Answer:
1.24 kJ is required to convert 14 g of liquid from 43.5°C to 128.2°C
Explanation:
This is a typical calorimetry problem:
We have to assume, no heat is lost to sourrounding.
First of all, we need to go from 43.5°C to 97.4°C, the boiling point.
Q = Ce . m . ΔT
We replace data, 1.18° J/g . 14 g . (97.4°C - 43.5°C)
Heat for the first stage is: 890.4 Joules
Now we have to change the state, and we need the ΔH. As we do not have latent heat, we can proceed like this:
1 mol release 30.1 kJ at vaporization.
We convert the mass to moles → 14 g. 1mol/ 67.44g = 0.207 mol
0.207 mol will release (0.207 . 30.1 kJ) = 6.25 kJ
Now, we are at gaseous phase.
Q = Ce . m . ΔT → 0.792 J/g°C . 14g . (128.2°C - 97.4°C)
Q = 341.5 Joules
To determine the amount of heat, we sum all the obtained values:
890.4 Joules + 6250 Joules + 341.5 Joules = 1238.2 J
We convert to kJ → 1238.2 J . 1kJ / 1000J = 1.24 kJ
The heat required to convert 14.0 g of an unknown liquid at 43.5 °C to gas at 128.2 °C is 7.48 kJ.
We want to calculate the heat required to convert 14.0 g of an unknown liquid at 43.5 °C to gas at 128.2 °C.
We can divide this process in 3 steps.
Heating of the liquid from 43.5 °C to 97.4 °C (normal boiling point).Vaporization of the liquid at 97.4 °C.Heating of the gas from 97.4 °C to 128.2 °C.1. Heating of the liquid from 43.5 °C to 97.4 °CWe will calculate the heat for this step (Q₁) using the following expression.
Q₁ = c(l) × m × ΔT
Q₁ = (1.18 J/g･°C) × 14.0 g × (97.4 °C - 43.5 °C) = 890 J = 0.890 kJ
where,
c(l) is the specific heat capacity of the liquid.m is the mass of the substance.ΔT is the change in the temperature.2. Vaporization of the liquid at 97.4 °C.We will calculate the heat for this step (Q₂) using the following expression.
Q₂ = (m/M) × ΔHvap
Q₂ = [14.0 g/(67.44 g/mol)] × 30.1 kJ/mol = 6.25 kJ
where,
m is the mass of the substance.M is the molar mass of the substance.ΔHvap is the enthalpy of vaporization of the substance.3. Heating of the gas from 97.4 °C to 128.2 °C.We will calculate the heat for this step (Q₃) using the following expression.
Q₃ = c(g) × m × ΔT
Q₃ = (0.792 J/g･°C) × 14.0 g × (128.2 °C - 97.4 °C) = 342 J = 0.342 kJ
where,
c(g) is the specific heat capacity of the gas.m is the mass of the substance.ΔT is the change in the temperature.4. Total amount of heat required (Q)Q = Q₁ + Q₂ + Q₃ = 0.890 kJ + 6.25 kJ + 0.342 kJ = 7.48 kJ
The heat required to convert 14.0 g of an unknown liquid at 43.5 °C to gas at 128.2 °C is 7.48 kJ.
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One mole of a metallic oxide reacts with one mole of hydrogen to produce two moles of the pure metal
and one mole of water. 5.00 g of the metallic oxide produces 2.32 g of the metal. What is the metallic
oxide? (Use molar masses)
Answer:
Formulas
3.2 Determining Empirical and Molecular Formulas
Learning Objectives
By the end of this section, you will be able to:
Compute the percent composition of a compound
Determine the empirical formula of a compound
Determine the molecular formula of a compound
The previous section discussed the relationship between the bulk mass of a substance and the number of atoms or molecules it contains (moles). Given the chemical formula of the substance, one may determine the amount of the substance (moles) from its mass, and vice versa. But what if the chemical formula of a substance is unknown? In this section, these same principles will be applied to derive the chemical formulas of unknown substances from experimental mass measurements.
