Answer:
4.46 mol of NH3
Explanation:
The equation of he reaction is given as;
2N + 3H2 --> 2NH3
From the stochiometry of the reaction, 1 mol of Nitrogen produces 2 mol of Ammonia.
Mass of Nitrogen = 31.2g
Molar mass of Nitrogen = 14g/mol
Number of moles = Mass / Molar mass = 31.2 / 14 = 2.23 mol
Since 1 mol of N = 2 mol of NH3;
2.23 mol of N2 would produce x
x = 2.23 * 2 = 4.46 mol of NH3
In the pictured cell, the side containing zinc is the Choose... and the side containing copper is the Choose... . The purpose of the N a 2 S O 4 NaX2SOX4 is to
Answer:
Zinc- anode
Copper- cathode
Sodium sulphate- salt bridge
Explanation:
A galvanic cell is an electrochemical cell in which electrical energy is produced by a spontaneous chemical reaction.
In the pictured galvanic cell, zinc is the anode since it looses electrons according to the reaction; Zn(s) -----> Zn^2+(aq) + 2e
Copper is the cathode as shown here; Cu^2+(aq) + 2e ----> Cu(s)
Sodium sulphate functions as the salt bridge. It keeps the both solutions neutral by ensuring charge balance in the both half cells.
Answer:
zinc=anode
copper=cathode
Explanation:
How are pH and pOH ?
A. pH = 14 + pOH
B. pOH = 14 - pH
C. pOH = 14 + pH
D. pH = 14 - pOH
Answer:
B. pOH = 14 - pH and D. pH = 14 - pOH.
Explanation:
Hello,
In this case, we must remember that pH and pOH are referred to a measure of acidity and basicity respectively, since pH accounts for the concentration of H⁺ and pOH for the concentration of OH⁻ in a solution. In such a way, since the maximum scale is 14, we say that the addition between the pH and pOH must be 14:
[tex]pH+pOH=14[/tex]
Therefore, the correct answers are B. pOH = 14 - pH and D. pH = 14 - pOH since the both of them are derived from the previous definition.
Best regards.
Answer:
D: by subtracting the pOH from 14.
Explanation:
The decomposition of nitramide in aqueous solution at 25 °C NH2NO2(aq)N2O(g) + H2O(l) is first order in NH2NO2 with a rate constant of 4.70×10-5 s-1. If an experiment is performed in which the initial concentration of NH2NO2 is 0.384 M, what is the concentration of NH2NO2 after 31642.0 s have passed? M
Answer:
[tex][NH_2NO_2]=0.0868M[/tex]
Explanation:
Hello,
In this case, for the given chemical reaction, the first-order rate law is:
[tex]r=\frac{d[NH_2NO_2]}{dt} =-k[NH_2NO_2][/tex]
Which integrated is:
[tex][NH_2NO_2]=[NH_2NO_2]_0exp(-kt)[/tex]
Thus, the concentration after 31642.0 s for a 0.384-M solution is:
[tex][NH_2NO_2]=0.384M*exp(-4.70x10^{-5}s^{-1}*31642.0s)\\[/tex]
[tex][NH_2NO_2]=0.0868M[/tex]
Best regards.
Answer:
[A] = 0.0868 M
Explanation:
Rate constant = 4.70×10-5 s-1
First order reaction
Initial concentration, [A]o = 0.384 M
Final concentration, [A] = ?
Time, t = 31642.0 s
All these variables are related by the following equation;
[A] = [A]o e^(-kt)
[A] = 0.384 e^(-4.70×10-5 x 31642.0)
[A] = 0.384 e^(-1.4872)
[A] = 0.384 * 0.2260
[A] = 0.0868 M
What's the name for the part of Earth made of rock?
