Answer:
B.bonds are broken and new bonds are formed
A substance is made up of slow-moving particles that have very little space between them. Based on this information, what can most likely be concluded about this substance? O It is not a gas because its particles do not move continuously. It is a gas because its particles move continuously in a straight line. 0 It is not a gas because its particles do not have large spaces between them. It is a gas because its particles move in many different directions.
Answer:
o
Explanation:
it is not a gas because the particles do not move freely it may be a liquid or a solid partly and mostly liquidized.
Assume you have 4 solids (A, B, C and D) of similar mass. Which of these requires the greatest energy input to melt?
polar covalent
covalent network
ionic compound
nonpolar covalent
The solid that require the greatest energy input to melt by mass is the option;
Covalent network
Reason for the above answer is as follows;
The elementary particles of a solid are held together by bonds that require
an input of energy to unlock, and once broken, the particles are then able
to change location within their containing vessels with less restrictions
Types of bonds
Polar covalent molecular solids have the following characteristics;a) Soluble in water b) Low melting point, b) Conduct electricity
Solids that are made up of a covalent network have the following characteristicsa) High melting point temperature b) Non conductive of electricity c) Not soluble in water
Solids of ionic compounds have the following characteristics;a) High melting point temperature b) The liquid state and solution
conducts electricity c) Soluble in water
Solids that have nonpolar covalent bonds have;a) Low melting point b) Normally in the gaseous or liquid state b) Not water soluble
Therefore, the covalent network, and the solids ionic compounds require the most energy to melt, however, the strength of the ionic bond in an ionic compound is a factor the charges present and the sizes of the atom, while
the covalent network solid, are combined to form essentially as a single
molecule and therefore require the greatest heat energy input break the bonds of the molecule down in order to melt
Learn more about the properties of the different types of bonds here;
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You need to make an aqueous solution of 0.182 M aluminum sulfate for an experiment in lab, using a 250 mL volumetric flask. How much solid aluminum sulfate should you add
Answer:
Hence, 15.99 g of solid Aluminum Sulfate should be added in 250 mL of Volumetric flask.
Explanation:
To make 0.187 M of Aluminum Sulfate solution in a 250 mL (0.250 L) Volumetric flask
The molar mass of Aluminum Sulfate = 342.15 g/mol
Using the molarity formula:-
Molarity = Number of moles/Volume of solution in a liter
Number of moles = Given weight/ molar mass
Molarity = (Given weight/ molar mass)/Volume of solution in liter
0.187 M = (Given weight/342.15 g/mol)/0.250 L
Given weight = 15.99 g
The fact that a beam of particles was deflected in the presence of an electric
or magnetic force led J.J. Thomson to conclude that the particles had a(n)
O A. large mass
B. electric charge
O C. negligible mass
O D. neutral charge
Answer:
electric charge
Explanation:
Charged particles are deflected in an electric or a magnetic field. The particles discovered by J.J. Thomson were charged particles.
When these charged particles are passed through electric and magnetic fields, deflection occurs depending on the nature of the charge.
A positive charge is deflected towards the negative part of an electric field or the south pole of a magnetic field.
A negative charge is selected towards the positive end of an electric field or the north pole of a magnetic field.
6) Hydrogen gas can be generated from the reaction between aluminum metal and hydrochloric acid:
2 Al(s) + 6 HCl(aq) + 2 AICI3, (aq) + 3 H2(g)
a. Suppose that 3.00 grams of Al are mixed with excess acid. If the hydrogen gas produced is directly collected
into a 850 mL glass flask at 24.0 °C, what is the pressure inside the flask (in atm)?
b. This hydrogen gas is then completely transferred from the flask to a balloon. To what volume (in L) will the
balloon inflate under STP conditions?
c. Suppose the balloon is released and rises up to an altitude where the temperature is 11.2 °C and the pressure is
438 mm Hg. What is the new volume of the balloon (in L)?
Stoichiometry refers to the relationship between the moles of reactants and products.
