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
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.
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
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
In recrystallization from boiling water of benzoic acid contaminated with acetanilide, you begin with an impure sample of 5 grams. If the % composition of the acetanilide impurity in the sample is 6.3 %, what is the minimum amount in mL of solvent (water) required for the recrystallization
This question is incomplete, the complete question is;
In recrystallization from boiling water of benzoic acid contaminated with acetanilide, you begin with an impure sample of 5 grams. If the % composition of the acetanilide impurity in the sample is 6.3 %, what is the minimum amount in mL of solvent (water) required for the recrystallization
Compound Solubility in water at 25°C Solubility in water at 100°C
Benzoic Acid 0.34 g/100mL 5.6 g/100mL
Acetanilide 0.53 g/100mL 5.5 g/100mL
Answer:
The minimum amount in mL of solvent (water) required for the recrystallization is 91 mL
Explanation:
Given the data in the question;
mass of sample = 5 g
percentage composition of the acetanilide impurity = 6.3%
mass of the acetanilide in impure sample will be;
⇒ 6.3% × 5 g = 0.315 g
Mass of benzoic acid in impure sample;
⇒ 5 g - 0.315 g = 4.685 g
now, solubility in water at 100°C for benzoic acid = 5.6 g/100mL
hence 4.685 g of benzoic acid is soluble in x mL
x = [ 100 mL × 4.685 g ] / 5.6 g
x = 83.66 ≈ 84 mL
Also, solubility in water at 100°C for acetanilide = 5.5 g/100mL
hence 0.315 g of benzoic acid is soluble in x mL
x = [ 100 mL × 0.315 g ] / 5.5 g
x = 5.727 ≈ 6 mL
So, the minimum amount in mL of solvent (water) required for the recrystallization will be;
⇒ 85 mL + 6 mL = 91 mL
The minimum amount in mL of solvent (water) required for the recrystallization is 91 mL
how many moles of oxygen are present in 16 g of oxygen gas
Answer:
Mole = molecular weight / molecular mass
Mole = 16/16
Mole= 1
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.
Learn more about chemical change,here:
https://brainly.com/question/2591189
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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.
1. What happens to global temperature averages that start an ice age?
Answer:
Around 46° F (7.8° C)
Explanation:
"Scientists have predicted that the global average temperature during the ice age was around 46 degrees Farenheit (7.8 degrees Celsius.) However, the polar regions were far colder, around 25 degrees Fahrenheit (14 degrees Celsius) colder than the global average."
Source: http://www.weforum.org
20. An oxide of osmium (symbol Os) is a pale yellow solid. If 2.89 g of the compound contains 2.16 g of osmium, what is its empirical formula?
The empirical formula is OsO₄ :
Explanation:
Osmium oxide contains osmium and oxygen only.
Thus, we shall determine the mass of oxygen in osmium oxide. This can be obtained as follow:
Mass of compound = 2.89 g
Mass of Os = 2.16 g
Mass of O =?Mass of O = (Mass of compound) – (Mass of Os)
Mass of O = 2.89 – 2.16
Mass of O = 0.73 g
Finally, we shall determine the empirical formula of the compound. This can be obtained as follow:
Mass of Os = 2.16 g
Mass of O = 0.73 g
Empirical formula =..?Os = 2.16 g
O = 0.73 g
Divide by their molar mass of
Os = 2.16 / 190 = 0.011
O = 0.73 / 16 = 0.046
Divide by the smallest
Os = 0.011 / 0.011 = 1
O = 0.046 / 0.011 = 4
Empirical formula = OsO₄Learn more: https://brainly.com/question/23629778
t the centers of some stars, fluorine-19 can undergo a fusion reaction with a proton (a hydrogen-1 nucleus) to produce two different nuclei. One of the products is an alpha particle (a helium-4 nucleus). What is the other
Answer:
Explanation:
¹⁹F₉ + ¹H₁ = ⁴He₂ + ¹⁶X₈
The atomic weight of fluorine is 19 and atomic weight of hydrogen is 1 . Total becomes 20 . Helium , the product has atomic weight of 4 . Therefore atomic weight of product X must be 16 . Similarly we can calculate atomic number of product X . It is found to be 8 .
Hence atomic weight and number of product will be 16 and 8 respectively.
It is similar to oxygen atom
Hence the other product will be oxygen.
tea contains approximately 2% caffeine by weight. assuming that you started with 18g of tea leaves, calculate your percent yield of extraced caffeine
Sulfur trioxide dissolves in water, producing H2SO4. How much sulfuric acid can be produced from 10.1 mL of water (d= 1.00 g/mL) and 23.9 g of SO3? How much of the reagent in excess is left over?
Answer:
29.2 g of H₂SO₄ are produced
0.263 moles of water remain after the reaction goes complete.
