If K is greater than 1, then products are favored
A compound with a molecular weight of about 64.47 g/mol was found to be 18.63 % of C, 1.56 % of H, 24.82 % of O, and 54.99 % of Cl by mass. Determine the molecular formula and draw the Lewis structure showing an accurate 3-D perspective. *Show your calculations
Answer:
See detailed explanation.
Explanation:
Hey there!
In this case, according to the given information, it turns out possible for us to solve this problem by firstly calculating the moles of each element, assuming those percentages are masses, so that we divide by their molar masses:
[tex]C=\frac{18.63}{12.01}=1.55\\\\H=\frac{1.56}{1.01} =1.55\\\\O=\frac{24.82}{16.00}=1.55\\\\Cl=\frac{54.99}{35.45}=1.55[/tex]
Then, we divide all of them by 1.55 to realize the empirical formula is:
[tex]CHOCl[/tex]
Whose molar mass is 64.47 g/mol, and therefore, since the molar mass of these two is the same, we infer the molecular formula is also CHOCl.
The Lewis structure is shown on the attached document, whereas, the central atom is C and it does complete its octet as well as both O and Cl.
Regards!
What is the molarity of a solution that contains 17g of NH₃ in 0.50L of solution?
Please explain as well!
Explanation:
Because molarity is mol/L, we'll have to convert 17g to mol.
After obtaining the mol, we'll divide that by the volume to obtain Molarity.
[tex]\\ \sf\longmapsto NH_3[/tex]
[tex]\\ \sf\longmapsto 14u+3(1u)[/tex]
[tex]\\ \sf\longmapsto 14u+3u[/tex]
[tex]\\ \sf\longmapsto 17u[/tex]
[tex]\\ \sf\longmapsto 17g/mol[/tex]
Moles of Ammonia:-
[tex]\boxed{\sf No\:of\:moles =\dfrac{Given\:mass}{Molar\:mass}}[/tex]
[tex]\\ \sf\longmapsto No\:of\:moles=\dfrac{17}{17}[/tex]
[tex]\\ \sf\longmapsto No\;of\:moles=1mol[/tex]
Volume of solution=0.50LWe know
[tex]\boxed{\sf Molarity=\dfrac{Moles\:of\:solute}{Volume\:of\: Solution\:in\:L}}[/tex]
[tex]\\ \sf\longmapsto Molarity=\dfrac{1}{0.50}[/tex]
[tex]\\ \sf\longmapsto Molarity=2M[/tex]
Which of the following ions is the less likely to be formed?
A) Li+3
B) Na+
C) I-
D) Sr2+
Ε) Η+
Answer:
Li^3+
Explanation:
The electronic configuration of lithium is ; 1s2 2s1. This means that lithium has one electron in its outermost shell and two core electrons.
We know that it is difficult to remove these core electrons during ionization. Lithium belongs to group 1 hence Li^+ is formed more easily.
It is very difficult to form Li^3+ because it involves loss of core electrons which requires a lot of energy.
A chunk of a metal alloy displaces 0.58 L of water and has a mass of 2.9 kg. What is the density of the alloy in g/cm3?
Answer:
5g/cm3
Explanation:
firstly convert the litres and kilograms to grams and centimeters.
1l is equivalent to 1000cm3
0.58×1000
580cm3
and 1kg is equivalent to 1000g
2.9×1000
2900
then find the density by using the formula
density=mass/volume
=2900g/580cm3
=5g/cm3
I hope this helps
A complex ion that forms in solution has a structure that:____.
a. can be determined simply by stoichiometry.
b. can be predicted on the basis of electrical charge.
c. can only be determined experimentally.
d. cannot be determined.
Answer:
can only be determined experimentally.
Explanation:
In the early days of inorganic chemistry, the structure of complex ions remained a mystery hence the name ''complex''.
These ions appear to have structures that defied accurate elucidation. However, by diligent laboratory investigation, Alfred Werner was able to accurately determine the structure of cobalt complexes. As a result of this, he is regarded as a pathfinder in coordination chemistry.
Hence, the structure of complex ions can only be determined experimentally.
Answer:
c. can only be determined experimentally
Explanation:
It is not possible to know for certain the structure of a complex ion on the basis of stoichiometry or by the electrical charges on the components. The structure of the resulting complex ion can only be known by experiment.