A. Geosphere
B. Atmosphere
C. Hydrosphere
D. Biosphere
SUBMIT
Answer:I think it's Geosphere
Explanation:
Answer:
A
Explanation:
Geo means rock, or earth. Hydro means water, Atmosphere is space, and Bio global ecosystem composed of living organisms
Comparing the 2-bromobutane + methoxide and 2-bromobutane + t-butoxide reactions, choose the statements that BEST describe the data and mechanism. a. the mechanism for this reaction is E2 b. an increase in 1-butene was observed when t-butoxide was used c. an increase in 1-butene was observed when methoxide was used d. the mechanism for this reaction is E1 e. no significant difference was observed
Answer:
an increase in 1-butene was observed when t-butoxide was used
Explanation:
When a base reacts with an alkyl halide, an elimination product is formed. This reaction is an E2 reaction.
Here we are to compare the reaction of two different bases with one substrate; 2-bromobutane. Both reactions occur by the E2 mechanism but follow different transition states due to the size of the base.
The Saytzeff product, 2-butene, is obtained when the methoxide is used while the non Saytzeff product, 1-butene, is obtained when t-butoxide is used.
The Saytzeff rule is reliable in predicting the major products of simple elimination reactions of alkyl halides given the fact that a small/strong bases is used for the elimination reaction. Therefore hydroxide, methoxide and ethoxide bases give similar results for the same alkyl halide substrate. Bulky bases such as tert-butoxide tend to yield a higher percentage of the non Saytzeff product and this is usually attributed to steric hindrance.
11mg of cyanide per kilogram of body weight is lethal for 50% of domestic chickens. If a chicken weighs 3kg, how many grams of cyanide would it need to ingest to kill 50% of domestic chickens?
Answer:
[tex]0.033g[/tex]
Explanation:
Hello,
In this case, since 11 mg per kilogram of body weight has the given lethality, the mg that turn out lethal for a chicken weighting 3 kg is computed by using a rule of three:
[tex]11mg\longrightarrow 1kg\\\\?\ \ \ \ \ \ \longrightarrow 3kg[/tex]
Thus, we obtain:
[tex]?=\frac{3kg*11mg}{1kg}\\ \\?=33mg[/tex]
That in grams is:
[tex]=33mg*\frac{1g}{1000mg} \\\\=0.033g[/tex]
Regards.
A 100.0 mL sample of 0.10 M NH3 is titrated with 0.10 M HNO3. Determine the pH of the solution before the addition of any HNO3. The Kb of NH3 is 1.8 × 10-5.
Answer:
[tex]pH=11.12[/tex]
Explanation:
Hello,
In this case, ammonia dissociation is:
[tex]NH_3(aq)+H_2O(l)\rightleftharpoons NH_4^+(aq)+OH^-(aq)[/tex]
So the equilibrium expression:
[tex]Kb=\frac{[NH_4^+][OH^-]}{[NH_3]}[/tex]
That in terms of the reaction extent and the initial concentration of ammonia is written as:
[tex]1.8x10^{-5}=\frac{x*x}{0.10M-x}[/tex]
Thus, solving by using solver or quadratic equation we find:
[tex]x=0.00133M[/tex]
Which actually equals the concentration of hydroxyl ion, therefore the pOH is computed:
[tex]pOH=-log([OH^-])=-log(0.00133)=2.88[/tex]
And the pH from the pOH is:
[tex]pH=14-pOH=14-2.88\\\\pH=11.12[/tex]
Best regards.
f a substance has a half-life of 8.10 hr, how many hours will it take for 75.0 g of the substance to be depleted to 3.90 g?
Answer:
35 hrs
Explanation:
half life of the substance [tex]t_{1/2 }[/tex] = 8.1 hr
initial amount [tex]N_{0}[/tex] = 75 g
The final amount [tex]N[/tex] = 3.9 g
The time elapsed [tex]t[/tex] = ?
we use the relationship
[tex]N[/tex] = [tex]N_{0}[/tex] [tex](\frac{1}{2} )^{\frac{t}{t_{1/2} } }[/tex]
substituting values, we have
3.9 = 75 x [tex]\frac{1}{2}^{\frac{t}{8.1} }[/tex]
0.052 = [tex]\frac{1}{2}^{\frac{t}{8.1} }[/tex]
take the log of both side
log 0.052 = log [tex]\frac{1}{2}^{\frac{t}{8.1} }[/tex]
log 0.052 = [tex]\frac{t}{8.1}[/tex] log 1/2
-1.284 = [tex]\frac{t}{8.1}[/tex] x -0.301
1.284 = 0.301t/8.1 =
1.284 = 0.0372t
t = 1.284/0.037 = 34.5 ≅ 35 hrs
What chemical bonds hold atoms?
place the following substances in Order of decreasing boiling point H20 N2 CO
Answer:
-195.8º < -191.5º < 100º
Explanation:
Water, or H20, starts boiling at 100ºC.