Percent Composition
The elemental makeup of a compound defines its chemical identity, and chemical formulas are the most succinct way of representing this elemental makeup. When a compound’s formula is unknown, measuring the mass of each of its constituent elements is often the first step in the process of determining the formula experimentally. The results of these measurements permit the calculation of the compound’s percent composition, defined as the percentage by mass of each element in the compound. For example, consider a gaseous compound composed solely of carbon and hydrogen. The percent composition of this compound could be represented as follows:
%H=mass Hmass compound×100%
%C=mass Cmass compound×100%
If analysis of a 10.0-g sample of this gas showed it to contain 2.5 g H and 7.5 g C, the percent composition would be calculated to be 25% H and 75% C:
%H=2.5g H10.0g compound×100%=25%
%C=7.5g C10.0g compound×100%=75%
EXAMPLE 3.9
Calculation of Percent Composition
Analysis of a 12.04-g sample of a liquid compound composed of carbon, hydrogen, and nitrogen showed it to contain 7.34 g C, 1.85 g H, and 2.85 g N. What is the percent composition of this compound?
Solution
To calculate percent composition, divide the experimentally derived mass of each element by the overall mass of the compound, and then convert to a percentage:
%C=7.34g C12.04g compound×100%=61.0%%H=1.85g H12.04g compound×100%=15.4%%N=2.85g N12.04g compound×100%=23.7%
The analysis results indicate that the compound is 61.0% C, 15.4% H, and 23.7% N by mass.
Check Your Learning
A 24.81-g sample of a gaseous compound containing only carbon, oxygen, and chlorine is determined to contain 3.01 g C, 4.00 g O, and 17.81 g Cl. What is this compound’s percent composition?
ANSWER:
12.1% C, 16.1% O, 71.8% Cl
Determining Percent Composition from Molecular or Empirical Formulas
Percent composition is also useful for evaluating the relative abundance of a given element in different compounds of known formulas. As one example, consider the common nitrogen-containing fertilizers ammonia (NH3), ammonium nitrate (NH4NO3), and urea (CH4N2O). The element nitrogen is the active ingredient for agricultural purposes, so the mass percentage of nitrogen in the compound is a practical and economic concern for consumers choosing among these fertilizers. For these sorts of applications, the percent composition of a compound is easily derived from its formula mass and the atomic masses of its constituent elements. A molecule of NH3 contains one N atom weighing 14.01 amu and three H atoms weighing a total of (3 × 1.008 amu) = 3.024 amu. The formula mass of ammonia is therefore (14.01 amu + 3.024 amu) = 17.03 amu, and its percent composition is:
%N=14.01amu N17.03amuNH3×100%=82.27%%H=3.024amu H17.03amuNH3×100%=17.76%
This same approach may be taken considering a pair of molecules, a dozen molecules, or a mole of molecules, etc. The latter amount is most convenient and would simply involve the use of molar masses instead of atomic and formula masses, as demonstrated Example 3.10. As long as the molecular or empirical formula of the compound in question is known, the percent composition may be derived from the atomic or molar masses of the
What is the mass of 4.67 mol of Sulfuric Acid (H2SO4)
458 g H₂SO₄
General Formulas and Concepts:Math
Pre-Algebra
Order of Operations: BPEMDAS
Brackets Parenthesis Exponents Multiplication Division Addition Subtraction Left to RightChemistry
Atomic Structure
Reading a Periodic TableStoichiometry
Using Dimensional AnalysisExplanation:Step 1: Define
4.67 mol H₂SO₄
Step 2: Identify Conversions
[PT] Molar Mass of H - 1.01 g/mol
[PT] Molar Mass of S - 32.07 g/mol
[PT] Molar Mass of O - 16.00 g/mol
Molar Mass of H₂SO₄ - 2(1.01) + 32.07 + 4(16.00) = 98.09 g/mol
Step 3: Convert
Set up: [tex]\displaystyle 4.67 \ mol \ H_2SO_4(\frac{98.09 \ g \ H_2SO_4}{1 \ mol \ H_2SO_4})[/tex]Multiply/Divide: [tex]\displaystyle 458.08 \ g \ H_2SO_4[/tex]Step 4: Check
Follow sig fig rules and round. We are given 3 sig figs.
458.08 g H₂SO₄ ≈ 458 g H₂SO₄
In Purdue's Chemistry department, the chemists have found that in a water based solution containing 1616 grams of certain undissolved chemicals, the rate of change of the amount of chemicals dissolved in the solution is proportional to the amount of the undissolved chemicals. Let Q(t)Q(t) (in grams) be the amount of dissolved chemicals at time tt and let kk be the positive proportionality constant. The differential equation describing the given situation is:
Answer:
The differential equation describing the given situation is dQ/dt = k( 16 - Q )
Explanation:
Given the data in the question;
Initially, the water based solution contains 16 grams of undissolved chemicals;
Assume Q(t) is the amount of dissolved chemical at time L
then the amount of undissolved chemicals at time t is ( 16 - Q)
The rate of change of amount of chemicals dissolved in the solution is proportional to the amount of undissolved chemicals
this means;
dQ/dt ∝ ( 16 - Q)
dQ/dt = k( 16 - Q )
where k is the positive proportionality constant.
