Answer:
A chemical bond is a lasting attraction between atoms, ions or molecules that enables the formation of chemical compounds. The bond may result from the electrostatic force of attraction between oppositely charged ions as in ionic bonds or through the sharing of electrons as in covalent bonds.
Explanation:
You have to put energy into a molecule to break its chemical bonds. The amount needed is called the bond energy. After all, molecules don't spontaneously break
What is the mass of carbon in 69.00 mg of co2
Answer:
18.82 mg
Explanation:
From the given information:
The molar mass of CO2 is calculated as follow
= (12 + (16 ×2))
= 44
The mass of carbon is determined by dividing the mass no of carbon from co2 by the molar mass of CO2, followed by multiplying it by 69.00 mg
= [tex](\dfrac{12}{44}\times 69 )[/tex]
=(0.2727 × 69 )
= 18.82 mg
Oxygen is composed of three isotopes: oxygen-16, oxygen-17 and oxygen-18 and has an average atomic mass of 15.9982 amu. Oxygen-17 has a mass of 16.988 amu and makes up 0.032% of oxygen. Oxygen-16 has a mass of 15.972 amu and oxygen-18 has a mass of 17.970 amu. What is the percent abundance of oxygen-18?
Answer:
The percent abundance of oxygen-18 is 1.9066%.
Explanation:
The average atomic mass of oxygen is given by:
[tex] m_{O} = m_{^{16}O}*\%_{16} + m_{^{17}O}*\%_{17} + m_{^{18}O}*\%_{18} [/tex]
Where:
m: is the atomic mass
%: is the percent abundance
Since the sum of the percent abundance of oxygen isotopes must be equal to 1, we have:
[tex] 1 = \%_{16} + \%_{17} + \%_{18} [/tex]
[tex] 1 = x + 3.2 \cdot 10^{-4} + \%_{18} [/tex]
[tex] \%_{18} = 1 - x - 3.2 \cdot 10^{-4} [/tex]
Hence, the percent abundance of O-18 is:
[tex] m_{O} = m_{^{16}O}*\%_{16} + m_{^{17}O}*\%_{17} + m_{^{18}O}*\%_{18} [/tex]
[tex]15.9982 = 15.972*x + 16.988*3.2 \cdot 10^{-4} + 17.970*(1 - 3.2 \cdot 10^{-4} - x)[/tex]
[tex] x = 0.980614 \times 100 = 98.0614 \% [/tex]
Hence, the percent abundance of oxygen-18 is:
[tex]\%_{18} = (1 - 3.2 \cdot 10^{-4} - 0.980614) \times 100 = 1.9066 \%[/tex]
Therefore, the percent abundance of oxygen-18 is 1.9066%.
I hope it helps you!
3. (07.05 LC)
When zinc reacts with hydrochloric acid, it produces hydrogen gas. As the reaction proceeds, why does the rate of production of hydrogen gas decrease? (3 points)
the rate would decrease because the reactants are being depleted.
According to an informal 1992 survey, the drinking water in about one-third of the homes in Chicago had lead levels of about 10 ppb. Dr. Koether lived in Chicago from 1996 to 1998. Assuming she drank 1.4 L of water a day, calculate the total amount of lead in mg (using one decimal place) that she was exposed to over the two years if she lived in a home that had such high levels of lead.
Answer:
10.2 mg
Explanation:
Step 1: Calculate the total amount of water she drank
1 year has 365 days and she lived in Chicago for 2 years = 2 × 365 days = 730 days.
If she drank 1.4 L of water per day, the total amount of water she drank is:
730 day × 1.4 L/day = 1022 L
Step 2: Calculate the amount of Pb in 1022 L of water
The concentration of Pb is 10 ppb (10 μg/L).
1022 L × 10 μg/L = 10220 μg
Step 3: Convert 10220 μg to milligrams
We will use the conversion factor 1 mg = 1000 μg.