Nitrogen, or N2, starts boiling at -195.8ºC.
Carbon monoxide, or C0, starts boiling at -191.5ºC.
When we place these in order from decreasing boiling point:
-195.8º goes first, then -191.5º, and 100º goes last.
Answer:
therefore, N2, CO, H20
Decreasing boiling point
Explanation:
the bond existing in H2O is hydrogen bond
bond existing in N2 is covalent bond, force existing is dipole-dipole-interaction
bond existing in CO is covalent bond , force existing between is induced -dipole- induced dipole-interaction
hydrogen bond is the strongest , followed by dipole-dipole-interaction and induced -dipole- induced dipole-interaction
the stronger the bond , the higher the boiling point
therefore, N2, CO, H20
-------------------------------------->
Decreasing boiling point
One gram is approximately the same as half the mass of a new U.S.
A) penny.
B) dime.
C) quarter.
D) dollar.
Answer:
b) dime
Explanation:
a dime is approximately 2.2g
half of this is 1.1g, which can be rounded down to one gram.
hope this helps
Beginning with Na, record the number of energy levels, number of protons, and atomic radius for each element in period 3.
Answer:
Sodium, magnesium, aluminium, silicon, phosphorus, sulfur, chlorine, and argon are the elements of third period.
Explanation:
There are three energy levels in sodium atom. It has 11 electrons revolving around the nucleus. the atomic radius of sodium atom is 227 ppm. Magnesium, aluminium, silicon, phosphorus, sulfur, chlorine, and argon has also three energy levels like sodium because all these elements belongs to third period. There are 12 electrons present in magnesium, 13 in aluminium, 14 in silicon, 15 in phosphorus, 16 in sulfur, 17 in chlorine, and 18 electrons in argon. The atomic radius of magnesium atom is 173 ppm. The atomic radius of aluminium atom is 143 ppm. The atomic radius of silicon atom is 111 ppm. The atomic radius of phosphorus atom is 98 ppm. The atomic radius of sulfur atom is 87 ppm. The atomic radius of chlorine atom is 79 ppm and the atomic radius of argon atom is 71 ppm.
To find the pH of a solution of NaNO2, one would have to construct an ICE chart using:
a. the Kb of NO−2 to find the hydroxide concentration.
b. the Kb of HNO2 to find the hydronium concentration.
c. the Kb of NO-2, to find the hydronium concentration.
d. the Kb of HNO2, to find the hydroxide concentration.
Answer:
a. the Kb of NO₂⁻ to find the hydroxide concentration.
Explanation:
When sodium nitrite is dissolved in water, it dissociates in sodium cation and nitrite anion according to the following equation.
NaNO₂(s) ⇒ Na⁺(aq) + NO₂⁻(aq)
Na⁺ comes from NaOH (strong base) so it doesn't react with water.
NO₂⁻ comes from HNO₂ (weak acid) so it reacts with water according to the following equation.
NO₂⁻(aq) + H₂O(l) ⇄ HNO₂(aq) + OH⁻(aq)
This is the basic reaction of nitrite ion, so we need the Kb of NO₂⁻ to find the hydroxide concentration.
Write a net ionic equation for the reaction that occurs when aqueous solutions of hydrofluoric acid and sodium hydroxide are combined. (Use the lowest possible coefficients. Use the pull-down boxes to specify states such as (aq) or (s). If a box is not needed, leave it blank. If needed, use H for the hydronium ion.)
