Answer:
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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.
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
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
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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name a factor tht affects the value of electron affinity
Answer:
Atomic sizeNuclear chargesymmetry of the electronic configurationFor 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
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
Compound X has a molar mass of 266.64 g/mol and the following composition: aluminum 20.24% chlorine 79.76% Write the molecular formula of X
Answer:
Explanation:
Assume we have 100g of this substance. That means we would have 20.24g of Cl and 79.76g of Al. Now we can find how many moles of each we have:
[tex]\frac{79.76 \:g}{35.45 \: g/mol}[/tex] = 2.25 mol of chlorine
[tex]\frac{20.24 \: g}{26.98 \: g/mol}[/tex] = 0.750 mol of Al.
To form a integer ratio, do 2.25/0.75 = 2.99999 ~= 3.
So the ratio is essentially Al : Cl => 1 : 3. To the compound is possibly [tex]AlCl_3[/tex].
However, it says it has a molar mass of 266.64 g/mol, and since AlCl3 has a molar mass of 133.32, it must be [tex]Al_2Cl_6[/tex].
Actually this molecule isn't exactly AlCl3 (which is ionic). Al2Cl6 forms a banana bond where Cl acts as a hapto-2 ligand. But that's a bit advanced. All you need to know is X = Al2Cl6
The molecular formula of the compound is Al₂Cl₆
To solve the question given above, we'll begin by obtaining the empirical formula of the compound. This can be obtained as follow:
Aluminum (Al) = 20.24%
Chlorine (Cl) = 79.76%
Empirical formula =?Al = 20.24%
Cl = 79.76%
Divide by their molar mass
Al = 20.24 / 27 = 0.75
Cl = 79.76 / 35.5 = 2.25
Divide by the smallest
Al = 0.75 / 0.75 = 1
Cl = 2.25 / 0.75 = 3
Thus, the empirical formula of the compound is AlCl₃
Finally, we shall determine the the molecular formula of the compound.
Molar mass of compound = 266.64 g/mol
Empirical formula = AlCl₃
Molecular formula =? Molecular formula = [AlCl₃]ₙ = molar mass of compound[AlCl₃]ₙ = 266.64
[27 + (3×35.5)]n = 266.64
[27 + 106.5]n = 266.64
133.5n = 266.64
Divide both side by 133.5
n = 266.64 / 133.5
n = 2Molecular formula = [AlCl₃]ₙ
Molecular formula = [AlCl₃]₂
Molecular formula = Al₂Cl₆Therefore, the molecular formula of the compound is Al₂Cl₆
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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.
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
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%
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
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
What is the mass of a piece of iron if its density is 1.98 g/mL and its volume is 2.45 mL?
0.80 g
4.858
1.248
5.998
2.71 g
Answer:
4.858 g
Explanation:
Start with the formula
density = [tex]\frac{mass}{volume}[/tex]
density = 1.98 g/mL
volume = 2.45 mL
mass = ??
rearrange the formula to solve for mass
(density) x (volume) = mass
Add in the substitutes and solve for mass
1.98 g/mL x 2.45 mL = 4.858 g
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
A major component of gasoline is octane (C8H8). When liquid octane is burned in air it reacts with oxygen (O2) gas to produce "0.050 mol" carbon dioxide gas and water vapor. Calculate the moles of octane needed to produce of carbon dioxide.
Answer:
0.0063 mol
Explanation:
Step 1: Write the balanced combustion equation
C₈H₁₈(l) + 12.5 O₂(g) ⇒ 8 CO₂(g) + 9 H₂O(g)
Step 2: Establish the appropriate molar ratio
According to the balanced equation, the molar ratio of C₈H₁₈ to CO₂ is 1:8.
Step 3: Calculate the number of moles of C₈H₁₈ needed to produce 0.050 moles of CO₂
0.050 mol CO₂ × 1 mol C₈H₁₈/8 mol CO₂ = 0.0063 mol C₈H₁₈
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:
MoO2A dehydration reaction starting with 3.0 g cyclohexanol produces 1.9 g cyclohexene. Calculate the theoretical yield for this reaction. Report your answer with two significant figures.
Answer:
77%
Explanation:
First we convert 3.0 g of cyclohexanol (C₆H₁₂O) to moles, using its molar mass:
Molar mass of C₆H₁₂O = 100.158 g/mol3.0 g ÷ 100.158 g/mol = 0.030 molThen we convert 1.9 g of cyclohexene (C₆H₁₀) to moles, using its molar mass:
Molar mass of C₆H₁₀ = 82.143 g/mol1.9 g ÷ 82.143 g/mol = 0.023 molFinally we calculate the theoretical yield:
0.023 mol / 0.030 mol * 100% = 77%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 is the molality of a glucose solution prepared by dissolving 16.7 g of glucose, C6H12O6, in 133.6 g of water
Answer:
0.696 m
Explanation:
We'll begin by calculating the number of mole in 16.7 g of C₆H₁₂O₆. This can be obtained as follow:
Mass of C₆H₁₂O₆ = 16.7 g
Molar mass of C₆H₁₂O₆ = (6×12) + (12×1) + (6×16)
= 72 + 12 + 96
= 180 g/mol
Mole of C₆H₁₂O₆ =?