Explanation:
We make the reaction in the first step:
Reactants are water and SO₃
H₂O + SO₃ → H₂SO₄
Let's determine moles of reactants:
23.9 g . 1 mol / 80.06g = 0.298 moles
We apply density, to determine mass of water:
D = m/ V so m = D . V
1 g/mL . 10.1 mL = 10.1 g
moles of water are: 10.1 g . 1 mol/ 18g = 0.561 moles
As ratio is 1:1, for 0.298 moles of SO₃ we need the same amount of water, and we have 0.561 moles. Then, water is the excess reagent and sulfur trioxide is the limiting.
0.561 - 0.298 = 0.263 moles of water that remain after the reaction goes complete.
As ratio is 1:1, again, 0.298 moles of SO₃ can produce 0.298 moles of acid.
We determine the mass: 0.298 mol . 98.06 g /mol = 29.2 g
In order to dry wet clothes, we spread them on a clothline. This is because (i) spreading increases surface area (ii) clothes become brighter when spread
Answer:
this is because spreading it makes more sunlight hit the cloth which results in it drying faster
The following pairs of soluble solutions can be mixed. In some cases, this leads to the formation of an insoluble precipitate. Decide, in each case, whether or not an insoluble precipitate is formed.
a. K2S and NH4Cl
b. CaCl2 and NH4CO3
c. Li2S and MnBr2
d. Ba(NO3)2 and Ag2SO4
e. RbCO3 and NaCl
Answer:
a) [tex]K_{2} S[/tex] and [tex]NH_{4} Cl[/tex] :
There are no insoluble precipitate forms.
b) [tex]Ca Cl_{2}[/tex] and [tex](NH_{4} )_{2} Co_{3}[/tex] :
There are the insoluble precipitates of [tex]CaCo_{3}[/tex] forms.
c) [tex]Li_{2}S[/tex] and [tex]MnBr_{2}[/tex] :
There are the insoluble precipitates of [tex]MnS[/tex] forms.
d) [tex]Ba(No_{3} )_{2}[/tex] and [tex]Ag_{2} So_{4}[/tex] :
As [tex]Ag_{2} So_{4}[/tex] is insoluble, therefore no precipitate forms.
e) [tex]Rb_{2}Co_{3}[/tex] and [tex]NaCl[/tex]:
There are no insoluble precipitates forms.
Explanation:
a)
Solubility rule suggests:- [tex]K_{2} S[/tex] ⇒ soluble, [tex]NH_{4} Cl[/tex] ⇒ soluble.
KCl ⇒ soluble, [tex](NH_{4})_{2} S[/tex] ⇒ soluble.
There are no insoluble precipitate forms.
b)
Solubility rule suggests:- [tex]Ca Cl_{2}[/tex] ⇒ soluble, [tex](NH_{4} )_{2} Co_{3}[/tex] ⇒ soluble.
[tex]CaCo_{3}[/tex] ⇒ insoluble, [tex]NH_{4} Cl[/tex] ⇒ soluble.
There are the insoluble precipitates of [tex]CaCo_{3}[/tex] forms.
c)
Solubility rule suggests:- [tex]Li_{2}S[/tex] ⇒ soluble, [tex]MnBr_{2}[/tex] ⇒ soluble.
[tex]LiBr[/tex] ⇒ soluble, [tex]MnS[/tex] ⇒ insoluble.
There are the insoluble precipitates of [tex]MnS[/tex] forms.
d)
Solubility rule suggests:- [tex]Ba(No_{3} )_{2}[/tex] ⇒ soluble, [tex]Ag_{2} So_{4}[/tex] ⇒insoluble.
As [tex]Ag_{2} So_{4}[/tex] is insoluble, therefore no precipitate forms.
e)
Solubility rule suggests:- [tex]Rb_{2}Co_{3}[/tex] ⇒ soluble, [tex]NaCl[/tex] ⇒ soluble.
[tex]RbCl[/tex] ⇒ soluble, [tex]Na_{2} Co_{3}[/tex] ⇒ soluble.
There are no insoluble precipitates forms.
the density of oxygen 1.43 gm/liter at 0°c and pressure 1.0 atm. if a 20 liter cylinder is filled with oxygen at pressure of 25 atm and temperature of 27°c. what is the mass of oxygen in the cylinder
Answer:
640 g
Explanation:
Step 1: Given and required data
Volume of the cylinder (V): 20 LPressure of the oxygen (P): 25 atmTemperature (T): 27 °C (300 K)Ideal gas constant (R): 0.082 atm.L/mol.KStep 2: Calculate the moles of oxygen gas
We will use the ideal gas equation
P × V = n × R × T
n = P × V / R × T
n = 25 atm × 20 L / (0.082 atm.L/mol.K) × 300 K = 20 mol
Step 3: Calculate the mass corresponding to 20 moles of oxygen
The molar mass of oxygen is 32.00 g/mol.
20 mol × 32.00 g/mol = 640 g
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
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
Find the oxidation number of:
A. sulfur in S032-
B. nickel in NiO2
c. iron in Fe(OH)2
Answer:
A. 4+.
B. 4+
C. 2+.
Explanation:
Hey there!
