Answer:
0.9933
(Text Text Text)
Triglycerides, waxes, and steroids are all _______ lipids because they contain only carbon, hydrogen, and oxygen.Question 20 options:A) long-chain compoundsB) methyl estersC) complexD) simple
Answer:
Explanation:
Here, we want to get the class of lipids that the given substances belong to
These substances are formed from glycerol and fatty acids
From their make up,we know that they are simple lipids
Thus, the answer here is simple
If 30.7 g of C₂H₅OH (MM = 46.07 g/mol) are added to a 500.0 mL volumetric flask, and water is added to fill the flask, what is the concentration of C₂H₅OH in the resulting solution?
The concentration of C₂H₅OH in the resulting solution is 1.332 M.
given that :
mass of C₂H₅OH = 30.7 g
molar mass =46.07 g/mol
volume = 500 ml = 0.5 L
number of moles can be calculated by the following formula :
number of moles = mass / molar mass
no. of moles = 30.7 g / 46.07 g/mol
no. of moles = 0.666 mol
now, the concentration of C₂H₅OH :
Molarity = no. of mole / volumes in l
Molarity = 0.666 / 0.5
Molarity = 1.332 M
Thus, If 30.7 g of C₂H₅OH (MM = 46.07 g/mol) are added to a 500.0 mL volumetric flask, and water is added to fill the flask, The concentration of C₂H₅OH in the resulting solution is 1.332 M.
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7. A given sample of gas is held in a container with the volume of 6.02 L with a temperature of 59.5℃ at a pressure of 1.20 atm. What is the final pressure when the sample of gas is administered to a new volume of 10.0 L at 20.2℃?
The final pressure of the sample gas based on the new volume and temperature would be 0.245 atm.
Combined gas lawThe combined gas law is expressed as:
[tex]\frac{p_1v_1}{t_1}[/tex] = [tex]\frac{p_2v_2}{t_2}[/tex]
Where
[tex]p_1[/tex]= initial pressure[tex]v_1[/tex]= initial volume[tex]t_1[/tex]= initial temperature[tex]p_2[/tex]= final pressure[tex]v_2[/tex]= final volume[tex]t_2[/tex]= final temperatureIn this case, we were given all the variables except the final pressure, [tex]p_2[/tex]:
[tex]p_1[/tex]= 1.20 atm[tex]v_1[/tex]= 6.02 L[tex]t_1[/tex]= 59.5℃[tex]v_2[/tex]= 10.0 L[tex]t_2[/tex]= 20.2℃Making p2 the subject of the formula in the combined gas equation, we have:
[tex]p_2[/tex]= [tex]p_1v_1t_2/t_1v_2[/tex]
Next, let's substitute the given variables:
[tex]p_2[/tex]= 1.2 x 6.02 x 20.2/59.5 x 10.0
= 145.9248/595
= 0.245 atm
In other words, the final pressure when the sample of gas is administered to a new volume of 10.0 L at 20.2℃ is 0.245 atm'
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Which statement is true about the formation of bonds!There is an overall energy threase when bonds formB. bond is formed as soms are solt apart from each otherO c There is an overall release of energy when bonds formO Bonds are always first-order
Answer:
A. There is an overall energy increase when bonds form.
Explanation:
Remember that when a chemical reaction occurs, molecular bonds are broken and other bonds are formed to make different molecules. For example, the bonds of two water molecules are broken to form hydrogen and oxygen. Energy is always required to break a bond, which is known as bond energy but we release energy when we form a bond.
Many atoms are unstable because they have to comply with the octet rule, so they form bonds with other atoms and the energy is released. So based on this logic, the answer would be A. There is an overall energy increase when bonds form.
How to balance ____CaCl2 —-> ____Ca+ ____ Cl
In order to properly balance an equation, we need to make sure that the same amount of elements on the reactants side matches the number of elements on the products side, we can do that by increasing the number in front of each molecule, the so called stoichiometric coefficient. In the reaction from the question we can properly balance by adding the following stoichiometric coefficients
For this question we have:
CaCl2 -> Ca + 2 Cl
Write a balanced equation for the decomposition reaction that occurred in Experiment 2. Include physical states.
