about trends in noble gases:
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Given the following reaction: 2O3---->3O2. DeltaH of reaction=-286kJ/mol
What is the Delta H reaction of the reaction 3O2----->2O3
The enthalpy of reaction (ΔH) of the above chemical reaction is equal to -143 kJ/mol.
What is a chemical reaction?A chemical reaction can be defined as a chemical process that involves the continuous transformation (rearrangement) of the ionic, atomic or molecular structure of a chemical element by breaking down and forming chemical bonds, in order to produce a new chemical compound while new bonds are formed.
What is a balanced equation?A balanced chemical equation can be defined as a chemical equation wherein the number of atoms on the reactant (left) side is equal to the number of atoms on the product (right) side.
Next, we would write the properly balanced chemical equation for this chemical reaction:
3/2O₂ -----> 2/2O₃
Therefore, the enthalpy of reaction (ΔH) of the above chemical reaction is half the enthalpy of reaction (ΔH) of the initial chemical reaction:
Enthalpy of reaction (ΔH) = ½ × -286 kJ/mol
Enthalpy of reaction (ΔH) = -143 kJ/mol.
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A piece of plastic has a net charge of 2. 00 μc. How many more protons than electrons does this piece of plastic have?
The piece of plastic has 1.256× 10¹³ protons more than that of electrons.
What are protons and electrons?A proton is a subatomic particle with a defined mass of 1 and a charge of +1 (positive charge). The symbol for the proton is either p or p+.
An electron is a subatomic particle that has a negative electric charge and that actively orbits the nucleus of the atom .
To calculate ,
given ,
magnitude of charge , q=+2μC=+2×10⁻⁶
Since, protons carry positive charge. Then, we know that standard number of proton in 1 Coulomb of charge is,
1C=6.28×10¹⁸ protons
Number of protons in 1 microcoulomb charge is,
=6.28×10¹⁸×10⁻⁶
=6.28×10¹²
Then for the net charge of q, the number of protons is given as,
np= q×6.28×10¹²
np= 2×6.28×10¹²
np=1.256×10¹³ protons
Thus, the piece of plastic has 1.256×10¹³ more protons than electrons.
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In general ethers have a much lower boiling point than their isomeric alcohols. why?
Ethers have a much lower boiling point than their isomeric alcohols because ether have not hydrogen bond that why ethers have lower boiling point .
Ethers will not have a hydrogen atom attached to the oxygen atom as alcohols do. Because ether molecules can not form intermolecular hydrogen bonds, their boiling points are much shorter than those of alcohol with such a comparable mass.
Owing to the combination of intermolecular H-bonding in alcohols though not in ethers, alcohols possess higher boiling temperatures than isomeric ethers.
Therefore, ethers have a much lower boiling point than their isomeric alcohols because ether have not hydrogen bond that why ethers have lower boiling point .
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A chemist reacted 0. 2 moles sodium benzoate with 0. 25 moles of hydrochloric acid. if she generated 22 g benzoic acid, what was her percent yield? (mw of benzoic acid = 122. 12 g mol-1)
A chemist reacted 0. 2 moles sodium benzoate with 0. 25 moles of hydrochloric acid. if she generated 22 g benzoic acid, the percent yield is 90%.
Reactant + HCl ------ Product
0.2 mol 0.25 22g
mol
Number of moles of reactant = 0.2mol
Molar mass of product = 122.12g
Number of moles of product = given mass/ molar mass
= 22g/ 122.12 g
= 0.180mol
Percent yield = (number of moles of product/ number of moles of reactant) × 100
= (0.180/0.2) × 100
= 90%
Thus we concluded that the percent yield of the given solution is 90%.
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What is the kinetic product obtained from the addition of 1 mole of bromine to 1,3-butadiene?
From the addition of 1 equivalent of HBr to 1,3-butadiene, 3-bromo-1-butene is obtained as a product.
What is moles?