8) Determine whether mixing each pair of the following results in a buffera. 100.0 mL of 0.10 M NH3 with 100.0 mL of 0.15 MNH4Cl b. 50.0 mL of 0.10 M HCL with 35.0 mL of 0.150 M NaOHc. 50.0 mL of 0.15 M HF with 20.0 mL of 0.15 M NaOHd. 175.0 mL of 0.10 M NH3 with 150.0 mL of 0.12 M NaOH
Answer:
a. 100.0 mL of 0.10 M NH₃ with 100.0 mL of 0.15 M NH₄Cl.
c. 50.0 mL of 0.15 M HF with 20.0 mL of 0.15 M NaOH.
Explanation:
A buffer system is formed in 1 of 2 ways:
A weak acid and its conjugate base.A weak base and its conjugate acid.Determine whether mixing each pair of the following results in a buffer.
a. 100.0 mL of 0.10 M NH₃ with 100.0 mL of 0.15 M NH₄Cl.
YES. NH₃ is a weak base and NH₄⁺ (from NH₄Cl ) is its conjugate base.
b. 50.0 mL of 0.10 M HCl with 35.0 mL of 0.150 M NaOH.
NO. HCl is a strong acid and NaOH is a strong base.
c. 50.0 mL of 0.15 M HF with 20.0 mL of 0.15 M NaOH.
YES. HF is a weak acid and it reacts with NaOH to form NaF, which contains F⁻ (its conjugate base).
d. 175.0 mL of 0.10 M NH₃ with 150.0 mL of 0.12 M NaOH.
NO. Both are bases.
Which of the following compounds would you expect to be an electrolyte?
N2
CH4
H2O
O2
КСІ
Answer:
N2 but i really didn't know
The compound that would be expected to be an electrolyte is : ( A ) N₂
What is an electrolyte
An electrolyte is any subsatnce which conducts electircity when dissolved in a solvent such as water. From the question the compound that can conduct electricty when dissolved in water is N₂
Hence we can conclude that The compound that would be expected to be an electrolyte is : ( A ) N₂
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Select all of the statements that are true about a buffer solution. A buffer solution always changes color when the pH changes. A buffer solution reacts with basic solutions. A buffer solution has a pH of 7. A buffer solution resists small changes in pH. A buffer solution reacts with acidic solutions. At what point on the titration curve for a weak acid is the solution a buffer
Answer: A buffer solution reacts with basic solutions.
A buffer solution resists small changes in pH.
A buffer solution reacts with acidic solutions.
Explanation:
A buffer solution simply refers to an aqueous solution that consist of a mixture of a weak acid and the conjugate. From the options given, the ones application to a buffer solution include:
• A buffer solution reacts with basic solutions.
• A buffer solution resists small changes in pH.
• A buffer solution reacts with acidic solutions.
Balance the redox reaction Al(s) + MnO4^- (aq) --> MnO2 (s) + Al(OH)4^- (aq) in aqueous basic solution
Answer:
Al + MnO4- + 2H2O → Al(OH)4- + MnO2
Explanation:
First of all, we out down the skeleton equation;
Al + MnO4- → MnO2 + Al(OH)4-
Secondly, we write the oxidation and reduction equation in basic medium;
Oxidation half equation:Al + 4H2O + 4OH- → Al(OH)4- + 4H2O + 3e-
Reduction half equation:MnO4- + 4H2O + 3e- → MnO2 + 2H2O + 4OH-
Thirdly, we add the two half reactions together to obtain:
Al + MnO4- + 8H2O + 4OH- + 3e- → Al(OH)4- + MnO2 + 6H2O + 3e- + 4OH-
Lastly, cancel out species that occur on both sides of the reaction equation;
Al + MnO4- + 8H2O→ Al(OH)4- + MnO2 + 6H2O
The simplified equation now becomes;
Al + MnO4- + 2H2O → Al(OH)4- + MnO2
Select the structure of a compound C6H14 with a base peak at m/z 43.
A) CH3CH2CH2CH2CH2CH3
B) (CH3CH2)2CHCH3
C) (CH3)3CCH2CH3
D) (CH3)2CHCH(CH3)2
E) None of these choices.
The structure of a compound C₆H₁₄ with a base peak at m/z 43 is none of these .
What is a compound?Compound is defined as a chemical substance made up of identical molecules containing atoms from more than one type of chemical element.
Molecule consisting atoms of only one element is not called compound.It is transformed into new substances during chemical reactions. There are four major types of compounds depending on chemical bonding present in them.They are:
1)Molecular compounds where in atoms are joined by covalent bonds.
2) ionic compounds where atoms are joined by ionic bond.
3)Inter-metallic compounds where atoms are held by metallic bonds
4) co-ordination complexes where atoms are held by co-ordinate bonds.
They have a unique chemical structure held together by chemical bonds Compounds have different properties as those of elements because when a compound is formed the properties of the substance are totally altered.