Answer:
The net ionic reaction is : H⁺ (aq) + OH⁻ (aq) ---> H₂O (l)
Explanation:
The reaction between aqueous solutions of hydrofluoric acid and sodium hydroxide is an example of a neutralization reaction.
A neutralization reaction is a reaction between and acid and an abase to produce salt and water only.
Hydrofluoric acid is the acid while sodium hydroxide is the base. during the reaction the hydrofluoric acid will produce hydrogen and fluoride ions, while sodium hydroxide will produce hydroxide and sodium ions. The hydroxide and hydrogen ions will combine to produce water while the sodium and fluoride ions remain in solution as ions.
The equation of the reaction is as follows:
H⁺F⁻ (aq) + Na⁺OH⁻ (aq) ----> Na⁺F⁻ (aq) + H₂O (l)
Since the sodium and fluoride ions appear on both sides of the equation, they are known as spectator ions and are cancelled out to give the net ionic equation.
The net ionic reaction is : H⁺ (aq) + OH⁻ (aq) ---> H₂O (l)
Consider the acid H3PO4. This acid will react with water by the following equation. H3PO4+H2O↽−−⇀H2PO−4+H3O+ What will be true of the resulting conjugate base H2PO−4? Select the correct answer below: H2PO−4 can act as an acid.
Answer:
H+/PO-4^-2
Explanation:
hydrogen has dissolved completely
In the given reaction conjugate base is H₂PO₄⁻, it also behave as a weak acid.
What is acid - conjugate base pair?
An acid and conjugate base pairs are those pairs in which they are differentiated by the one atom of hydrogen atom.
Given chemical reaction is:
H₃PO₄ + H₂O → H₂PO₄⁻ + H₃O⁺
In the above reaction H₃PO₄ is an acid as it gives H⁺ ion to the solution and formed H₂PO₄⁻, which is a conjugate base of H₃PO₄ acid. H₂PO₄⁻ will also behave as an acid because it have H⁺ ion to gives in the solution but nature of this acid is weak as they not readily dissociates.
Hence, H₂PO₄⁻ is a conjugate base.
To know more about acid-base pair, visit the below link:
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The front curve of a spectacle lens is called?
Answer:
Corrective lense or just lens.
Explanation:
21. What are the two main ways of working with clay?
Answer:
Diferentes tipos de arcilla
ARCILLA DE LADRILLOS. Contiene muchas impurezas. ...
ARCILLA DE ALFARERO. Llamada también barro rojo y utilizada en alfarería y para modelar. ...
ARCILLA DE GRES. Es una arcilla con gran contenido de feldespato. ...
ARCILLAS “BALL CLAY” O DE BOLA. ...
CAOLIN. ...
ARCILLA REFRACTARIA. ...
BENTONITA.
Explanation:
Answer:
Coil method and the slab method.
Explanation:
What happens to the rate of dissolution as the temperature is increased in a gas solution?
A.
The rate stays the same.
B.
The rate decreases.
C.
The rate increases.
D.
There is no way to tell.
Answer:
The rate decreases
Explanation:
When we dissolve a gas in a water, the process is exothermic. This implies that heat is evolved upon dissolution of a gas in water.
Recall from Le Chateliers principle that for exothermic reactions, an increase in temperature favours the reverse reaction. The implication of these is that when the temperature of the gas is increased, less gas will dissolve in water.
Hence increase in temperature decreases the rate of solubility of a gas in water.
Answer:
B.
The rate decreases.
Explanation:
Divers often inflate heavy duty balloons attached to salvage items on the sea floor. If a balloon is filled to a volume of 1.20 L at a pressure of 6.25 atm, what is the volume of the balloon when it reaches the surface?
Answer:
7.50 L
Explanation:
The balloon has a volume of 1.20 L (V₁) when the pressure at the sea floor is 6.25 atm (P₁). When it reaches the surface, the pressure is that of the atmosphere, that is, 1.00 atm (P₂). If we consider the gas to behave as an ideal gas and the temperature to be constant, we can calculate the final volume (V₂) using Boyle's law.
