A certain liquid has a normal freezing point of and a freezing point depression constant . Calculate the freezing point of a solution made of of iron(III) chloride () dissolved in of . Round your answer to significant digits.
The question is incomplete, the complete question is:
A certain liquid X has a normal freezing point of [tex]0.80^oC[/tex] and a freezing point depression constant [tex]K_f=7.82^oC.kg/mol[/tex] . Calculate the freezing point of a solution made of 81.1 g of iron(III) chloride () dissolved in 850. g of X. Round your answer to significant digits.
Answer: The freezing point of the solution is [tex]-17.6^oC[/tex]
Explanation:
Depression in the freezing point is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution.
The expression for the calculation of depression in freezing point is:
[tex]\text{Freezing point of pure solvent}-\text{freezing point of solution}=i\times K_f\times m[/tex]
OR
[tex]\text{Freezing point of pure solvent}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times w_{solvent}\text{(in g)}}[/tex] ......(1)
where,
Freezing point of pure solvent = [tex]0.80^oC[/tex]
Freezing point of solution = [tex]?^oC[/tex]
i = Vant Hoff factor = 4 (for iron (III) chloride as 4 ions are produced in the reaction)
[tex]K_f[/tex] = freezing point depression constant = [tex]7.82^oC/m[/tex]
[tex]m_{solute}[/tex] = Given mass of solute (iron (III) chloride) = 81.1 g
[tex]M_{solute}[/tex] = Molar mass of solute (iron (III) chloride) = 162.2 g/mol
[tex]w_{solvent}[/tex] = Mass of solvent (X) = 850. g
Putting values in equation 1, we get:
[tex]0.8-(\text{Freezing point of solution})=4\times 7.82\times \frac{81.1\times 1000}{162.2\times 850}\\\\\text{Freezing point of solution}=[0.8-18.4]^oC\\\\\text{Freezing point of solution}=-17.6^oC[/tex]
Hence, the freezing point of the solution is [tex]-17.6^oC[/tex]
An atom that ______ electrons is called a positive ion. A. has 0 B. has 8 C. loses D. gains
Answer:
Gains
Explanation:
It gets more electrons
which of the following molecules would you expect to have a dipole moment of zero? a,CH2 Ch3
bH2C=0
cCH2cl
dNH3
Answer: The molecule [tex]CH_{3}-CH_{3}[/tex] is expected to have a dipole moment of zero.
Explanation:
The product of magnitude of the charge calculated in electrostatic units is called dipole moment.
Formula for dipole moment is as follows.
Dipole moment = Charge (in esu) [tex]\times[/tex] distance (in cm)
Non-polar molecules have zero dipole moment.
For example, [tex]CH_{3}-CH_{3}[/tex] is a non-polar molecule so its dipole moment is zero.
[tex]H_{2}C=O[/tex] is a polar molecule so it will have dipole moment.
[tex]CH_{2}Cl_{2}[/tex] is a polar molecule so it will have dipole moment.
[tex]NH_{3}[/tex] has nitrogen atom as more electronegative than hydrogen atom. So, net dipole moment will be in the direction of nitrogen atom.
Thus, we can conclude that the molecule [tex]CH_{3}-CH_{3}[/tex] is expected to have a dipole moment of zero.
Select all the correct answers
When two generalizations can be made based on what you know about cycles of matter in a closed system?
New matter is added, and old matter is destroyed.
Matter changes its physical form, allowing it to return to its original state.
The amount of matter within the system remains the same
Matter and energy can cross the boundaries of the system.
The cycle has a well-defined starting and Stopping point
Answer:
A
Explanation:
If the electromagnet in the PhET simulation is disconnected from the battery, the compass needle will
A. Not move
B. Flip directions
C. Point north
D. Point south
What is the concentration of a solution in which 15 grams of sugar is dissolved in 0.2 L of water?
Answer:
0.2 M
Explanation:
Step 1: Given data
Mass of sugar (sucrose): 15 gVolume of water: 0.2 L (we will assume it is the volume of the solution)There are different ways to express the concentration of a solution. We will calculate molarity, which is one of the most used.
