Answer: Volume occupied by given neon sample at standard condition is 123.84 L.
Explanation:
Given: [tex]V_{1}[/tex] = 105 L, [tex]T_{1} = 27^{o}C = (27 + 273) K = 300 K[/tex], [tex]P_{1}[/tex] = 985 torr
At standard conditions,
[tex]T_{2}[/tex] = 273 K, [tex]P_{2}[/tex] = 760 K, [tex]V_{2}[/tex] = ?
Formula used to calculate the volume is as follows.
[tex]\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}[/tex]
Substitute the values into above formula as follows.
[tex]\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\frac{985 torr \times 105 L}{300 K} = \frac{760 torr \times V_{2}}{273 K}\\V_{2} = \frac{94116.75}{760} L\\= 123.84 L[/tex]
Thus, we can conclude that volume occupied by given neon sample at standard condition is 123.84 L.
How are all compounds similar?
A. They are all made up of ions that are held together by attractions.
B. They are all made up of the same few elements.
C. They are all made up of atoms of two or more different elements.
D. They are all made up of atoms that share electrons.
Answer:
the answer is C
Explanation:
a molecule can be made up of two atoms of the same kind, as when two oxygen atoms bind together to make an oxygen molecule
I'd really appreciate a brainleast
balance equation of potassium sulphate+ water
Answer:
2KHCO
3
+H
2
SO
4
→K
2
SO
4
+2CO
2
+2H
2
O
Cathodic protection of iron involves using another more reactive metal as a sacrificial anode. Classify each of the following metals by whether they would or would not act as a sacrificial anode to iron under standard conditions.
a. Ag
b. Mg
c. Cu
d. Pb
e. Sn
f. Zn
g. Au
Answer:
a. Ag ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.
b. Mg ---> can serve as a sacrificial anode for iron because it is higher than iron in the reactivity series. Hence, it is more reactive than iron.
c. Cu ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.
d. Pb ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.
e. Sn ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.
f. Zn ---> can serve as a sacrificial anode for iron because it is higher than iron in the reactivity series. Hence, it is more reactive than iron.
g. Au ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.
Explanation:
Cathodic protection of iron involves using another more reactive metal as a sacrificial anode. The reactivity series of metals arranges metals based on decreasing order of reactivity. The more reactive metals are found higher up in the series while the least reactive metals are found at the lower ends of the series. Thus, metals above iron in the reactivity series can serve as sacrificial anodes by protecting against corrosion, while those lower than iron cannot.
Based on the reactivity series, the following metals can be classified as either a sacrificial anode for iron or not:
a. Ag ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.
b. Mg ---> can serve as a sacrificial anode for iron because it is higher than iron in the reactivity series. Hence, it is more reactive than iron.
c. Cu ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.
d. Pb ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.
e. Sn ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.
f. Zn ---> can serve as a sacrificial anode for iron because it is higher than iron in the reactivity series. Hence, it is more reactive than iron.
g. Au ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.
what is the charge on the Mn ions in Mn2o3? 1+, 2+, 3+,3-,4+?
How many mL of 0.200M KI would contain 0.0500 moles of KI?
Please explain and show work.
Answer:
250ml
Explanation:
call it V
V*0.2=0.05 (moles)
so V=0.05/0.2 = 0.25l = 250ml
We know
[tex]\boxed{\Large{\sf Molarity=\dfrac{No\:of\:moles\:of\:solute}{Volume\:of\:solution\:in\;\ell}}}[/tex]
[tex]\\ \Large\sf\longmapsto Volume\:of\:KI=\dfrac{0.05}{0.2}[/tex]
[tex]\\ \Large\sf\longmapsto Volume\:of\:KI=0.25L[/tex]
[tex]\\ \Large\sf\longmapsto Volume\:of\:KI=250mL[/tex]
how many moles of KF are present in 46.5 grams of KF
Explanation:
here's the answer to your question
Answer:
0.8017
Explanation:
Find the molar Mass of KF
K = 39
F = 19
Total = 58
Note: these numbers are approximate. Use your periodic table to get the exact numbers.
mols = given mass / molar mass
given mass = 46.5
molar mass = 58
mols = 46.5 / 58
mols = 0.8017
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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An atom's first 2 energy levels are filled and there are 2 electrons in the third energy
level. It's atomic number is:
Answer:
12
Explanation:
2+8+2=12
atomic no is the No of protons
Answer:
Atomic number is 12.