P₁ × V₁ = P₂ × V₂
V₂ = P₁ × V₁ / P₂
V₂ = 6.25 atm × 1.20 L / 1.00 atm
V₂ = 7.50 L
Explain your reasoning. Match each explanation to the appropriate blanks in the sentences on the right.
1. the atomic radius decreases
2. the number of gas molecules decreases
3. molar mass and structure complexity decreases
4. structure complexity decreases
5. molar mass decreases
6. each phase (gas, liquid, solid) becomes more ordered
A (I_2(g), Br_2 (g), Cl_2 (g), F_2 (B): The ranking can best be explained by the trend entropy decreases as______.
B (H_2O_2 (g), H_2S(g), H_2O(g): The ranking can best be explained by the decreases a trend entropy decreases as_______.
C. (C(s, amorphous), C(s, graphite), C(s, diamond): The ranking can best be explained by the trend entropy decreases as_______.
Answer:
A (I_2(g), Br_2 (g), Cl_2 (g), F_2 (B): The ranking can best be explained by the trend entropy decreases as 5. molar mass decreases.
B (H_2O_2 (g), H_2S(g), H_2O(g): The ranking can best be explained by the decreases a trend entropy decreases as 3. molar mass and structure complexity decreases.
C. (C(s, amorphous), C(s, graphite), C(s, diamond): The ranking can best be explained by the trend entropy decreases as 4. structure complexity decreases.
Explanation:
Hello.
In this case, we can understand a higher entropy when more disorder is present and a lower entropy when less disorder is present, thus:
A (I_2(g), Br_2 (g), Cl_2 (g), F_2 (B): The ranking can best be explained by the trend entropy decreases as 5. molar mass decreases since iodine has the greatest molar mass (254 g/mol) and fluorine the least molar mass (38 g/mol).
B (H_2O_2 (g), H_2S(g), H_2O(g): The ranking can best be explained by the decreases a trend entropy decreases as 3. molar mass and structure complexity decreases since hydrogen peroxide weights 34 g/mol as well as hydrogen sulfide but the peroxide has more bonds (more complex, higher entropy).
C. (C(s, amorphous), C(s, graphite), C(s, diamond): The ranking can best be explained by the trend entropy decreases as 4. structure complexity decreases since diamond has a well-ordered structure and amorphous carbon has a very disordered one.
Best regards.
Ammonia, methane, and phosphorus trihydride are three different compounds with three different boiling points. Rank their boiling points in order from lowest to highest.
A. CH4< NH3 < PH3
B. NH3 < PH3< CH4
C. CH4 < PH3 < NH3
D. NH3 < CH4< PH3
E. PH3< NH3 < CH4
Answer:
B. NH3 < PH3< CH4
Explanation:
Hello,
In this case, taking into account that the boiling point of ammonia, methane and phosphorous trihydrate are -33.34 °C , -161.5 °C and -87.7 °C , clearly, methane has the lowest boiling point (most negative) and ammonia the greatest boiling point (least negative), therefore, ranking is:
B. NH3 < PH3< CH4
Best regards.
The reaction of butadiene gas (C4H6) with itself produces C8H12 gas as follows: The reaction is second order with a rate constant equal to 5.76 × 10−2 L/mol/min under certain conditions. If the initial concentration of butadiene is 0.200 M, what is the concentration in molarity remaining after 10.0 min? Report your answer to 3 decimal places.
Answer:
[tex]C_{C_4H_6}=0.179M[/tex]
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
[tex]2C_4H_6\rightarrow C_8H_{12}[/tex]
And the rate law is:
[tex]\frac{dC_{C_4H_6}}{dt}=kC_{C_4H_6}^2[/tex]
Which integrated is:
[tex]\frac{1}{C_{C_4H_6}} =\frac{1}{C_{C_4H_6}^0}+kt[/tex]
In such a way, the concentration after 10.0 min is:
[tex]\frac{1}{C_{C_4H_6}} =\frac{1}{0.200M}}+5.76x10^{-2}\frac{L}{mol*min}*10.0min\\ \\\frac{1}{C_{C_4H_6}}=5.58\frac{L}{mol} \\\\C_{C_4H_6}=\frac{1}{5.58\frac{L}{mol} } \\\\C_{C_4H_6}=0.179M[/tex]
Regards.