10220 μg × 1 mg/1000 μg = 10.2 mg
) The C O bond dissociation energy in CO2 is 799 kJ/mol. The maximum wavelength of electromagnetic radiation required to rupture this bond is ________.
Answer:
λ = 150 nm
Explanation:
For C-O bond rupture:
The required energy to rupture C-O bond = bond energy of C-O bond
= 799 kJ/mol
[tex]\mathsf{= 799 \ kJ/mol \times ( \dfrac{1 \ mol }{6.023 \times 10^{23} \ C-O \ bonds })}[/tex]
[tex]\mathsf{= 1.3265 \times 10^{-21} \ kJ/ C-O \ bond}[/tex]
[tex]\mathsf{= 1.33 \times 10^{-18} \ J/C-O \ bond}[/tex]
Recall that the wavelength associated with energy and frequency is expressed as:
[tex]E = \dfrac{hc}{\lambda}[/tex]
[tex]\lambda = \dfrac{hc}{E}[/tex]
[tex]\lambda = \dfrac{(6.626 \times 10^{-34} \ J.s^{-1}) \times (3.0 \times 10^8 \ ms^{-1})}{ 1.33 \times 10^{-18} \ J/C-O \ bond}}[/tex]
[tex]\mathsf{\lambda = 1.50 \times 10^{-7} \ m}[/tex]
λ = 150 nm
The density of an aqueous solution containing 25.0 percent of ethanol (C2H5OH) by mass is 0.950 g/mL. (a) Calculate the molality of this solution. m (b) Calculate its molarity. M (c) What volume of the solution would contain 0.275 mole of ethanol
Answer:
a. 7.24m
b. 5.15M
c. 53.4mL of the solution would contain this amount of ethanol.
Explanation:
Molality, m, is defined as the moles of solute (ethanol, in this case) per kg of solvent.
Molarity, M, are the moles of solute per kg of solvent
To solve this question we need to find the moles of solute in 100g of solution and the volume using its density as follows:
a. Moles ethanol -Molar mass: 46.07g/mol-:
25g ethanol * (1mol/46.07g) = 0.54265 moles ethanol
kg solvent:
100g solution - 25g solute = 75g solvent * (1kg / 1000g) = 0.075kg
Molality:
0.54265 moles ethanol / 0.075kg = 7.24m
b. Liters solution:
100g solution * (1mL / 0.950g) = 105.3mL * (1L / 1000mL) = 0.1053L
Molarity:
0.54265 moles ethanol / 0.1053L = 5.15M
c. 0.275 moles ethanol * (1L / 5.15moles Ethanol) = 0.0534L =
53.4mL of the solution would contain this amount of ethanol
All of the following are TRUE for activities and activity coefficients, except: A) activity for a chemical species is the product of concentration and activity coefficient. B) the activity coefficient corrects for non-ideal behavior due to ionic strength. C) as ionic strength increases, the value of the activity coefficient increases. D) for ions, the activity coefficient approaches unity as the ionic strength approaches 0. E) the activity coefficient for neutral molecules is approximately unity.
Answer:
C) as ionic strength increases, the value of the activity coefficient increases.
Explanation:
The effective concentration of ions available for reactions is known as the activity of the ion.
The activity coefficient important in chemistry because it accounts for the deviation of a solution from ideal behaviour.
The activity of a chemical species is defines as the product of concentration and activity coefficient.
Following the Debye–Hückel limiting law; log γ = −0.509z2I1/2. The ionic strength of a solution tends to increase as the activity coefficient (γ) of the ion decreases.
Congratulations! You are now the head biologist at the local "Cells and Bells" research lab! It has come to other cell biologists' attention recently that some cells are too small to contain all of the organelles inside of them. They decide that it's best to get rid of an organelle, but they're not sure which one. In the first process of this decision, they need to know "which organelle is the most important?"
Your job, as the head cell biologist, is to decide which organelle the cell cannot live without.