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Predict the reactants of this chemical reaction. That is, fill in the left side of the chemical equation. Be sure the equation you submit is balanced. (You can edit both sides of the equation to balance it, if you need to.)
Note: you are writing the molecular, and not the net ionic equation.
NaClO3(aq) + H2O(l)
Answer:
HClO₃(aq) + NaOH(aq) ⇒ NaClO₃(aq) + H₂O(l)
Explanation:
We have the products of a reaction and we have to predict the reactants. Since the products are salt and water, this must be a neutralization reaction. In a neutralization reaction, an acid reacts with a base. To form NaClO₃, the acid must be HClO₃(aq) and the base NaOH(aq). The balanced chemical equation is:
HClO₃(aq) + NaOH(aq) ⇒ NaClO₃(aq) + H₂O(l)
A laboratory utilizes a mixture of 10% dimethyl sulfoxide (DMSO) in the freezing and long-term storage of embryonic stem cells. If DMSO has a specific gravity of 1.1004, calculate the specific gravity, to four decimal places, of the mixture (assume water to be the 90% portion).
Answer:
The correct answer is "1.0100".
Explanation:
Let the volume of mixture be 100 ml.
then,
The volume of DMSO will be 10 mL as well as that of water will be 90 mL.
DMSO will be:
= [tex]10\times 1.1004[/tex]
= [tex]11.004 \ g[/tex]
The total mass of mixture will be:
= [tex]90+11.004[/tex]
= [tex]101.004 \ g[/tex]
Density of mixture will be:
= [tex]\frac{Mass}{Volume}[/tex]
= [tex]\frac{101.004}{100}[/tex]
= [tex]1.01004 \ g/mL[/tex]
hence,
Specific gravity of mixture will be:
= [tex]\frac{Density \ of \ mixture}{Density \ of \ water}[/tex]
= [tex]\frac{1.01004}{1}[/tex]
= [tex]1.0100[/tex]
385 x 42.13 x 0.079 is (consider significant figures):
385 x 42.13 x 0.079 = 1281.38395
A chemist determines by measurements that moles of bromine liquid participate in a chemical reaction. Calculate the mass of bromine liquid that participates. Round your answer to significant digits.
Answer:
5.20 grams of Br₂
Explanation:
From our previous knowledge;
We understand that:
The number of moles of a given element = mass of the element divided by its molar mass.
Mathematically:
[tex]\mathbf{no \ of \ moles =\dfrac{ mass}{ molar \ mass}}[/tex]
From the given information, let's assume that the 0.065 moles of liquid -bromine partake in the reaction.
From the periodic table, the molar mass of Bromine is = 79.9 g/mol
As such, the mass of liquid that partakes is calculated as:
0.065 mol = mass/ 79.9 g/mol
mass = 0.065 mol × 79.9 g/mol
mass of liquid that partakes in the reaction = 5.20 grams of Br₂
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
Consider an equilibrium (K1) that is established after 10 mL of compound A and 10 mL of compound B are mixed. Now, imagine the equilibrium (K2) where 1 mL of compound A is added to 100 mL of compound B. How are K1 and K2 related algebraically (read this question VERY carefully, at least one more time)
The equilibrium constant K₁ = Equilbrium constant K₂.
The equilibrium constant, K, of a reaction, is defined as:
"The ratio between concentration of products powered to their reaction quotient and concentration of reactants powered to thier reaction quotient".
For the reaction:
aA + bB ⇄ cC + dD
The equilibrium constant, K, is:
[tex]K = \frac{[C]^c[D]^d}{[A]^a[B]^b}[/tex]
Now, assuming the reaction of the problem is 1:1:
A + B ⇄ C + D
[tex]K = \frac{[C][D]}{[A][B]}[/tex]
The concentrations of the reactants are directly proportional to the volume added. Thus, we can assume that concentration = Volume. Replacing for K₁ and K₂:
[tex]K_1 = \frac{[C][D]}{[10mL][10mL]} = K_1 = \frac{[C][D]}{100mL^2}[/tex]
In the same way:
[tex]K_2 = \frac{[C][D]}{[1mL][100mL]} = K_2 = \frac{[C][D]}{100mL^2}[/tex]
Thus, we can say:
K₁ = K₂Learn more about chemical equilibrium in:
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What functional group is found in an alcohol?
A. Ester
B. Amino
C. Carbonyl
D. Hydroxyl
Answer:
an alcohol is a Hydroxyl group due to the OH~ that is associated with it's molecules
The functional group found in an alcohol is Hydroxyl . Therefore, the correct option is option D.
