Answer:
4.92×10²³ atoms of oxygen.
Explanation:
We'll begin by calculating the number of mole of oxygen in 1 mole of CO₂. This can be obtained as follow:
1 mole of CO₂ contains 2 mole of oxygen.
1 mole of CO₂ = 12 + (2×16)
= 12 + 32
= 44 g
Thus,
44 g of CO₂ contains 2 moles of oxygen.
Next, we shall determine the number of mole of oxygen in 18 g of CO₂. This can be obtained as follow:
44 g of CO₂ contains 2 moles of oxygen.
Therefore, 18 g of CO₂ will contain
= (18 × 2)/ 44 = 0.818 mole of oxygen.
Finally, we shall determine the number of atoms in 0.818 mole of oxygen. This can be obtained as follow:
From Avogadro's hypothesis,
1 mole = 6.02×10²³ atoms.
Therefore,
0.818 mole = 0.818 × 6.02×10²³ atoms.
0.818 mole of oxygen = 4.92×10²³ atoms.
Thus, 18 g of CO₂ contains 4.92×10²³ atoms of oxygen.
A 12.37 g sample of Mo2O3(s) is converted completely to another molybdenum oxide by adding oxygen. The new oxide has a mass of 13.197 g. Identify the empirical formula of the new oxide
Answer:
MoO2
Explanation:
The empirical formula is defined as the simplest whole number ratio of atoms present in a molecule.
To solve this question we need to find the moles of Mo2O3. Twice these moles = Moles of Mo. With the moles of Mo we can find its mass.
The difference in masses between mass of new oxide and mass of Mo = Mass of oxygen. With the mass of oxygen we can find its moles and the empirical formula as follows:
Moles Mo2O3 -Molar mass: 239.9g/mol-
12.37g * (1mol / 239.9g) = 0.05156 moles Mo2O3 * (2mol Mo / 1mol Mo2O3) = 0.1031 moles of Mo
Mass Mo -95.95g/mol-:
0.1031 moles of Mo * (95.95g/mol) = 9.895g of Mo
Mass oxygen in the oxide:
13.197 - 9.895g = 3.302g Oxygen
Moles oxygen -Molar mass: 16g/mol-:
3.302g Oxygen * (1mol / 16g) = 0.206 moles O
Now, the ratio of moles O / moles Mo is:
0.206 moles O / 0.1031 moles Mo = 2
That means there are 2 moles of O per mole of Mo and the empirical formula of the new oxide is:
MoO2What is the molarity of a solution that contains 0.75 mol Naci in 3.0 L of solution? Select one: O a. 4.0 M O b. 2.3 M O d. 3.8 M O d. 0.25 M Clear my choice
Answer:
[tex]\boxed {\boxed {\sf D. \ 0.25 \ M}}[/tex]
Explanation:
Molarity is a measure of concentration in moles per liter.
[tex]molarity= \frac{moles \ of \ solute}{ liters \ of \ solution}[/tex]
The solution contains 0.75 moles of sodium chloride and has a volume of 3.0 liters.
moles of solute = 0.75 mol NaCl liters of solution = 3.0 LSubstitute these values into the formula.
[tex]molarity= \frac{ 0.75 \ mol \ NaCl}{3.0 \ L}[/tex]
[tex]molarity= 0.25 \ mol \ NaCl/L[/tex]
Molarity has the molar (M) as its unit. 1 molar is equal to 1 mole per liter.
[tex]molarity= 0.25 \ M \[/tex]
The molarity of the solution is 0.25 Molar and Choice D is correct.
calculate the volume of 20.5g of oxygen occupied at standard temperature and pressure.what the volume
Answer :
volume of a gas = weight * 22.4 l / gram molecular weight
volume of o2 = ?
weight given = 20.5 g
gram molecular weight of oxygen = 32 (because of 2 oxygen atoms )
volume of oxygen = 20.5 * 22.4 / 32
volume of oxygen = 14.35 liters
Explanation:
hope this helps you
if wrong just correct me
Another method for creating a buffer, in situ, is to add an appropriate amount of a strong base, e.g., NaOH, to a weak acid OR add an appropriate amount of a strong acid, e.g., HNO3, to a weak base. As an example, mixing 1.0 mol of acetic acid with 0.5 mol of NaOH will result in a buffer solution with 0.5 mol of acetic acid and 0.5 mol of acetate. The acetate is created by the reaction of acetic acid and the strong base, hydroxide. Given this information, which of the following, when mixed with the appropriate amount of HCl, would create a buffer solution?
a. HNO3
b. HClO2
c. LiCl
d. NH3
Answer:
As an example, mixing 1.0 mol of acetic acid with 0.5 mol of NaOH will result in a buffer solution with 0.5 mol of acetic acid and 0.5 mol of acetate. The acetate is created by the reaction of acetic acid and the strong base, hydroxide.