This question must be solved using both stoichiometry and the gas laws
The reaction equation is;
2 Al(s) + 6 HCl(aq) --------> 2 AICI3, (aq) + 3 H2(g)
Using stoichiometryNumber of moles of Al = 3g/27g/mol = 0.11 moles
According to the reaction equation;
2 moles of Al yields 3 moles of H2
0.11 moles of Al yields 0.11 * 3/2 = 0.165 moles
Using the gas lawsFrom the ideal gas equation;
PV=nRT
P = ?
n= 0.165 moles
V = 0.85 L
T = 297 K
R = 0.082 atmLK-1mol-1
P= nRT/V
P = 0.165 * 0.082 * 297/0.85
P= 4.73 atm
Under STP conditions;P1 = 4.73 atm
T1 = 297 K
V1 = 0.85 L
P2 = 1 atm
T2 =273 K
V2 =?
From the general gas equation;P1V1/T1 = P2V2/T2
P1V1T2 = P2V2T1
V2 = P1V1T2/P2T1
V2 = 4.73 * 0.85 * 273/1 * 297
V2 = 3.69 L
P1 = 760 mmHg
T1 = 273 K
V1 = 3.69
P2 = 438 mm Hg
T2 = 284.2 K
V2 =?
P1V1/T1 = P2V2/T2
P1V1T2 = P2V2T1
V2 = P1V1T2/P2T1
V2 = 760 * 3.69 * 284.2/438 *273
V2 = 797010.48/119574
V2= 6.67 L
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how is the molecule of substance formed
Answer:
When atoms approach one another closely, the electron clouds interact with each other and with the nuclei. If this interaction is such that the total energy of the system is lowered, then the atoms bond together to form a molecule.
Explanation:
Which of the following is a reduction half-reaction?
Solution : An oxidation reduction (redox) reaction is a type of chemical reaction that involves a transfer of electrons between tow species an oxidaion reductin reaction is any chemical reaction in which the oxidation number of a molecule atom or ion changes by gaining or losing an electron
The reversible reaction 2H2 CO <------> CH3OH heat is carried out by mixing carbon monoxide and hydrogen gases is a closed vessel under high pressure with a suitable catalyst . After equilibrium is established at high temperature and pressure, all three substances are present. If the pressure on the system is lower, with the temperature kept constant, what will be the result
Answer:
The amount of CH3OH present in the mixture would decrease
Explanation:
According to Le Cha-telier's principle, when a reaction is in equilibrium and one of the constraints that influence the rate of reactions is applied, the equilibrium would shift so as to neutralize the effects of the constraint.
In this case, looking at the equation of the reaction:
2H2 + CO <------> CH3OH + heat
the total number of moles on the reactant's (left hand) side is 3 (2+1) while on the product's (right hand) side, it is 1. If the pressure on the system is increased, more CH3OH (and less of H2 and CO) will be produced because its side has the lower number of moles out of the two sides.
If the pressure on the system is otherwise lowered, more of H2 and CO would be produced while the amount of CH3OH present would gradually decrease.
why is it difficult to undergo nucleophilic substitution in haloarene?
Answer:
In Haloarenes the C atom to which the X group is attached is SP2 hybridized thus it is become difficult to replace it by the Nucleophile. Since arenes and Vinyl halides are electron rich molecules due to presenceof n bonds, they repel Nucleophile attacking them.
Question:-
Why it is difficult for haloarenes to undergo nucleoplhilic subsituⁿ reaction?
Answer:-
Haloarenes are less reactive towards the nucleoplhilic substitution rxⁿ . This is due to following reasons :-
[tex]\red{\bigstar}\underline{\textsf{ Reason 1 :- Partial double bond character .}}[/tex]
Halogen atom has one lone pair, and due to presence of π - σ - lp , resonance is established in the compound ( see attachment) . Due to resonance there is a partial double bond character in the carbon halogen bond , so it is difficult to break a double bond than a single bond.
[tex]\rule{200}2[/tex]
[tex]\red{\bigstar}\underline{\textsf{ Reason 2 :- $\pi$ cloud .}}[/tex]
When a nucleoplhile comes to attack , it is repelled by the π-cloud of the benzene ring.