Write a research paper (intro, body, and conclusion) on which organelle is the most important and why.
Phosphorylation of enzymes:_______.
a. always increases their activity.
b. generally occurs on Ser, Thr, and/or Tyr side chains and to a lesser extent on the His side chain.
c. is irreversible.
d. is one of only five known covalent forms of regulation.
Answer:
generally occurs on Ser, Thr, and/or Tyr side chains and to a lesser extent on the His side chain
What volume of each solution contains 0.14 mol of KCl? Express your answer using two significant figures.
1.8 M KCl
Answer:
Solution given:
1 mole of KCl[tex]\rightarrow [/tex]22.4l
1 mole of KCl[tex]\rightarrow [/tex]74.55g
we have
0.14 mole of KCl[tex]\rightarrow [/tex]74.55*0.14=10.347g
74.55g of KCl[tex]\rightarrow [/tex]22.4l
10.347 g of KCl[tex]\rightarrow [/tex]22.4/74.55*10.347=3.11litre
volume of each solution contains 0.14 mol of KCl contain 3.11litre.
1 mol of any gas contains 22.4L of volume at STP
Moles of KCl=0.14Volume of KCl:-
0.14(22.4)3.14LCritique this statement: Promotion of electrons is accompanied by a release of energy
Answer: Promotion of electrons is accompanied by a release of energy because of absorption of photon.
Explanation:
Promotion of electrons occurs when an electron accepts or absorbs a photon which leads to it's movement from a lower energy level orbital to a higher energy orbital.
According to Bohr, the electrons was restricted to certain energy levels and was thought to move along certain circular orbits around the nucleus. These energy levels were identified by means of principal quantum number, n. The wave mechanics model of atom does not restrict the electrons to a certain energy levels only. Instead it describes a region around the Nucleus called orbitals where there is a possibility of finding an electron with a certain amount of ENERGY.
The energy levels are composed of one or more orbitals and the distribution of electrons around the nucleus is determined by the number and kind of energy levels that are occupied.
Bohr made an assumption that an electron emits energy in the form of radiation when it moves from a higher to a lower permitted orbit, this produces a line in the atomic emission spectrum. Since the energies of the higher and lower orbits are fixed, the line will be of a particular energy and frequency.
An aqueous solution contains 0.374 M ammonia (NH3). How many mL of 0.276 M nitric acid would have to be added to 125 mL of this solution in order to prepare a buffer with a pH of 8.970.
Answer:
111.95mL of HNO3 are needed to prepare the buffer
Explanation:
We can solve this equation using H-H equation for bases:
pOH = pKb + log [HA+] / [A]
Where pOH is the pOH of the solution
pOH = 14 - pH = 14 - 8.970 = 5.03
pKb is the pKb of NH3 = 4.74
[HA+] could be taken as moles of NH4+
[A] as moles of NH3
The NH3 reacts with nitric acid, HNO3, as follows:
NH3 + HNO3 → NH4+ + NO3-
That means the moles of HNO3 added = X = Moles of NH4+ produced
And moles of NH3 are initial moles NH3 - X
Initial moles of NH3 are:
0.125L * (0.374mol/L) = 0.04675 moles NH3
Replacing in H-H equation:
pOH = pKb + log [HA+] / [A]
5.03 = 4.74 + log [X] / [0.04675-X]
0.29 = log [X] / [0.04675-X]
1.95 = [X] / [0.04675-X]
0.0912 - 1.95X = X
0.0912 = 2.95X
X = 0.0309 moles
We need to add 0.0309 moles of HNO3. From a solution that is 0.276M:
0.0309 moles of HNO3 * (1L / 0.276moles) = 0.11195L of HNO3 are needed
In mL:
111.95mL of HNO3 are needed to prepare the buffer
A student has accidentally spilled 100.0 mL of 3.0 mol/L nitric acid onto the lab bench. What mass of sodium bicarbonate would the teacher need to sprinkle on this spill to neutralize and clean it up?