In this case, according to the given substances, it turns out possible for us to find the oxidation number of each element by applying the concept of charge balance in all of them as shown below:
A. sulfur in S032- : overall charge is 2- and the oxidation number of oxygen is 2-, thus:
[tex]x-6=-2\\\\x=6-2\\\\x=4+[/tex]
B. nickel in NiO2 : overall charge is 0 and the oxidation number of oxygen is 2-, thus:
[tex]x-4=0\\\\x=4+[/tex]
C. iron in Fe(OH)2: overall charge is 0 and the oxidation state of the OH ion is 1-, thus:
[tex]x-2=0\\\\x=2+[/tex]
Regards!
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:
MoO2here 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.
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.
Please help answering 46)
Answer:
CO2 is a trigonal planar.
PLEASE HELP ME
a)
b)
c)
or d)?
Answer:
D / 15.0 g
Explanation:
3 % volume thus shows that there are 3 g of an solute in every 100mL of solutions
.. there will be 3 × 5000÷ 100 of H2O2 in a 500 mL bottle
Aqueous sulfuric acid will react with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . Suppose 2.9 g of sulfuric acid is mixed with 3.53 g of sodium hydroxide. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to significant digits.
Answer:
1.07 g of water.
Explanation:
A reaction between an acid and a base makes water and a salt as product.
Our reaction is:
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
Reactants are the acid and the base. Which is the limiting?
2.9 g . 1mol /98 g = 0.0296 moles of acid
3.53 g . 1mol / 40 g = 0.088 moles of base
2 moles of base react to 1 mol of acid
0.088 moles may react to (0.088 . 1)/2 = 0.044 moles of acid
And we only have 0.0296, sulfuric acid is the limiting
Ratio is 1:2. 1 mol of acid can produce 2 moles of water.
Our 0.0296 moles may produce (0.0296 . 2) /1 = 0.0592 moles of water.
We convert moles to mass:
0.0592 mol . 18g /mol = 1.07 g
Give the formula of each coordination compound. Include square brackets around the coordination complex. Do not include the oxidation state on the metal. Use parentheses only around polyatomic ligands.
a) potassium tetracyanonickelate(II)
b) sodium diamminedicarbonatoruthenate(III)
c) diamminedichloroplatinum(II)
Answer:
a) K2[Ni(CN)4]
b) Na3[Ru(NH3)2(CO3)2]
c) Pt(NH3)2Cl2
Explanation:
Coordination compounds are named in accordance with IUPAC nomenclature.
According to this nomenclature, negative ligands end with the suffix ''ato'' while neutral ligands have no special ending.
The ions written outside the coordination sphere are counter ions. Given the names of the coordination compounds as written in the question, their formulas are provided above.
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
A sample of a compound is analyzed and found to contain 0.420 g nitrogen, 0.480g oxygen, 0.540 g carbon and 0.135 g hydrogen. What is the empirical formula of the compound? a. C2H5NO b. CH3NO c. C3H9N2O2 d. C4HN3O4 e. C4H13N3O3
Answer:
c. C3H9N2O2
Explanation:
The empirical formula of a compound is defined as the simplest whole number ratio of atoms present in a molecule. To solve this question we need to convert the mass of each atom to moles. With the moles we can find the ratio as follows:
Moles N -Molar mass: 14.01g/mol-
0.420g N * (1mol/14.01g) = 0.0300 moles N
Moles O -Molar mass: 16g/mol-
0.480g O * (1mol/16g) = 0.0300 moles O
Moles C -Molar mass: 12.01g/mol-
0.540g C * (1mol/12.01g) = 0.0450 moles C
Moles H -Molar mass: 1.0g/mol-
0.135g H * (1mol/1g) = 0.135moles H
Dividing in the moles of N (Lower number of moles) the ratio of atoms is:
N = 0.0300 moles N / 0.0300 moles N = 1
O = 0.0300 moles O / 0.0300 moles N = 1
C = 0.0450 moles C / 0.0300 moles N = 1.5
H = 0.135 moles H / 0.0300 moles N = 4.5
As the empirical formula requires whole numbers, multiplying each ratio twice:
N = 2, O = 2, C = 3 and H = 9
And the empirical formula is:
c. C3H9N2O2
The second law of thermodynamics requires that spontaneous processes generate entropy, either in the system or in the surroundings. What is the thermodynamic driving force for dissolving a solid in a liquid if it is an endothermic process (which reduces the entropy of the surroundings)
Answer:
See explanation
Explanation:
Truly, the second law of thermodynamics requires that spontaneous processes generate entropy, either in the system or in the surroundings.
When a solid is dissolved in a liquid, the solid dissociates into ions. These ions increases the number of particles and hence the entropy of the system thereby making the process spontaneous.
Hence, the dissolution of a substance via an endothermic process is spontaneous because of increase in the number of particles which in turn increases the entropy of the system.