Therefore, The differential equation describing the given situation is dQ/dt = k( 16 - Q )
What is the chemical formula of tin(IV) chloride pentahydrate?
Answer:
SnCl4 * 5H2O
Explanation:
An electric circuit has an expected current of 80 amps An electrician measures the current in the circuit at 120 amps. Which
statement provides a possible explanation for this observation?
There is not enough voltage in the circuit.
The circuit has an extra resistor
A resistor in the circuit is broken
There is not enough electrical energy in the circuit.
From where do the placenta and umbilical cord develop?
Answer:
it develops from the womb
Answer:
outer cells of the blastocyst
Explanation:
on edg:)
Pls give a detailed explanation about what are enzyme mutations
Answer:
Enzyme mutations can lead to serious or fatal human disorders and are the consequence of inherited abnormalities in the affected individual's DNA. The mutation may be at a specific position in an enzyme encoded by a mutated gene, just like a single abnormal amino acid residue.
Explanation:
Consider the following unknown solution analysis:
Flame test: red flame Solutions reactions: ammonium carbonate - white precipitate (ppt.) ammonium phosphate - white ppt. Halide test: purple hexane layer The complete chemical name for the unknown compound is______. The correct chemical formula of the compound is_______ .
Answer:
The answer is "Strontium iodide and [tex]SrI_2[/tex]".
Explanation:
In the flames test, two ionic metals give red fire. It was calcium (Ca, which gives a red brick flame) and strontium (Sr, which gives a persistent red flame). Both Ca and Sr respond to insoluble carbonate with ammonium carbonate. Ca and Sr produce insoluble ammonium phosphate phosphates. Even then, the water is reactive with calcium sulfate (CaSO4) whereas the strontium sulfate (SrSO4) is illiquid. Because the metal ion formed 3 precipitates, they can't get Ca, and we have Sr as its metal.
The hexane of its halide gives a violet color which would be typical of iodide (I-).
Strontium iodide is the chemical name of the unknown ionic compound.
Strontium iodide's chemical formula is [tex]SrI_2[/tex].
Hello, could someone help me with this?
Answer:
c. bigger planets have more gravity and smaller planets have less gravity. i hope this helped :))
Calculate the kinetic energy of an electron ejected from a piece of sodium (Φ = 4.41x10–19 J) that is illuminated with 265 nm light.
in Joules
Answer:
Explanation:
Energy of falling radiation having wavelength of 265 nm
= h c / λ where h is plank's constant , c is velocity of light and λ is wavelength of radiation . Putting the values
Energy of light photon = 6.6 x 10⁻³⁴ x 3 x 10⁸ / 265 x 10⁻⁹
= .0747 x 10⁻¹⁷
= 7.47 x 10⁻¹⁹ J .
Work function of sodium is 4.41 x 10⁻¹⁹
So kinetic energy of ejected electron = energy of falling photon - work function
= 7.47 x 10⁻¹⁹ - 4.41 x 10⁻¹⁹
= 3.06 x 10⁻¹⁹ eV .
Predict how many H1 NMR signals (individual resonances, not counting splitting) are expected for the compound.
Answer:
3 H1 NMR signals
Explanation:
NB: kindly check the diagram of the chemical compound in the attached picture.
This particular Question is based on the part of chemistry which is known as spectroscopy. Spectroscopy is used in the Determination or in identifying chemical compounds. H'NMR works on the principle of nuclear magnetic resonance.
In order to solve this question, one has to count the number of hydrogen in unique location. The diagram in the attached show how hydrogen is been counted.
The numbers of signals is the number of different chemical environments in which hydrogen atoms are located.
NB: signals is also the same as peak in H'NMR.
Hence, the number of H1 NMR signals in this chemical compound is 3.
What is the molecular formula of the molecule that has an empirical formula of C2H40 and a molar mass of 176.21 g/mol?
Type your answer using the following format:
CuCl2 for CuCl2.
Answer:
C8H16O4
Explanation:
C2H4O= 24+4+16
44
n=molar mass/empirical formula
n=176.21/44
=4
Therefore
Molar Formula= (C2H4O)4=C8H16O4
What is the density of an object with a mass of 8.7 g and a volume of 8.6 cm??