When 91.96g of Na reacts with 32.o g of O2 how many grams of NaO2 are produced
Answer:
123.96 g Na₂O
Explanation:
4 Na + O₂ ⇒ 2 Na₂O
You first need to find the limiting reagent. Convert the reactants to moles and see which produces the least amount of product using the mole ratios in the chemical equation.
(91.96 g Na)/(22.99 g/mol Na) = 4 mol Na
(4 mol Na) × (2 mol Na₂O/4 mol Na) = 2 mol Na₂O
(32.0 g O₂)/(32.0 g/mol) = 1 mol O₂
(1 mol O₂) × (2 mol Na₂O/1 mol O₂) = 2 mol Na₂O
Since they both produce the same amount of product, you don't need to pick a limiting reagent. Now, convert moles of Na₂O to grams.
(2 mol Na₂O) × (61.98 g/mol Na₂O) = 123.96 g Na₂O
The ΔHvap of nitrous oxide is 16.53 kJ · mol−1 and its ΔSvap is 89.51 J · mol−1 · K−1. What it the boiling point of nitrous oxide?
Answer:
[tex]T_b=-88.48\°C[/tex]
Explanation:
Hello,
In this case, since the entropy of vaporization is defined in terms of the enthalpy of vaporization and the boiling point of the given substance, nitrous oxide, as shown below:
[tex]\Delta _{vap}S=\frac{\Delta _{vap}}{T_b}[/tex]
Solving for the boiling point of nitrous oxide, we obtain:
[tex]T_b=\frac{\Delta _{vap}H}{\Delta _{vap}S}=\frac{16.53\frac{kJ}{mol}*\frac{1000J}{1kJ} }{89.51\frac{J}{mol} } \\ \\T_b=184.67K[/tex]
Which in degree Celsius is also:
[tex]Tb=184.67-273.15\\\\T_b=-88.48\°C[/tex]
Best regards.
a. Name a chemical or product that was once considered safe but is now considered
harmful. (1 point)
-
Answer:
Bisphenol A (BPA)
Explanation:
Bisphenol A (BPA) is a chemical additive commonly found in resins and plastics, such as water bottles or food containers. It can also be found in household electronics, medical devices, dental fillings and sales receipts, just to name a few other applications.
An ice cube at 0.00C with a mass of 8.32g is placed Into 55g of water, initially at 25C. If no heat is lost to the surroundings, what is the final temperature of the entire water sample after all the ice is melted (answer must be in 3 sig figs)
Answer:
The final temperature of the entire water sample after all the ice is melted, is 12,9°C. We should realize that if there is no loss of heat in our system, the sum of lost or gained heat is 0. It is logical to say that the temperature has decreased because the ice gave the water "heat" and cooled it
Thats all i know
A solution containing a unknown ionic compound, vigorously bubbles when hydrochloric acid (HCl) is added to the solution. This might indicate that the solution contains which anion?
Answer:
CO3^2-
Explanation:
In qualitative analysis, we try to use chemical reactions to determine the composition of an unknown substance. The addition of certain reagents to the unknown solution gives certain results that show the presence or absence of certain species from the unknown sample.
When dilute HCl is added to an unknown sample and effervescence is observed, then the unknown sample must contain CO3^2- or HCO3^-. The presence of these species is confirmed if the gas evolved is passed through limewater and the gas turns limewater milky.
Calculate the energy required to heat 566.0mg of graphite from 5.2°C to 23.2°C. Assume the specific heat capacity of graphite under these conditions is ·0.710J·g−1K−1 . Be sure your answer has the correct number of significant digits.
Answer:
7.23 J
Explanation:
Step 1: Given data
Mass of graphite (m): 566.0 mgInitial temperature: 5.2 °CFinal temperature: 23.2 °CSpecific heat capacity of graphite (c): 0.710J·g⁻¹K⁻¹Step 2: Calculate the energy required (Q)
We will use the following expression.