Answer:
25 g
Explanation:
Step 1: Write the balanced equation
HNO₃ + NaHCO₃ ⇒ NaNO₃ + H₂O + CO₂
Step 2: Calculate the reacting moles of HNO₃
100.0 mL of 3.0 mol/L HNO₃ reacted.
0.1000 L × 3.0 mol/L = 0.30 mol
Step 3: Calculate the reacting moles of NaHCO₃
The molar ratio of HNO₃ to NaHCO₃ is 1:1. The reacting moles of NaHCO₃ are 1/1 × 0.30 mol = 0.30 mol.
Step 4: Calculate the mass corresponding to 0.30 moles of NaHCO₃
The molar mass of NaHCO₃ is 84.01 g/mol.
0.30 mol × 84.01 g/mol = 25 g
If 1 mol of ferric oxide reacts with 3 moles of carbon monoxide to yield 2 mols of iron and 3 mols of carbon dioxide, how much CO will be needed to completely react with 50.26 g of ferric oxide?
Answer:
26.4g
Explanation:
The balanced chemical equation as stated in this question is given as follows:
Fe2O3 + 3CO → 2Fe + 3CO2
According to this balanced equation, 3 moles of carbon monoxide (CO) will react with 1 mole of Ferric oxide (Fe2O3).
We need to change 50.26 g of ferric oxide to moles by using the formula;
mole = mass/molar mass
Molar mass of Fe2O3 = 56(2) + 16(3)
= 112 + 48
= 160g/mol
mole = 50.26/160
mole = 0.314mol of Fe2O3
If 3 moles of carbon monoxide (CO) will react with 1 mole of Ferric oxide (Fe2O3).
Hence, 0.314 mol of Fe2O3 will completely react with (0.314 × 3) mol of CO
0.314 × 3 = 0.94 mol of CO
molar mass of CO = 12 + 16 = 28g/mol
mole = mass/molar mass
mass = mole × M.M
mass = 0.94 × 28
mass = 26.4g of CO
Each 5-ml teaspoon of Extra Strength Maalox Plus contains 450 mg of magnesium hydroxide and 500 mg of aluminum hydroxide. How many moles of hydronium ions H3O are neutralized by 1 teaspoon of antacid product?
Answer:
0.0347 moles of hydronium ions
Explanation:
The equation of the neutralization reaction between hydroxide and hydronium ions is given below:
H₃O+ (aq) + OH- (aq) ----> 2 H₂O (l)
From the equation above, 1 mole of hydroxide ions will neutralize one mole hydronium ions.
The moles of hydroxide ions present in 1 teaspoon or 5 mL of antacid product is calculated as follows:
Number of moles = mass / molar mass
Molar mass of Magnesium hydroxide, Mg(OH)₂ = 58 g/mol
Molar mass of aluminium hydroxide, Al(OH)₃ = 78 g/mol
Mass of magnesium hydroxide = 450 g = 0.45 g
Mass of aluminium hydroxide = 500 mg = 0.5 g
Moles of magnesium hydroxide = (0.45/58) moles
Moles of aluminium hydroxide = (0.5/78) moles
Equation of the ionization of magnesium hydroxide and aluminium hydroxide is given below:
Mg(OH)₂ (aq) ----> Mg²+ (aq) + 2 OH- (aq)
Al(OH)₃ (aq) ---> Al³+ (aq) + 3 OH- (aq)
Number of moles of hydroxide ions present in (0.45/58) moles of magnesium hydroxide = 2 × (0.45/58) moles = 0.0155 moles
Number of moles of hydroxide ions present in (0.5/78) moles of aluminium hydroxide = 3 × (0.5/78) moles = 0.0192 moles
Total moles of hydroxide ions = 0.0155 + 0.0192 = 0.0347 moles hydroxide ions
Therefore, 0.0347 moles of hydroxide ions will neutralize 0.0347 moles of hydronium ions.