When HClO2 is mixed with the appropriate amount of HCl it would create a buffer solution. That is option B.
Methods used to form buffer solutionA buffer solution is the solution that resists a change in pH of a solution when acid or base is added because it is made up of weak acid and the conjugate base or weak base and the conjugate acid.
The methods that can be used to form a buffer solution include:
Adding a strong base to a weak acid: For example, mixing 1.0 mol of acetic acid with 0.5 mol of NaOH will result in a buffer solution with 0.5 mol of acetic acid and 0.5 mol of acetate.Adding a weak acid to a conjugate base: For example HCl is a strong acid which will react with a conjugate base such as HClO2.Although HCl is a strong acid, it can be converted to a weak acid through dilution with water. It is in this context that it can be used to form a buffer solution.
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Select the choice that best completes the following sentence: When cooled slowly, transformations near the melting temperature tend to yield ______ grains due to the formation of ______ nucleation sites followed by ______ grain growth.
Question Completion with Options:
O coarse...few...rapid
O fine...few...slow
O fine...multiple...rapid
O coarse...few...slow
O fine...multiple...slow
Answer:
The choice that best completes the sentence is:
O coarse...few...slow
Explanation:
Transformations near the melting temperature develop coarse grains because few nucleation sites are formed and the rate of the grain growth is usually slow. This is because of the process that starts with recrystallization, recovery, and nucleation before growth can occur. While recrystallization enables the grain to increase in size at high temperature, nucleation gives the grain the energy to irreversibly grow into larger-sized nucleus.
11 Explain how you would obtain solid lead carbonate from a mixture of lead carbonate and sodium chloride
Explanation:
Add water, Na2CO3 dissolves, filter, PbCO3 stays in the paper and dissolved Na2CO3 goes through as the solution. Dry the PbCO3 and you have the dry solid.
OR
Add water to dissolve then filter to obtain PbCo3 as you're residue and Na2Co3 as the filtrate. Dry the insoluble PbCo3 between filter papers and you obtain solid PbCo3
4.
Ammonia gas occupies a volume of 450. mL at a pressure of 720 mm Hg. What volume in
liters will the gas occupy at standard atmospheric pressure?
Answer:
[tex]\boxed {\boxed {\sf 426 \ mL}}[/tex]
Explanation:
We are asked to find the volume of ammonia gas given a change in pressure. We will use Boyle's Law, which states the volume of a gas is inversely proportional to the pressure of a gas. The formula is:
[tex]P_1V_1= P_2V_2[/tex]
The ammonia gas originally occupies a volume of 450 milliliters at a pressure of 720 millimeters of mercury. Substitute the values into the formula.
[tex]450 \ mL * 720 \ mm \ Hg = P_2V_2[/tex]
The pressure is changed to standard atmospheric pressure, which is 760 millimeters of mercury. The new volume is unknown.
[tex]450 \ mL * 720 \ mm \ Hg = 760 \ mm \ Hg * V_2[/tex]
We are solving for the volume at standard pressure. We will need to isolate the variable V₂. It is being multiplied by 760 millimeters of mercury. The inverse of multiplication is division. Divide both sides of the equation by 760 mm Hg.
[tex]\frac {450 \ mL * 720 \ mm \ Hg }{760 \ mm \ Hg}= \frac{760 \ mm \ Hg * V_2}{760 \ mm \ Hg}[/tex]
[tex]\frac {450 \ mL * 720 \ mm \ Hg }{760 \ mm \ Hg}= V_2[/tex]
The units of millimeters of mercury (mm Hg) cancel.