Moles are defined as the smallest base unit that is equivalent to [tex]6.02214 X 10^{23}[/tex] units.
Moles are chemical entities, which are used to measure the mass or molecular mass of the compound, molecule, or atom. The moles are, for example can be explained as equal to the mass of the atom. For example, there is one mole in 12 grams of carboncarbon.
Thus the kinetic product formed is 3-bromo-1-butene.
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help me with this question please like right now
Using the gas laws, the volume of nitrogen produced is 6.6 L.
What is the product of a reaction?We know that the product of a reaction is obtained from the balanced reaction equation. The reactants combine together to yield the products hence we have; N2 + 3H2 ----> 2NH3
We know that 1 mole of hydrogen occupies 22.4 L
x moles of hydrogen occupies 25500 L
x = 1 mole * 25500 L/ 22.4 L
x = 1138.4 moles
If 3 mole of hydrogen reacts with 1 mole of nitrogen
1138.4 moles reacts with 1138.4 moles * 1 mole/ 3 moles = 379.5 moles
Mass of nitrogen = 379.5 moles * 28 g/mol = 10626 g
Now 3 moles of hydrogen produces 44.8 L of NH3
1138.4 moles produces 1138.4 moles * 44.8 L/ 3 moles = 17000 L
Now;
P1V1/T1 = P2V2/T2
P1V1T2 = P2V2T1
V2 = P1V1T2/P2T1
V2 = 1 * 17000 * 848/5.5 * 273
V2 =9601 L
2) 2NH4NNO3(s) ---> 4H2O(g) + 2N2(g) + O2(g)
Number of moles = 12g/ 80 g/mol = 0.15 moles
Now;
2 moles of NH4NNO3 produced 2 moles of nitrogen hence 0.15 moles of N2 was produced.
1 mole of N2 occupies 22.4 L
0.15 moles of N2 occupies 0.15 moles * 22.4 L/1 mole = 3.36 L
P1 = 760 torr
V1 = 3.36 L
T1 = 273 K
P2 = 745 torr
T2 = 527 degrees or 800 K
V2 = ?
V2 = P1V1T2/P2T1
V2 = 760 * 3.36 * 527/745 * 273
V2 = 6.6 L
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What is the theoretical yield (mass of Al2O3 formed) for this reaction?
(answer is not 204)
Answer:
102 g Al₂O₃
Explanation:
The theoretical yield is the mass of product calculated via the molar masses and balanced chemical equation.
The limiting reagent is the reactant that is completely reacted before the other reactant(s) are used up. Since Al produces the smaller amount of product, it appears that Al is the limiting reactant. You can only make as much product as the limiting reactant allows. As such, the theoretical yield is 102 grams Al₂O₃.
If the sodium oxide (Na2O) in soda-lime glass was replaced by potassium oxide (K2O), this would result in . . .
How many milliliters of a 9. 0 mh2so4 solution are needed to make 0. 45 l of a 3. 5 m solution?
There are 1157.14 milliliters of a 9. 0 M [tex]H_{2} SO_4}[/tex] solution are needed to make 0. 45 L of a 3. 5 M solution.
Calculation,
The formula for the dilution of a solution is given as:
[tex]M_{1} V_{1}[/tex] = [tex]M_{2} V_{2}[/tex]
Where M is molarity and V is the volume of the solution in liters ( L ).
Given data,
[tex]M_{1}[/tex] = 9. 0 M
[tex]V_{1}[/tex] = 0. 45 L
[tex]M_{2}[/tex] = 3. 5 M
[tex]V_{2}[/tex] = ?
9. 0 M ×0. 45 = 3. 5 M × [tex]V_{2}[/tex]
[tex]V_{2}[/tex] in Lit = 9. 0 M ×0. 45/ 3. 5 M = 1.15 L
[tex]V_{2}[/tex] in the ml = 1.15 × 1000ml = 1157.14 ml
Volume in milliliters is 1157.14 .
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