What is functional group?A functional group in organic chemistry is a substituent and moiety inside a molecule that triggers the molecule's distinctive chemical processes. No matter how the rest of a molecule is made up, the very same functional group would experience the same or a similar set of chemical events.
This permits the design of synthetic chemistry as well as the methodical forecasting of chemical reactions as well as the behaviour of chemical molecules. Other functional groups close by can affect a functional group's reactivity. Retrosynthetic analysis can be used to design organic synthesis by using functional group interconversion. The functional group found in an alcohol is Hydroxyl .
Therefore, the correct option is option D.
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Aqueous hydrobromic acid HBr will react with solid sodium hydroxide NaOH to produce aqueous sodium bromide NaBr and liquid water H2O . Suppose 4.9 g of hydrobromic acid is mixed with 3.86 g of sodium hydroxide. Calculate the maximum mass of sodium bromide that could be produced by the chemical reaction. Round your answer to 2 significant digits.
Answer:
6.2g of NaBr are produced
Explanation:
The reaction of HBr with NaOH occurs as follows:
HBr + NaOH → NaBr + H2O
Where 1 mole of each reactant produce 1 mole of NaBr
To solve this question we need to find the moles of each reactant using their molar mass. With moles we can find limiting reactant and the moles (And mass) of NaBr produced, as follows:
Moles HBr -Molar mass: 80.9119g/mol)-
4.9g * (1mol/80.9119g) = 0.0606 moles HBr
Moles NaOH -Molar mass: 40g/mol-
3.86g * (1mol/40g) = 0.0965 moles NaOH
As the reaction is 1:1 and the moles of HBr < Moles NaOH, the limiting reactant is HBr and moles of NaBr produced are 0.0606 moles.
The mass of NaBr (Molar mass: 102.894g/mol) is:
0.0606 moles * (102.894g/mol) =
6.2g of NaBr are producedg When 2.50 g of methane (CH4) burns in oxygen, 125 kJ of heat is produced. What is the enthalpy of combustion (in kJ) per mole of methane under these conditions
Answer:
-800 kJ/mol
Explanation:
To solve the problem, we have to express the enthalpy of combustion (ΔHc) in kJ per mole (kJ/mol).
First, we have to calculate the moles of methane (CH₄) there are in 2.50 g of substance. For this, we divide the mass into the molecular weight Mw) of CH₄:
Mw(CH₄) = 12 g/mol C + (1 g/mol H x 4) = 16 g/mol
moles CH₄ = mass CH₄/Mw(CH₄)= 2.50 g/(16 g/mol) = 0.15625 mol CH₄
Now, we divide the heat released into the moles of CH₄ to obtain the enthalpy per mole of CH₄:
ΔHc = heat/mol CH₄ = 125 kJ/(0.15625 mol) = 800 kJ/mol
Therefore, the enthalpy of combustion of methane is -800 kJ/mol (the minus sign indicated that the heat is released).
what vent system nitrogen vessel used to?
it's helpful
you can try this answer
Determine the value of the equilibrium constant for the following reaction, where the following amounts of each species are present at equilibrium in a 5.00 L container: 1.34 mol HCl, 4.30 mol O2, 30 g H2O, and 2.42 mol Cl2.
4 HCl(g) O2(g) ----> 2 H2O(l) 2 Cl2(g)
Explanation:
here's the answer to your question about
A mixture of methane and carbon dioxide gases contains methane at a partial pressure of 431 mm Hg and carbon dioxide at a
partial pressure of 504 mm Hg. What is the mole fraction of each gas in the mixture?
XCHA
Xc02
Answer:
XCH₄ = 0.461
XCO₂ = 0.539
Explanation:
Step 1: Given data
Partial pressure of methane (pCH₄): 431 mmHgPartial pressure of carbon dioxide (pCO₂): 504 mmHgStep 2: Calculate the total pressure in the container
We will sum both partial pressures.
P = pCH₄ + pCO₂
P = 431 mmHg + 504 mmHg = 935 mmHg
Step 3: Calculate the mole fraction of each gas
We will use the following expression.
Xi = pi / P
XCH₄ = pCH₄/P = 431 mmHg/935 mmHg = 0.461
XCO₂ = pCO₂/P = 504 mmHg/935 mmHg = 0.539
(d) 40g of sulphur
Calculate the number of moles of 40g of sulphur
Answer:
It is 1.25 moles
Explanation:
Molar mass of sulphur = 32 g
[tex]{ \bf{moles = \frac{given \: mass}{molar \: mass} }} [/tex]
Substitute:
[tex]{ \sf{moles = \frac{40}{32} }} \\ { \sf{ = 1.25 \: moles}}[/tex]
2FePO4+3 Na2so4--->Fe2(so4) + 2 Na3Po4 if i ise 25grams of iron iii phosphate 18.5 g iton what is my percent yeild?