[tex]\frac {450 \ mL * 720 }{760} = V_2[/tex]
[tex]\frac {324,000}{760} \ mL = V_2[/tex]
[tex]426.3157895 \ mL =V_2[/tex]
The original values of volume and pressure have 3 significant figures. Our answer must have the same. For the number we calculated, that is the ones place. The 3 in the tenths place tells us to leave the 6 in the ones place.
[tex]426 \ mL \approx V_2[/tex]
The volume at standard atmospheric pressure is approximately 426 milliliters.
A gas at 273K temperature has a pressure of 590 MM Hg. What will be the pressure if you change the temperature to 273K? 
Explanation:
here's the answer to your question
here is the question
Answer:
1. Nitrate ions, NaNO3 - Sodium nitrate.
2. Sulphide ions, K2S - Potassium sulphide.
3. Sulphate ions, CaSO4 - Calcium sulphate.
4. Hydrogensulphite ions, NaHSO3 - Sodium hydrogensulphite.
5. Carbonate ions, CaCO3 - Calcium carbonate.
6. Hydrogencarbonate ions, KHCO3 - Potassium hydrogencarbonate.
7. Phosphite ions, PH3 - Hydrogen phosphite.
8. Nitride ions, NH3 - Hydrogen nitride ( ammonia ).
9. Ethanoate ions, CH3COONa - Sodium ethanoate.
10. Methanoate ions, HCOONa - Sodium methanoate.
11. Fluoride ions, HF - Hydrogen fluoride.
12. Chloride ions, KCl - Potassium chloride.
13. Bromide ions, HBr - Hydrogen bromide.
14. Iodide ions, NaI - Sodium iodide.
15. Phosphate ions, K3PO3 - potassium phosphate.
what are the properety of covalent bond
Explanation:
1. boiling and melting point
2. electrical conductivity
3. Bond strength
4. bond length
A covalent bond consists of negative electrons that are shared in between atoms. Because of this bond, they possess and manifest physical abilities, including electrical pressure/conductivity and lower melting points compared to ionic compounds.
How many grams of magnesium chloride can be produced from 2.30 moles of chlorine gas reacting w excess magnesium Mg(s)+Cl2(g)->MgCl2(s)
The mass of magnesium chloride produced from 2.30 moles of chlorine gas is 218.99 grams.
How to calculate moles in stoichiometry?Stoichiometry refers to the study and calculation of quantitative (measurable) relationships of the reactants and products in chemical reactions.
According to this question, magnesium reacts with chlorine gas to form magnesium chloride as follows:
Mg + Cl₂ → MgCl₂
Based on the above chemical equation, 1 mole of chlorine gas forms 1 mole of magnesium chloride.
This means that 2.30 moles of chlorine gas will 2.30 moles of magnesium chloride.
Next, we convert moles of magnesium chloride to mass as follows:
molar mass of magnesium chloride = 95.211g/mol
mass of magnesium chloride = 95.211 × 2.30 = 218.99 grams.
Therefore, 218.99 grams of magnesium chloride will be formed.
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Consider the following reaction:
Cr(NO3)3 (aq) + 2NaF (aq) --> 3NaNO3 (aq) + CrF3 (s)
If 21.0 grams of NaF are needed to precipitate all of the Cr+3 ions present in 0.125L of a solution of Cr(NO3)3, what is the molarity of the Cr(NO3)3 solution?
Your answer should be to 2 decimal places.
Answer:
2.01
Explanation:
First, let's convert grams to moles
(Na) 22.99 + (F) 18.998 = 41.988
Every mole of NaF is 41.988 grams
21/41.988 = 0.500143 moles of NaF
For every Cr+3, we will need 2 NaF, so Cr+3 will be half of NaF
0.500143/2 = 0.250071
molarity = moles/liters
0.250071/0.125 = 2.0057 M
if an element has an atomic number of 9 what is the electronic structure of the same element
9 is the element Florine
Florine has 9 electrons as well as the 9 protons that determine its atomic number.
The ground state configuration is the lowest energy configuration.
name a factor tht affects the value of electron affinity
Answer:
Atomic sizeNuclear chargesymmetry of the electronic configurationtea contains approximately 2% caffeine by weight. assuming that you started with 18g of tea leaves, calculate your percent yield of extraced caffeine
For the titration of 50. mL of 0.10 M ammonia with 0.10 M HCl, calculate the pH. For ammonia, NH3, Kb
Answer:
11.12 → pH
Explanation:
This is a titration of a weak base and a strong acid.