Mole = mass / molar mass
Mole of C₆H₁₂O₆ = 16.7 / 180
Mole of C₆H₁₂O₆ = 0.093 mole
Next, we shall convert 133.6 g of water to Kg. This can be obtained as follow:
1000 g = 1 Kg
Therefore,
133.6 g = 133.6 g × 1 Kg / 1000 g
133.6 g = 0.1336 Kg
Thus, 133.6 g is equivalent to 0.1336 Kg.
Finally, we shall determine the molality of the solution. This can be obtained as illustrated below:
Mole of C₆H₁₂O₆ = 0.093 mole
Mass of water = 0.1336 Kg
Molality =?
Molality = mole / mass of water (in Kg)
Molality = 0.093 / 0.1336
Molality = 0.696 m
Therefore, the molality of the solution is 0.696 m
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.
Please help thank you
Answer:
[tex]K=1.7x10^{-3}[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to solve this problem by firstly setting up the equilibrium expression for the given reaction, in agreement to the law of mass action:
[tex]K=\frac{[NO]^2}{[N_2][O_2]}[/tex]
Next, we plug in the given concentrations on the data table to obtain:
[tex]K=\frac{(0.034)^2}{(0.69)(0.98)}\\\\K=1.7x10^{-3}[/tex]
Regards!
tea contains approximately 2% caffeine by weight. assuming that you started with 18g of tea leaves, calculate your percent yield of extraced caffeine
Which of the following would tell you that a weak base is present?
A. If the pH of the solution is close to 14.
B. If the solution is a good conductor of electricity.
C. An indicator added to the solution turns green-blue.
D. If there is a high reaction rate.
Answer:
An indicator added to the solution turns green-blue.
Explanation:
An indicator is a substance which shows the degree of acidity or alkalinity of a solution by change in colour of the solution.
The universal indicator changes colour as the pH of the solution changes. Looking at the colour ranges for the universal indicator, the green-blue colour indicates a weak base.
Stalactites and stalagmites form as ________ precipitates out of the water evaporating in underground caves.
Stalactites and stalagmites form as ________ precipitates out of the water evaporating in underground caves.
Group of answer choices
hydrochloric acid
sodium bicarbonate
calcium carbonate
sodium chloride
sodium hydroxide
Answer:
calcium carbonate
Explanation:
A stalactite is an icicle-looking mould that is formed by the precipitation of natural minerals as a result of water dripping from the ceiling, hanging from a cave.
A stalagmites in the other hand, grows upwards and is also a mound that is formed by the deposits of minerals gotten by the water dripping on the floor of a cave.
Therefore, stalactites and stalagmites form as calcium carbonate precipitates out of the water evaporating in underground caves.
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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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
The Tollen's test is the reaction of aldehydes with silver(l) ions in basic solution to form silver metal and a carboxylate. 2 Ag+ + + 3 OH- - HR 2 Ag +_ W + 2 H2O ÃR Which species is being oxidized in the reaction? Choose... Which species is being reduced in the reaction? Choose... Which species is the visual indicator of a positive test? v Choose... Carboxylate ion Aldehyde Silver metal Water Silver(1) ion Hydroxide ion
Answer:
Explanation:
When Tollen's test is done by aldehyde , silver ion is converted into silver which forms a layer which looks like a mirror.
Ag⁺ + e = Ag
It is a reduction process where silver(1) ion is reduced to metallic silver.
Aldehyde is oxidized to carboxylate ion.
CH₃CHO + 2 OH⁻ = CH₃COO⁻ + H₂O + H⁺ + 2e
Visual indicator is silver metal which forms silver mirror at the bottom of test tube .
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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Melanie has completed the analysis of her data for the reaction of KMnO4 with malonic acid and data for a reaction of KMnO4 with tartaric acid. She compared the activation energies, Ea, she calculated for the two reactions and found the Ea for the malonic acid reaction to be greater than the Ea for the tartaric acid reaction.
Required:
What does this mean about the magnitude of the rate constant, k, and the rate of the reaction?
Answer:
See explanation
Explanation:
The relationship between the activation energy and rate of reaction is best captured by the Arrhenius equation;
k= Ae^-Ea/RT
Where;
k= rate constant
A= pre-exponential factor
Ea=activation energy
R= gas constant
T= temperature
We can see from the foregoing that, as the activation energy increases, the rate of reaction decreases and vice versa. reactions that have a very high activation energy are markedly slow.
Since the activation energy for the malonic acid reaction is found to be greater than the activation energy for the tartaric acid reaction, then the rate of the malonic acid reaction(k) will be slower than that of the tartaric acid reaction.
The study of chemistry and bonds is called chemistry. There are two types of elements metal and nonmetals.
The correct answer is mentioned below.
What is the Arrhenius equation?The relationship between the activation energy and rate of reaction is best captured by the Arrhenius equationThe equation is as follows:-
[tex]k= Ae^{-Ea/RT[/tex] Where;
k= rate constantA= pre-exponential factorEa=activation energyR= gas constantT= temperatureWe can see from the foregoing that, as the activation energy increases, the rate of reaction decreases and vice versa. reactions that have very high activation energy are markedly slow. Since the activation energy for the malonic acid reaction is found to be greater than the activation energy for the tartaric acid reaction, then the rate of the malonic acid reaction(k) will be slower than that of the tartaric acid reaction.
Hence, the correct answer is mentioned above.
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https://brainly.com/question/1388366