Explanation:
Atomic number = electrons in filled shells + outermost electrons
= 2 + 8 +2
= 12
A sample of gas occupies 12 L under a pressure of 1.2 atm. What
would its volume be if the pressure were increased to 3.6 atm?
(assume temp is constant)
Explanation:
here's the answer to your question
Balance the following skeleton reaction and identify the oxidizing and reducing agents: Include the states of all reactants and products in your balanced equation. You do not need to include the states with the identities of the oxidizing and reducing agents.
NO_2(g) rightarrow NO_3^-(aq) +NO_2^- (aq) [basic]
The oxidizing agent is:______.
The reducing agent is:_______.
Answer:
a. 2NO₂ (g) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻ (aq) + H₂O (l)
b. i. NO₂⁻ is the oxidizing agent
ii. NO₃⁻ is the reducing agent.
Explanation:
a. Balance the following skeleton reaction
The reaction is
NO₂ (g) → NO₃⁻ (aq) + NO₂⁻ (aq)
The half reactions are
NO₂ (g) → NO₃⁻ (aq) (1) and
NO₂ (g) → NO₂⁻ (aq) (2)
We balance the number of oxygen atoms in equation(1) by adding one H₂O molecule to the left side.
So, NO₂ (g) + H₂O (l) → NO₃⁻ (aq)
We now add two hydrogen ions 2H⁺ on the right hand side to balance the number of hydrogen atoms
NO₂ (g) + H₂O (l) → NO₃⁻ (aq) + 2H⁺ (aq)
The charge on the left hand side is zero while the total charge on the right hand side is -1 + 2 = +1. To balance the charge on both sides, we add one electron to the right hand side.
So, NO₂ (g) + H₂O (l) → NO₃⁻ (aq) + 2H⁺ (aq) + e⁻ (4)
Since the number of atoms in equation two are balanced, we balance the charge since the charge on the left hand side is zero and that on the right hand side is -1. So, we add one electron to the left hand side.
So, NO₂ (g) + e⁻ → NO₂⁻ (aq) (5)
We now add equation (4) and (5)
So, NO₂ (g) + H₂O (l) → NO₃⁻ (aq) + 2H⁺ (aq) + e⁻ (4)
+ NO₂ (g) + e⁻ → NO₂⁻ (aq) (5)
2NO₂ (g) + H₂O (l) + e⁻ → NO₃⁻ (aq) + NO₂⁻ (aq) + 2H⁺ (aq) + e⁻ (4)
2NO₂ (g) + H₂O (l) → NO₃⁻ (aq) + NO₂⁻ (aq) + 2H⁺ (aq)
We now add two hydroxide ions to both sides of the equation.
So, 2NO₂ (g) + H₂O (l) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻ (aq) + 2H⁺ (aq) + 2OH⁻ (aq)
The hydrogen ion and the hydroxide ion become a water molecule
2NO₂ (g) + H₂O (l) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻ (aq) + 2H₂O (l)
2NO₂ (g) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻ (aq) + H₂O (l)
So, the required reaction is
2NO₂ (g) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻ (aq) + H₂O (l)
b. Identify the oxidizing agent and reducing agent
Since the oxidation number of oxygen in NO₂ is -2. Since the oxidation number of NO₂ is zero, we let x be the oxidation number of N.
So, x + 2 × (oxidation number of oxygen) = 0
x + 2(-2) = 0
x - 4 = 0
x = 4
Since the oxidation number of oxygen in NO₂⁻ is -1. Since the oxidation number of NO₂⁻ is -1, we let x be the oxidation number of N.
So, x + 2 × (oxidation number of oxygen) = 0
x + 2(-2) = -1
x - 4 = -1
x = 4 - 1
x = 3
Also, the oxidation number of oxygen in NO₃⁻ is -1. Since the oxidation number of NO₃⁻ is -1, we let x be the oxidation number of N.
