Answer:
(a) 282 kJ
(b) 67.4 Calories
Explanation:
(a) The molar enthalpy, ΔH = −2802.5 kJ/mol, means that the heat produced by the reaction is 2802.5 kJ per mol of glucose.
We can multiply the enthalpy by the number of moles of glucose to get the heat produced by the metabolism. Grams of glucose will be converted to moles using the molar mass of glucose (180.156 g/mol):
(18.1 g)(mol/180.156g)(2802.5 kJ/mol) = 282 kJ
(b) Using the result we obtained above, kJ will be converted to Calories using the conversion factor of 4.184J = 1 cal. Calorie with a capital C is the same as a kilocalorie.
(282 kJ)(1 cal/4.184J) = 67.4 kcal = 67.4 Calories
Answer:
For a: The amount of heat produced for given amount of glucose is -283.05 kJ
For b: The amount of heat produced for given amount of glucose is -67648.9 Cal
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass of glucose = 18.1 g
Molar mass of glucose = 180.16 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of glucose}=\frac{18.1g}{180.16g/mol}=0.101mol[/tex]
The given chemical reaction follows:
[tex]C_6H_{12}O_6(s)+6O_2(g)\rightarrow 6CO_2(g)+6H_2O(l);\Delta H=-2802.5kJ/mol[/tex]
For a:By Stoichiometry of the reaction:
When 1 mole of glucose is reacted, the amount of heat released is 2802.5 kJ
So, when 0.101 moles of glucose is reacted, the amount of heat released is [tex]\frac{2802.5}{1}\times 0.101=283.05kJ[/tex]
Hence, the amount of heat produced for given amount of glucose is -283.05 kJ
For b:To convert the heat produced in kilo joules to calories, we use the conversion factor:
1 kJ = 239 Cal
So, [tex]-283.05kJ\times (\frac{239Cal}{1kJ})=-67648.9Cal[/tex]
Hence, the amount of heat produced for given amount of glucose is -67648.9 Cal
Acetylene torches are used for welding. These torches use a mixture of acetylene gas, C2H2, and oxygen gas, O2 to produce the following combustion reaction: 2 C2H2 (g) + 5 O2 (g) → 4 CO2 (g) + 2 H2O (g) Imagine that you have a 5 L gas tank and a 3.5 L gas tank. You need to fill one tank with oxygen and the other with acetylene to use in conjunction with your welding torch. If you fill the larger tank with oxygen to a pressure of 127 atm , to what pressure should you fill the acetylene tank to ensure that you run out of each gas at the same time? Assume ideal behavior for all gases.
Answer:
72.6 atm should be the pressure at which acetylene tank.
Explanation:
[tex]2C_2H_2 (g)+5O_2 (g)\rightarrow 4CO_2 (g)+2H_2O[/tex]
Let the temperature of the both tanks be same as T.
Volume of the tank in which oxygen is filled = [tex]V_1=5L[/tex]
Pressure of the oxygen in tank =[tex]P_1[/tex]= 127 atm
According to reaction 5 moles of oxygen reacts with 2 moles of acetylene.
[tex]n_1= 5 moles[/tex]
[tex]P_1V_1=n_1RT[/tex]
[tex]T=\frac{P_1V_1}{n_1}[/tex]..[1]
[tex]n_2=2 mol[/tex]
Volume of the tank in which acetylene is filled = [tex]V_2=3.5L[/tex]
Pressure of the acetylene in tank =[tex]P_2[/tex]= ?
[tex]T=\frac{P_2V_2}{n_2}[/tex]..[2]
[1] = [2]
[tex]\frac{P_1V_1}{n_1}=\frac{P_2V_2}{n_2}[/tex]
[tex]P_2=\frac{P_1\times V_1\times n_2}{n_1\times V_2}[/tex]
[tex]=\frac{127 atm\times 5 L \times 2 mol}{5 mol\times 3.5 L}[/tex]
[tex]P_2=72.6 atm[/tex]
72.6 atm should be the pressure at which acetylene tank.
Which one of the following statements is not correct?
Select one:
a. During allotropic transformation, the number of atoms in the material changes.
b. During allotropic transformation the volume of the material changes.
c. Isotopes of the same element have the same number of protons.
d. In an element, the size of its anion is larger than its atomic size
Answer:
option a, During allotropic transformation, the number of atoms in the material changes.
Explanation:
Allotrops are different crystalline forms of the same element. So, two allotrops of an element have same number of atoms.
for example diamond, graphite, graphene, fullerene are allotropic forms of carbon.
Allpotops differ in physical and chemical properties. So, volume changes during alloptropic transformation. So, statement b is correct.
Allotropic transformation is the transformation of one allotropic form to other therefore, number of atom does not change during allotropic transformation.
Chemical species having same atomic number or same no. of protons are called isotopes. so statement c is correct.
When an electron is added to the neutral element, electronic repulsion increases which lead to the increase in atomic size, So, statement d is also correct.
So among given, option a is incorrect
Heather just drank 40.0 grams of water (H20). How many moles of water did she just drink? O a. 2.22 moles b. 45 moles O c. 40.0 moles O d. 0.45 moles O e.720 moles
Answer:
This question begins with something, you should know: molar mass from water is aproximately 18 g/m, so if 18 grams of water are contained in 1 mole, the 40 grams occuped 2.22 moles. As you see, opcion a is the best!
Explanation:
Which of the following statements about material bonding is correct? C O a. Ionic bonds are formed by the sharing of valence electrons among two or more atoms b. Van der Waals bonds are formed by Van der Waals forces in which molecules or atoms have either an induced or permanent dipole moment to attract each other C c. Metallic elements with metallic bonds have atoms that donate valence electrons to other atoms, thus filling the outer energy shells of these other atoms d. Covalent bonds are formed by atoms that donate their valence electrons to form a "sea" of electrons surrounding the atoms
Answer: Option (b) is the correct answer.
Explanation:
In material bonding, there occurs Vander waal foces between the molecules in which their is either an induced or permanent dipole moment that attract molecules towards each other.
And, due to these forces the molecules are held together.
On the other hand, in a ionic bond there will always be transfer of electrons from one atom to another. This is because on atom which loses its valence electrons acquires a positive charge and another atom which gains the electrons acquires a negative charge.
Hence, these opposite charges strongly gets attracted towards each other forming a strong bond.
Whereas in a covalent bond, there will be sharing of electrons between the combining atoms.
In a metallic bond, there occurs a sea of electrons which is uniformly distributed throughout the solid substance or material.
Thus, we can conclude that the statement, Van der Waals bonds are formed by Van der Waals forces in which molecules or atoms have either an induced or permanent dipole moment to attract each other, about material bonding is correct.
Final answer:
The correct statement is that Van der Waals bonds are formed by Van der Waals forces due to induced or permanent dipoles, unlike ionic or covalent bonds.
Explanation:
The correct statement about material bonding is option b: Van der Waals bonds are formed by Van der Waals forces in which molecules or atoms have either an induced or permanent dipole moment to attract each other. Unlike ionic and covalent bonds, Van der Waals bonds involve weaker, more temporary attractions. In contrast, ionic bonds occur when electrons are transferred from one atom to another, usually between metals and nonmetals, creating a positive and a negative ion that attract each other.
Finally, covalent bonds involve the sharing of electron pairs between atoms, and in metallic bonds, valence electrons are not donated to other atoms; instead, they form a 'sea of electrons' that surrounds the metal ions.
have an infinite number of significant figures
Exact numbers have an infinite number of significant figures because the number of significant figures in a value indicates the level of uncertainty associated with that value, and exact numbers have no associated uncertainty.
What are exact numbers ?
Exact numbers are those whose values are known with absolute certainty, devoid of any measurement uncertainty or approximation. They are typically derived from precise definitions, counting procedures, or other mathematical operations that guarantee their accuracy.
Due to their certainty, exact numbers are considered to possess an infinite number of significant figures. This is because the concept of significant figures reflects the level of uncertainty in a value, and exact numbers, being devoid of uncertainty, transcend this limitation.
Complete question:
Why do exact numbers have an infinite number of significant figures?
Calculate the density of CO2 in g/cm3 at room temperature(25 degrees Celsuis) and pressure(1 atm) assuming it acts as an ideal gas
Answer:
[tex]density=1.8x10^{-3}g/mL[/tex]
Explanation:
Hello,
Considering the ideal equation of state:
[tex]PV=nRT[/tex]
The moles are defined in terms of mass as follows:
[tex]n=\frac{m}{M}[/tex]
Whereas [tex]M[/tex] the gas' molar mass, thus:
[tex]PV=\frac{mRT}{M}[/tex]
Now, since the density is defined as the quotient between the mass and the volume, we get:
[tex]P=\frac{m}{V} \frac{RT}{M}[/tex]
Solving for [tex]m/V[/tex]:
[tex]density= m/V=\frac{PM}{RT}[/tex]
Thus, the result is given by:
[tex]density=\frac{(1atm)(44g/mol)}{[0.082atm*L/(mol*K)]*298.15K} \\density=1.8g/L=1.8x10^{-3}g/mL[/tex]
Best regards.
Methyl isocyanate, H3C-N=C=O, is used in the industrial synthesis of a type of pesticide and herbicide known as a carbamate. As a historical note, an industrial accident in Bhopal, India in 1984 resulted in leakage of an unknown quantity of this chemical into the air. An estimated 200,000 persons were exposed to its vapors and over 2000 of these people died.
(a) Draw the Lewis structure for methyl isocyanate.
Explicitly draw all H atoms.
Include all valence lone pairs in your answer.
(b) What is the hybridization of the carbonyl carbon? _________sp²spsp³
What is the hybridization of the nitrogen? _________sp²spsp³
Answer:
(a) The lewis structure for methylisocyanate is in the attached.
(b) The carbonyl carbon have an sp² hybridization
(c) The nitrogen have an sp² hybridization?
Explanation:
(a) The lewis structure for methylisocyanate has the nitrogen with one lone pair and the oxygen with two lone pairs.
(b) The carbonyl carbon form double bond with the oxygen causing to form three hybrid orbitals sp².
The Nitrogen also forms a double bond with the carbon having an sp² hybridization too.
Question:Dimethylhydrazine, the
fuel used in the Apollo lunar descentmodule, has a molar mass of
60.10 g/mol. It is made up of carbon,hydrogen, and nitrogen atoms.
The combustion of 2.859g of the fuelin excess oxygen yields 4.190g
of carbon dioxideand 3.428g ofwater. What are the simplest and
molecular formulas fordimethylhydrazine?
Answer:
Explanation:
Given parameters:
Molar mass of compound = 60.1g/mol
Mass of fuel used in the combustion process = 2.859g
Mass of carbon dioxide produced = 4.190g
Mass of oxygen produced= 3.428g
Unknown parameters;
Empirical and molecular formula of the compound
Solution
The empirical formula of a compound is its simplest formula. The molecular formula is the actual formula of the compound showing the proportions of the atoms.
We can derive the empirical formula of a compound from its molecular formula and vice versa.
Here, we have to work from empirical formula to molecular formula.
Solving
We know that the compound contains C, H and N atoms. We need to first find the masses of these atoms in the compoundFor Carbon, we can determine the mass from the amount of carbon dioxide produced:
mass of carbon in compound = [tex]\frac{12}{44}[/tex] x 4.19g = 1.14g
For Hydrogen, we can determine the mass from the amount of water produced:
mass of hydrogen in compound = [tex]\frac{2}{18}[/tex] x 3.428g = 0.38g
To determine the mass of N in the compound:
mass of compound = mass of C + mass of H + mass of N
mass of N = mass of compound - (mass of C + mass of H)
mass of N = 2.859g - (1.14g + 0.38g) = 1.34g
2. we now proceed to find the empirical formula using the process below:
Elements C H N
mass of the
elements 1.14 0.38 1.34
Atomic mass
of elements 12 1 14
Number of
moles 1.14/12 0.38/1 1.34/14
0.095 0.38 0.095
Dividing by
the smallest 0.095/0.095 0.38/0.095 0.095/0.095
1 4 1
The empirical formula of the compound is CH₄N
To obtain the molecular formula, we need to find the number of times the empirical formula must have repeated itself in the original form.
Molar mass of CH₄N = 12 + 4 + 14 = 30g/mol
Ratio = [tex]\frac{molar mass of molecular formula}{molar mass of empirical formula}[/tex] = [tex]\frac{60.1}{30}[/tex] = 2
Molecular formula = 2(CH₄N) = C₂H₈N₂
For IR radiation with û = 1,130 cm 1, v=__THz
Answer: The frequency of the radiation is 33.9 THz
Explanation:
We are given:
Wave number of the radiation, [tex]\bar{\nu}=1130cm^{-1}[/tex]
Wave number is defined as the number of wavelengths per unit length.
Mathematically,
[tex]\bar{\nu}=\frac{1}{\lambda}[/tex]
where,
[tex]\bar{\nu}[/tex] = wave number = [tex]1130cm^{-1}[/tex]
[tex]\lambda[/tex] = wavelength of the radiation = ?
Putting values in above equation, we get:
[tex]1130cm^{-1}=\farc{1}{\lambda}\\\\\lambda=\frac{1}{1130cm^{-1}}=8.850\times 10^{-4}cm[/tex]
Converting this into meters, we use the conversion factor:
1 m = 100 cm
So, [tex]8.850\times 10^{-4}cm=8.850\times 10^{-4}\times 10^{-2}=8.850\times 10^{-6}m[/tex]
The relation between frequency and wavelength is given as:[tex]\nu=\frac{c}{\lambda}[/tex]
where,
c = the speed of light = [tex]3\times 10^8m/s[/tex]
[tex]\nu[/tex] = frequency of the radiation = ?
Putting values in above equation, we get:
[tex]\nu=\frac{3\times 10^8m/s}{8.850\times 10^{-4}m}[/tex]
[tex]\nu=0.339\times 10^{14}Hz[/tex]
Converting this into tera Hertz, we use the conversion factor:
[tex]1THz=1\times 10^{12}Hz[/tex]
So, [tex]0.339\times 10^{14}Hz\times \frac{1THz}{1\times 10^{12}Hz}=33.9THz[/tex]
Hence, the frequency of the radiation is 33.9 THz
Radon-219 decays to radon-218 by releasing... a. a positron b.a muon O c. a neutron O d. an electron O e. a proton
Final answer:
Radon-219 decays into radon-218 by emitting an alpha particle, a process that reduces its atomic mass and number, leading to a new isotope.
Explanation:
Radon-219 decays to radon-218 by releasing an alpha particle (Helium nucleus). When radium-226 undergoes alpha decay, it forms radon-222 and an alpha particle. The process involves the nucleus of an atom releasing two protons and two neutrons, which together form an alpha particle. In the case of radon decay, the alpha particle is emitted, decreasing the atomic mass by 4 units and atomic number by 2, resulting in a new isotope. Therefore, the correct answer is b. a muon. The emission of an alpha particle is a common mode of decay for heavy, unstable nuclei such as radon-219.
Lewis base is a(n) a. producer of OH ions. b. proton acceptor. c. electron-pair donor. d. electron-pair acceptor. In the reaction represented by the equation Ag (aq) +2NH3(aq)-[Ag[ a. Bronsted-Lowry acid. b. Lewis acid
Answer:
c. electron-pair donor.
Explanation:
Lewis base -
A Lewis base is an electron rich species , which is available for donation .
Hence , Lewis base is a electron - pair donor .
The indication of a Lewis base , is it has a negative charge or lone pairs of electrons .
Hence , the species with a lone pair or negative charge act as a Lewis base .
for example , OH ⁻ is a Lewis base , due to its negative charge which is available for donation .
H₂O is also a Lewis base , due to its lone pairs of electrons , that are available for donation .
Describe the molecular structure of water and explain why the water molecule is polar
Explanation:
Hybridization of O in [tex]H_2O = sp^3[/tex]
So, water molecule has four hybrid orbitals.
Two hybrid orbitals form 2 sigma bond with two H atoms.
Remaining two hybrid orbitals are occupied by two lone pairs.
Because of lone pair-lone pair repulsion, shape of [tex]H_2O[/tex] becomes bent.
Water molecule is polar because of difference in eletronegativities of O and H.
O is more electronegative as comapared to hydrogen. So bonding electrons get attracted towards O atom which results in the development of partial negative charge on O atom and partial positive charge on H atoms.
Because of development of partial negative and partial positive charge, water molecule becomes polar.
Water's molecular structure is a bent shape with two hydrogen atoms bonded to one oxygen atom. The electronegativity difference between oxygen and hydrogen results in a polar molecule, leading to properties like high boiling point and solubility.
Explanation:The molecular structure of water, referred to as H2O, comprises two hydrogen atoms bonded to one oxygen atom. Each hydrogen atom forms a single covalent bond with the oxygen atom, creating a bent structure. Yet, oxygen is more electronegative than hydrogen, meaning it pulls shared electrons closer to itself. This results in oxygen having a partial negative charge, and hydrogen having a partial positive charge, making water a polar molecule.
Water's polarity contributes to its unique properties, such as high boiling point and ability to dissolve many substances. Water molecules can form hydrogen bonds - attractions between the positively charged hydrogen of one molecule, and the negatively charged oxygen of another - due to their polarity, which makes the water molecule extremely cohesive and leads to a higher than expected boiling point.
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Select the cations responsible for water hardness.1) Ca2+ 2) CO3−2 3) CaCO3 4) Mg2+
Answer:
The correct options are: 1. Ca²⁺ and 4. Mg²⁺
Explanation:
Hard water is the water with high mineral content. Temporary hardness and permanent hardness are the two types of hardness of water.
Temporary hardness is due to the presence of dissolved bicarbonate minerals. These minerals present in the water, dissociate to give multivalent calcium cations (Ca²⁺) and magnesium cations (Mg²⁺).
Therefore, the presence of metal cations such as calcium cations (Ca²⁺) and magnesium cations (Mg²⁺) makes the water hard.
A solution is to be prepared with a sodium ion concentrationof
0.148 mol/L. What mass of sodium sulfate (g) is needed toprepare
2.29 liters of such a solution?
Answer : The mass of sodium sulfate needed is 5.7085 grams.
Explanation : Given,
Concentration of sodium ion = 0.148 mol/L
Volume of solution = 2.29 L
Molar mass of sodium sulfate = 142 g/mole
First we have to determine the moles of sodium ion.
[tex]\text{Concentration of sodium ion}=\frac{\text{Moles of sodium ion}}{\text{Volume of solution}}[/tex]
[tex]0.184mol/L=\frac{\text{Moles of sodium ion}}{2.29L}[/tex]
[tex]\text{Moles of sodium ion}=0.08035mole[/tex]
Now we have to calculate the moles of sodium sulfate.
The balanced chemical reaction will be,
[tex]Na_2SO_4\rightarrow 2Na^++SO_4^{2-}[/tex]
As, 2 moles of sodium ion produced from 1 moles of [tex]Na_2SO_4[/tex]
So, 0.08035 moles of sodium ion produced from [tex]\frac{0.08035}{2}=0.040175[/tex] moles of [tex]Na_2SO_4[/tex]
Now we have to calculate the mass of sodium sulfate.
[tex]\text{Mass of }Na_2SO_4=\text{Moles of }Na_2SO_4\times \text{Molar mass of }Na_2SO_4[/tex]
[tex]\text{Mass of }Na_2SO_4=0.040175mole\times 142g/mole=5.7085g[/tex]
Therefore, the mass of sodium sulfate needed is 5.7085 grams.
How many kilojoules of heat will be released when exactly 1 mole of manganese, Mn, is burned to form Mn3O4 at standard state condaition?
The heat released (ΔH) when one mole of manganese is burned to form Mn3O4 can be derived from a thermochemical equation and molar enthalpies. For specific values, consult a standard enthalpy of formation table.
Explanation:The calculation for the heat released when one mole of manganese is burned to form Mn3O4 at standard state condition involves understanding of chemistry concepts such as molar enthalpies and thermochemical equations. Unfortunately, without specifying the thermochemical equation for the formation of Mn3O4 from manganese or the molar enthalpy of this specific reaction, an exact value cannot be given.
However, as a general guide, the heat released (also known as enthalpy change, ΔH) can be found from the formula ΔH = -q, where q represents the heat absorbed. A negative value indicates heat is being released. In thermochemical equations, the ΔH value is often given per mole of a substance involved in the reaction, so you would usually directly obtain the heat released when one mole of a substance is involved from the molar enthalpy.
For specific values, refer to a standard enthalpy of formation table, a resource often found in chemistry textbooks or scientific literature, to find the molar enthalpy for the formation of Mn3O4 from manganese.
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Define ecology, environmental science and environmentalism.
Explanation:
Ecology -
It is the branch of biology , which studies the interactions between the organisms and their biophysical environment , including both abiotic and biotic components .
Environmental science -
It is an interdisciplinary field , of information science , physical , biological science , which helps to study the environment and the environmental problems .
Environmentalism -
It is the rights regarding the concern for the protection of the environment and the to improve the health of the environment .
The main focus on environment and nature-related aspects of green ideology and politics .
.Answer:
Explanation:
Environmentalism -
It is the rights regarding the concern for the protection of the environment and the to improve the health of the environment .
The main focus on environment and nature-related aspects of green ideology and politics.
Ecology -
It is the branch of biology , which studies the interactions between the organisms and their biophysical environment , including both abiotic and biotic components .
Environmental science -
It is an interdisciplinary field , of information science , physical , biological science , which helps to study the environment and the environmental problems .
(a) Consider a carbon atom in its ground state. Would such an atom offer a satisfactory model for the carbon of methane? If not, why not? (Hint: Consider whether a ground state carbon atom could be tetravalent, and consider the bond angles that would result if it were to combine with hydrogen atoms.)
(b) Consider a carbon atom in the excited state. Would such an atom offer a satisfactory model for the carbon of methane? If not, why not?
Answer:
(a) No. Ground-state carbon has only 2 half-filled orbitals that could be used for bonding.
(b) No. The bond angles would be incorrect as the p-orbitals are all perpendicular to each other (90°).
Explanation:
See attachment for the ground-state and excited-state electron orbital diagrams of carbon.
A methane molecule has all four CH bonds the same length and at 109.5° from each other. Hybridization of the s and p orbitals to sp³ orbitals is necessary.
If the concentration of an analyte in a solution prepared by digesting a 2.05 g solid sample and making up to 100 mL, is found to be 21.1 mg L-1. What was the mass percentage of the analyte in the original sample?
Answer:
Mass percentage of analyte = 0.10%
Explanation:
The mass of analyte in the solution is calculated as follows:
n = CV = (21.1 mgL⁻¹)(100 mL)(1L/1000mL) = 2.11 mg
The mass percentage of the analyte is calculated as follows:
(g analyte)/(g sample) x 100%
(2.11 mg)(1g/1000g) / (2.05g) x 100% = 0.10%
in the fali, the relative humidity is high in early morning and as the temperature increases, the relative humidity decreases because (3 pts) A. saturation pressure of water increases B. water condenses C. water evaporates D. saturation pressure of water decreases
Answer:
The correct option is: A. saturation pressure of water increases
Explanation:
Relative humidity, at a given temperature, is defined as ratio of partial pressure of the water vapor preset in the air mixture and equilibrium vapor pressure above pure water.
Relative humidity of a system is dependent upon the pressure and temperature. When the amount of water vapor in a system is constant, the higher relative humidity is higher at low temperature and lower at high temperature. This is because at high temperatures, the air capacity and the saturation pressure increases.
Therefore, in the morning when the temperature is low in fali, the relative humidity is high. As the temperature increases, the relative humidity in fali decreases.
Answer the following and round to the correct number of significant figures (don't forget units).
a) 0.02123 m + 1.12 m + 0.00123 m =
b) 2.3 cm - 1.23 cm + 120 cm =
c) 25,430 km - 3,500 km + 200 km =
d) (1.21 x 105 ) x (2.6 x 103 ) =
e) 7.13 mm x 9.1 mm =
f) 3.0 cm x 8.222 cm =
g) 4.1 g ÷ 0.121 cm =
h) 0.413 ÷ (9.212 x 103 ) =
i) (12.1 cm - 4.15 cm) / 35.64 g =
j) (11.00 m - 3.356 m) x 45.1 kg /35.64 s =
k) 73.0 x 1.340 x (25.31 – 1.6) =
l) (418.7 x 31.8) / (19.27 – 18.98) =
Answer:
a) 0.02123 m + 1.12 m + 0.00123 m = 1.14246m it can be round off as 1.1425m
b) 2.3 cm - 1.23 cm + 120 cm = 121.07 cm it can be round off as 121.1 cm
c) 25,430 km - 3,500 km + 200 km = 22130 km
d) (1.21 x 105 ) x (2.6 x 103 ) = 34023.99 it can be round off as 34024
e) 7.13 mm x 9.1 mm = 64.883 mm it can be round off as 65 mm
f) 3.0 cm x 8.222 cm = 24.666 cm it can be round off as 25 cm
g) 4.1 g ÷ 0.121 cm = 33.8842 g/cm it can round off as 34 g/cm
h) 0.413 ÷ (9.212 x 103 ) = 0.00043527
i) (12.1 cm - 4.15 cm) / 35.64 g = 0.223063973 cm/g
j) (11.00 m - 3.356 m) x 45.1 kg /35.64 s = 9.672962963 m kg/s
k) 73.0 x 1.340 x (25.31 – 1.6) = 2319.3122
l) (418.7 x 31.8) / (19.27 – 18.98) = 45912.62069 45912.6207
The question requests computations involving addition, subtraction, multiplication, and division while carefully applying the rules of significant figures. The answers are expressed in various physical units such as meters, centimeters, km, mm, g, m/s, and their calculated results are in accordance with significant figures rules.
Explanation:This question involves numerical calculations with significant figures focusing on measurement and precision, a concept prevalent in Physics.
a) 0.02123 m + 1.12 m + 0.00123 m = 1.14 mb) 2.3 cm - 1.23 cm + 120 cm = 121 cmc) 25,430 km - 3,500 km + 200 km = 22,100 kmd) (1.21 x 105 ) x (2.6 x 103 ) = 3.14 x 108e) 7.13 mm x 9.1 mm = 65 mm2f) 3.0 cm x 8.222 cm = 25 cm2g) 4.1 g ÷ 0.121 cm = 33.9 g/cmh) 0.413 ÷ (9.212 x 103 ) = 4.48 x 10-5i) (12.1 cm - 4.15 cm) / 35.64 g = 0.223 g/cmj) (11.00 m - 3.356 m) x 45.1 kg /35.64 s = 9.60 kg.m/sk) 73.0 x 1.340 x (25.31 – 1.6) = 2,300l) (418.7 x 31.8) / (19.27 – 18.98) = 71,900Learn more about Significant Figures here:https://brainly.com/question/37022020
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15 g of anhydrous calcium chloride is dissolved in 185 mL of water. What is molarity of the prepared solution? 81.1 M O0.73 M 0.081 M O0.073 M 1.07 M
Answer:
When you prepared the solution, you will find that the molarity is 0.73M
Explanation:
First of all you should get by the periodic table, molar mass in the anhydrouds calcium.
CaCl2 · 0H20 = 110.98 g/m
Now we have to find out how many mols are 15 g.
So 15g / 110.98 g/m = 0.135 moles
This moles are in 185 ml of water. Molarity as you should know are moles of solute in 1 L of solution (either 1000 ml)
185 ml ______ 0.135 moles
1000 ml _____ x x = (1000*0.135) /185 = 0.730M
Evaporation of Cane Sugar Solutions. An evaporator is used to concentrate cane sugar solutions. A feed of 10 000 kg/d of a solution containing 38 wt % sugar is evaporated, producing a 74 wt % solution. Calculate the weight of solution produced and amount of water removed.
Answer:
Weight of solution produced = 5135 kg
Amount of water removed = 4865 kg
Explanation:
For the balance of mass, the incoming mass of sugar must be equal to the outgoing mass. So, the incoming mass (mi) is 38% of 10000 kg
mi = 0.38x10000 = 3800 kg
The outgoing mass (mo) must be 3800 kg, and it is 74% of the total mass (mt)
mo = 0.74xmt
0.74xmt = 3800
mt = 3800/0.74
mt = 5135 kg
This is the mass of solution produced.
The amount of water removed (wr) is the amount of water incoming (wi) less the amount of water outgoing (wo). Both will be the total mass less the mass of sugar :
wi = 10000 - 3800 = 6200 kg
wo = 5135 - 3800 = 1335 kg
wr = wi - wo
wr = 6200 - 1335
wr = 4865 kg
From the data provided and the concept of balance of mass,, the mass of solution produced is 5135 kg and the water removed is 4865 kg
What is evaporation?
Evaporation is the process by which molecules of a liquid turn to gas.
To calculate the weight of solution produced and amount of water removed:
Using the concept of balance of mass, the incoming mass of sugar must be equal to the outgoing mass.
Incoming mass (Mi) = 38% of 10000 kg
Mi = 3800 kg
Therefore, the outgoing mass (Mo)= 3800 kg
Mo= 74% of the total mass, Mt
Mo = 0.74 x Mt
Mt = 3800/0.74
Mt = 5135 kg
Thus, the mass of solution produced is 5135 kg
The amount of water removed (Wr) = water incoming (Wi) - water outgoing (wo).
Wi = 10000 - 3800 = 6200 kg
Wo = 5135 - 3800 = 1335 kg
Wr = Wi - Wo
Wr = 6200 - 1335
Wr = 4865 kg
Therefore, the water removed is 4865 kg
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A sample of lake water was analyzed to determine the amount of metals found in the lake. The standard deviation of the sampling method was found to be ±6.0%±6.0% . The standard deviation of the analytical method used to determine the amount of metals in the sample was determined to be ±2.4%±2.4% . What is the overall standard deviation?
Answer:
The overall standard deviation, s = 6.46 %
Given:
Sampling variance, [tex]s_{b} = \pm 6.0% = 0.06[/tex]
Analytical variance, [tex]s_{a} = \pm 2.4% = 0.024[/tex]
Solution:
Variance additive is given by:
[tex]s^{2} = s_{a}^{2} + s_{b}^{2}[/tex] (1)
where
s = overall variance
Also, we know that:
Standard Deviation, [tex]\sigma = \sqrt{variance}[/tex]
Therefore the standard deviation of the sampling, analytical and overall sampling is given by taking the square root of eqn (1) on both the sides:
[tex]s = \sqrt{s_{a}^{2} + s_{b}^{2}}[/tex]
[tex]s = \sqrt{0.024^{2} + 0.06^{2}} = 0.0646[/tex]
s = 6.46 %
The overall standard deviation considering both the sampling method (±6.0%) and analytical method (±2.4%) is calculated by squaring each to get variances, summing these, and taking the square root of the sum, resulting in an overall standard deviation of ±6.464%.
To calculate the overall standard deviation when given two separate standard deviations from different sources (sampling and analytical method), you should first square each standard deviation to find the variances. Then, add the variances together, and finally, take the square root of the total to find the overall standard deviation.
The standard deviation of the sampling method is ±6.0%, and this value squared gives 36.0%. The analytical method's standard deviation is ±2.4%, which squared gives 5.76%. Adding these two variances together, we get 36.0% + 5.76% = 41.76%. The overall standard deviation is then the square root of 41.76%, which is approximately ±6.464%.
Therefore, the combined or overall standard deviation considering both the sampling and analytical methods is ±6.464%.
A buffer, consisting ofH2PO4-
andHPO42-, helps control the pH of
physiologicalfluids. Many carbonated soft drinks also use this
buffer system.What is the pH of a soft drink in which the major
bufferingredients are 6.70 g ofNaH2PO4 and
6.50g of Na2HPO4 per 355 mL
ofsolution?
Answer:
The pH of the drink is 7.12
Explanation:
First, we calculate the concentration of NaH₂PO₄ and Na₂HPO₄, using their molecular weight and the volume in L (355 mL= 0.355 L):
[NaH₂PO₄] = [tex]\frac{6.70g}{0.355L*120g/mol}= 0.1573 M[/tex]
[Na₂HPO₄] = [tex]\frac{6.50g}{0.355L*142g/mol} = 0.1289 M[/tex]
Now we calculate the pH of the solution, keeping in mind the equilibrium:
H₂PO₄⁻ ↔ HPO₄⁻² + H⁺From literature, we know that the pka for the previous equilibrium is 7.21
The equation that gives us the pH of a buffer solution is the Henderson–Hasselbalch equation:
pH = pka + [tex]log\frac{[HPO4^{-2} ]}{[H2PO4^{-} ]}[/tex]
Replacing in the equation the data we know gives us:
[tex]pH=7.21+log\frac{0.1289M}{0.1573M} \\pH=7.12[/tex]
33.56 g of fructose (C6H,206) and 18.88 g of water are mixed to obtain a 40.00 ml solution a. What is this solution's density? b. What is the mole fraction of fructose in this solution? c. What is the solution's average molar mass? d. What is the specific molar volume of the solution?
Explanation:
Mass of fructose = 33.56 g
Mass of water = 18.88 g
Total mass of the solution = Mass of fructose + Mass of water = M
M = 33.56 g + 18.88 g =52.44 g
Volume of the solution = V = 40.00 mL
Density =[tex]\frac{Mass}{Volume}[/tex]
a) Density of the solution:
[tex]\frac{M}{V}=\frac{52.44 g}{40.00 mL}=1.311 g/mL[/tex]
b) Molar mass of fructose = 180.16 g/mol
Moles of fructose = [tex]n_1=\frac{ 33.56 g}{180.16 g/mol}=0.1863 mol[/tex]
Molar mass of water = 18.02 g/mol
Moles of water= [tex]n_2=\frac{ 18.88 g}{18.02 g/mol}=1.0477 mol[/tex]
Mole fraction of fructose in this solution:[tex]\chi_1[/tex]
[tex]\chi_1=\frac{n_1}{n_1+n_2}=\frac{0.1863 mol}{0.1863 mol+1.0477 mol}[/tex]
[tex]\chi_1=0.1510[/tex]
Mole fraction of water = [tex]\chi_2=1-\chi_1=0.8490[/tex]
c) Average molar mass of of the solution:
=[tex]\chi_1\times 180.16 g/mol+\chi_2\times 18.02 g/mol[/tex]
[tex]=0.1510\times 180.16 g/mol+0.8490\times 18.02 g/mol=42.50 g/mol[/tex]
d) Mass of 1 mole of solution = 42.50 g/mol
Density of the solution = 1.311 g/mL
d) Specific molar volume of the solution:
[tex]\frac{\text{Average molar mass}}{\text{Density of the mass}}[/tex]
[tex]=\frac{42.50 g/mol}{1.311 g/mL}=32.42 mL/mol[/tex]
Name the type of bond in organic chemistry that corresponds to a glycoside bond
Answer:
covalent bond
Explanation:
The bond which is most common in the organic molecules is the covalent bond which involves sharing of the electrons between the two atoms.
Glycosidic bond, also known as glycosidic linkage is type of the covalent bond which joins carbohydrate molecule to other group that may not or may be a carbohydrate.
Glycosidic bond is the bond which is formed between hemiacetal or hemiketal group of the saccharide and hydroxyl group of compounds like alcohol.
CH4 with pressure 1 atm and volume 10 liter at 27°C is passed into a reactor with 20% excess oxygen, how many moles of oxygen is left in the products?
Answer : The moles of [tex]O_2[/tex] left in the products are 0.16 moles.
Explanation :
First we have to calculate the moles of [tex]CH_4[/tex].
Using ideal gas equation:
[tex]PV=nRT[/tex]
where,
P = pressure of gas = 1 atm
V = volume of gas = 10 L
T = temperature of gas = [tex]27^oC=273+27=300K[/tex]
n = number of moles of gas = ?
R = gas constant = 0.0821 L.atm/mol.K
Now put all the given values in the ideal gas equation, we get:
[tex](1atm)\times (10L)=n\times (0.0821L.atm/mol.K)\times (300K)[/tex]
[tex]n=0.406mole[/tex]
Now we have to calculate the moles of [tex]O_2[/tex].
The balanced chemical reaction will be:
[tex]CH_4+2O_2\rightarrow CO_2+2H_2O[/tex]
From the balanced reaction we conclude that,
As, 1 mole of [tex]CH_4[/tex] react with 2 moles of [tex]O_2[/tex]
So, 0.406 mole of [tex]CH_4[/tex] react with [tex]2\times 0.406=0.812[/tex] moles of [tex]O_2[/tex]
Now we have to calculate the excess moles of [tex]O_2[/tex].
[tex]O_2[/tex] is 20 % excess. That means,
Excess moles of [tex]O_2[/tex] = [tex]\frac{(100 + 20)}{100}[/tex] × Required moles of [tex]O_2[/tex]
Excess moles of [tex]O_2[/tex] = 1.2 × Required moles of [tex]O_2[/tex]
Excess moles of [tex]O_2[/tex] = 1.2 × 0.812 = 0.97 mole
Now we have to calculate the moles of [tex]O_2[/tex] left in the products.
Moles of [tex]O_2[/tex] left in the products = Excess moles of [tex]O_2[/tex] - Required moles of [tex]O_2[/tex]
Moles of [tex]O_2[/tex] left in the products = 0.97 - 0.812 = 0.16 mole
Therefore, the moles of [tex]O_2[/tex] left in the products are 0.16 moles.
Answer the following using the following information: ∆Hfus=6.02 kJ/mol; ∆Hvap= 40.7 kJ/mol; specific heat of water is 4.184 J/g∙˚C; specific heat of ice is 2.06 J/g∙˚C; specific heat of water vapor is 2.03 J/g∙˚C.
A. How much heat is required to vaporize 25 g of water at 100˚C?
B. How much heat is required to convert 25 g of ice at -4.0 ˚C to water vapor at 105 ˚C (report your answer to three significant figures)?
C. An ice cube at 0.00 ˚C with a mass of 8.32 g is placed into 55 g of water, initially at 25 ˚C. If no heat is lost to the surroundings, what is the final temperature of the entire water sample after all the ice is melted (report your answer to three significant figures)?
Answer:
A. 56 kJ
B. 75.8 kJ
C. 11 ˚C
Explanation:
A. The heat of vaporization, ∆Hvap = 40.7 kJ/mol, gives the amount of energy per mole of water required to vaporize water to steam. The molar mass of water is 18.02 g/mol.
Q = M·∆Hvap = (25 g)(mol/18.02g)(40.7 kJ/mol) = 56 kJ
B. Five steps are necessary in this process. First, the ice will be warmed to 0 °C, then melted to water. The water will be heated to 100 °C, then vaporized. Finally, the vapor will be heated from 100 °C to 105 °C.
We calculate the heat required to warm the ice from -4.0 °C to 0 °C:
Q₁ = mcΔt = (25 g)(2.06 J∙g⁻¹˚C⁻¹)(0 °C - (-4.0 °C)) = 206 J
Then we calculate the heat required to melt the ice to water:
Q₂ = M∙∆Hfus = (25 g)(mol/18.02 g)(6.02 kJ/mol) = 8.35 kJ
Then, we calculate the heat required to warm the water from 0 °C to 100 °C.
Q₃ = mcΔt = (25 g)(4.184 J∙g⁻¹˚C⁻)(100 °C - 0 °C) = 10460 J
Then we calculate the heat required to vaporize the water:
Q₄ = M∙∆Hvap = (25 g)(mol/18.02 g)(40.7 kJ/mol) = 56.5 kJ
Finally, the vapor is heated from 100 °C to 105 °C.
Q₅ = mcΔt = (25 g)(2.03 J∙g⁻¹˚C⁻)(105 °C - 100 °C) = 254 J
The total heat required is the sum of Q₁ through Q₅
Qtotal = Q₁ + Q₂ + Q₃ + Q₄ + Q₅
Qtotal = (206 J)(1 kJ/1000J) + 8.35 kJ + (10460 J)(1 kJ/1000J) + 56.5 kJ + (254 J)(1 kJ/1000J)
Qtotal = 75.8 kJ
C. The heat required to melt the ice is provided by the water as it decreases in temperature.
First, we calculate the energy required to melt ice to water
Q = M∙∆Hfus = (8.32 g)(mol/18.02 g)(6.02 kJ/mol) = 2.779 kJ
There are at least two ways to solve this problem. Here, we will calculate the heat lost when all the water is brought to a temperature of 0 °C:
Q = mc∆t = (55 g)(4.184 J∙g⁻¹˚C⁻¹)(25 °C - 0°C) = 5753 J
We see that the water has enough energy to melt all of the ice. The residual heat energy of the water after melting all the ice is:
5753 J - (2.779 kJ)(1000J/kJ) = 2974 J
Now the problem becomes that we have (8.32 g + 55 g) = 63.32 g of water at 0 °C that will be raised to some final temperature by the residual heat of 2974 J:
Q = mcΔt ⇒ Δt = Q/(mc)
Δt = (2974 J) / (63.32 g)(4.184 J∙g⁻¹˚C⁻¹) = 11 ˚C
T(final) - T(inital) = 11 ˚C
T(final) = 11 ˚C + T(inital) = 11 ˚C + 0 ˚C = 11 ˚C
Thus, the final temperature will be 11 ˚C.
The heat required to vaporize 25g of water at 100˚C is 56.529 kJ. To convert 25g of ice at -4.0˚C to water vapor at 105˚C, approximately 83.8 kJ of heat is required. The final temperature of the water sample after the melting of an 8.32g ice cube is about 24.856˚C.
Explanation:A. To find the heat required to vaporize 25 g of water at 100˚C, we need to use the given heat of vaporization (∆Hvap) which is 40.7 kJ/mol. However, we need the molar mass of water which is approximately 18.015 g/mol. The calculation is as follows: (25 g) / (18.015 g/mol) * (40.7 kJ/mol) = 56.529 kJ. Therefore, the heat needed is 56.529 kJ.
B. To convert 25 g of ice at -4.0˚C to water vapor at 105˚C we first calculate the heat needed to melt the ice to water at 0˚C, then to heat the water from 0˚C to 100 ˚C, and finally to vaporize the water. When all of these are added up, the total heat can be calculated as approximately 83.8 kJ.
C. To find the final temperature of the entire water sample after the ice is melted, we set the heat gained by the ice equal to the heat lost by the water. Solving gives a final temperature of about 24.856˚C.
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A 41-g sample of potassium completely reacts with chlorine to form 78 g of potassium chloride. How many grams of chlorine must have reacted?
Answer:
37.275g must react
Explanation:
Step 1: Balance the reaction
The reaction is as follow:
2 K + Cl2---------> 2KCl
This means that for 2 moles K we have 1 mole Cl2 ( and 2 moles KCl)
Step 2: Calculate moles
Moles of potassium: moles = mass / Molar mass
Moles of potassium = 41g / 39.1g/mol
Moles of potassium = 1.05 moles
for each 2 moles potassium we have 1 mole Cl2 ( and 2 moles KCl)
Moles of Chlorine : 1.05 / 2 = 0.525 moles ( and 1.05 moles of KCl)
Step 3: Calculate mass of chlorine
mass chlorine = moles x Molar mass = 0.525 x 35.5 x 2
= 37.275g Chlorine
The work function for metallic caesium is 2.14 eV. Calculate the
kinetic energy and the speed of the electrons ejected by light of
wavelength
(a) 750 nm,
(b) 250 nm.
The question involves calculating the kinetic energy and speed of photoelectrons ejected from metallic caesium due to incident light using the photoelectric effect. The photoelectron kinetic energy is found using the difference between the photon energy and the work function, and the speed is calculated from the kinetic energy.
Explanation:The question asks to calculate the kinetic energy and speed of electrons ejected from metallic caesium when exposed to light of different wavelengths, applying the principles of the photoelectric effect. The work function given for caesium is 2.14 eV.
To find the kinetic energy of the ejected electrons, we use the equation:
KEmax = E - φwhere KEmax is the maximum kinetic energy of the photoelectrons, E is the energy of the incident photons, and φ is the work function. The energy of the photons (E) can be calculated using:
E = (hc) / λwhere h is Planck's constant (4.135667696 × 10-15 eV·s), c is the speed of light (≈ 3.00 × 108 m/s), and λ is the wavelength of the incident light.
After finding KEmax, we can find the speed of the electrons using the equation:
v = √(2 * KE / m)where v is the speed of the electron, m is the mass of the electron (9.10938356 × 10-31 kg), and KE is the kinetic energy in joules. Remember to convert electron volts to joules (1 eV = 1.602 × 10-19 J) to use SI units.
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For 750 nm light, no electrons are ejected. For 250 nm light, the ejected electrons have a kinetic energy of 4.52 × [tex]10^{-19[/tex] J and a speed of 996 km/s.
Let's solve this problem step by step for both parts (a) and (b) using the photoelectric effect equation:
hf = Φ + KE
Where:
h = Planck's constant = 6.626 × [tex]10^{-34[/tex] J·sf = frequency of lightΦ = work function = 2.14 eVKE = kinetic energy of ejected electrons
convert eV to Joules: 1 eV = 1.602 × [tex]10^{-19[/tex] J
So, Φ = 2.14 eV = 2.14 × (1.602 × [tex]10^{-19[/tex] ) J = 3.42828 × [tex]10^{-19[/tex] J
Also, use c = λf, where c is the speed of light (3 × [tex]10^{8[/tex] m/s)
Part A) For λ = 750 nm = 750 × [tex]10^{-19[/tex] m1. Calculate frequency:
f = [tex]\frac{C}{\lambda}[/tex] = (3 × [tex]10^{8[/tex] m/s) ÷ (750 × [tex]10^{-19[/tex]) = 4 × [tex]10^{14[/tex] Hz
2. Calculate energy of photon:
E = hf = (6.626 × [tex]10^{-34[/tex])(4 ×[tex]10^{14[/tex]) = 2.6504 × [tex]10^{-19[/tex] J
3. Calculate kinetic energy:
KE = E - Φ = 2.6504 × [tex]10^{-19[/tex] - 3.42828 × [tex]10^{-19[/tex] = -0.77788 × [tex]10^{-19[/tex] J
The negative KE means no electrons are ejected for this wavelength.
Part B) For λ = 250 nm = 250 × [tex]10^{-9[/tex] m1. Calculate frequency:
f = [tex]\frac{C}{\lambda}[/tex] = (3 × [tex]10^{8[/tex]) ÷ (250 × [tex]10^{-9[/tex]) = 1.2 × [tex]10^{15[/tex] Hz
2. Calculate energy of photon:
E = hf = (6.626 × [tex]10^{-34[/tex])(1.2 × [tex]10^{15[/tex]) = 7.9512 × [tex]10^{-19[/tex] J
3. Calculate kinetic energy:
KE = E - Φ = 7.9512 × [tex]10^{-19[/tex] - 3.42828 × [tex]10^{-19[/tex] = 4.52292 × [tex]10^{-19[/tex]
Therefore Kinetic energy is 4.52 × [tex]10^{-19[/tex] J.
4. Calculate speed:
KE = [tex]\frac{1}{2} \times m v^2[/tex] v = [tex]\sqrt{\frac{2kE}{m}[/tex], where m is mass of electron = 9.1093 ×[tex]10^{-31[/tex] kg v = [tex]\sqrt{\frac{2 \times 4.52292 \times 10^{-19}}{9.1093 \times 10^{-31}}[/tex]= 995,999.7 m/s ≈ 996 km/sTherefore the speed is approximately 996 km/s.