Step 2: Calculate the moles of sucrose
The molar mass of sucrose is 342.3 g/mol.
15 g × 1 mol/342.3 g = 0.044 mol
Step 3: Calculate the molarity of the solution
Molarity is equal to the moles of solute divided by the liters of solution.
M = 0.044 mol/0.2 L = 0.2 M
f. . A metal cylinder has a mass of 100.00 g is heated to 95.50 celcius and then put in 245.5 g of water whose initial temperature is 22.50 Celsius. The final temperature of the mixture is 24.17 Celsius what is the specific heat of the metal.
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A sample of gas contains 0.1800 mol of CO(g) and 0.1800 mol of NO(g) and occupies a volume of 23.2 L. The following reaction takes place:
2CO(g) + 2NO(g 2Co2(g) +N2(g)
Calculate the volume of the sample after the reaction takes place, assuming that the temperature and the pressure remain constant.
Answer:
The volume of the sample is 17.4L
Explanation:
The reaction that occurs requires the same amount of CO and NO. As the moles added of both reactants are the same you don't have any limiting reactant. The only thing we need is the reaction where 4 moles of gases (2mol CO + 2mol NO) produce 3 moles of gases (2mol CO2 + 1mol N2). The moles produced are:
0.1800mol + 0.1800mol reactants =
0.3600mol reactant * (3mol products / 4mol reactants) = 0.2700 moles products.
Using Avogadro's law (States the moles of a gas are directly proportional to its pressure under constant temperature and pressure) we can find the volume of the products:
V1n2 = V2n1
Where V is volume and n moles of 1, initial state and 2, final state of the gas
Replacing:
V1 = 23.2L
n2 = 0.2700 moles
V2 = ??
n1 = 0.3600 moles
23.2L*0.2700mol = V2*0.3600moles
17.4L = V2
The volume of the sample is 17.4LThe specific rate constant, k, for radioactive beryllium–11 is 0.049 s–1. What mass of a 0.500 mg sample of beryllium–11 remains after 28 seconds? This reaction was found to be first order.
Answer: The mass of sample that remained is 0.127 mg
Explanation:
The integrated rate law equation for first-order kinetics:
[tex]k=\frac{2.303}{t}\log \frac{a}{a-x}[/tex] ......(1)
Given values:
a = initial concentration of reactant = 0.500 mg
a - x = concentration of reactant left after time 't' = ?mg
t = time period = 28 s
k = rate constant = [tex]0.049s^{-1}[/tex]
Putting values in equation 1:
[tex]0.049s^{-1}=\frac{2.303}{28s}\log (\frac{0.500}{(a-x)})\\\\\log (\frac{0.500}{(a-x)})=\frac{0.049\times 28}{2.303}\\\\\frac{0.500}{a-x}=10^{0.5957}\\\\frac{0.500}{a-x}=3.94\\\\a-x=\frac{0.500}{3.942}=0.127mg[/tex]
Hence, the mass of sample that remained is 0.127 mg
2.67 Determine the density (g/mL) for each of the following:
a. A 20.0-mL sample of a salt solution has a mass of 24.0 g.
The density (g/mL) for a 20.0-mL sample of a salt solution has a mass of 24.0 g is 1.2 g/ml.
What is density?Density is the mass per unit volume. Density is a scalar quantity. It is denoted by d and the symbol for density is given as rho, a Greek symbol. Density is calculated as mass divided by volume.
Mass is the quantity of matter in a physical body. The product of the molar mass of the compound and the moles of the substance are defined as mass.
Volume is the space occupied by a three-dimensional object. Volume is calculated by dividing mass by density.
Given, the 20.0-mL sample of a salt solution, which is the volume.
The mass of the solution is 24.0 g
To calculate the density
mass/volume
24.0 / 20.0 = 1.2 g/ml
Thus, the density of the given salt solution is 1.2 g/ml.
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Calculate the mass of water produced when 9.57 g of butane reacts with excess oxygen.
Express your answer to three significant figures and include the appropriate units.