Q = c × m × ΔT
Q = 0.710J·g⁻¹K⁻¹ × 0.5660 g × (23.2°C-5.2°C)
Q = 7.23 J
A student sets up the following equation to convert a measurement. The (?) Stands for a number the student is going to calculate. Fill in the missing part of this equation. (0.030 cm^3) x ? =m^3
Answer:
\text{0.30 cm}^{3} \times \left (\dfrac{10^{-2}\text{ m}}{\text{1 cm}}\right )^{3} = 3.0 \times 10^{-7} \text{ m}^{3}
Explanation:
0.030 cm³ × ? = x m³
You want to convert cubic centimetres to cubic metres, so you multiply the cubic centimetres by a conversion factor.
For example, you know that centi means "× 10⁻²", so
1 cm = 10⁻² m
If we divide each side by 1 cm, we get 1 = (10⁻² m/1 cm).
If we divide each side by 10⁻² m, we get (1 cm/10⁻² m) = 1.
So, we can use either (10⁻² m/1 cm) or (1 cm/10⁻² m) as a conversion factor, because each fraction equals one.
We choose the former because it has the desired units on top.
The "cm" is cubed, so we must cube the conversion factor.
The calculation becomes
[tex]\text{0.30 cm}^{3} \times \left (\dfrac{10^{-2}\text{ m}}{\text{1 cm}}\right )^{3} = 0.30 \times 10^{-6}\text{ m}^{3} = \mathbf{3.0 \times 10^{-7}} \textbf{ m}^{\mathbf{3}}\\\\\textbf{0.30 cm}^{\mathbf{3}} \times \left (\dfrac{\mathbf{10^{-2}}\textbf{ m}}{\textbf{1 cm}}\right )^{\mathbf{3}} = \mathbf{3.0 \times 10^{-7}} \textbf{ m}^{\mathbf{3}}[/tex]
The intermolecular forces present in CH 3NH 2 include which of the following? I. dipole-dipole II. ion-dipole III. dispersion IV. hydrogen bonding
Answer:
I. dipole-dipole
III. dispersion
IV. hydrogen bonding
Explanation:
Intermolecular forces are weak attraction force joining nonpolar and polar molecules together.
London Dispersion Forces are weak attraction force joining non-polar and polar molecules together. e.g O₂, H₂,N₂,Cl₂ and noble gases. The attractions here can be attributed to the fact that a non -polar molecule sometimes becomes polar because the constant motion of its electrons may lead to an uneven charge distribution at an instant.
Dispersion forces are the weakest of all electrical forces that act between atoms and molecules. The force is responsible for liquefaction or solidification of non-polar substances such as noble gas an halogen at low temperatures.
Dipole-Dipole Attractions are forces of attraction existing between polar molecules ( unsymmetrical molecules) i.e molecules that have permanent dipoles such as HCl, CH3NH2 . Such molecules line up such that the positive pole of one molecule attracts the negative pole of another.
Dipole - Dipole attractions are more stronger than the London dispersion forces but weaker than the attraction between full charges carried by ions in ionic crystal lattice.
Hydrogen Bonding is a dipole-dipole intermolecular attraction which occurs when hydrogen is covalently bonded to highly electronegative elements such as nitrogen, oxygen or fluorine. The highly electronegative elements have very strong affinity for electrons. Hence, they attracts the shared pair of electrons in the covalent bonds towards themselves, leaving a partial positive charge on the hydrogen atom and a partial negative charge on the electronegative atom ( nitrogen in the case of CH3NH2 ) . This attractive force is know as hydrogen bonding.
Answer:
The intermolecular forces present in CH_3NH_2 includes
II. (ion-dipole) and IV. (hydrogen bonding)Explanation:
The intermolecular forces present in CH_3NH_2 includes II. (ion-dipole) and IV. (hydrogen bonding)
It is a polar molecule due to NH polar bond and it can form Hydrogen bond also due to NH bond.
Interaction will be dipole- dipole and Hydrogen dispersion forces can always be taken into account.
For more information on intermolecular forces, visit
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