Answer:
56%
Explanation:
If I use 25 grams of iron (III) phosphate and obtain 18.5 g of iron (III) sulfate, what is my percent yield?
Step 1: Write the balanced equation
2 FePO₄ + 3 Na₂SO₄ ⇒ Fe₂(SO₄)₃ + 2 Na₃PO₄
Step 2: Calculate the theoretical yield of Fe₂(SO₄)₃
The mass ratio of FePO₄ to Fe₂(SO₄)₃ is 301.64:399.88.
25 g FePO₄ × 399.88 g Fe₂(SO₄)₃/301.64 g FePO₄ = 33 g Fe₂(SO₄)₃
Step 3: Calculate the percent yield of Fe₂(SO₄)₃
We will use the following expression.
%yield = (experimental yield/theoretical yield) × 100%
%yield = (18.5 g/33 g) × 100% = 56%
Select all that are True.
a. For an isoelectronic series, the species with the most negative charge has the smallest first ionization energy.
b. The removal of an electron from a neutral atom results in a release of energy in the form of heat.
c. For an isoelectronic series, the species with the most positive charge has the smallest first ionization energy.
Given the following formula for calculating the ionization energy of one-electron species such as Li2+, He+, and H, calculate the ionization energy (in J/mol) for B4+. Use scientific notation in answers (ex: 1E10, 3.20E-6)
Answer:
The answer is "[tex]32819.9 \ \frac{J}{mol}\\\\[/tex]"
Explanation:
[tex]Boron: 5^{B}\to 1s^2 2s^2 2p^1[/tex]
[tex]\Delta E=-2.18\times 10^{-18}\ \frac{J}{atom}\ (\frac{1}{\infity^2}-\frac{1}{n^2_{initial}})(z^2) (6.022\times 10^{23}\ \frac{atom}{mol})\\\\[/tex]
[tex]=-2.18\times 10^{-18}\ \frac{J}{atom}\ (0-\frac{1}{1})(5^2) (6.022\times 10^{23}\ \frac{atom}{mol})\\\\ =2.18\times 10^{-18}\times 25 \times 6.022\times 10^{23}\ (\frac{J}{mol})\\\\ =328.199 \times 10^{5}\ (\frac{J}{mol})\\\\ =32819 \times 10^{3}\ (\frac{J}{mol})\\\\ =32819.9 \ (\frac{J}{mol})\\\\[/tex]
what is food nutrients
Answer:
Nutrients arw compounds in foods essential to life and heath
Answer: In simple terms nutrients are the energy that you get from food certain foods give more nutrients and others give close to none. That is what nutrients in your food is
Explanation:
When should a line graph be used
Answer:
Line graphs are used to track changes over short and long periods of time. When smaller changes exist, line graphs are better to use than bar graphs. Line graphs can also be used to compare changes over the same period of time for more than one group.
Calculating the expected pH of the buffer solution: Given that the pKa for Acetic Acid is 4.77, calculate the expected pH of the buffer solutions using the Henderson-Hasselbalch equation and the concentrations of Acetic Acid and Acetate added to the 250 ml Erlenmeyer flask: pH
Answer:
[tex]pH=4.77[/tex]
Explanation:
From the question we are told that:
pKa for Acetic Acid [tex]pK_a= 4.77[/tex]
Therefore
For Equal Concentration of acetic acid and acetatic ion
[tex]CH_3COOH=CH_3COO^-[/tex]
Generally the Henderson's equation for pH value is mathematically given by
[tex]pH=pK_a+log\frac{base}{acid}[/tex]
[tex]pH=4.77+log\frac{CH_3COO^-}{CH_3COOH}[/tex]
[tex]pH=4.77+log1[/tex]
[tex]pH=4.77[/tex]
The chemical formula for strontium sulfide is SrS . A chemist measured the amount of strontium sulfide produced during an experiment. She finds that 199.g of strontium sulfide is produced. Calculate the number of moles of strontium sulfide produced. Be sure your answer has the correct number of significant digits.
Answer:
The number of moles of strontium sulfide produced is:
= 1.663.
Explanation:
Chemical formula for strontium sulfide = SrS
Production of strontium sulfide = 199g
1 mole = 1 moles Strontium Sulfide, which is equal to 119.685 grams
The number of moles of strontium sulfide produced = 1.663 (199/119.685)
The number of moles of strontium sulfide produced is the dividend of the amount of strontium sulfide produced during the experiment divided by the mass of 1 mole.