In the reaction, 2NaOH + H2SO4 —> Na2SO4 + H2O, 40.0 g NaOH reacts with 60.0 g H2SO4. Which is the limiting reactant
Step 1
The reaction must be written, completed, and balanced:
2 NaOH + H2SO4 => Na2SO4 + H2O
-----------
Step 2
Information provided:
Mass of NaOH = 40.0 g
Mass of H2SO4 = 60.0 g
--
Information needed:
The molar masses:
NaOH) 40.0 g/mol approx.
H2SO4) 98.0 g/mol approx.
-----------
Step 3
The limiting reactant?
By stoichiometry:
1 mole NaOH = 40.0 g
1 mole H2SO4 = 98.0 g
2 NaOH + H2SO4 => Na2SO4 + H2O
2 x 40.0 g NaOH ----------- 98.0 g H2SO4
40.0 g NaOH ----------- X
X = 40.0 g NaOH x 98.0 g H2SO4/2 x 40.0 g NaOH = 49.0 g H2SO4
For 40.0 g of NaOH, 49.0 g of H2SO4 is needed but is provided 60.0 g of H2SO4. Therefore, the excess is the H2SO4, and the limiting reactant is the NaOH.
Answer: the limiting reactant is NaOH
How many atoms of O are there in 7.00 g FeSO4 ?
Answer:
1.11 x 10²³ atoms O
Explanation:
To find the number of oxygen atoms in FeSO₄, you need to (1) convert FeSO₄ from grams to moles (using the molar mass), then (2) convert moles FeSO₄ to moles O (using the mole-to-mole ratio of FeSO₄), and then (3) convert moles O to atoms O (using Avogadro's Number). It is important to arrange the conversions/ratios in a way that allows for the cancellation of units (the desired unit should be in the numerator). The final answer should have 3 sig figs like the given number (7.00 = 3 sig figs).
Atomic Mass (Fe): 55.845 g/mol
Atomic Mass (S): 32.065 g/mol
Atomic Mass (O): 15.999 g/mol
Molar Mass (FeSO₄): 55.845 g/mol + 32.065 g/mol + 4(15.999 g/mol)
Molar Mass (FeSO₄): 151.906 g/mol
1 mole FeSO₄: 1 mole Fe, 2 mole S, 4 moles O
Avogadro's Number:
6.022 x 10²³ atoms = 1 mole
7.00 g FeSO₄ 1 mole 4 moles O 6.022 x 10²³ atoms
----------------------- x ------------------- x ----------------------- x ----------------------------
151.906 g 1 mole FeSO₄ 1 mole
= 1.11 x 10²³ atoms O
A radioactive substance is found to register 5000 counts per second on a Geiger counter. Twenty-four hours later it registers 1250 counts per second. What is its half-life?
Answer
The half-life of the radioactive substance = 12 hours.
Explanation
Given that:
R₀ = 5000 counts/sec
R₁ = 1250 counts/sec
t = 24 hr
What to find:
The half-life of the radioactive substance.
Step-by-step solution:
The half-life of life of the radioactive substance can be calculated using the formula below:
[tex]\frac{0.693t}{T_{1\text{/}2}}=ln(\frac{R_0}{R_1})[/tex]Putting the values of the parameters into the formula, we have:
[tex]\begin{gathered} \frac{0.693\times24}{T_{1\text{/}2}}=ln(\frac{5000}{1250}) \\ \\ \frac{16.632}{T_{1\text{/}2}}=ln(4) \\ \\ \frac{16.632}{T_{1\text{/}2}}=1.3863 \\ \\ T_{1\text{/}2}=\frac{16.632}{1.3863} \\ \\ T_{1\text{/}2}=11.997\text{ }hr \\ \\ T_{1\text{/}2}\approx12\text{ }hr \end{gathered}[/tex]Thus, the half-life of the radioactive substance is 12 hours.
Calculate the pH of the resulting solution when 25.0mL of 0.30M HClO4 is added to 60.0ml of 0.35 M CH3NH2. Kb for CH3NH2 = 4.4*10^-4
The pH of the resulting solution is 10.9 .