What is the energy of a photon emitted with a wavelength of 654 nm?
O A. 3.04 x 10^-19 J
O B. 1.01 * 10^-27 J
O C. 1.30 x 10^-22 J
O D. 4.33 * 10^-22 J
Answer:
A. 3.04×10^-19J
Explanation:
Hope this will help you.
How can beta particles be dangerous to living cells?
A. They move fast and penetrate the skin.
B. They are very high in energy and can travel through most
materials.
C. They move slowly but are very large.
D. They are very low in energy but remain in the body for a long time.
14. What is the oxidation number of oxygen in HSO4 -
Answer:
2
Explanation:
i did this
the ability for carbon to form long chain or rings is
which primitive organic molecule was essential to form lipid bilayer?
a)protenoid
b)phospholipid
c)autocatalytic RNA
d)aminoacids
Answer:
c) autocatalytic RNA is the primitive organic molecules was essential to form lipid bilayer.
balance equation of aluminium chloride+ hydrogen
[tex]\boxed{\sf {AlCl_3\atop Aluminium\:Chloride}+{H_2\atop Hydrogen}\longrightarrow {Al\atop Aluminium}+{HCl\atop Hydrochloric\:acid}}[/tex]
Balanced Equation:-
[tex]\boxed{\sf {2AlCl_3\atop Aluminium\:Chloride}+{3H_2\atop Hydrogen}\longrightarrow {2Al\atop Aluminium}+{6HCl\atop Hydrochloric\:acid}}[/tex]
When naming organic compounds, there are strict rules regarding punctuation.
1. A comma is used to separate two numbers.
2. A hyphen is used to separate a number from a letter.
Rewrite the name of this compound using hyphens and commas as appropriate.
The question is incomplete, the complete question is shown in the image attached to this answer
Answer:
2,3,3-trimethylhexane
Explanation:
IUPAC nomenclature provides a universally acceptable method of naming organic compounds from its structure.
According to this system of nomenclature;
A comma is used to separate two numbers.
A hyphen is used to separate a number from a letter.
Applying these rules, the name of the compound shown in the question should be written as 2,3,3-trimethylhexane.
Phosphagens are a group of molecules that includes creatine phosphate (in vertebrates), and arginine phosphate, lombricine, and phosphoopheline (in invertebrates). These molecules have similar functions in different organisms.
a. True
b. False
Answer: True
Explanation:
Phosphagens are high energy storage compounds that are usually found in the tissue of animals.
Based on the question, the molecules have similar functions in different organisms such as the fact that they can accept phosphoryl groups from ATP in a situation where the ATP is in excess. Also, they donate phosphoryl groups to ADP in order for the regeneration of ATP.
Tech A says that the PCM monitors the pre-cat and post-cat oxygen sensors to determine catalytic converter efficiency. Tech B says that a catalytic converter can be tested by graphing the oxygen sensor readings on a scan tool or lab scope and comparing them. Who is correct
Answer:
Both Tech A and Tech B.
Explanation:
Catalyst is an element used to start chemical reaction but is not used in the reaction. Catalysts material used in catalytic converter include Rhodium, Palladium and platinum. The pre cat and post cat oxygen sensors helps determine converter efficiency.
Answer:
Explanation:
B
Read the scales of this balance.
The unknown sample has a mass of:
11.2 g
01.012 kg
1.220 g
O 1.200 g
Answer:
and I'll call you when the party's over
quiet when I'm come in home
when I'm all alone
Answer:
Explanation:
Don't you know too much already?
I'll only hurt you if you let me
Call me friend but keep me closer (call me back)
And I'll call you when the party's over
Using the following equation for the combustion of octane calculate the heat associated with the formation of 100.0 g of carbon dioxide. The molar mass of octane is 114.33 g/mole.