So, x + 2 × (oxidation number of oxygen) = -1
x + 3(-2) = -1
x - 6 = -1
x = 6 - 1
x = 5
i. The oxidizing agent
The oxidation number of N changes from +4 in NO₂ to +3 in NO₂⁻. So, Nitrogen is reduced and thus NO₂⁻ is the oxidizing agent
ii. The reducing agent
The oxidation number of N changes from +4 in NO₂ to +5 in NO₃⁻. So, Nitrogen is oxidized and thus and NO₃⁻ is the reducing agent.
Nicotine is a toxic substance present in tobacco leaves. There are two lone pairs in the structure of nicotine. In general, localized lone pairs are much more reactive than delocalized lone pairs. With this information in mind, do you expect both lone pairs in nicotine to be reactive?
A. Both lone pairs are delocalized and, therefore, both are expected to have the same reactivity.
B.Lone pair in pyrrolidine ring is localized and, therefore, is expected to be more reactive.
C. Both lone pairs are localized and, therefore, both are expected to be reactive.
D. Lone pair in pyridine ring is localized and, therefore, is expected to be more reactive.
Answer:
B.Lone pair in pyrrolidine ring is localized and, therefore, is expected to be more reactive.
Explanation:
There are two nitrogen atoms bearing lone pairs of electrons in the structure of nicotine as shown in the image attached.
One nitrogen atom is found in the pyrrolidine ring. The lone pair on this nitrogen atom is localized hence it is more reactive than the lone pair of electrons found on the nitrogen atom in the pyridine ring which is delocalized a shown in the image attached to this answer.
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
All of the following statements concerning crystal field theory are true EXCEPT Group of answer choices in an isolated atom or ion, the five d orbitals have identical energy. low-spin complexes contain the maximum number of unpaired electrons. in low-spin complexes, electrons are concentrated in the dxy, dyz, and dxz orbitals. the energy difference between d orbitals often corresponds to an energy of visible light. the crystal field splitting is larger in low-spin complexes than high-spin complexes.
Answer:
low-spin complexes contain the maximum number of unpaired electrons.
Explanation:
In the crystal field theory, the magnitude of crystal field splitting and the pairing energy determines whether a complex will be low spin or high spin.
Low spin complexes often have greater magnitude of crystal field splitting energy than low spin complexes.
High spin complexes have maximum number of unpaired electrons(most of the electrons are unpaired) while low spin complexes have a minimum number of unpaired electrons in a complex(most of the electrons are paired).
The structure of the compound CuI is best described as a cubic closest packed array of iodide ions with the copper ions in tetrahedral holes. What percent of the tetrahedral holes are occupied in this solid
Answer:
12.5 %
Explanation:
In CCP, the effective number of anion is 4
That is there are 4 I- present in 1 unit cell
Number of tetrahedral void = 2*effective number of anion
= 2*4
= 8
In ZrI4, for every 4 anion, there are only 1 Zr atom.
So, one tetrahedral void is occupied per unit cell out of 8
% tetrahedral void occupied = 1*100/8
= 12.5 %
Answer: 12.5 %
Which of the following events takes place in the Kreb entry phase (acetyl COA from pyruvate)?
A). Only CO2 output
B). NAD is reduced, CO2 is released
C). NADH is oxidized, CO2 is released
D). Only NADH is oxidized
E). Only NAD is reduceed
Answer:
Alphabet C :NADH is oxidized,CO2 is reduced
In the reoxidation of QH2 by purified ubiquinone-cytochrome c reductase (Complex III) from heart muscle, the overall stoichiometry of the reaction requires 2 mol of cytochrome c per mole of QH2 because:
Answer: Options related to your question is missing below are the missing options
a. cytochrome c is a one-electron acceptor, whereas QH2 is a two-electron donor.
b. cytochrome c is a two-electron acceptor, whereas QH2 is a one-electron donor.
c. cytochrome c is water soluble and operates between the inner and outer mitochondrial membranes
d. heart muscle has a high rate of oxidative metabolism, and therefore requires twice as much cytochrome c as QH2 for electron transfer to proceed normally.
e. two molecules of cytochrome c must first combine physically before they are catalytically active.
