ILL GIVE YOU BRAINLIST !!
Which are the quantitative data in the example above? Explain.
Answer:
helicona
throated carib
eulampis jugularis
Answer:
The percentages
Explanation:
Quantitative data are numbers, while qualitative data are characteristics. So I think that the numbers are your answer.
Hope that helps!
Gold (Au) has a density of 19.3 g/ml. A gold filling falls out of your teeth and you bring it to chemistry class. When you drop it into a graduated cylinder the water rises from 9.27 ml to 12.03 ml.
What is the mass of your gold filling?
Answer:
53.27 g.
Explanation:
From the question given above, the following data were obtained:
Density of gold = 19.3 g/mL
Volume of water = 9.27 mL
Volume of water + gold = 12.03 mL
Mass of gold =.?
Next, we shall determine the volume of the gold filling. This can be obtained as follow:
Volume of water = 9.27 mL
Volume of water + gold = 12.03 mL
Volume of gold =?
Volume of gold = (Volume of water + gold) – (Volume of water)
Volume of gold = 12.03 – 9.27
Volume of gold = 2.76 mL
Finally, we shall determine the mass of the gold filling as follow:
Density of gold = 19.3 g/mL
Volume of gold = 2.76 mL
Mass of gold =.?
Density = mass/volume
19.3 = mass of gold /2.76
Cross multiply
Mass of gold = 19.3 × 2.76
Mass of gold = 53.27 g
The, the mass of the gold filling is 53.27 g.
Which solution below would have the greatest buffering capacity? The solution was prepared from a weak acid and the salt of its conjugate base. 0.160 M C6H5OHCOOH and 0.160 M C6H5OHCOONa, Ka = 1.05e-3 0.0892 M H2NCH2COOH and 0.0892 M H2NCH2COOK, Ka = 4.50e-3 0.0725 M H2NCH2COOH and 0.0725 M H2NCH2COONa, Ka = 4.50e-3 0.1360 M C6H5OHCOOH and 0.1360 M C6H5OHCOOK, Ka = 1.05e-3
Answer:
0.1360 M C₆H₅OHCOOH and 0.1360 M C₆H₅OHCOOK,
Ka = 1.05x 10⁻³ .
Explanation:
The first mixture
0.160 M C₆H₅OHCOOH and 0.160 M C₆H₅OHCOONa,
Ka = 1.05x 10⁻³ .
The second mixture
0.0892 M H₂NCH₂COOH and 0.0892 M H₂NCH₂COOK,
Ka = 4.5 x 10⁻³ .
The third mixture
.0725 M H₂NCH₂COOH and 0.0725 M H₂NCH₂COONa,
Ka = 4.5 x 10⁻³
fourth mixture
0.1360 M C₆H₅OHCOOH and 0.1360 M C₆H₅OHCOOK,
Ka = 1.05x 10⁻³ .
In all the mixtures the ratio of acid and its salt are same and equal to one so this ratio will not determine their relative buffering capacity .
Now we know that weak acid has more buffering capacity so mixture having acid of less Ka will have more buffering capacity .
Ka is less if
Ka = 1.05 x 10⁻³ .
Dilute acids have greater buffering capacity
So ultimate answer is
0.1360 M C₆H₅OHCOOH and 0.1360 M C₆H₅OHCOOK,
Ka = 1.05x 10⁻³ .
One possible source of error in this experiment is not completely drying the NaCl. Effect of Percent yield ? And other questions
Answer:
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12.9 Theoretical Yield and Percent Yield
FlexBooks® 2.0 > CK-12 Chemistry For High School > Theoretical Yield and Percent Yield
It is best to have high yields for chemical reactions
Can we save some money?
The world of pharmaceutical production is an expensive one. Many drugs have several steps in their synthesis and use costly chemicals. A great deal of research takes place to develop better ways to make drugs faster and more efficiently. Studying how much of a compound is produced in any given reaction is an important part of cost control.
Percent Yield
Chemical reactions in the real world don’t always go exactly as planned on paper. In the course of an experiment, many things will contribute to the formation of less product than would be predicted. Besides spills and other experimental errors, there are usually losses due to an incomplete reaction, undesirable side reactions, etc. Chemists need a measurement that indicates how successful a reaction has been. This measurement is called the percent yield.
To compute the percent yield, it is first necessary to determine how much of the product should be formed based on stoichiometry. This is called the theoretical yield, the maximum amount of product that could be formed from the given amounts of reactants. The actual yield is the amount of product that is actually formed when the reaction is carried out in the laboratory. The percent yield is the ratio of the actual yield to the theoretical yield, expressed as a percentage.
Percent yield is very important in the manufacture of products. Much time and money is spent improving the percent yield for chemical production. When complex chemicals are synthesized by many different reactions, one step with a low percent yield can quickly cause a large waste of reactants and unnecessary expense.
Typically, percent yields are understandably less than 100% because of the reasons indicated earlier. However, percent yields greater than 100% are possible if the measured product of the reaction contains impurities that cause its mass to be greater than it actually would be if the product was pure. When a chemist synthesizes a desired chemical, he or she is always careful to purify the products of the reaction.
Sample Problem: Calculating the Theoretical Yield and the Percent Yield
Potassium chlorate decomposes upon slight heating in the presence of a catalyst according to the reaction below:
In a certain experiment, 40.0 g KClO3 is heated until it completely decomposes. What is the theoretical yield of oxygen gas? The experiment is performed and the oxygen gas is collected and its mass is found to be 14.9 g. What is the percent yield for the reaction?
First, we will calculate the theoretical yield based on the stoichiometry.
Step 1: List the known quantities and plan the problem.
Known
given: mass of KClO3 = 40.0 g
molar mass KClO3 = 122.55 g/mol
molar mass O2 = 32.00 g/mol
Unknown
theoretical yield O2 = ? g
Apply stoichiometry to convert from the mass of a reactant to the mass of a product:
Step 2: Solve.
The theoretical yield of O2 is 15.7 g.
Step 3: Think about your result.
The mass of oxygen gas must be less than the 40.0 g of potassium chlorate that was decomposed.
Now, we use the actual yield and the theoretical yield to calculate the percent yield.
Step 1: List the known quantities and plan the problem.
Known
Actual yield = 14.9 g
Theoretical yield = 15.7 g (from Part 12.11A)
Unknown
Percent yield = ? %
Use the percent yield equation above.
Step 2: Solve.
Step 3: Think about your result.
Since the actual yield is slightly less than the theoretical yield, the percent yield is just under 100%.
Summary
Theoretical yield is calculated based on the stoichiometry of the chemical equation.
The actual yield is experimentally determined.
The percent yield is determined by calculating the ratio of actual yield/theoretical yield.
Review
What do we need in order to calculate theoretical yield?
If I spill some of the product before I weigh it, how will that affect the actual yield?
How will spilling some of the product affect the percent yield?
I make a product and weigh it before it is dry. How will that affect the actual yield?
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Explanation:
A glass of water weighs 40g and on adding ice to it weighs 250g. How much ice is added to glass.
Answer:
[tex]\huge\boxed{\sf Ice\ Added = 210\ g}[/tex]
Explanation:
Weight of glass of water before adding ice = 40 g
Weight of glass of water after adding ice = 250 g
Ice Added = 250 g - 40 g
Ice Added = 210 g
What is the Temperature of the suns most middle layer the chromosphere
The temperature in the chromosphere varies between about 4000 K at the bottom (the so-called temperature minimum) and 8000 K at the top (6700 and 14,000 degrees F, 3700 and 7700 degrees C), so in this layer (and higher layers) it actually gets hotter if you go further away from the Sun
Calculate the mole fraction of nitric acid of a(n) 15.7% (by mass) aqueous solution of nitric acid. Calculate the mole fraction of nitric acid of a(n) 15.7% (by mass) aqueous solution of nitric acid. 2.56×10−2 0.102 5.33×10−2 5.11×10−2 The density of the solution is needed to solve the problem.
Answer:
0.0505
Explanation:
Step 1: Calculate the masses of solute and solvent
We have a 15.7% by mass nitric acid solution, that is, there are 15.7 g of nitric acid (solute) per 100 g of solution. The mass of water (solvent) is:
m(solution) = m(solute) + m(solvent)
m(solvent) = m(solution) - m(solute)
m(solvent) = 100 g - 15.7 g = 84.3 g
Step 2: Calculate the moles of nitric acid
The molar mass of nitric acid is 63.01 g/mol.
15.7 g × (1 mol/63.01 g) = 0.249 mol
Step 3: Calculate the moles of water
The molar mass of water is 18.02 g/mol.
84.3 g × (1 mol/18.02 g) = 4.68 mol
Step 4: Calculate the mole fraction of nitric acid
[tex]X(HNO_3) = \frac{nHNO_3}{nHNO_3+nH_2O} = \frac{0.249mol}{0.249mol+4.68mol} = 0.0505[/tex]
Thinking about simple organic molecules made with only carbon and hydrogen and
no other elements (hydrocarbons), what are four different ways to make new
molecules? (Hint: Two versions of the same "way" only count as one thing!)
Answer:
Explanation:
Simple organic molecules having only carbon and hydrogen are alkanes, alkenes and alkynes.
Ways of making these molecules are referred to as synthesis
Synthesis of
(1) alkanes can be achieved by halogenation of an alkene or alkyne as shown in the reaction below
C₂H₄ + H₂ ⇒ C₂H₆
The reaction above shows an alkene (ethene) undergoing halogenation to form an alkane (ethane). This reaction occurs in the presence of a platinum or nickel catalyst
C₂H₂ + 2H₂ ⇒ C₂H₆
The reaction above shows an alkyne (ethyne) undergoing halogenation to form an alkane (ethane). This reaction also occurs in the presence of a nickel or platinum catalyst
(2) Alkanes can be achieved by Wurtz reaction; treating alkyl halides with sodium metal as shown below
CH₃CH₂Br + 2Na + BrCH₂CH₃ ⇒ CH₃CH₂CH₂CH₃ + 2NaBr
This reaction proceed in the presence of a dry ether.
(3) alkanes from Kolbe's electrolytic method; this involves the synthesis of alkanes through the electrolysis of sodium or potassium salt of a carboxylic acid
2CH₃COO⁻Na⁺ + 2H₂O ⇒ CH₃CH₃ + 2CO₂ + H₂ + 2NaOH
(4) alkenes from dehydration of alcohol, as shown in the reaction below
C₂H₅OH ⇒ C₂H₄ (-H₂O)
The reaction above proceeds by heating the alcohol (ethanol) at high temperature in the presence of strong acid such as sulphuric acid.
How do the chemical characteristics of carbon affect the characteristics of organic molecules?
Answer:
The chemical characteristics of carbon affect the characteristics of organic molecules due to its tetravalent nature. It has four valence electrons in which it shares with other elements in order to form an octet configuration.
Carbon atoms are also capable of forming double and triple bonds with other atoms. These properties help determine the functional group present and gives us a knowledge of the chemical features such as polarity, melting and boiling present in the compound.
Cell theory states:
A. All living things need glucose to survive
B. All living things Must have a nucleus surrounding DNA
C. All living things are composed of cells
D. The basic unit of life is a cell
Group of answer choices
C & D
A & D
B, C & D
B only
Answer:
C&D all living things are composed of cells
The basic unit of life is a cell
Convert 45 kL to mL (must show work)
Answer:
1 kiloliter = 1,000,000 milliliters
45 × 1,000,000
45 KL = 45,000,000 ML
CHML 1046 general chemistryExperiment "Buffers and Buffer Capacity" 1. What (NH3/ NH4+) ratio is required for a buffer solution that has pH = 7.00? Кь = 1.8x10-5 for NH3 and Ka = 56x10-10 for NH4.2. Why is a mixture of NHs and NH.CI a poor choice for a buffer having pH 7?
Answer:
Explanation:
For pH of a buffer solution , the equation is
pH = pKa + log [ A⁻ ] / [ HA ]
Putting the values
7 = - log 56 x 10⁻¹⁰ + log [ NH₃ ] / [ NH₄⁺]
7 = 10 - log 56 + log [ NH₃ ] / [ NH₄⁺]
- 3 + 1.748 = log [ NH₃ ] / [ NH₄⁺]
log [ NH₃ ] / [ NH₄⁺] = - 1.25
[ NH₃ ] / [ NH₄⁺] = [tex]10^{-1.25}[/tex]
Required ratio = [tex]10^{-1.25}[/tex]
2 )
In the formation of NH₄⁺ , generally HCl acid is required . A little bit of excess HCl will change the pH required . So , this mixture is a poor choice for getting pH 7 for a buffer solution .
what is the average speed in miles per hour of a sprinter who who does thehundred yard dash in 11.3 seconds
Answer:
18.1 mi/h
Explanation:
Step 1: Given data
Distance traveled (d): 100 yd
Time elapsed (t): 11.3 s
Step 2: Convert "d" to miles
We will use the relationship 1 mi = 1,760 yd.
100 yd × (1 mi/1,760 yd) = 0.0568 mi
Step 3: Convert "t" to hours
We will use the relationship 1 h = 3,600 s.
11.3 s × (1 h/3,600 s) = 3.14 × 10⁻³ h
Step 4: Calculate the average speed (s)
We will use the following expression.
s = d / t
s = 0.0568 mi / 3.14 × 10⁻³ h
s = 18.1 mi/h
4. Which statement gives you enough information to say that the atom is electrically neutral?
O A. The atom has 15 neutrons and 15 electrons.
OB. The atom has 4 neutrons and 4 protons.
O C. The atom has 7 protons and 7 electrons.
O D. The atom has 19 electrons and 19 neutrons.
Answer:
O C. The atom has 7 protons and 7 electrons.
If the proton amd electron of atoms are equal it is said to be electrically nuetral
Which of the following is not one of the four groups of organic compounds
The following that is not one of the four groups of organic compounds is a monosaccharide. The correct option is D.
What are organic compounds?Any of a vast group of chemical compounds known as organic compounds contain one or more carbon atoms that are covalently connected to atoms. Organic compounds are made up of carbon and hydrogen, and oxygen.
Carbohydrates, lipids, proteins, nucleic acids, and nucleotides are organic compounds. Due to their presence of both carbon and hydrogen, these substances are referred to as organic.
Monosaccharides are the single unit of carbohydrates. They join to form polymers of carbohydrates. They are glucose, fructose, etc. It is a single unit of carbohydrates. It is not one of the organic compound.
Thus, the correct option is D) monosaccharides.
To learn more about organic compounds, refer to the below link:
https://brainly.com/question/5994723
#SPJ2
The question is incomplete. Your most probably complete question is given below:
A) carbohydrates D) monosaccharides B) lipids E) proteins C) nucleic acids
What can be observed when concentrated nitric acid is added to iron two sulfate solution?
Answer:
It will produce iron(III) sulphate ,nitrogen dioxide and water
Explanation:
hope it helps you☺️
if I'm correct do mark my answer as brainliest
Choose the best description for the indicated bond H-CEC-CH3 A) This is a triple bond between two carbons in which three pi bonds hold the atoms together. B) This is a single bond between two carbons formed by the overlap of an sp orbital on each carbon. Extra electrons on each carbon exist as lone pairs in p orbitals. C) This is a triple bond between two carbons in which a sigma bond arises from sp- sp overlap and two pi bonds are formed by side-to-side overlap of unhybridized p orbitals on each carbon. D) This is a triple bond between two carbons in which there are three sigma bonds between the atoms, which are each spa hybridized
Answer:
(Chemical compound and indicated bond in attachment).
C) This is a triple bond between two carbons in which a sigma bond arises from sp- sp overlap and two pi bonds are formed by side-to-side overlap of unhybridized p orbitals on each carbon.
Explanation:
Alkynes are hydrocarbons that contain carbon-carbon triple bonds. Each carbon atom is bound to the other two atoms, and there are no free valence electrons. Each carbon atom needs two hybrid orbitals to form the skeleton of sigma bonds. The hybridization of the s orbital with a p orbital generates two hybrid orbitals, oriented at 180º of separation for each carbon atom. The overlap of these hybrid orbitals sp with the s orbitals from the hydrogen produces the skeleton of sigma bonds.
The result of the overlap of the two remaining unhybridized p orbitals of each atom of carbon is the formation of two pi bonds. These orbitals overlap in right angles with each other, forming a pi bond with electronic density above and below the sigma bond and the other with electronic density in front of and behind the sigma bond. The form of these pi bonds is such that they combine to form a cylinder of electron density that "wraps" the sigma bond between the two carbon atoms.
PLEASE HELP ASAP!!!!!!!!!!! WITH PAGE 2 FOR QUESTION # 3 WITH THE GRAPHING THANK YOU. THE QUESTION IS AS FOLLOWED: GRAPH THE VAPOR PRESSURE OF WATER USING THE FOLLOWING TEMPERATURES. Temperature (k) Vapor Pressure (Torr) (dependent variable) 273 4.6 293 17.5 313 55.0 333 149.2 353 355.5 373 760.0
Answer:
a
Explanation:
Which of the following molecules is achiral? A) (2R,3R)-2,3-Dichloropentane B) (2R,3S)-2,3-Dichloropentane C) (2S,3S)-2,3-Dichlorobutane D) (2R,3S)-2,3-Dichlorobutane E) None of these
Answer:
E) None of these
Explanation:
For this question, we must remember the definition of a chiral carbon. In the chiral carbons, we have four different groups around the carbon.
So, an achiral molecule would be a molecule that does not have chiral carbons. That is, in an achiral molecule in all carbons we have at least one repeating group.
In all the molecules, which we have in the question, we have absolute configuration. That is, we have chiral carbons. In molecule A we have two chiral carbons (R and R), in molecule b we have two chiral carbons (R and S), in molecule c we have two chiral carbons (S and S) and in molecule d we have two chiral carbons (R and S). Therefore in all molecules, we have chiral carbons.
Now, if we analyze molecules a and c, we have a plane of symmetry. If we have a plane of symmetry, even though there are chiral carbons, there will be no optical activity. That is, these molecules do not have the ability to deflect polarized light despite having chiral carbons. Therefore molecules a and c are meso compounds.
But optical activity is not related to chirality.
See figure 1 to further explanations
I hope it helps!
An unknown substance is measured. It has a mass of 0.221 g and a volume of 2.25 mL. What is its density?
Answer:
d ≈ 0.098 g/mL
Explanation:
The density of a substance can be found by dividing the mass by the volume.
d=m/v
The mass of the substance is 0.221 grams and the volume is 2.25 milliliters.
m= 0.221 g
v= 2.25 mL
Substitute the values into the formula.
d= 0.221 g / 2.25 mL
Divide
d= 0.098222222 g/mL
Let’s round to the nearest thousandth. The 2 in the ten thousandths tells us to keep the 8 in the thousandth place.
d ≈ 0.098 g/mL
The density of the substance is about 0.098 grams per milliliter.
Answer:
Explanation:5.70
An iceberg has a volume of 783 m3. If the density of ice = 0.917 g/cm3, how much mass does the iceberg have in kg?
Answer:
7.18 × 10⁵ kg
Explanation:
Step 1: Given data
Volume of the iceberg (V): 783 m³
Density of ice (ρ): 0.917 g/cm³
Step 2: Convert "V" to cm³
We will use the relationship 1 m³ = 10⁶ cm³.
783 m³ × (10⁶ cm³/1 m³) = 7.83 × 10⁸ cm³
Step 3: Calculate the mass (m) of the iceberg
We will use the following expression.
ρ = m/V
m = ρ × V
m = 0.917 g/cm³ × 7.83 × 10⁸ cm³
m = 7.18 × 10⁸ g
Step 4: Convert "m" to kg
We will use the relationship 1 kg = 10³ g.
7.18 × 10⁸ g × (1 kg/10³ g) = 7.18 × 10⁵ kg
What is the volume of an object that has a mass of 18 g and a density of 0.5 g/cm3 ?
437 mm converted to m
Answer:
0.437
Explanation:
Divide the length value by 1000 to get the answer.
Draw all of the constitutional isomers of the molecule with formula C3H5Br. Ignore geometric and stereoisomers.
Answer:
-) 3-bromoprop-1-ene
-) 2-bromoprop-1-ene
-) 1-bromoprop-1-ene
-) bromocyclopropane
Explanation:
In this question, we can start with the I.D.H (hydrogen deficiency index):
[tex]I.D.H~=~\frac{(2C)+2+(N)-(H)-(X)}{2}[/tex]
In the formula we have 3 carbons, 5 hydrogens, and 1 Br, so:
[tex]I.D.H~=~\frac{(2*3)+2+(0)-(5)-(1)}{2}~=~1[/tex]
We have an I.D.H value of one. This indicates that we can have a cyclic structure or a double bond.
We can start with a linear structure with 3 carbon with a double bond in the first carbon and the Br atom also in the first carbon (1-bromoprop-1-ene). In the second structure, we can move the Br atom to the second carbon (2-bromoprop-1-ene), in the third structure we can move the Br to carbon 3 (3-bromoprop-1-ene). Finally, we can have a cyclic structure with a Br atom (bromocyclopropane).
See figure 1
I hope it helps!
Hydrogen reacts with nitrogen monoxide to form dinitrogen monoxide (aughing gas) according to the equation: H2 (g) + 2NO(g) → N2O(g) + H2O(g)Based on the following data: Trail 1 Trail 2 Trail 3H2 (M) 0.30 0.60 0.60NO (M) 0.35 0.35 0.70Rate (M*s) 2.835 x 10^-3 1.134 x 10^-2 2.268 x 10^-2Required:a. What is the order with respect to H2? b. What is the order with respect to NO? c. What is the rate equation for this reaction? d. Calculate the rate constant for the reaction.
Explanation:
H2 (g) + 2NO(g) → N2O(g) + H2O(g)
Trial 1 Trial 2 Trial 3
H2 (M) 0.30 0.60 0.60
NO (M) 0.35 0.35 0.70
Rate (M*s) 2.835 x 10^-3 1.134 x 10^-2 2.268 x 10^-2
a. What is the order with respect to H2?
Comparing trial 1 and 2, the conc of H2 is doubled, the rate of the reaction increased by a factor f 4. This means the rate is in second order with respect to H2.
b. What is the order with respect to NO?
Comparing Trial 2 and 3, the concentration of NO is doubled, the rate of the reaction increased by a factor of 2. This means the rate is in first order with respect to NO.
c. What is the rate equation for this reaction?
The rate equation is given as;
rate = k [H2]²[NO]
d. Calculate the rate constant for the reaction.
Taking trial 1;
2.835 x 10^-3 = k (0.30)²(0.35)
k = 2.835 x 10^-3 / 0.0315 = 90 x 10^-3
k = 0.09 L2 mol-2 s-1
10 grams of compound J is found to have a mass ratio of 8:2 of lithium (Li): Hydrogen (H). How many grams
of hydrogen would found in 42.0 grams of compound J?
Answer:
Mass of hydrogen in 42 gram of J = 8.4
Explanation:
Given:
Amount of compound J = 10 gram
Mass ratio[lithium (Li): Hydrogen (H)] = 8:2
Find:
Mass of hydrogen in 42 gram of J
Computation:
Mass of hydrogen in 10 gram of J = [2 / 10] 10
Mass of hydrogen in 10 gram of J = 2 gram
Mass of hydrogen in 42 gram of J = [2 / 10] 42
Mass of hydrogen in 42 gram of J = 8.4
For which reaction, carried out at standard conditions, wouldboth the enthalpy and entropy changes drive the reaction in thesame direction? Please EXPLAIN.
A. 2H2(g) + O2(g) --->2H2O(l) ΔH = -571.1 kJ
B. 2Na(s) + Cl2(g)---->2NaCl(s) ΔH = -822.0kJ
C. N2(g) + 2O2(g) ---->2NO2(g) ΔH= +67.7 kJ
D. 2NH3(g) ----> N2(g) +3H2(g) ΔH= +92.4 kJ
Answer:
2NH3(g) ----> N2(g) +3H2(g) ΔH= +92.4 kJ
Explanation:
Entropy increases with increase in the number of particles from left to right in a reaction. Hence, the reaction; 2NH3(g) ----> N2(g) +3H2(g) ΔH= +92.4 kJ is favoured by increase in entropy.
Similarly, the enthalpy change for the reaction is only +92.4 KJ. Hence for this reaction, both enthalpy and entropy changes drive the reaction in the same direction.
The specific heat of water is 4.184 J/ g°C, the specific heat of wood is 1.760 J/g°C, and the specific heat of coal is 1.26 J/g°C. Which of these materials requires the most heat to raise its temperature by one degree Celsius?
A. All materials need the same amount of energy.
B. Wood
C. Coal
D. Water
Answer:
D. Water
Explanation:
Specific heat is defined as the amount of heat you need to add to a substance per unit of mass to increase its temperature in 1°C.
That means if you have the same mass of substances, the one with the higherst specific heat is the one that requires the most amount of heat to increase its temperature by one degree Celsius.
As you can see, water is the material with the highest specific heat doing the material that requires the most heat.
Right answer:
D. WaterAnswer: D
Explanation: I took the test
What is the relationship between the bases displayed when the arrow is pointed to the left versus when it is pointed to the right?
Answer:
Options
Explanation:
If you lived in Flint, Michigan why would it be important for you to understand chemistry
Answer:
If I lived in Flint, Michigan, it would be important for me to understand chemistry because of the high contamination levels of water in the city, in order to know how to help to solve this issue and to avoid any potential harm to my health.
There is an ongoing drinking water issue since 2014 in Flint, Michigan. In April 2014, the Flint authorities changed the source of water from the Detroit Water and Sewage Department, which takes it from Lake Huron and the Detroit River, to the Flint River for cost-effective reasons. This led to a series of water quality problems culminating in the presence of lead, creating a serious threat to public health.