In the first step we did not add any acid, so our solution is totally ammonia.
Equation of neutralization is:
NH₃ + HCl → NH₄Cl
Equilibrium for ammonia is:
NH₃ + H₂O ⇄ NH₄⁺ + OH⁻ Kb = 1.8×10⁻⁵
Initially we have 50 mL . 0.10M = 5 mmoles of ammonia
Our molar concentration is 0.1 M
X amount has reacted.
In the equilibrium we have (0.1 - x) moles of ammonia and we produced x amount of ammonium and hydroxides.
Expression for Kb is : x² / (0.1 - x) = 1.8×10⁻⁵
As Kb is so small, we can avoid the x to solve a quadratic equation.
1.8×10⁻⁵ = x² / 0.1
1.8×10⁻⁵ . 0.1 = x²
1.8×10⁻⁶ = x²
√1.8×10⁻⁶ = x → 1.34×10⁻³
That's the value for [OH⁻] so:
1×10⁻¹⁴ = [OH⁻] . [H⁺]
1×10⁻¹⁴ / 1.34×10⁻³ = [H⁺] → 7.45×10⁻¹²
- log [H⁺] = pH
- log 7.45×10⁻¹² = 11.12 → pH
If the starting material has no stereogenic centers, when carbonyl compounds are reduced with a reagent such as LiAlH4 or NaBH4 and a new stereogenic center is formed, what will the composition of the product mixture be?
A) Forms a racemic mixture of the two possible enantiomers.
B) Forms more of one enantiomer than another because of steric reasons around the carbonyl.
C) Forms more of one enantiomer than another depending on the temperature of the reaction.
D) Forms different products depending on the solvent used.
Answer:
A) Forms a racemic mixture of the two possible enantiomers
When carbonyl compounds are reduced with a reagent such as LiAlH₄ or NaBH₄ and new stereogenic center is formed chemical change will lead to products that form a racemic mixture of the two possible enantiomers.
What is a chemical change?
Chemical changes are defined as changes which occur when a substance combines with another substance to form a new substance.Alternatively, when a substance breaks down or decomposes to give new substances it is also considered to be a chemical change.
There are several characteristics of chemical changes like change in color, change in state , change in odor and change in composition . During chemical change there is also formation of precipitate an insoluble mass of substance or even evolution of gases.
There are three types of chemical changes:
1) inorganic changes
2)organic changes
3) biochemical changes
During chemical changes atoms are rearranged and changes are accompanied by an energy change as new substances are formed.
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Chromium-51 is a radioisotope that is used to assess the lifetime of red blood cells The half-life of chromium-51 is 27.7 days. If you begin with 39.7 mg of this isotope, what mass remains after 48.2 days have passed?
Answer:
11.9g remains after 48.2 days
Explanation:
All isotope decay follows the equation:
ln [A] = -kt + ln [A]₀
Where [A] is actual amount of the isotope after time t, k is decay constant and [A]₀ the initial amount of the isotope
We can find k from half-life as follows:
k = ln 2 / Half-Life
k = ln2 / 27.7 days
k = 0.025 days⁻¹
t = 48.2 days
[A] = ?
[A]₀ = 39.7mg
ln [A] = -0.025 days⁻¹*48.2 days + ln [39.7mg]
ln[A] = 2.476
[A] = 11.9g remains after 48.2 days
Van der Waals forces hold molecules together by: A. moving electrons from one molecule to another. B. attracting a lone pair of electrons to the positive charge of a hydrogen. C. inducing temporary dipoles that attract each other. D. sharing electrons between atoms.
Van der Waals forces hold molecules together by inducing temporary dipoles that attract each other. That is option C
Van Der Waals forces are example of those intermolecular forces which are weaker than ionic and covalent bonds that exists between molecules.
Van Der Waals forces was postulated by a Dutch physicist known as Van Der Waals. He postulated the existence of weak, short-range forces of attraction, which are independent of normal bonding forces, between non-polar molecules. He came to this conclusion after studying the of real gases at low temperatures and high pressures that:
electrons in a non-polar molecule such as hydrogen are close to one nucleus as to the other, although momentary concentration at one end of the molecule may occur, this momentary concentration of electron cloud on one side create a temporary dipole in the hydrogen molecule, that is, one side of the molecule acquires a partial negative charge while the other side acquires a partial positive charge of equal magnitude, the temporary dipole induces a similar dipole in an adjacent behavior molecule, this results in a temporary dipole-induced dipole attraction between the positive and negative ends of the adjacent molecules.This is how weak Van Der Waals forces are set up. Therefore, option C is CORRECT
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Calculate the percent error in the atomic weight if the mass of a Cu electrode increased by 0.4391 g and 6.238x10-3 moles of Cu was produced. Select the response with the correct Significant figures. You may assume the molar mass of elemental copper is 63.546 g/mol. Refer to Appendix D as a guide for this calculation.
Answer:
10.77%
Explanation:
Molar mass of Cu = mass deposited/number of moles of Cu
Molar mass of Cu = 0.4391 g/6.238x10^-3 moles
Molar mass of Cu = 70.391 g/mol
%error = 70.391 g/mol - 63.546 g/mol/63.546 g/mol × 100
%error = 10.77%
Considering a fish breeder decided to breed small fishes which needs a pH between 6,0 to 7,0 to stay alive. He needs to adjust the water's pH that is 5,0 to a value of 6.5, having available only calcium carbonate. The mass in mg added to 5L of water is about:
A)2,5
B)5,5
C)6,5
D)7,5
E)9,5
which of the following is indicated by the ph value of a solution?
a- it's hydrogen ion concentration
b- its ammonium ion concentration
c- ability to undergo chemical reaction
d- its ratio of solute amount to solvent volume
Answer:
c- ability to undergo chemical reaction
How many moles of gas occupy a volume of 101.3L?
Answer:
V= n VmV: gas volume , n : The number of moles of gas , Vm : molar volume*The molar volume of any gas at standard conditions of temperature and pressure is 22.4 L/mol
V= 101.3 L , n=? , Vm = 22.4 L/mol
V=n Vm101.3 = n × 22.4
n=101.3 / 22.4
n = 4.52 mol
I hope I helped you ^_^
Answer:
[tex]\boxed {\boxed {\sf 4.522 \ mol}}[/tex]
Explanation:
We are asked to find how many moles of gas occupy a volume of 101.3 liters.
1 mole of any gas at STP (standard temperature and pressure) has a volume of 22.4 liters. We can use this information to make a proportion.
[tex]\frac {1 \ mol}{22.4 \ L}[/tex]
We are converting 101.3 liters to moles, so we multiply the proportion by that value.
[tex]101.3 \ L *\frac {1 \ mol}{22.4 \ L}[/tex]
The units of liters (L) cancel.
[tex]101.3 *\frac {1 \ mol}{22.4}[/tex]
[tex]\frac {101.3}{22.4} \ mol[/tex]
Divide.
[tex]4.52232143 \ mol[/tex]
The original value of liters (101.3 L) has 4 significant figures, so our answer must have the same. For the number we calculated, that is the thousandths place. The 3 in the ten-thousandths place to the right tells us to leave the 2 in the thousandths place.
[tex]4.522 \ mol[/tex]
101.3 liters of gas is equal to approximately 4.522 moles of gas.
what is the difference between 25ml and 25.00ml
Answer:
There is no difference between the two.
Explanation:
They both show the same volume. But, adding decimal places shows the least count of the instrument used and is more acceptable when recording values in scientific experiments
1. Draw the condensed structural formula of sodium benzoate showing all charges, atoms including any lone pairs in the side chain functional group, and all sigma and pi bonds.
2. Draw the condensed structural formula of benzoic acid showing all atoms including any lone pairs in the side chain functional group, and all sigma and pi bonds. Indicate the acidic hydrogen.
3. Draw the condensed structural formula of tetrahydrofuran (THF) showing all heteroatoms plus their lone pairs and all sigma and pi bonds.
The structures are shown in the image attached.
A structural formula is the representation of the molecule in which all atoms and bonds in the molecule are shown.
Since the question requires that all the lone pairs, formal charges and sigma and pi bonds should be shown, then the simple condensed structural formula becomes insufficient in this case.
I have attached images of the structural formula of sodium benzoate (image 1), benzoic acid (image 2) and tetrahydrofuran (image 3).