[tex]\rule{200}2[/tex]
[tex]\red{\bigstar}\underline{\textsf{ Reason 3 :- Phenyl cation .}}[/tex]
If somehow the halogen atoms leaves the benzene ring ,being more electronegative than carbon , it takes away the electron , thus a positive charge is left on benzene ring and the phenyl cation so formed is very unstable .
[tex]\rule{200}2[/tex]
The standard entropy change of a reaction has a positive value. This reaction results in: Select the correct answer below: a decrease in entropy. an increase in entropy. no entropy change. neither an entropy increase nor decrease.
Explanation:
The standard entropy change of a reaction has a positive value. This reaction results in an increase in entropy.
Positive entropy means the system has increased its degree of disorderness.
2. XC12 is the chloride of metal X. The formulae of its sulphate is
Answer:
XSO₄
Explanation:
XCl₂ is the chloride of metal X. The sum of the charges of the cation and the anion must be zero because the salt is electrically neutral. The charge of the cation of X is:
1 × X + 2 × Cl = 0
1 × X + 2 × (-1) = 0
X = +2
X has a charge +2 and sulphate (SO₄²⁻) a charge -2. The neutral salt they form is XSO₄.
If Sterling silver is 90.0% silver and 10.0% copper, what is the maximum amount of Sterling silver that can be made if you have 48.3 grams of silver metal and an unlimited amount of copper
Answer:
[tex]x=54g[/tex]
Explanation:
From the question we are told that:
Content of Sterling silver:
Let x be sterling silver
Silver [tex]S=0.9x[/tex]
Copper [tex]C=0.1x[/tex]
Total silver available [tex]T=48.3[/tex]
Generally the equation for Total amount to be made is mathematically given by
[tex]T=\frac{x*90}{100}[/tex]
[tex]x=\frac{48.3*100}{90}[/tex]
[tex]x=54g[/tex]
If a buffer is composed of 23.34 mL of 0.147 M acetic acid and 33.66 mL of 0.185 M sodium acetate, how many mL of 0.100 M NaOH can be added before the buffer capacity is reached
Answer:
25.5mL of 0.100M NaOH are needed to reach buffer capacity.
Explanation:
The buffer capacity is reached when the ratio between moles of conjugate base (Sodium acetate) and moles of weak acid (Acetic acid) is 10:
Moles sodium acetate / Moles Acetic acid = 10
The reaction of acetic acid, HA, with NaOH, to produce sodium acetate, NaA is:
HA + NaOH → H2O + NaA
That means the moles of NaOH added = Moles of HA that are being subtracted and moles of NaA that are been produced.
The initial moles of each species is:
Acetic acid:
23.34mL = 0.02334L * (0.147mol / L) = 0.00343 moles Acetic Acid
Sodium Acetate:
33.66mL = 0.03366L * (0.185mol / L) = 0.00623 moles Sodium Acetate
We can write the moles of each species when NaOH is added as:
Moles sodium acetate / Moles Acetic acid = 10
0.00623 moles + X / 0.00343 moles - X = 10
Where X are moles of NaOH added
Solving for X:
0.00623 moles + X = 0.0343 moles - 10X
11X = 0.0281
X = 0.00255 moles of NaOH are needed
In Liters:
0.0255mol NaOH * (1L / 0.100mol) = 0.0255L of 0.100M NaOH are needed =
25.5mL of 0.100M NaOH are needed to reach buffer capacity
Of these gases, which has the fastest-moving molecules (on average) at a given temperature?
-N2
-They all have the same average speed.
-Cl2
-HCl
Which gas molecules have the highest average kinetic energy at a given temperature?
-They all have the same average kinetic energy.
-Cl2
-HCl
-N2
Answer:
a) N2
b) They all have the same average kinetic energy.
Explanation:
At a given temperature, the speed of a gas molecule depends on its relative molecular mass. The heavier the gas, the lesser its average velocity at a given temperature. On that basis, N2 molecules are the fastest moving gas molecules.
At a particular temperature, all gases have the same average kinetic energy.
Lewis Structures are used to describe the covalent bonding in molecules and ions. Draw a Lewis structure for NO3- and answer the following questions based on your drawing.
1. For the central nitrogen atom:
The number of lone pairs = ________
The number of single bonds=_______
The number of double bonds= ______
2. The central nitrogen atom :
Answer:
The lewis structure for NO₃⁻ is shown in the attachment below
For the central nitrogen atom:
The number of lone pairs = 0
The number of single bonds = 2
The number of double bonds= 1
Explanation:
The lewis structure for NO₃⁻ is shown in the attachment below.
From the Lewis structure
For the central nitrogen atom:
The number of lone pairs = 0
The number of single bonds = 2
The number of double bonds= 1
A solution is prepared by dissolving 6.60 g of an nonelectrolyte in water to make 550 mL of solution. The osmotic pressure of the solution is 1.84 atm at 25 °C. The molecular weight of the nonelectrolyte is ________ g/mol.
Answer:
160 g/mol
Explanation:
Step 1: Calculate the molarity of the solution
We will use the following expression.
π = M × R × T
where,
π: osmotic pressure of a nonelectrolyteM: molarityR: ideal gas constantT: absolute temperature (25 °C = 298 K)M = π / R × T
M = 1.84 atm / (0.0821 atm.L/mol.K) × 298 K = 0.0752 mol/L
Step 2: Calculate the moles of solute in 550 mL (0.550 L)
0.550 L × 0.0752 mol/L = 0.0413 mol
Step 3: Calculate the molecular weight of the nonelectrolyte
0.0413 moles weigh 6.60 g.
6.60 g/0.0413 mol = 160 g/mol
Consider the reaction: A(aq) + 2B (aq) === C (aq). Initially 1.00 mol A and 1.80 mol B
were placed in a 5.00-liter container. The mole of B at equilibrium was determined to
be 1.00 mol. Calculate K value.
0.060
5.1
25
17
Ugh
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
What is the correct ratio of carbon to hydrogen to oxygen in glucose (CH1206)?
•12:12:6
•2:1:1
•1:2:1
•6:6:12
Answer:
1:2:1 is the correct ratio of carbon hydrogen to oxygen in glucose.
The Lewis dot model of a molecule is shown.
Based on the model, which of the following is true?
Each carbon has three lone pairs of electrons on it.
The octet of carbon atom remains incomplete in the molecule.
The two carbon atoms share a total of six electrons with each other.
The difference between the electronegativity of carbon and hydrogen is greater than 1.7.
Answer:
The two carbon atoms share a total of six electrons with each other.
Explanation:
Looking at the structure of the molecule H-C≡C-H as shown in the question, we will notice that there exists a triple bond between the two carbon atoms.
Each bond between the two carbon atoms represents two electrons shared. Since there are three bonds between the two carbon atoms, then a total of six electrons were shared between the two carbon atoms hence the answer chosen above.
Hydrogen fluoride will react with glass (silicon dioxide. S_iO_2) according to the equation below What mass of HF is required to react completely with 16 0 g of glass? (Molar masses; Si 28 09 g/mol F 19 00 g/mol H. 1.01 g/mol O, 16.00 g/mol]
4HF(g) + SiO_2(s) --> SiF_4(g) + 2H_2O(g)
A. 5.33 g
B. 21.3 g
C. 64.0 g
D. 80.0 g
E. 320 g
Answer:
21.3 g. Option B
Explanation:
The reaction is:
4HF(g) + SiO₂(s) → SiF₄(g) + 2H₂O(g)
We analysed it and it is correctly balanced.
4 moles of hydrogen fluoride react to 1 mol of silicon dioxide in order to produce 1 mol of silicon fluoride and 2 moles of water vapor.