Answer:
14.9 g
Explanation:
Step 1: Write the balanced equation
C₄H₁₀ + 6.5 O₂ ⇒ 4 CO₂ + 5 H₂O
Step 2: Calculate the moles corresponding to 9.57 g of C₄H₁₀
The molar mass of C₄H₁₀ is 58.12 g/mol.
9.57 g × 1 mol/58.12 g = 0.165 mol
Step 3: Calculate the moles of H₂O produced from 0.165 moles of C₄H₁₀
0.165 mol C₄H₁₀ × 5 mol H₂O/1 mol C₄H₁₀ = 0.825 mol H₂O
Step 4: Calculate the mass corresponding to 0.825 mol of H₂O
The molar mass of H₂O is 18.02 g/mol.
0.825 mol × 18.02 g/mol = 14.9 g
Hydrogen gas can be prepared in the laboratory by a sin- gle-displacement reaction in which solid zinc reacts with hydrochloric acid. How much zinc in grams is required to make 14.5 g of hydrogen gas through this reaction
Answer:
941 g
Explanation:
Step 1: Write the balanced equation
Zn + 2 HCl ⇒ ZnCl₂ + H₂
Step 2: Calculate the moles corresponding to 14.5 g of H₂
The molar mass of H₂ is 1.01 g/mol.
14.5 g × 1 mol/1.01 g = 14.4 mol
Step 3: Calculate the number of moles of Zn required to form 14.4 moles of H₂
The molar ratio of Zn to H₂ is 1:1. The moles of Zn required are 1/1 × 14.4 mol = 14.4 mol.
Step 4: Calculate the mass corresponding to 14.4 moles of Zn
The molar mass of Zn is 65.38 g/mol.
14.4 mol × 65.38 g/mol = 941 g
how many atoms of one formula unit in Fe2O3
Answer:
5 atoms form one formula unit of Fe2O3
Explanation:
2 atoms of Fe (Iron)
and 3 atoms of O ( Oxygen)
so total = 3 + 2
= 5
Answer:
fe203 the right answer is thus
what is the chemical fomula for water
Answer:
H2O.....................
Suppose you need to prepare 21.0 mL of formate buffer with a ratio of 4 of [sodium formate]/[formic acid] by mixing 0.10 M formic acid and 0.10 M sodium formate. How many milliliters of sodium formate do you need to measure to make this buffer (assuming the rest is formic acid)
Answer: A volume of 20.49 milliliters of sodium formate do you need to measure to make this buffer (assuming the rest is formic acid).
Explanation:
Given: Total volume of the buffer = 21.0 mL
[tex]\frac{[HCOONa]}{[HCOOH]} = 4[/tex] ... (1)
It is assumed that the volume of HCOONa is x. Hence, volume of HCOOH is (21.0 - x) mL.
Hence,
[HCOONa] = Molarity [tex]\times[/tex] Volume
= 0.10 [tex]\times[/tex] x
= 0.1x mmol
Similarly, [HCOOH] = Molarity [tex]\times[/tex] Volume
= 0.10 [tex]\times[/tex] (21.0 - x) mmol
Using equation (1),
[tex]\frac{[HCOONa]}{[HCOOH]} = 4\\\frac{0.1x}{(21.0 - x)} = 4\\0.1x = 84.0 - 4x\\4.1x = 84.0\\x = 20.49 mL[/tex]
As x is the volume of sodium formate. Hence, 20.49 mL of sodium formate is required to make the buffer.
Thus, we can conclude that a volume of 20.49 milliliters of sodium formate do you need to measure to make this buffer (assuming the rest is formic acid).
The combustion of ethylene proceeds by the reaction
C2H4 (g) + 3O2 (g) → 2CO2 (g) + 2H2O (g)
When the rate of disappearance of C2H4 is 0.13 M s-1, the rate of appearance of CO2 is ________ M s-1.
Which of the following choices is a source of groundwater pollution?
O sewage
very warm water
O silt
O All of these choices are correct.
Answer:
very warm water consuving
why ionic compound are good conduct of electricity in their molten state ??