answer:
cytochrome c is a one-electron acceptor, whereas QH2 is a two-electron donor. ( A )
Explanation:
The overall stoichiometry of the reaction requires 2 mol of cytochrome per mole of QH2 because a cytochrome is simply a one-electron acceptor while QH2 is not a one-electron donor ( i.e. it is a two-electron donor )
An electron donor in a reaction is considered a reducing agent because it donates its electrons to another compound thereby self oxidizing itself in the process.
pls help name any of these compounds
Answer:
D. Propanol
Explanation:
C3H7OH the presence of alcohol functional group makes it propanol
What mass of NaNO3 must be dissolved to make 838mL of a 1.25 M solution
Answer:
89.04 g of NaNO₃.
Explanation:
We'll begin by converting 838 mL to L. This can be obtained as follow:
1000 mL = 1 L
Therefore,
838 mL = 838 mL × 1 L / 1000 mL
838 mL = 0.838 L
Next, we shall determine the number of mole of NaNO₃ in the solution. This can be obtained as follow:
Volume = 0.838 L
Molarity = 1.25 M
Mole of NaNO₃ =?
Mole = Molarity × volume
Mole of NaNO₃ = 1.25 × 0.838
Mole of NaNO₃ = 1.0475 mole
Finally, we shall determine the mass of NaNO₃ needed to prepare the solution. This can be obtained as follow:
Mole of NaNO₃ = 1.0475 mole
Molar mass of NaNO₃ = 23 + 14 + (16×3)
= 23 + 14 + 48
= 85 g/mol
Mass of NaNO₃ =?
Mass = mole × molar mass
Mass of NaNO₃ = 1.0475 × 85
Mass of NaNO₃ = 89.04 g
Therefore, 89.04 g of NaNO₃ is needed to prepare the solution.
Determine the highest level of protein structure described by each item.
a. Primary structure
b. Secondary structure
c. Tertiary structure
d. Quaternary structure
1. order of amino acids
2. overall macromolecule structure containing more than one polypeptide chain
From each of the protein structures listed, the option with the highest level of protein structure as regards with order of amino acids and overall macromolecule structure is quaternary structure. That is option D.
The protein one of the essential nutrients that is found in and consumed by mammals.
There are different types of proteins and their functions depends on their shape, structure or conformation.
The structure of proteins include:
Primary structure: This is the simplest shape of proteins. This is because, the amino acids of a polypeptide is arranged in a linear form.Secondary structure: This is the local folded structures that form within a polypeptide due to interactions between atoms of the backbone.Tertiary structure: These are three dimensional structures of proteins that occurs as a result of the interactions between the R groups of the amino acids that make up the protein.Quaternary structure: This protein structure contains multiple polypeptide chains also called subunits.Therefore, the option with the highest level of protein structure as regards with order of amino acids and overall macromolecule structure is quaternary structure.
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The metal sample suspected of being aluminum is warmed and then submerged into water, which is near room temperature. The final temperature of the water and the metal is given below. The specific heat capacity of water is 4.18 J/g.oC. Calculate the specific heat capacity of the metal based on the data below. Remember heat lost = heat gained.
Type of metal used:
Trial 1 Trial 2 Trial 3
Mass of metal, g 2.746 g 2.750 g 2.900 g
Mass of water, g 15.200 g 15.206 g 15.201 g
Initial Temp. of Water, oC 24.7 oC 24.6 oC 24.5 oC
Initial Temp. of Metal, oC 72.1 oC 72.2 oC 71.9 oC
Final Temp of Water & Metal,oC 26.3 oC 26.2 oC 24.7 oC
ΔT for water, oC ______ ______ ______
ΔT for metal, oC ______ ______ ______
Specific heat capacity of metal, J/g.oC ______ ______ ______
Average specific heat capacity, J/g .oC ______ (use two significant figures due to ΔT of water)
Answer:
Average specific heat capacity of metal = 0.57 J/g°C
Explanation:
Heat lost = Heat gained
Heat energy gained or lost, H = mcΔT
where m = mass of substance, c = specific heat capacity, ΔT = temperature change
Trial 1:
Heat lost by metal = -[2.746 g × c × ΔT]
ΔT = (26.3 - 72.1) °C = -45.8 °C
Heat lost by metal = -[2.746 g × c × (-45.8 °C)] = c × (125.7688)g°C
Heat gained by water = 15.200 × 4.18 × ΔT
ΔT = (26.3 - 24.7) = 1.6 °C
Heat gained by water = 15.200 × 4.18 × 1.6 = 101.6576 J
From Heat lost = Heat gained
c × (125.7688)g°C = 101.6576 J
c = 101.6576 J / 125.7688 g°C
c = 0.8083 J/g°C
Trial 2:
Heat lost by metal = -[2.750 g × c × ΔT]
ΔT = (26.2 - 72.2)°C] = - 46 °C
Heat lost by metal = -[2.750 g × c × (-46 °C)
Heat lost by metal = c × (126.5) g°C
Heat gained by water = 15.206 × 4.18 × ΔT
ΔT = (26.2 - 24.6) = 1.6 °C
Heat gained by water = 15.206 × 4.18 × 1.6 = 101.697728 J
From Heat lost = Heat gained
c × (126.5)g°C = 101.6977 J
c = 101.697728 J / 126.5 g°C
c = 0.8039 J/g°C
Trial 3:
Heat lost by metal = -[2.900 g × c × ΔT]
ΔT = (24.7 - 71.9)°C] = - 47.2 °C
Heat lost by metal = -[2.900 g × c × (- 47.2 °C)
Heat lost by metal = -[2.900 g × c × (- 47.2)°C] = c × (136.88)g°C
Heat gained by water = 15.201 × 4.18 × ΔT
ΔT = (24.7 - 24.5) = 0.2 °C
Heat gained by water = 15.201 × 4.18 × 0.2 = 12.708036 J
From Heat lost = Heat gained
c × (136.88)g°C = 12.708036 J
c = 12.708036 J / 136.88 g°C
c = 0.0928 J/g°C
Average specific heat capacity of metal = (0.8083 + 0.8039 + 0.0928) J/g°C / 3
Average specific heat capacity of metal = 0.57 J/g°C
what is the IUPAC name of 2NaOH(s)
Answer:
NaoH= sodium hydroxide
Define pressure. Group of answer choices force exerted by solids to the surrounding area force used to compress a gas force used to melt a solid force exerted per unit area by gas particles as they strike the surfaces around them force applied to a gas to condense it
Answer:
force exerted per unit area by gas particles as they strike the surfaces around them
Explanation:
According to the kinetic molecular theory, a gas is composed of molecules. The molecules of a gas are in constant random motion and collide frequently with each other as well as with the walls of the container.
Pressure is defined as force per unit area. The pressure of a gas is the force exerted per unit area by gas particles as they strike the surfaces around them hence the answer above.
Oxygen and hydrogen are compressed into two cubical boxes of the same
size at a temperature of 28 K. What do these gases have in common
according to the kinetic theory?
Explanation:
Following are the kinetic theory of gases postulates:
1) Space-volume to molecules ratio is negligible.
2)There is no force of attraction between the molecules at normal temperature and pressure. The force of attraction between the molecules build when the temperature decreases and the pressure increases.
3) There is large space between the molecules resulting in continuous motion.
4) The free movement of molecules results in collision which is perfectly elastic.
5) The molecules have kinetic energy due to random movement. But the average kinetic energy of these molecules differs with temperature.
6) Molecules exert pressure on the walls of the container.
100 mL of 0.2 mol/L sodium carbonate solution and 200 mL of 0.1 mol/L calcium nitrate solution are mixed together. Calculate the mass of calcium carbonate that would precipitate and the concentration of the sodium nitrate solution that will be produced.
Answer:
Explanation:
Na2CO3+Ca(NO3)2=CaCO3+2NaNO3
nNa2CO3=0.02
nCa(NO3)2=0.02
mCaCO3=0.02*100=2 gram
nNaNo3=0.04
Cm=2/15
From the calculation, the mass of the product is 2 g.
What is a reaction?A chemical reaction occurs when two more substances are mixed together. In this case, the reaction is shown by; Ca(NO3)2 + Na2CO3 ----> CaCO3(s) + 2NaNO3.