given that :
for HClO₄
volume = 25 mL = 0.025 L
M = 0.30 M
no. of moles = 0.025 × 0.30
= 0.0075
for CH₃NH₂
no. of moles = 0.35 × 0.06
= 0.021
as a resultant no. of moles of CH₃NH₂ that remain in solution
= 0.021-0.0075
= 0.0135
total volume = 0.025 + 0.060
= 0.085
the concentration will be :
concentration for HClO₄ = 0.0075 / 0.085
= 0.088 M
concentration for CH₃NH₂ = 0.0135 / 0.088
= 0.153
kb = 4.4 × 10⁻⁴
pkb = - log kb
pkb = 3.3
the pOH formula is given as :
POH = pKb + log [ acid ] / [ base ]
pOH = 3.3 + log [ 0.088 ] / [ 0.153 ]
pOH = 3.06
pH + pOH = 14
pH = 14 - 3.06
pH = 10.9
Thus, The pH of the resulting solution when 25.0mL of 0.30M HClO₄ is added to 60.0ml of 0.35 M CH₃NH₂. Kb forCH₃NH₂ = 4.4 × 10^-4 . pH is 10.9.
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Calculate the AHrxn from the AH of formation for the following reaction. C2H4(g) + 302(g) 2C02(g) + 2H20(1). Formation AH values for C2H4(g) = 52.30 kJ/mol, for 02(g) = 0 kJ/mol, for CO2(g) = -393.5 kJ/mol and for H20(1) =-285.8kJ/mol.A. -1305kJB. 1350kJC. 1411kJD. -1411kJ
Answer
A. -1305 kJ
Explanation
Given:
Equation: C2H4(g) + 302(g) ---- > 2C02(g) + 2H20(l).
Formation ΔH values:
for C2H4(g) = 52.30 kJ/mol,
for 02(g) = 0 kJ/mol,
for CO2(g) = -393.5 kJ/mol, and
for H20(1) = -285.8kJ/mol.
What to find:
The ΔHrxn from the ΔH of formation for the given reaction.
Step-by-step solution:
ΔHrxn = (Sum of ΔH formation for the product) - (Sum of ΔH formation for the reactants).
ΔHrxn = (ΔH for 2CO2(g) + ΔH for 2H2O(l)) + (ΔH for C2H4(g) + ΔH for 3O2(g))
ΔHrxn = [(2 x -393.5) + (2 x -285.8)] + [52.30 + (3 x 0)]
ΔHrxn =(-787.0 - 571.6) + (52.30 + 0)
ΔHrxn = -1358.6 + 52.30
ΔHrxn = -1306.3 kJ
so the closest answer is A. -1305 kJ
I’m getting certain numbers but I want to be sure I’m doing this right
To write the numbers with the specific number of significant figures, we need to remember that zeros to the left don't count as a significant figure and also that we can add zeros to the right to add significant figures to the numbers.
1) 1.5408 to 3 significant figures, we start counting from left to right, so we will maintain the numbers until the second decimal place, the "8". Since the next is "0", the rounding is "8", so the number is 1.54.
2) Here is similar, but since we don't have decimal places, we need to put the zeros in place of the numbers we are rounding. Since we want 2 significant figures, we will maintain only the first two. However, since the third is greater than 5, we need to round up, so the number is 4400.
3) To get this, we will maintain the zeros to the elft, but they don't count, so we start counting from the "1". Since the third number is less than 5, we round down, so the number is 0.019.
4) Now, here we have an example we need to add zeros. The way it is, 0.5 has only 1 significant figure, so we need to add 3 more zeros to the right to get to 4 significant figures. The number is 0.5000.
5) Here, is the same as item 3, but we want 3 significant figures and since the fourth is greater than 5 we round it up. So, the number is 0.066.
Which of the following best describes what happens when salt dissolves in water?A.) The polar solvent molecules surround the positive sodium ions.B.) The polar solvent molecules surround the positive sodium ions and the negativechloride ions.C.) The solute and solvent molecules form a crystalline structure.D.) The solute and solvent molecules do not interact.
When salt dissolves in water, dissociation occurs. This dissociation forms two ions, one positive and one negative. Water is polar and its molecules has poles negatives and positives. So positive poles of water will surround the negative ions. While negative poles will surround the positive ions.
Answer: Alternative "B"
How many milliliters of a 2.59 M H2SO4 solution are needed to neutralize 47.50 mL of a 0.827 M KOH solution?