We determine molar mass of each reactant:
HF → 1.01 g/mol + 19 g/mol = 20.01 g/mol
SiO₂ → 16 g/mol . 2 + 28.09 g/mol = 60.09 g/mol
We convert mass to moles: 16 g . 1 mol /60.09g = 0.266 moles of glass
Ratio is 1:4. 1 mol of glass react to 4 moles of HF
Our 0.266 moles may react to (0.266 . 4) / 1 = 1.07 moles of gas
We convert moles to mass: 1.07 mol . 20.01 g/mol = 21.3 g
Nitric acid can be formed in two steps from the atmospheric gases nitrogen and oxygen, plus hydrogen prepared by reforming natural gas. In the first step, nitrogen and hydrogen react to form ammonia: N2 (g) + 3H2 (g) â 2NH3 (g) =ÎHâ92.kJ In the second step, ammonia and oxygen react to form nitric acid and water:
NH3 (g) + 2O2 (g) â HNO3 (g) + H2O (g) =ÎHâ330.kJ
Required:
Calculate the net change in enthalpy for the formation of one mole of nitric acid from nitrogen, hydrogen and oxygen from these reactions.
Answer:
-376 kJ
Explanation:
The first step equation:
[tex]\mathsf{N_{2(g)} + 3H_2{(g)} \to 2NH_3{(g)} \ \ \ \Delta H = -92\ kJ}[/tex] ---- (1)
The second step equation:
[tex]\mathsf{NH_{3(g)} + 2O_2{(g)} \to HNO_3{(g)} +H_2O_{(g)} \ \ \ \Delta H = -330\ kJ}[/tex] ---- (2)
To determine the enthalpy of formation for 1 mole of HNO₃ (nitric acid), we have the following.
From the above equations; let multiply equation (1) by 1 and equation (2) by 2.
[tex]\mathsf{N_{2(g)} + 3H_2{(g)} \to 2NH_3{(g)} \ \ \ \Delta H = -92\ kJ}[/tex] ---- (3)
[tex]\mathsf{2NH_{3(g)} + 4O_2{(g)} \to 2HNO_3{(g)} +2H_2O_{(g)} \ \ \ \Delta H = 2(-330)\ kJ}[/tex] ----- (4)
adding the above two equations, we have:
[tex]\mathsf{N_{2(g)} + 3H_2{(g)}+ 2NH_{3(g)} + 4O_{2(g)} \to 2HNO_{3(g)} + 2NH_3{(g)} +2H_2O_{(g)} \ \ \ \Delta H = (-660 \ kJ -92\ kJ)}[/tex][tex]\mathsf{N_{2(g)} + 3H_2{(g)} + 4O_{2(g)} \to 2HNO_{3(g)} +2H_2O_{(g)} \ \ \ \Delta H = (-752 \ kJ)}[/tex]
Now, from the recent equation, we have:
2 moles of nitric acid = -752 kJ
∴
1 mole of nitric acid will be: = (1 mole × (-752 kJ)) ÷ 2 moles
1 mole of nitric acid will be: = -376 kJ
Consider the reaction below. How much heat is absorbed if 5.00 moles of nitrogen react
with excess oxygen?
2 N2 (8) + O2(g) → 2 N20 (8) AHrxn- +163.2 kJ
Explanation:
The given chemical reaction is:
[tex]2 N_2 (g) + O_2(g) -> 2 N_20 (g) delta Hrxn= +163.2 kJ[/tex]
When two moles of nitrogen reacts with oxygen, it requires 163.2kJ of energy.
When 5.00 mol of nitrogen requires how much energy?
[tex]5.00 mol x \frac{163.2 kJ }{2 mol} \\=408 kJ[/tex]
Hence, the answer is 408 kJ of heat energy is required.
A hot pot of water is set on the counter to cool. After a few minutes it has lost 495 J of heat energy. How much heat energy has the surrounding air gained?
_____unit_____
Answer:
495 J
Explanation:
When the hot pot was set on the counter to cool, heat energy was lost from the pot. Note that according to the first law of thermodynamics, heat is neither created nor destroyed.