Answer:
1.01 grams/ cm^3
Explanation:
because that's just how it is
Can anyone help me? Plsss
Which of the following elements has the largest atomic radius?
A. Ba
B.F
C. Ga
D. P
Answer:
A. Ba
Explanation:
Atomic radius increases as you go down and decreases as you go right.
How many grams of water at 0 degree c can be frozen into ice at zero degree c if 55 kj of heat is removed
Answer:
what is that's a question
According to specific heat capacity, 13.09 grams of water at 0 degree Celsius can be frozen into ice at zero degree c if 55 kJ of heat is removed.
What is specific heat capacity?Specific heat capacity is defined as the amount of energy required to raise the temperature of one gram of substance by one degree Celsius. It has units of calories or joules per gram per degree Celsius.
It varies with temperature and is different for each state of matter. Water in the liquid form has the highest specific heat capacity among all common substances .Specific heat capacity of a substance is infinite as it undergoes phase transition ,it is highest for gases and can rise if the gas is allowed to expand.
It is given by the formula ,
Q=mcΔT, substitution in formula gives Q/c=m which is m= 55/4.2=13.09 grams.
Thus, 13.09 grams of water at 0 degree Celsius can be frozen into ice at zero degree c if 55 kJ of heat is removed.
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2. What is the difference between an investigation and a demonstration?
Answer:
An investigation is about collecting information, while an experiment is testing something based off an investigation
Explanation:
Hope this helps
An experiment tests something based on an investigation, whereas an investigation pertains to gathering knowledge.
What are an investigation and a demonstration?The operation of something is demonstrated. There are independent and dependent variables in an experiment. The control of variables distinguishes an experiment from a demonstration. Add an independent as well as a dependent variable to the project in order to convert a demonstration into an experiment.
An information resource's value to a particular professional group and its influence on the procedures and results of healthcare are just a few of the problems that demonstration studies address. An experiment tests something based on an investigation, whereas an investigation pertains to gathering knowledge.
Therefore, an experiment tests something based on an investigation, whereas an investigation pertains to gathering knowledge.
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Is the chemical formula below an Element, Molecule or Compound?
H2SO4
[tex]{ \boxed{ \bold{ \blue{Question}}}}[/tex]
Is H2SO4 an element, compound or molecule?
[tex]{ \boxed{ \bold{ \blue{Answer}}}}[/tex]
H2SO4 (sulphuric acid) is a compound. It is made up of 2 hydrogen atoms, 1 sulphur atom, and 4 oxygen atoms.
_______________________________
[tex] \underline \bold \orange{hope \: it \: helps}[/tex]
A chemical compound is any substance that consists of two or more different elements combined together.
I think that H2SO4 fits this definition of a compound .
HOPE IT HELPS
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Can somebody PLEASE tell me the empirical formula for br2o6
Answer:
BrO3
Explanation:
The empirical formula is the smallest whole-number ratio, you find the greatest common factor (which is 2, in this case), then divide the subscripts by it.
So:
Br2 / 2 = Br1
O6 / 2 = O3
How many molecules of NaOH are there in 22 moles?
Answer:
i hope my answer help u :))
Explanation:
no , it will be 39.99711
Convert a speed of 141 mi/h to units of feet per minute show work.
Answer:
12408 feet per minute
Explanation:
Given: Speed is 141 mi/h
To find: speed in units of feet per minute
Solution:
Use the following units to convert the given speed into feet per minute.
1 mile = 5280 foot
1 h = 60 minutes
Therefore,
141 mi/h [tex]=\frac{141(5280)}{60} =12408[/tex] feet per minute.
WILL MARK BRANLIEST FOR CORRECT ANSWER! Given the following equation, write the expression for its relative rate.
2N2O(g) — 2N2(g) + O2(9)
[tex]\tt -\dfrac{1}{2}\dfrac{d[N_2O]}{dt}=\dfrac{1}{2}\dfrac{d[N_2]}{dt}=\dfrac{1}{1}\dfrac{d[O_2]}{dt}[/tex]
Further explanationReaction
2N2O(g) — 2N2(g) + O2(g)
Required
relative rate
Solution
The reaction rate (v) shows the change in the concentration of the substance (changes in addition to concentrations for reaction products or changes in concentration reduction for reactants) per unit time.
so the relative rates for the reaction above are :
[tex]\tt -\dfrac{1}{2}\dfrac{d[N_2O]}{dt}=\dfrac{1}{2}\dfrac{d[N_2]}{dt}=\dfrac{1}{1}\dfrac{d[O_2]}{dt}[/tex]
It pulls everything down towards earth
Binary compounds are formed by ............... ............... elements.