Answer:
Ionic compounds conduct electricity when molten (liquid) or in aqueous solution (dissolved in water), because their ions are free to move from place to place. Ionic compounds cannot conduct electricity when solid, as their ions are held in fixed positions and cannot move.
Explanation:
because their ions are free to move from place to place.
3. Calculate the answers to the appropriate number of significant figures. e) 43.678 x 64.1 = f) 1.678/0.42 =
The pH of a certain orange juice is 3.33.Calculate the +ion concentration.
Answer:
[tex]4.67\times 10^{-4}[/tex]
Explanation:
Given that,
The pH of a certain orange juice is 3.33.
We need to find the +ion concentration.
We know that,
[tex]pH=-log[H^+][/tex]
So,
[tex]3.33=-log[H^+]\\\\\[H^+=10^{-3.33}\\\\=4.67\times 10^{-4}[/tex]
So, the +ion concentraion is equal to [tex]4.67\times 10^{-4}[/tex].
Name the following compound: Cuzs
O sulfur copperide (ll)
O sulfur copperide (1)
O copper(I) sulfide
copper(ll) sulfide
Answer:
THE ANSWER IS: copper(I) sulfide.
hope this helped <3
Explanation:
How many moles are present in a sample if it consists of 5.61x1022 particles? Report your answer to 3 decimal places. Do not include units.
Answer:
The mole is defined as a collection of 6.022 × 1023 particles.
The atomic mass given on a periodic table that is given in grams is the mass of
one mole (6.022 × 1023 particles) of that element
Explanation:
In a titration to find the concentration of 30ml of a H2SO4 solution, a student found that 40ml of 0.2M KOH solution was needed to reach the endpoint. What's the concentration of the H2SO4?
Question 21 options:
A) 0.27M
B) 0.53M
C) 0.4M
D) 1.1M
Answer:
it's B
Explanation:
Write the balanced equation: H2SO4 + 2KOH → K2SO4 +2H2O. So 2(moles KOH) = (moles H2SO4); 2(volume KOH)(concentration KOH) = (volume H2SO4)(concentration H2SO4); 2(40ml)(0.2M) = (30ml)(x); x = 0.53M
The concentration of H₂SO₄ solution is equal to 0.133 M.
What is a neutralization reaction?A neutralization reaction can be described as a chemical reaction in which an acid and base react together to form respective salt and water. When a strong acid such as HCl will react with a strong base such as NaOH the salt can be neither acidic nor basic.
When H₂SO₄ (a strong acid) reacts with KOH, the resulting salt will be K₂SO₄ and water.
H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O
Given, the concentration of KOH solution = 0.2 M
The volume of the KOH solution = 40 ml = 0.040 ml
The number of moles of KOH, n = M × V = 0.2 × 0.04 = 0.008 mol
The volume of the H₂SO₄ = 30 ml = 0.03 L
The number of moles of H₂SO₄, n = 0.008/2 = 0.004 mol
The concentration of H₂SO₄ solution = 0.004/0.03 = 0.133 M
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Identify the indicated protons in the following molecules as unrelated, homotopic, enantiotopic, or diastereotopic. a) Methyls a & b: _________ b) Ha & Hc: _________
Answer:
Identify the indicated protons in the following molecules as unrelated, homotopic, enantiotopic, or diastereotopic. a) Methyls a & b: _________ b) Ha & Hc: ________
Explanation:
Homotopic hydrogens:
Consider two hydrogens in the given molecule and replace one by one with a different atom say for example deuterium, then if the two molecules formed by replacing hydrogens are the same then the two hydrogens are called homotopic hydrogens.
After replacing the two hydrogens with a different atom then, enantiomers are formed then, the two hydrogens are called enantiotopic hydrogens.
After replacing the two hydrogens with a different atom then, diastereomers are formed then, the two hydrogens are called diastereotopic hydrogens.
In the methyl group, select two hydrogens and replace one hydrogen atom with a D-atom name the compound.
Again replace another hydrogen atom with D-atom.
Name it.
If both are the same then, the hydrogens are homotopic and they are shown below:
Hence, they are homotopic protons.