2C8H18 + 25O2 → 16 CO2 + 18 H2O
ΔH°rxn = -11018 kJ
Answer:
The right solution is "-602.69 KJ heat".
Explanation:
According to the question,
The 100.0 g of carbon dioxide:
= [tex]\frac{100.0 \ g}{114.33\ g/mole}[/tex]
= [tex]0.8747 \ moles[/tex]
We know that 16 moles of [tex]CO_2[/tex] formation associates with -11018 kJ of heat, then
0.8747 moles [tex]CO_2[/tex] formation associates with,
= [tex]-\frac{0.8747}{16}\times 11018 \ KJ \ of \ heat[/tex]
= [tex]-0.0547\times 11018[/tex]
= [tex]-602.69 \ KJ \ heat[/tex]
NCEPT The mass of NaClcontaining the Avogadro Number of particles is.
Answer:
one mole of water (6.022 x 10 23 molecules) has a mass of 18.02 g. One mol of NaCl (6.02 x1023 formulas) has a mass of 58.44 g.
Explanation:
• The mole (or mol) represents a certain number of objects.
• SI def.: the amount of a substance that contains the same
number of entities as there are atoms in 12 g of carbon-12.
• Exactly 12 g of carbon-12 contains 6.022 x 10 23 atoms.
• One mole of H 2O molecules
contains 6.022 x 10 23 molecules.
• 1 mole contains 6.022 x 10 23 entities (Avogadro’s number)
• One mole of NaCl contains 6.022 x 10 23 NaCl formula units.
• Use the mole quantity to count formulas by weighing them.
• Mass of a mole of particles = mass of 1 particle x 6.022 x 1023
Mass of 1 H atom: 1.008 amu x 1.661 x10-24 g/amu = 1.674 x10-24 g
Mass of 1 mole of H atoms:
1.674 x10-24g/H atom x 6.022 x1023H atoms = 1.008 g
• The mass of an atom in amu is numerically the same
as the mass of one mole of atoms of the element in grams.
• One atom of sulfur has a mass of 32.07 amu;
one mole of S atoms has a mass of 32.07 g
PLEASE HELP ASAP
Use the equation below to answer the following questions.
2Al(s) + 3Cu(NO3)2(aq) 3Cu(s) + 2Al(NO3)3(aq)
Determine the oxidation state of the atoms in the equation's reactants and products: (6 points)
Oxidation state of Al in reactant:
in product:
Oxidation state of Cu in reactant:
in product:
Oxidation state of N in reactant:
in product:
Oxidation state of O in reactant:
in product:
Explain why this is a redox reaction.
Thank you!
Answer:
hlo.......................,
g consider the following pair of aqueous solutions. which pair will result in the formation of a precipitate? give the formula for the precipitate in the blank. write none if no precipitate forms. a) libr and nh4no3 b) kcl and pb(ch3coo)2
Answer:
kcl and pb(ch3coo)2
The precipitate is PbCl2
Explanation:
Let us take the options provided one after the other;
In the first case, we have;
LiBr(aq) + NH4NO3(aq) ----> LiNO3(aq) + NH4Br(aq)
You can see that no precipitate is formed here.
In the second case;
2KCl(aq) + Pb(CH3COO)2(aq) ----> PbCl2(s) + 2CH3COOK(aq)
The precipitate here is PbCl2.
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) Send a detailed email Send a detailed text message Oc) Book a one-hour meeting for tomorrow morning O d) Call with a quick update
Kevin can effectively deliver an update by sending a detailed EMAIL to Jill
Email, which means electronic mail is a technological advanced way of passing information from persons to persons without physical contact. Sending emails are also official ways of passing vital information regarding business, work to and fro. According to this question, Jill is a very busy supervisor who hardly. The best way for Kevin to deliver any update concerning the store he is managing is to send Jill an updated email that can even be assessed outside work hours. Learn more: https://brainly.com/question/7098974