This implies that, the heat energy lost from the pot must be gained by the surrounding air. Therefore, if 495 J of energy is lost from the pot, then 495 J of energy is gained by the surrounding air.
plzzz guys help it urgent
Answer:
requiring immediate action or attention.
word similar to urgent: Acute
During the reaction of 2-methyl-2-butanol with the nucleophile-solvent mixture two layers are formed after shaking the reaction for 5 minutes. After removing the aqueous layer with a Pasteur pipette the organic layer is diluted with 1 mL dichloromethane. The organic phase is washed with 1 mL water. Two layers are obtained.
a. Top layer is Aqueous (H20/ H2SO4/NH4CI)
b. Top layer is Organic (CH2Cl2 and product)
c. Bottom layer is Organic (CH2Cl and product)
d. Top layer is Aqueous (H20)
Answer:
Top layer is Organic (CH2Cl2 and product)
Explanation:
In a solvent mixture, there are usually two phases, the organic phase and the aqueous phase.
It is usual that the organic phase is almost always less dense than the aqueous phase hence the organic phase tend to remain on top of the aqueous phase.
Hence, the top layer is expected to be the organic CH2Cl2 and product.
write the chemistry of Epsom salt
In practice, the second law of thermodynamics means that:
a. Systems move from ordered behavior to more random behavior.
b. Systems move from random behavior to more ordered behavior.
c. Systems move between ordered and random behavior patterns based on temperature.
d. Systems are constantly striving to reach equilibrium.
Answer:
Systems move from ordered behavior to more random behavior.
Explanation:
Entropy refers to the degree of disorderliness in a system. The second law of thermodynamics can be restated in terms of entropy as follows; “any spontaneous process in any isolated system always results in an increase in the entropy of that system.''(science direct)
According to this law, systems tend towards a more disorderly behaviour (increase in entropy) hence the answer given above.
Consider a hypothetical metal that has a density of 10.6 g/cm3, an atomic weight of 176.8 g/mol, and an atomic radius of 0.130 nm. Compute the atomic packing factor if the unit cell has tetragonal symmetry, values for the a and c lattice parameters are 0.570 and 0.341, respectively.
Answer:
0.3323
Explanation:
GIven that:
Density of the metal = 10.6 g/cm^3
atomic weight = 176.8 g/mol
atmic radius = 0.130 nm
values of a and c = 0.570 nm and 0.341 nm respectively
For us to determine the atomic packing factor, we need to first determine the volume of all spheres (Vs) and the volume of unit cell (Vc).
However, the number of atoms in the unit cell (n) can be computed as:
[tex]n = \dfrac{\rho * V_c *N_A}{A} \\ \\ n = \dfrac{(10.6) * (5.7)^2 (3.41)*(10^{-24}) *(6.022*10^{23})}{176.8}[/tex]
n = 4.0
Thus, the number of atoms in the unit cell is 4
∴
The atomic paking factor (APF) is calculated by using the formula:
[tex]\dfrac{Vs}{Vc} = \dfrac{4 * \dfrac{4}{3}\pi *R^3 }{a^2 *c} \\ \\ \\ \dfrac{Vs}{Vc} = \dfrac{4 * \dfrac{4}{3}\pi *(1.30*10^{-8})^3 }{(5.70*10^{-8})^2 *(3.41*10^{-8})}[/tex]
= 0.3323
When hydrogen gas reacts with oxygen gas, water vapour is formed according to the
reaction 2H2 + O2 2H2O. If 3.00 mol of hydrogen gas react with 3.00 mol
of oxygen gas, which reactant will be the reactant in excess?
Explanation:
here's the answer to the question
During electrophilic aromatic substitution, a resonance-stabilized cation intermediate is formed. Groups, already present on the benzene ring, that direct ortho/para further stabilize this intermediate by participating in the resonance delocalization of the positive charge. Assume that the following group is present on a benzene ring at position 1 and that you are brominating the ring at positon 4. Draw the structure of the resonance contributor that shows this group actively participating in the charge delocalization.
-----OCH3
Answer:
See explanation and image attached
Explanation:
Aromatic compounds undergo electrophilic aromatic substitution reactions in which the aromatic ring is maintained.
Substituted benzenes may be more or less reactive towards electrophilic aromatic substitution than benzene depending on the nature of the substituent present in the ring.
Substituents that activate the ring towards electrophilic substitution such as -OCH3 are ortho-para directing.
The major products of the bromination of anisole are p-bromoanisole and o-bromoanisole. The resonance structures leading to these products are shown in the image attached.