All the formal charges, lone pairs as well as sigma and pi bonds are fully shown.
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how many moles of oxygen are present in 16 g of oxygen gas
Answer:
Mole = molecular weight / molecular mass
Mole = 16/16
Mole= 1
The standard enthalpies of combustion of fumaric acid and maleic acid (to form carbon dioxide and water) are - 1336.0 kJ moJ-1 and - 1359.2 kJ moJ-1, respectively. Calculate the enthalpy of the following isomerization process:
maleic acid ----> fumaric acid
Answer:
Explanation:
maleic acid ⇒ fumaric acid
ΔHreaction = ΔHproduct - ΔHreactant
ΔHproduct = -1336.0 kJ mol⁻¹
ΔHreactant = - 1359.2 kJ mol⁻¹.
ΔHreaction = -1336.0 kJ mol⁻¹ - ( - 1359.2 kJ mol⁻¹.)
= 1359.2 kJ mol⁻¹ -1336.0 kJ mol⁻¹
= 23.2 kJ mol⁻¹ .
Enthalpy of isomerization from maleic to fumaric acid is 23.2 kJ per mol.
What type of bonding is occuring in the compound below?
A. Covalent polar
B. Metallic
C. Ionic
D. Covalent nonpolar
Answer:
(B). it's metallic bonding
Which subshells are found in each of the following shells
electron subshell - M shell
Answer:
3
Explanation:
The electron shells are labelled as K,L,M,N,O,P, and Q or 1,2,3,4,5,6, and 7.
As we go from innermost shell outwards, this number denotes the number of subshell in the shell. Electrons in outer shells have higher average energy and travel farther from the nucleus than those in inner shells.
Hence, M shell contains s,p and d subshells.
How do I do this? What are the answers to the 5 questions shown?
Answer:
1. C₃H₆O₃
2. C₆H₁₂
3. C₆H₂₄O₆
4. C₆H₆
5. N₂O₄
Explanation:
1. Determination of the molecular formula.
Empirical formula => CH₂O
Mass of compound = 90 g
Molecular formula =?
Molecular formula = n × Empirical formula = mass of compound
[CH₂O]ₙ = 90
[12 + (2×1) + 16]n = 90
[12 + 2 + 16]n = 90
30n = 90
Divide both side by 30
n = 90/30
n = 3
Molecular formula = [CH₂O]ₙ
Molecular formula = [CH₂O]₃
Molecular formula = C₃H₆O₃
2. Determination of the molecular formula.
Empirical formula => CH₂
Mass of compound = 84 g
Molecular formula =?
Molecular formula = n × Empirical formula = mass of compound
[CH₂]ₙ = 84
[12 + (2×1)]n = 84
[12 + 2]n = 84
14n = 84
Divide both side by 14
n = 84/14
n = 6
Molecular formula = [CH₂]ₙ
Molecular formula = [CH₂]₆
Molecular formula = C₆H₁₂
3. Determination of the molecular formula.
Empirical formula => CH₄O
Mass of compound = 192 g
Molecular formula =?
Molecular formula = n × Empirical formula = mass of compound
[CH₄O]ₙ = 192
[12 + (4×1) + 16]n = 192
[12 + 4 + 16]n = 192
32n = 192
Divide both side by 32
n = 192/32
n = 6
Molecular formula = [CH₄O]ₙ
Molecular formula = [CH₄O]₆
Molecular formula = C₆H₂₄O₆
4. Determination of the molecular formula.
Empirical formula => CH
Mass of compound = 78 g
Molecular formula =?
Molecular formula = n × Empirical formula = mass of compound
[CH]ₙ = 78
[12 + 1]n = 78
13n = 78
Divide both side by 13
n = 78/13
n = 6
Molecular formula = [CH]ₙ
Molecular formula = [CH]₆
Molecular formula = C₆H₆
5. Determination of the molecular formula.
Empirical formula => NO₂
Mass of compound = 92 g
Molecular formula =?
Molecular formula = n × Empirical formula = mass of compound
[NO₂]ₙ = 92
[14 + (2×16)]n = 92
[14 + 32]n = 92
46n = 92
Divide both side by 46
n = 92/46
n = 2
Molecular formula = [NO₂]ₙ
Molecular formula = [NO₂]₂
Molecular formula = N₂O₄