Answer: i think its A diatomic compound..
Explanation: hope i helped! sorry if im wrong!
how many grams of water produced if we react 3 moles of hydrogen with 3 moles of oxygen
18. What is one of the three things that cause the surface currents of the oceans?
A.differences in salinity
B.temperature differences
C. density differences
D. Coriolis effect
Answer:
b. temperature difference
Asapp please
The percent composition of an unknown substance is 75.42% Carbon, 6.63 % Hydrogen,
8.38 % Nitrogen, and 9.57 % Oxygen. If its molar mass is 334.0 g/mol what is its empirical and
molecular formula?
The empirical and molecular formula : C₂₁H₂₂N₂O₂
Further explanationGiven
The percent composition of an unknown substance is 75.42% Carbon, 6.63 % Hydrogen, 8.38 % Nitrogen, and 9.57 % Oxygen
Required
The empirical and molecular formula
Solution
C : 75.42 : 12 = 6.285
H : 6.63 : 1 = 6.63
N : 8.38 : 14 = 0.599
O : 9.57 : 16 = 0.598
Divide by 0.598
C : H : N : O = 10.5 : 11 : 1 : 1 = 21 : 22 : 2 : 2
The empirical formula : C₂₁H₂₂N₂O₂
(C₂₁H₂₂N₂O₂)n = 334
(334)n=334
n = 1
The simplest chemical substance are composed of only what types of atom
Here, we are required to determine what types of atoms are present in the simplest chemical substance.
The simplest chemical substances are composed of only the atoms of one kind of Element.Chemical substances are generally classified as either;
ElementsCompoundsMixtures.Definition of terms;
Elements and Compounds are collectively termed Pure Substances.
Elements: An element is a chemical substance that is composed of a specific kind of atom and cannot be broken down or transformed by means of a chemical reaction.Elements: An element is a chemical substance that is composed of a specific kind of atom and cannot be broken down or transformed by means of a chemical reaction.Compound: A compound is a chemical substance formed by the combination of two or more elements or group of elements by means of a chemical reaction.Elements: An element is a chemical substance that is composed of a specific kind of atom and cannot be broken down or transformed by means of a chemical reaction.Compound: A compound is a chemical substance formed by the combination of two or more elements or group of elements by means of a chemical reaction.Mixture: A mixture put simply is a chemical substance that consists of diverse, nonbonded elements, molecules or compounds as the case may be.Therefore, inference drawn from the premises above point to the fact that the simplest chemical substances are composed of the atoms of one type of element.
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What is the energy of an electron in the third energy level of hydrogen?
Answer:
Electrons in a hydrogen atom must be in one of the allowed energy levels. If an electron is in the first energy level, it must have exactly -13.6 eV of energy.
...
Energy Levels of Electrons.
Energy Level Energy
1 -13.6 eV
2 -3.4 eV
3 -1.51 eV
4 -.85 eV
Open the Molecule Shapes interactive and select Real Molecules. Check the box to Show Bond Angles. Classify each molecule by whether its real bond angles are the same as or different than its model (ideal) bond angles. In other words, do the bond angles change when you switch between Real and Model mode at the top of the page? Same (angles do not change) Different (angles change) Answer Bank H0 00, 50, XP, BF, CIF, NH, CH, SP, XF, BF, PCI, SF,
Answer:
Same bond angle-CO2, CH4, SF6, PCl5, BF3, BeF2, XeF4,XeF2
Different bond angle- H2O, SO2, NH3, ClF3,SF4
Explanation:
We know that the shape of molecules and and bond angles between atoms in molecules are predicted on the basis of the valence shell electron pair repulsion theory.
Most times, certain molecules deviate from the expected bond angles for different reasons. The most common reason is the presence of lone pairs on the molecule. This is the case in the molecules; H2O, SO2, NH3, ClF3 and SF4.
However, in XeF4 and XeF2, the lone pairs orient themselves in such a way that they do not distort the expected bond angle. For instance, in XeF2, the three lone pairs occupy equatorial positions while the two bond pairs occupy apical positions. In XeF4, the bond pairs are directed at the corners of a square while the two lone pairs are positioned above and below the plane of the square.