Given Data:
Concentration of acid, H2SO4 = 2.59 M
For, H2SO4, n-factor = 2, i.e., the number of [tex]H^{+}[/tex] ions on dissociation
Thus, the concentration in normality, N1 = 2.59 x 2 = 5.18 N
Concentration of base, KOH = 0.827 M
For, KOH, n-factor = 1, i.e., the number of [tex]OH^{-}[/tex] ions on dissociation
Thus, the concentration in normality, N2 = 0.827 x 1 = 0.827 N
Volume of base, V2 = 47.50 mL
Using the formula, N1 x V1 = N2 x V2, where, V1 = volume of acid
Thus, volume of acid required to neutralize, V1 = [tex]\frac{N_{2}X V_{2} }{N_{1} }[/tex] = [tex]\frac{0.827 X 47.50}{5.18}[/tex]
= 7.58 mL.
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What type of blood vessels can you find in most mammals?
A) tendons, ligaments, and capillaries
B) nerves and veins
C) aorta and ventricle
D) arteries, capillaries, veins
Answer:
Explanation:
the answer is D
Answer:
D
Explanation:
capillaries veins and arteries
A gas has a volume of 1.82 L at-30°C and 131 kPa. At what temperature would the gasoccupy 1.3 L at 233 kPa?Answer in units of°C.
To solve this problem we can use the Ideal gas law:
[tex]\frac{P_1\cdot V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]The problem give us de following information:
P1= 131kPa
V1=1.82L
T1=-30°C= 243.15 K
P2= 233 kPa
V2= 1.3L
Then we just have to solve for T2 and use the information provided:
[tex]T_2=\frac{P_2\cdot V_2}{P_1\cdot V_1}\cdot T_1=\frac{233\text{ kPa}\cdot1.3L}{131kPa\cdot1.82L}\cdot243.15K=308.91K=35.76°C[/tex]Then the answer is T2=35.76°C
Suppose 2.68 g of barium acetate is dissolved in 300. mL of a 45.0 m Maqueous solution of ammonium sulfate.
Calculate the final molarity of acetate anion in the solution. You can assume the volume of the solution doesn't change when the barium
acetate is dissolved in it.
Round your answer to 3 significant digits.
Explanation
Final Molarity of acetate ion is 0.0394 M. The number of moles of solute per liter of solution is referred to as molarity.
What is Molarity?The amount of a substance in a given volume of solution is measured in molarity (M). Molarity is defined as the number of moles of a solute in one liter of solution. A solution's molarity is also known as its molar concentration.
Therefore,
We first assume that the reaction completes and that the volume of the reaction remains constant.
The balanced stoichiometric equation is as follows:
Ba(CH₃COO)₂ + (NH₄)₂SO₄ → BaSO₄ + 2NH₄(CH₃COO)
First, we must determine which reactant is in excess and which has been completely consumed by the reaction. To do so, we count the number of moles of each reactant at the beginning of the reaction.
For Ba(CH₃COO)₂
The number of moles = (Mass)/(Molar Mass)
Barium acetate mass = 2.68 g
Barium acetate molar mass = 255.43 g/mol
The number of moles = (2.68/255.43) = 0.0104 moles.
For (NH₄)₂SO₄,
Number of moles = (Concentration in mol/L) × (Volume in L)
Concentration of Ammonium surface in mol/L = 45 M
Volume in L = (300/1000) = 0.3 L
Number of moles = 45 × 0.3 = 13.5 moles
From the stoichiometric balance of the reaction,
1 mole of Ba(CH₃COO)₂ responds with 1 mole of (NH₄)₂SO₄
As a result, it is clear that Ba(CH₃COO)₂ is the limiting reagent; the chemical specie that is depleted during the reaction and determines the number of other reactants and products formed.
1 mole of Ba(CH₃COO)₂ provide 2 moles of NH₄(CH₃COO)
0.0591 moles of Ba(CH₃COO)₂ will give 2 × 0.0591 moles of NH₄(CH₃COO); 0.1182 moles of NH₄(CH₃COO).
The molarity of NH₄(CH₃COO) exist then provide as (number of moles) ÷ (Vol in L)
The number of moles = 0.1182 moles
L volume of the solution = 0.3 L
Molarity of NH₄(CH₃COO) = (0.01182/0.3)
= 0.0394 M
Note that 1 mole of NH₄(CH₃COO) contains 1 mole of acetate ion,
Hence, 0.0394 M of NH₄(CH₃COO) also contains 0.0394 M of acetate ion.
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HELP PLEASEEEEE!!!!!!