Number of moles of Na2CO3 = 100/1000 L * 0.2 mol/L = 0.02 moles
Number of moles of Ca(NO3)2 = 200/1000 L * 0.1 mol/L = 0.02 moles
Since the reaction is equimolar, amount of the product = 0.02 moles * 100 g/mol = 2 g
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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.
Benzoyl chloride undergoes hydrolysis when heated with water to make benzoic acid. Reaction scheme of benzoyl chloride with water and heat over the arrow, and benzoic acid and hydrochloric acid as products. Calculate the molar mass of the reactant and product. Report molar masses to 1 decimal place.
Answer:
The molar mass of benzoic acid is 122.1 g/mol
The molar mass of hydrochloric acid = 36.5 g/mol
Explanation:
Benzoyl chloride is an organic compound with the molecular formula C₆H₅COCl. It is an acyl chloride since is it an organic derivative of a carboxylic acid. Acyl chlorides have the general molecular formula, R-COCl, where R is a side chain.
The R group of benzoyl chloride is the benzyl group C₆H₅. It reacts with water (hydrolysis) to produce hydrochloric acid and benzoic acid. The equation of the reaction is given below:
C₆H₅COCl + H₂O → C₆H₅CO₂H + HCl
The molar mass of benzoic acid as well as of hydrochloric acid is calculated from the sum of the masses of the atoms of the elements present in the compound thus:
Molar mass of carbon = 12.0107 g
Molar mass of hydrogen = 1.00784 g
Molar mass of oxygen = 15.999 g
Molar mass of chlorine = 35.453 g
Molar mass of benzoic acid, C₆H₅CO₂H containing 7 moles of atoms of carbon, 6 moles of atoms of hydrogen and 2 moles of atoms of oxygen = 7 × 12.0107 + 6 × 1.00784 + 2 × 15.999 = 122.1 g
Therefore, the molar mass of benzoic acid is 122.1 g/mol
Molar mass of hydrochloric acid, HCl, containing 1 mole of atoms of hydrogen and 1 mole of atoms of chlorine = 1 × 1.00784 + 1 × 35.453 = 36.5 g
Therefore, the molar mass of hydrochloric acid = 36.5 g/mol
importance of hematology
Answer:
Haematology is the specialty important for the diagnosis and management of a wide range of benign and malignant disorders of the red and white blood cells, platelets and the coagulation system in adults and children.
If the balance were not tared prior to weighing out the KHP.... how would you expect this to affect the molarity of NaOH calculated? What type of error is this?
Answer:
Following are the response to the given question:
Explanation:
In the given scenario, When the balance has never been tainted before the KHP is weighted, which can affect the molar concentration of NaOH because its molarity is directly proportional to the weight including its substance. In this question it is the mistake is systemic because it may be corrected by modifying balancing parameters.
Gallium is produced by the electrolysis of a solution made by dissolving gallium oxide in concentrated NaOH(aq). Calculate the amount of Ga(s) that can be deposited from a Ga(III) solution using a current of 0.680 A that flows for 80.0 min.
Answer:
Mass gallium (Ga°(s)) produced ≅ 0.800 grams (1 sig. fig.)
Explanation:
Ga(OH)₃ => Ga⁺³ + 3OH⁻
Ga⁺³ + 3e⁻ => Ga°(s)
? grams Ga°(s) = 0.680 Amps x 1 mole e⁻/1 Faraday x 1 Faraday/96,500 Amp·sec x 1 mole Ga°/3 moles e⁻ x 69.723 grams Ga°/mole Ga° x 60 sec/1 min x 80 min = [(0.680)(69.723)(60)(80)/(96,500)(3)] grams Ga° = 0.786099731 grams Ga° (calc. ans.) ≅ 0.800 grams Ga° (1 sig. fig.)
Which of the given statements best represent what to do in the event of a spill of concentrated sulfuric acid.
A. First, rinse the affected area with copious amount of water.
B. First, rinse the affected area with copious amounts of sodium hydroxide.
C. Second, treat the area with aqueous sodium bicarbonate solution.
D. Second, add sand to absorb the remaining acid.