Epinephrine (adrenaline) is a hormone secreted into the bloodstream in times of danger and stress. It is 59.0% carbon, 7.15% hydrogen, 26.20% oxygen, and 7.65% nitrogen by mass and has a molar mass of 183 g/mol. Determine the empirical formula for Epinephrine.
Answer: The empirical and molecular formula for the given organic compound is [tex]C_9H_{13}O_3N[/tex]
Explanation:
Let the mass of the compound be 100 g
Given values:
% of C = 59.0%
% of H = 7.15%
% of O = 26.20%
% of N = 7.65%
Mass of C = 59.0 g
Mass of H = 7.15 g
Mass of O = 26.20 g
Mass of N = 7.65 g
The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)
To formulate the empirical formula, we need to follow some steps:
Step 1: Converting the given masses into moles.Molar mass of C = 12 g/mol
Molar mass of H = 1 g/mol
Molar mass of O = 16 g/mol
Molar mass of N = 14 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of C}=\frac{59.0g}{12g/mol}=4.917 mol[/tex]
[tex]\text{Moles of H}=\frac{7.15g}{1g/mol}=7.15 mol[/tex]
[tex]\text{Moles of O}=\frac{26.20g}{16g/mol}=1.6375 mol[/tex]
[tex]\text{Moles of N}=\frac{7.65g}{14g/mol}=0.546 mol[/tex]
Step 2: Calculating the mole ratio of the given elements.Calculating the mole fraction of each element by dividing the calculated moles by the least calculated number of moles that is 0.546 moles
[tex]\text{Mole fraction of C}=\frac{4.917}{0.546 }=9[/tex]
[tex]\text{Mole fraction of H}=\frac{7.15}{0.546 }=13[/tex]
[tex]\text{Mole fraction of O}=\frac{1.6375}{0.546 }=2.99\approx 3[/tex]
[tex]\text{Mole fraction of N}=\frac{0.546}{0.546 }=1[/tex]
Step 3: Taking the mole ratio as their subscripts.The ratio of C : H : O : N = 9 : 13 : 3 : 1
The empirical formula of the compound becomes [tex]C_9H_{13}O_3N_1=C_9H_{13}O_3N[/tex]
To calculate the molecular formula, the number of atoms of the empirical formula is multiplied by a factor known as valency that is represented by the symbol, 'n'.
[tex]n =\frac{\text{Molecular mass}}{\text{Empirical mass}}[/tex] .....(2)
We are given:
Mass of molecular formula = 183 g/mol
Mass of empirical formula = 183 g/mol
Putting values in equation 2, we get:
[tex]n=\frac{183g/mol}{183g/mol}=1[/tex]
Multiplying this valency by the subscript of every element of empirical formula, we get:
[tex]C_{1\times 9}H_{1\times 13}O_{1\times 3}N_{1\times 1}=C_9H_{13}O_3N[/tex]
Hence, the empirical and molecular formula for the given organic compound is [tex]C_9H_{13}O_3N[/tex]
Write a formula for the ionic compound that forms from magnesium
and oxygen.
Answer:
MgO
Explanation:
how has society influenced our opinions on lithium mining
Answer:
LIBs have had a huge impact on our society. They enabled modern portable electronics such as laptops and mobile phones. And they are now enabling clean and low-carbon transport, be it via electric cars or even flying taxis, and grid-scale storage of renewable energy
Explanation:
One of the most common causes of inaccurate melting point ranges is rapid heating of the compound. Under these circumstances, how will the observed MP range compare to the true MP range
Answer:
INCREASE in the difference between the melting point measured and the true melting temperature.
Explanation:
Melting point of a compound is defined as the temperature at which the soils compound changes into liquid at the atmospheric pressure. There are different circumstances that can lead to inaccurate melting point. These include:
--> presence of impurities in the compound,
--> Molecular composition,
--> Force of attraction, and
--> Rapid heating of the compound.
Under the circumstances of rapid heating of the compound, there would be an increase in the melting point range when compared with the true melting point range of the compound.