Organism – Squid
Number of chromosomes after meiosis: 26
Number of chromosomes in the body cells: 52
Number of chromosomes in gametes: 26
Number of chromosomes after mitosis: 52
Organism – Sunflower
Number of chromosomes after meiosis: 19
Number of chromosomes in the body cells: 38
Number of chromosomes in gametes: 19
Number of chromosomes after mitosis: 38
Organism – Pea plant
Number of chromosomes after meiosis: 7
Number of chromosomes in the body cells: 14
Number of chromosomes in gametes: 7
Number of chromosomes after mitosis: 14
Organism – Cow
Number of chromosomes after meiosis: 30
Number of chromosomes in the body cells: 60
Number of chromosomes in gametes: 30
Number of chromosomes after mitosis: 60
Chromosome are made up of genetic material called DNA and it looks like a thread. It is present in the inside of nucleus of a cell.
During meiosis number of chromosomes would be reduced to half whereas in mitosis number of chromosomes remains the same.
The number of chromosomes in body cells is the chromosomal number of an organism and in gametes it would be reduced to half.
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Given the following equation, if I make 18.5G of iron (III) sulphate, what is the percent yield? I picked B but I’m not sure if I’m correct
Answer:
B. The percent yield is 55.9%.
Explanation:
From the balanced reaction we know that 2 moles of FePO4 produce 1 mole of Fe2(SO4)3.
1st) It necessary to convert moles into grams, using the molar mass of FePO4 (151g/mol) and the molar mass of Fe2(SO4)3 (400g/mol):
- FePO4 conversion:
[tex]2moles*\frac{151g}{1mole}=302g[/tex]- Fe2(SO4)3 conversion:
In this case, it is not necessary the conversion, because there is 1 mole, so it is equal to 400g.
Now we know that 302g of FePO4 produce 400g of Fe2(SO4)3.
2nd) Now we have to calculate the Theoretical yield of Iron (III) sulfate, that is, the amount of Iron (III) sulfate that we must produce from 25g of Iron (III) phosphate:
[tex]\begin{gathered} 302gFePO_4-400gFe_2(SO_4)_3 \\ 25gFePO_4-x=\frac{25gFePO_4*400gFe_2(SO_4)_3}{302gFePO_4} \\ x=33.11gFe_2(SO_4)_3 \end{gathered}[/tex]So, the Theoretical yield is 33.11g.
3rd) Finally, we can calculate the Percent yield using the Theoretical yield (33.11g) and the Actual yield (18.5g):
[tex]\begin{gathered} PercentYield=\frac{ActualYield}{TheoreticalYield}*100\% \\ PercentYield=\frac{18.5g}{33.11g}*100\% \\ PercentYield=55.9\% \end{gathered}[/tex]So, the percent yield is 55.9%.
How many electrons can have the quantum numbers n = 3, l = 2, ml = -2 ?
1
2
5
10
There are only two electrons that could posses the quantum numbers as shown.
What are quantum numbers?We now that it is often common to seek to describe the probability of finding the electrons in the atom. While it is know that the electron is not a given point in the atom. The probability of finding the electron in the atom is given by the help of the quantum numbers that can be used to describe the position of the electron as shown.
In tis case, we have an electron that has the quantum numbers; n = 3, l = 2, ml = -2. This describes an orbital and there are two electrons in a given orbital.
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After conditions changed to a volume of a sample of helium at 15.56 mL, 138.7°C and 334.6 kPa. What was its initial volyme at 63.2 °C and 57.3 kPa?O a. 74.2O b. 41.4O c. 2.18O d. 111
Answer
a. 74.2
Explanation
Given that:
The initial temperature, T₁ = 63.2 °C + 273 = 336.2 K
Initial pressure, P₁ = 57.3 kPa
The final volume, V₂ = 15.56 mL
Final temperature, T₂ = 138.7°C + 273 = 411.7 K
Final pressure, P₂ = 334.6 kPa
What to find:
The initial volume, V₁.
Step-by-step solution:
The initial volume, V₁ can be calculated using the combined gas law equation.
[tex]\begin{gathered} \frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2} \\ \\ V_1=\frac{P_2V_2T_1}{P_1T_2}=\frac{334.6kPa\times15.56mL\times336.2K}{57.3kPa\times411.7K} \\ \\ V_1=\frac{1750383.611\text{ }mL}{23590.41}=74.2\text{ }mL \end{gathered}[/tex]Hence, its initial volume at 63.2 °C and 57.3 kPa is 74.2 mL
7. A given sample of gas is held in a container with the volume of 6.02 L with a temperature of 59.5℃ at a pressure of 1.20 atm. What is the final pressure when the sample of gas is administered to a new volume of 10.0 L at 20.2℃?
The final pressure of the gas when the sample gas is administered to a new volume of 10.0 L at 20.2℃ would be 0.245 atm.
Combined gas lawThe problem here has to do with the combined gas law. The law is mathematically expressed as:
[tex]p_1v_1/t_1[/tex] =[tex]p_2v_2/t_2[/tex]
Where [tex]p_1[/tex] = initial pressure, [tex]v_1[/tex] = initial volume, [tex]t_1[/tex] = initial temperature, [tex]p_2[/tex] = final pressure, [tex]v_2[/tex] = final volume, and [tex]t_2[/tex] = final temperature.
In this case,
[tex]p_1[/tex]= 1.20 atm[tex]v_1[/tex] = 6.02 L[tex]t_1[/tex] = 59.5℃[tex]v_2[/tex] = 10.0 L[tex]t_2[/tex] = 20.2℃Rearranging the combined gas equation:
[tex]p_2[/tex] = [tex]p_1v_1t_2/t_1v_2[/tex]
Substituting the different variables:
[tex]p_2[/tex] = 1.2 x 6.02 x 20.2/59.5 x 10.0
= 145.9248/595
= 0.245 atm
Thus, the final pressure of the sample gas will be 0.245 atm.
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According to Newton's Second Law, which of the following is TRUE?
Answer: the second one
Explanation:
Answer:
there's no answer choices?
If a volcano erupts and ejects 12.5 moles of sulfur into the atmosphere how many atoms of sulfur is this
In this question, we have to find the number of atoms in 12.5 moles of Sulfur, and the way to do it is by using the Avogadro's number, which is 6.02*10^23 atoms, this value represents how many atoms we have in 1 single mol of any element. Therefore in 12.5 moles we will have:
1 mol = 6.02*10^23 atoms
12.5 moles = x atoms
x = 7.52*10^24 atoms of Sulfur in 12.5 moles
Predict the shape and bond angles of the following molecules:
H2S
CF4
HCN
NF3
BCl3
NH2Cl
OF2
The shape and bond angle of the molecule will be, H2S =shape= bent, Bond angle =less than 109 degrees
2) CH4 = shape = tetrahedral
bond angle-=109 degrees
1)H2S =shape= bent
Bond angle =less than 109 degrees
2) CH4 = shape = tetrahedral
bond angle-=109 degrees
3) HCN = shape = Linear
Bond angle = 180 degrees
4) NF3 = shape = trigonal planar
Bond angle = less than 109 degrees
5) BCl3 = shape = trigonal planar
Bond angle = 120 degrees
6) NH2Cl = shape = trigonal pyramidal
Bond angle = 107 degrees
7) OF2 = shape = linear
Bond angle = 109 degrees
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Name the functional group in thefollowing molecule:NH2A. aldehydeB. amineC. amideD. thiol
The functional group of the given molecule is located at one of the extremes of it, it is:
This functional group is a carbonyl derived group known as amide. We can recognized because of the carbon that has a double bond with the oxygen and at the same time has a simple bond with nitrogen.
It means that the correct answer is amide.
9. Determine the percent compositions for each of the following compounds. a. Ba(OH)2 b. copper (I) oxide C. Fe(C2H3O2)3 d. iron (III) nitrate hexahydrate (*see notes on backside to help with the hexahydrate part)
Answer:
Iron 24.03%
Carbon 30.90%
Hydrogen 3.86%
Oxygen 41.20 %
Explanation:
Here, we want to get the percentage composition of the given compound
What this means is that we want to get the percentage composition of each of the elements present in the compound
Firstly, we calculate the molar mass of the given compound
We can calculate the molar mass by adding the atomic masses of the elements and their multiplicities
Mathematically, we have that as:
[tex]\begin{gathered} 56\text{ + 3\lparen2\lparen12\rparen + 3\lparen1\rparen +2\lparen16\rparen\rparen} \\ =\text{ 56 + 3\lparen24+3+32\rparen} \\ 56\text{ + 177 = 233 g/mol} \end{gathered}[/tex]Now, let us get the percentage compositions
For Iron, we have:
[tex]\frac{56}{233}\times\text{ 100 \% = 24.03 \%}[/tex]For Carbon, we have:
[tex]\frac{72}{233}\times\text{ 100 \% = 30.90 \%}[/tex]For Hydrogen, we have:
[tex]\frac{9}{233}\times\text{ 100 \% = 3.86 \%}[/tex]For Oxygen, we have:
[tex]\frac{96}{233}\times\text{ 100 \% = 41.20\%}[/tex]What is the total energy required to break all the bonds in 1 mol hexane C6H14?A) 760 kJB) 5780 kJC) 1740 kJD) 7520 kJ
The total energy to break the hexane bonds will be equal to the sum of each of the energy to break each bond.
So the first thing we're going to do is determine how many and what type of bonds hexane has. Hexane is an alkane, meaning the C-C bonds are all single bonds. We also have C-H bonds. According to the structure of hexane, we will count each of the bonds in the following figure:
The number of C-H bonds is written in blue, we see that there are 14 C-H bonds, which coincides, being an alkane, with the number of hydrogen atoms.
The number of C-C bonds is written in orange. The number of C-C bonds is equal to 5.
Now, we multiply the energy to break each bond by the number of bonds and add the energy of the C-H and C-C bonds.
[tex]TotalEnergy=14\times Energy(C-H)+5\times Energy(C-C)[/tex][tex]\begin{gathered} TotalEnergy=14\times413kJ/mol+5\times347kJ/mol \\ TotalEnergy=7517kJ/mol \end{gathered}[/tex]So, for 1 mol of hexane the energy. required to break all the bonds will be 7514kJ
Answer: The closest value to the answer is 7520. D) 7520 kJ
Question 7 of 10What makes up an ionic compound?O A. A positive cation and a negative anionB. A negative anion and a negative anionOC. A positive anion and a negative cationOD. A positive cation and a positive anionаSUBMIT
A ionic compound is made of two ions, one with positive charge and the other one with a negative charge.
The ion with positive charge is known as cation and the one with negative charge is known a anion.
It means that the correct answer is A. A positive cation and a negative anion.
Chloroform, CHCl3, has a vapour pressure of 120 mmHg at a certain temperature. Whatis the vapour pressure of a 0,200 m solution of a non-volatile, non-electrolyte solute inchloroform at the same temperature?
Explanation:
Raoult's law states that the vapor pressure of a solvent that is a over a solution is equal to the vapor pressure of the pure solvent at the same temperature scaled by the mole fraction of the solvent.
Psolution = Xsolvent * P°solvent
Where Psolution is the vapor pressure of the solution, X is the molar fraction of the solvent and P°solvent is the vapor pressure of the pure solvent.
Our solvent is chloroform, we know its vapor pressure.
P°solvent = 120 mmHg
We still have to find the molar fraction of the solvent. The molar fraction of the solvent will be defined like:
Xsolvent = moles of solvent/(total number of moles)
We know that the concentration of the solution is 0.200 m. That means that we have 0.200 moles of solute in each kg of solvent.
In order to find the mole fraction we have to suppose that the solution has 1 kg of solvent.
mass of solvent = 1 kg
molality = 0.200 m
molality = moles of solute/(mass of solvent in kg)
0.200 m = moles of solute/1 kg
moles of solute = 0.200 m * 1 kg
moles of solute = 0.200 moles
So our solution has 0.200 moles of solute in 1 kg of solvent. The solvent is chloroform (CHCl₃). We can convert the mass of solvent into moles using the molar mass of it.
molar mass of CHCl₃ = 119.38 g/mol
mass of solvent = 1 kg = 1 kg * 1000 g/kg
mass of solvent = 1000 g
moles of solvent = 1000 g * 1 moles/(119.38 g)
moles of solvent = 8.377 moles
Our solution has 8.377 moles of solvent and 0.200 moles the non-volatile solute.
total number of moles = moles of solute + moles of solvent
total number of moles = 8.377 moles + 0.200 moles
total number of moles = 8.577 moles
Xsolvent = moles of solvent/(total number of moles)
Xsolvent = 8.377 moles/(8.577 moles)
Xsolvent = 0.977
Finally we found the mole fraction of the solvent and we can find the answer to our problem.
Psolution = Xsolvent * P°solvent
Psolution = 0.977 * 120 mmHg
Psolution = 117.2 mmHg
Answer: The vapor pressure of the solution is 117.2 mmHg.