The higher the heating rate, the more rapid the rise in oven temperature, increasing the difference between the melting point measured and the true melting temperature.
Label each formula and name pair as correct or incorrect.
Formula Name Correct/Incorrect
Aluminum tribromide
Sulfur dioxide
Beryllium hydride
Magnesium(II) oxide
Copper(II) oxide
Calcium sulfate
Nitric acid
Answer:
Aluminum tribromide: AlBr₃, however, it should be just aluminum bromide.
Sulfur dioxide: SO₂.
Beryllium hydride: BeH₂
Magnesium(II) oxide: MgO; however the roman numeral is not used in Mg as it just has one oxidation number.
Copper(II) oxide: CuO.
Calcium sulfate: CaSO₄
Nitric acid: HNO₃.
Explanation:
Hello there!
In this case, it seems that the formulas were not given, however, we can write the correct one for each given compound according to the widely used nomenclature rules as shown below:
Aluminum tribromide: AlBr₃, however, it should be just aluminum bromide.
Sulfur dioxide: SO₂.
Beryllium hydride: BeH₂
Magnesium(II) oxide: MgO; however the roman numeral is not used in Mg as it just has one oxidation number.
Copper(II) oxide: CuO.
Calcium sulfate: CaSO₄
Nitric acid: HNO₃.
Regards!
Please help me name these organic compounds
Answer:
Aldehydes and Ketones
Both aldehydes and ketones contain a carbonyl group, a functional group with a carbon-oxygen double bond. The names for aldehyde and ketone compounds are derived using similar nomenclature rules as for alkanes and alcohols, and include the class-identifying suffixes -al and -one, respectively:

In an aldehyde, the carbonyl group is bonded to at least one hydrogen atom. In a ketone, the carbonyl group is bonded to two carbon atoms:


As text, an aldehyde group is represented as –CHO; a ketone is represented as –C(O)– or –CO–.
In both aldehydes and ketones, the geometry around the carbon atom in the carbonyl group is trigonal planar; the carbon atom exhibits sp2 hybridization. Two of the sp2 orbitals on the carbon atom in the carbonyl group are used to form σ bonds to the other carbon or hydrogen atoms in a molecule. The remaining sp2 hybrid orbital forms a σ bond to the oxygen atom. The unhybridized p orbital on the carbon atom in the carbonyl group overlaps a p orbital on the oxygen atom to form the π bond in the double bond.
Like the C=OC=O bond in carbon dioxide, the C=OC=O bond of a carbonyl group is polar (recall that oxygen is significantly more electronegative than carbon, and the shared electrons are pulled toward the oxygen atom and away from the carbon atom). Many of the reactions of aldehydes and ketones start with the reaction between a Lewis base and the carbon atom at the positive end of the polar C=OC=O bond to yield an unstable intermediate that subsequently undergoes one or more structural rearrangements to form the final product (Figure 1).
Figure 1. The carbonyl group is polar, and the geometry of the bonds around the central carbon is trigonal planar.
The importance of molecular structure in the reactivity of organic compounds is illustrated by the reactions that produce aldehydes and ketones. We can prepare a carbonyl group by oxidation of an alcohol—for organic molecules, oxidation of a carbon atom is said to occur when a carbon-hydrogen bond is replaced by a carbon-oxygen bond. The reverse reaction—replacing a carbon-oxygen bond by a carbon-hydrogen bond—is a reduction of that carbon atom. Recall that oxygen is generally assigned a –2 oxidation number unless it is elemental or attached to a fluorine. Hydrogen is generally assigned an oxidation number of +1 unless it is attached to a metal. Since carbon does not have a specific rule, its oxidation number is determined algebraically by factoring the atoms it is attached to and the overall charge of the molecule or ion. In general, a carbon atom attached to an oxygen atom will have a more positive oxidation number and a carbon atom attached to a hydrogen atom will have a more negative oxidation number. This should fit nicely with your understanding of the polarity of C–O and C–H bonds. The other reagents and possible products of these reactions are beyond the scope of this chapter, so we will focus only on the changes to the carbon atoms:
