Answer:
Explanation:
Carbohydrates are basically composed of Carbon, Hydrogen and Oxygen having the general formula CnH2nOn.
There are 3 types of carbohydrates which are dependent on the number of "n"
Monosaccharides which are n>3 (Triose) are the aldose and ketose.
They are the simpleat and smallest form and they are Glucose, fructose and galactose
Disaccharides are structure of the combination of the monosaccharides by glycosidic bond and they are sucrose, lactose, maltose etc
Polysaccharides are the largest and insoluble form of carbohydrates. They are cellulose, starch, glycogen etc.
Lipids(triglycerides) are solid fats or series of repeated fats at room temperature, they are insoluble in water both soluble in some organic solvents. They are also composed of glycerides (3 molecules). Its structure is composed of two parts, the soluble part composing of the -COOH group and the insoluble part that can be saturated or unsaturated hydrocarbon chain
Saturated fats - CH3(CH2)nCOOH
Their types are phospholipids, glycolipids etc
Proteins are polymers of peptides called polypeptides. The bond linking the structure together is called a peptide bond (-CONH-). They form chains of amino acid.
There are 4 levels of protein structures and they are
The primary structure defines the basic straight chain structure of an amino acid. They form the basis of genetic mutation.
Secondary structure involves the folding of this chain into alpha helix or beta pleated.
Tertiary structure is a 3-D structure that involves the hydrophobic and Hydrophilic parts pf the structure. The hydrophobic part apreads outwards while the hydrophyllic parts curve inwards by the action of van der waals forces.
Tertiary structure in this case is the example of the Haemoglobin
Nucleic acids is the building block for RNA and DNA (ribo- and Deoxyribonucleic acid). This is composed of a nitrogenous base which can either be purine or pyrimidine bases, a ribose sugar (5- Carbon sugar and phosphate group
The bond holding the nucleotides together is called phosphodiester bond.
Carbohydrates, lipids, proteins, and nucleic acids, are composed of monosaccharides, fatty acids, amino acids, and nucleotides respectively. They serve as energy sources, structural components, and information carriers. Each has a unique composition but all are vital for life.
Explanation:The building blocks of carbohydrates, lipids, proteins, and nucleic acids are monosaccharides, fatty acids, amino acids, and nucleotides, respectively. Carbohydrates are energy-generating compounds, where lipids are used mainly for long-term energy storage and insulating the body from cold. Proteins function in everything from immune response to structural support to chemical catalysis. Nucleic acids store and transmit genetic information.
Carbohydrates are molecules composed of carbon, hydrogen, and oxygen in a ratio of 1:2:1. Monosaccharides, are simple sugars like glucose and fructose. They serve as an immediate source of energy for organisms.
Lipids are composed of glycerol and fatty acids. They can be saturated or unsaturated, solid or liquid at room temperature. They function as a long-term energy storage molecule, provide insulation, and can serve as a metabolic water source.
Proteins are composed of amino acids linked by peptide bonds. They have a complex structure going from primary to quaternary. They serve multiple functions including enzymatic activity, transport, structural support, storage, and immune response.
Nucleic Acids, DNA and RNA, are made of nucleotides consisting of a pentose sugar, phosphate group, and nitrogenous base. They store genetic information and conduct protein synthesis.
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While driving your rental car on your vacation in Europe, you find that you are getting 8.8 km/L of gasoline. What does this value correspond to in miles per gallon?
Answer : The value correspond to in miles per gallon is, 20.6976 mile/gallon
Explanation :
The conversion used to convert kilometer to miles is:
1 km = 0.6214 miles
The conversion used to convert liter to gallon is:
1 L = 0.2642 gallons
Thus,
1 km/L = [tex]\frac{0.6214}{0.2642}mile/gallon[/tex]
1 km/L = 2.352 mile/gallon
As we are given that 8.8 km/L of gasoline. Now we have to convert it into mile/gallon.
As, 1 km/L = 2.352 mile/gallon
So, 8.8 km/L = [tex]\frac{8.8km/L}{1km/L}\times 2.352mile/gallon[/tex]
= 20.6976 mile/gallon
Thus, the value correspond to in miles per gallon is, 20.6976 mile/gallon
To convert the fuel economy from km/L to mpg, multiply the km/L value by 2.8248.
Explanation:To convert the fuel economy from kilometers per liter (km/L) to miles per gallon (mpg), we can use the conversion factor:
1 km/L = 2.8248 mpg
So, to find the fuel economy in miles per gallon, we multiply 8.8 km/L by 2.8248:
8.8 km/L × 2.8248 mpg = 24.89904 mpg
Therefore, the fuel economy of the rental car is approximately 24.9 miles per gallon.
Aluminum is more reactive than iron is, but iron corrodes more quickly. This occurs because aluminum is reduced in the presence of oxygen while iron is oxidized. aluminum develops a coating of aluminum oxide, which protects it from further reaction. iron has a lower density than aluminum does. iron is exposed to both oxygen and water, while aluminum is rarely exposed to water.
Answer:aluminum develops a coating of aluminum oxide,
Explanation:
Aluminium has an electrode potential value of -1.66V while iron has an electrode potential value of -0.44V for iron II and +0.77 V for iron III. Clearly, aluminum has a more negative value of electrode potential and ought to be more reactive. However, a protective coating formed on aluminium surface prevents corrosion of the metal.
MgCl2 + NH4NO3 --> Mg NO3 + NH4Cl is this a double replacement reaction?
Answer:
The answer to your question is Yes, it is a double replacement reaction
Explanation:
General equation of a double replacement reaction
AB + CD ⇒ AD + BC
In a double replacement reaction, the cation of a compound is interchanged with the cation and another compound and also the anions are interchanged. The reaction given has the characteristics of a double replacement reaction.
Describe a sigma bond.A) overlap of two f orbitalsB) end to end overlap of p orbitalsC) s orbital overlapping with the side of a p orbitalD) side by side overlap of p orbitalsE) p orbital overlapping with a d orbital
Answer:
B - End to end overlap of p overlap
Explanation:
Sigma bonds (σ bonds) are the strongest type of covalent chemical bondSigma bonds are formed by end-to-end overlapping. Sigma bonds can occur between any kind of atomic orbitals;The combination of overlapping orbitals can be s-s, s-pz or pz-pz.
Silver has two naturally occurring isotopes with the following isotopic masses: 47Ag 107 – 106.90509 47Ag 109 – 108.9047 The average atomic mass of silver is 107.8682 amu. The abundance of the lighter of the two isotopes is __________. 3/1
A) 0.2422B) 0.4816C) 0.5184D) 0.7578E) 0.9047
Answer:
c) .51835
Explanation:
Let the relative abundance of the lighter of the two isotopes be X we have
Then the relative abundance of the heavier isotope is then (1-X)
Whereby we have that in nature the amount of the lighter silver found in proportion is X and the heavier isotope of silver is present as (1-X) proportion in nature.
To calculate the relative atomic mass of silver, we have
(Mass of light weight silver)×X + (mass of heavier isotope of silver×(1-X) = relative atomic mass of silver
106.90509(X) + 108.9047(1-X)
108.9-108.9(x)+106.9(x) = 107.87
-2x-1.03 = 0.517450902926
Closest answer is c
c) .5184
The relative atomic mass of isotopes is the weighted average by the mole-fraction of abundance of these isotopes which gives the atomic weight that is listed for that element on the periodic table.
Given the following reaction:
[tex]C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O[/tex]
How many moles of CO₂ will be produced from 79.0 g of C₃H₈, assuming O₂ is available in excess?
Answer:
5.38 moles of CO₂ are produced
Explanation:
This is the reaction:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
First of all, let's convert the mass of C₃H₈ to moles (mass / molar mass)
79 g / 44 g/mol = 1.79 moles
So ratio is 1:3.
1 mol of C₃H₈ is needed to produce 3 moles of CO₂
1.79 moles of C₃H₈ would produce (1.79 .3) /1 = 5.38 moles
I need help with organic chemistry!!!
Answer:
CHO2- ion
Explanation:
We have the lewis structure of a formate-ion here
This is CHO2-.
The carbon atom is the central atom in the structure. It's the least electronegative atom (C). Carbon has 4 valence electrons. Oxygen has 6 valence electrons.
The carbon will bind with 1 hydrogen atom, this will form 1 single bond, because hydrogen has 1 valence electron.
The carbon will bind with oxygen via a double bond.
Since carbon has only 4 valence electrons, it can only form 1 bond with the other oxygen atom.
There will formed 1 double bond between C and O and 1 single bond between C and O resulting in a negative charged O-atom.
This means there are two resonance structures. for the CHO2- ion
Organic chemistry is a branch of chemistry studying carbon-containing compounds. While it traditionally focused on naturally occurring compounds, it now includes human-made substances. Applications range from pharmaceuticals to plastics.
Explanation:Organic chemistry is a sub-discipline of chemistry that involves the study of the structure, properties, composition, reactions, and preparation of carbon-containing compounds, which includes not only hydrocarbons but also compounds with any number of other elements, including hydrogen (most compounds contain at least one carbon-hydrogen bond), nitrogen, oxygen, halogens, phosphorus, silicon, and sulfur.
This branch of chemistry was traditionally limited to compounds produced by living organisms but has been broadened to include human-made substances such as plastics. The range of application of organic compounds is enormous and also includes, but is not limited to, pharmaceuticals, dyes, coatings, plastics, and many more.
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When an ionic compound dissolves in water:_________.
a. the negative ends of water molecules surround the positive ions.
b. the negative ends of water molecules surround both the negative and the positive ions.
c. the positive ends of water molecules surround the positive ions.
d. the negative ends of water molecules surround the negative ions.
Answer:
a. the negative ends of water molecules surround the positive ions.
Explanation:
Ionization reaction is defined as the reaction in which an ionic compound dissociates into its ions when dissolved in aqueous solution.
Thus,
[tex]NaCl_{(s)}\rightarrow Na^+_{(aq)}+Cl^-_{(aq)}[/tex]
When the ionic compound dissociates, the cation which is positively charged gets attracted to the oxygen (Negative ends) which is present in the water. The anion which is negatively charged gets attracted to the hydrogen (Positive ends) of the water.
Hence, the correct option is:- a. the negative ends of water molecules surround the positive ions.
When an ionic compound dissolves in water, the positive end of the water molecule surrounds the negative ions, and the negative end of the water molecule surrounds the positive ions.
Explanation:When an ionic compound dissolves in water, the water molecules surround the individual ions and pull them apart through a process known as dissolving or hydration. Specifically, the positive end of the water molecule (which is the hydrogen end) surrounds the negative ions, while the negative end of the water molecule (which is the oxygen end) surrounds the positive ions. So, the correct answer is: (a) the negative ends of the water molecules surround the positive ions and (c) the positive ends of water molecules surround the negative ions. Therefore, water is described as a polar solvent, due to its capability to dissolve many ionic substances.
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An aluminum wire having a cross-sectional area equal to 5.40 10-6 m2 carries a current of 5.50 A. The density of aluminum is 2.70 g/cm3. Assume each aluminum atom supplies one conduction electron per atom. Find the drift speed of the electrons in the wire.
Answer:
Drift speed of electrons will be 1.056x10^-4 m/s
ExplanationGiven Data:
A(area)= 5.4 x 10^-6 [tex]m^{2}[/tex]
I(current)= 5.5 A
Density= 2.7 [tex]g/cm^{3}[/tex]
Calculation:The equation for drift velocity is:
[tex]v(drift)=I/nqA[/tex]
In this case 'q' will be charge of electron which is= 1.6 x 10-19
As each atoms supplies one conduction electron, so number of conduction electrons will be equal to number of atoms.
Hence,
n= no. of conduction electrons/[tex]m^{3}[/tex] = no. of atoms/[tex]m^{3}[/tex]
To find 'n' we can use following equation:
[tex]n= (mass/cm^{3} *atoms/mol)/(mass/mol)[/tex]
We know atoms/mol is equal to Avogadro`s number i.e 6.02 x 10^23
and molar mass of aluminium is 26.982 g.
Now,
[tex]n=(2.7g/cm^{3} * 6.02*10^{23} )/25.982g[/tex] (putting values in above equation)
[tex]n=6.024*10^{22} electrons/cm^{3}[/tex]
[tex]n= 6.024*10^{22} *10^{6} electrons/m^{3}[/tex] (converting electrons/cm3 to electrons/m3)
[tex]n= 6.024*10^{28} electrons/m^{3}[/tex]
To find drift velocity, we will use equations mention before:
[tex]v(drift)=I/nqA[/tex]
[tex]v(drift)=5.5A/(6.024*10^{28}electrons/m^{3} *1.6*10^{-19}C* 5.4*10^{-6}m^{2} )[/tex]
[tex]v(drift)= 1.056*10^{-4} m/s[/tex]
The drift velocity of electrons in the aluminum wire is 1.353 × 10^-3 m/s.
Explanation:The drift velocity of electrons in a wire can be calculated using the formula Vd = I / (nqA), where Vd is the drift velocity, I is the current, n is the number of free electrons per unit volume, q is the charge of an electron, and A is the cross-sectional area of the wire.
In this case, the cross-sectional area of the aluminum wire is given as 5.40 × 10-6 m2 and the current is 5.50 A. The charge of an electron is -1.60 × 10-19 C. To calculate the number of free electrons per unit volume, we need to know the density of aluminum and the molar mass of aluminum.
Using the given data, the drift velocity of the electrons in the aluminum wire can be calculated as follows:
Vd = (5.50 A) / [(8.44 × 1028 m-3) × (-1.60 × 10-19 C) × (5.40 × 10-6 m2)] = 1.353 × 10-3 m/s.
Using complete subshell notation (not abbreviations, 1s 22s 22p 6 , and so forth), predict the electron configuration of each of the following atoms: (a) C (b) P (c) V (d) Sb (e) Sm
Answer: The electronic configuration of the elements are written below.
Explanation:
Electronic configuration is defined as the representation of electrons around the nucleus of an atom.
Number of electrons in an atom is determined by the atomic number of that atom.
For the given options:
Option a: Carbon (C)Carbon is the 6th element of the periodic table. The number of electrons in carbon atom are 6.
The electronic configuration of carbon is [tex]1s^22s^22p^2[/tex]
Option b: Phosphorus (P)Phosphorus is the 15th element of the periodic table. The number of electrons in phosphorus atom are 15.
The electronic configuration of phosphorus is [tex]1s^22s^22p^63s^23p^3[/tex]
Option c: Vanadium (V)Vanadium is the 23rd element of the periodic table. The number of electrons in vanadium atom are 23.
The electronic configuration of vanadium is [tex]1s^22s^22p^63s^23p^64s^23d^3[/tex]
Option d: Antimony (Sb)Antimony is the 51st element of the periodic table. The number of electrons in antimony atom are 51.
The electronic configuration of antimony is [tex]1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^24d^{10}5p^3[/tex]
Option e: Samarium (Sm)Samarium is the 62nd element of the periodic table. The number of electrons in samarium atom are 62.
The electronic configuration of samarium is [tex]1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^24d^{10}5p^66s^24f^6[/tex]
Hence, the electronic configuration of the elements are written above.
The electrons configurations of the elements C, P, V, Sb, and Sm were determined using complete subshell notation. It depicts the filling of electrons into s, p, d, and f orbitals at different energy levels according to the Aufbau Principle.
Explanation:The electron configurations in complete subshell notation for the elements specified (C, P, V, Sb, and Sm) would be as following:
C (Carbon): 1s² 2s² 2p²: Carbon has 6 electrons. The two 1s electrons fill the first energy level, and of its four remaining electrons, two fill the second energy level's s subshell (2s) and the other two fill two of the six available p orbitals (2p). P (Phosphorus): 1s² 2s² 2p6 3s² 3p3: Phosphorus has 15 electrons. The 1s, 2s, and 2p subshells are filled just like in Carbon. The remaining five electrons fill the third level's s and p subshells. V (Vanadium): 1s² 2s² 2p6 3s² 3p6 4s² 3d3: Vanadium has 23 electrons, which fill the same subshells filled by Phosphorus and also extend into the fourth level's s subshell and the third level's d subshell. Sb (Antimony): 1s² 2s² 2p6 3s² 3p6 4s² 3d10 4p6 5s² 4d10 5p3. Antimony with 51 electrons extend into further energy levels. Sm (Samarium): 1s² 2s² 2p6 3s² 3p6 4s² 3d10 4p6 5s² 4d10 5p6 6s² 4f6. Samarium, a lanthanide, has electrons filling into the f orbital.Learn more about Electron Configuration here:
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A tanker truck carrying 6.05×103 kg of concentrated sulfuric acid solution tips over and spills its load. The sulfuric acid solution is 95.0%H2SO4 by mass and has a density of 1.84 g/mL.
Part A
Sodium carbonate (Na2CO3) is used to neutralize the sulfuric acid spill. How many kilograms of sodium carbonate must be added to neutralize 6.05×103 kg of sulfuric acid solution?
Express your answer with the appropriate units
Answer:
6,216.684 kilograms of sodium carbonate must be added to neutralize [tex]6.05\times 10^3 kg[/tex] of sulfuric acid solution.
Explanation:
Mass of sulfuric acid solution = [tex]6.05\times 10^3 kg=6.05\times 10^6 g[/tex]
[tex]1 kg = 10^3 g[/tex]
Percentage mass of sulfuric acid = 95.0%
Mass of sulfuric acid = [tex]\frac{95.0}{100}\times 6.05\times 10^6 g[/tex]
[tex]=5,747,500 g[/tex]
Moles of sulfuric acid = [tex]\frac{5,747,500 g}{98 g/mol}=58,647.96 mol[/tex]
[tex]H_2SO_4+Na_2CO_3\rightarrow Na_2SO_4+CO_2+H_2O[/tex]
According to reaction , 1 mole of sulfuric acid is neutralized by 1 mole of sodium carbonate.
Then 58,647.96 moles of sulfuric acisd will be neutralized by :
[tex]\frac{1}{1}\times 58,647.96 mol=58,647.96 mol[/tex] of sodium carbonate
Mass of 58,647.96 moles of sodium carbonate :
[tex]106 g/mol\times 58,647.96 mol=6,216,683.76 g[/tex]
6,216,683.76 g = 6,216,683.76 × 0.001 kg = 6,216.684 kg
6,216.684 kilograms of sodium carbonate must be added to neutralize [tex]6.05\times 10^3 kg[/tex] of sulfuric acid solution.
To neutralize 6.05×10^3 kg of sulfuric acid solution, we would need approximately 6.21×10^3 kg of sodium carbonate, based on the stoichiometry of the chemical reaction between these two compounds.
Explanation:The problem involves a chemical reaction between sulfuric acid (H2SO4) and sodium carbonate (Na2CO3). The balanced equation for this reaction is: H2SO4 + Na2CO3 -> Na2SO4 + H2O + CO2. From this equation, we can see that 1 mole of H2SO4 reacts with 1 mole of Na2CO3. Thus, the mass of Na2CO3 needed can be obtained by first determining the mass of H2SO4 present. Then, we convert the mass of H2SO4 to moles using its molar mass, and finally use the molar mass of Na2CO3 to find the mass of Na2CO3 needed.
First, we calculate the mass of H2SO4 in the solution: mass_H2SO4 = 0.95 * (6.05×10^3 kg of H2SO4 solution) = 5747.5 kg of H2SO4.
To convert this mass into moles, we divide by the molar mass of H2SO4, which is approximately 98.08 g/mol (or 0.09808 kg/mol). So, moles_H2SO4 = mass_H2SO4 / molar_mass_H2SO4 = 5747.5 kg / 0.09808 kg/mol = 58.6×10^3 mol.
From the balanced chemical equation, we know that 1 mole of H2SO4 neutralizes 1 mole of Na2CO3. Thus, to neutralize all the H2SO4, we need 58.6×10^3 mol of Na2CO3. Multiplying this number of moles by the molar mass of Na2CO3 (approximately 105.988 g/mol or 0.105988 kg/mol), we obtain the mass of Na2CO3 required: mass_Na2CO3 = moles_Na2CO3 * molar_mass_Na2CO3 = 58.6×10^3 mol * 0.105988 kg/mol = 6.21×10^3 kg of Na2CO3.
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A 22.25 g piece of metal is dropped into a graduated cylinder with 126.3 mL of water. If the density of the metal is 4.43 g/cm3, what is the total volume in the graduated cylinder in mL with the correct number of significant digits?
Answer:
131.2 mL would be the total volume in the graduated cylinder.
Explanation:
Let's work with the density of metal to find out its volume
Metal density = Metal mass / metal volume
4.46 g/mL = 22.25 g / metal volume
Metal volume = 22.25 g / 4.46 g/mL = 4.98 mL
If we suppose aditive volume, total volume in the graduated cylinder will be:
126.3mL + 4.9mL = 131.2 mL
An unknown compound contains only C, H, and O. Combustion of 6.10 g of this compound produced 14.9 g of CO₂ and 6.10 g of H₂O. What is the empirical formula of the unknown compound?
Answer:
The answer to your question is C₃H₃O
Explanation:
Data
Combustion of a compound C, H, O
mass = 6.10 g
mass CO2 = 14.9 g
mass of water = 6.10 g
Reaction
Cx Hy Oz + O2 ⇒ CO2 + H2O
Process
1.- Calculate the moles of carbon
44 g of CO2 -------------- 12 g of carbon
14.9 g of CO2 ------------- x
x = (14.9 x 12) / 44
x = 4.06 g
12 g of C ------------------ 1 mol
4.06 g ------------------- x
x = (4.06 x 1) / 12
x = 0.34 moles
2.- Calculate the moles of hydrogen
18 g of water ------------- 1 g of hydrogen
6.10 g of water ---------- x
x = (6.10 x 1) / 18
x = 0.33 g
1 g of H ---------------- 1 mol of H
0.33 g ---------------- x
x = (0.33 x 1) / 1
x = 0.33 moles of H
3.- Calculate the mass of oxygen
mass of Oxygen = 6.10 - 4.06 - 0.33
= 1.71 g
16 g of O --------------- 1 mol of O
1.71 g of O ------------- x
x = (1.71 x 1) / 16
x = 0.11 moles
4.- Divide by the lowest number of moles
Carbon = 0.34 / 0.11 = 3
Hydrogen = 0.33 / 0.11 = 3
Oxygen = 0.11 /0.11 = 1
5.- Write the empirical formula
C₃H₃O
Yeast and other organisms can convert glucose (C6H12O6) to Ethanol (CH3CH2OH) through a process called Alcoholic Fermentation. The Net reaction is:
C6H12O6------2C2H5OH(l) +2CO2(g)
Calculate the mass of glucose required to produce 2.25L of CO2 measured at P=1 atm and T=295k
Answer:
The answer to your question is 8.28 g of glucose
Explanation:
Data
Glucose (C₆H₁₂O₆) = ?
Ethanol (CH₃CH₂OH)
Carbon dioxide (CO₂) = 2.25 l
Pressure = 1 atm
T = 295°K
Reaction
C₆H₁₂O₆ ⇒ 2C₂H₅OH(l) +2CO₂(g)
- Calculate the number of moles
PV = nRT
Solve for n
[tex]n = \frac{PV}{RT}[/tex]
Substitution
[tex]n = \frac{(1)(2.25)}{(0.082)(295)}[/tex]
Simplification
n = 0.092
- Calculate the mass of glucose
1 mol of glucose --------------- 2 moles of carbon dioxide
x --------------- 0.092 moles
x = (0.092 x 1) / 2
x = 0.046 moles of glucose
Molecular weight of glucose = 180 g
180 g of glucose --------------- 1 mol
x g ---------------0.046 moles
x = (0.046 x 180) / 1
x = 8.28 g of glucose
the mass of potassium in a one cup serving of this cereal is determined to be 172 mg show a numerical setup for calculating the percent error for the mass of potassium to this serving.
Answer:
[(172 - 170)mg/170 mg]*100% = 2*100%/170 = 1.18%
Explanation:
Generally percent error is calculated by dividing the error by the actual value of the variable at standard conditions. In the problem above, the acceptable or actual mass of potassium in a cup of the cereal is 170 mg and the estimated mass of potassium in a cup of the cereal is 172 mg. Therefore, the percent error = [(172 - 170)mg/170 mg]*100% = 2*100%/170 = 1.18%
A helium balloon containing 0.100 mol of gas occupies a volume of 2.4 L at 25 C and 1.0 atm. how many moles have we added if we inflate it to 5.6 L?
Answer: 0.13mol
Explanation:Please see attachment for explanation
Write a balanced equation for the dissolution of sodium carbonate (Na2CO3) in water. Find the number of moles of Na+ produced when 0.207 mol of sodium carbonate dissolves.
Answer:
in the presence of water H2O
Na2CO3 (S) --> 2Na+ (aq)+ (CO3)2-(aq)
One mole of sodium carbonate produces two moles of Na+ ions
Therefore 0.207 moles produces 0.414 moles of Na+ ions
= 0.414 moles of Na+ ions
Explanation:
In water
Na2CO3 --> 2Na+ (aq)+ (CO3)2-(aq)
In a limited reaction, the carbonate ion reacts with the water molecules as follows
(CO3)2-(aq) + H2O←→HCO3-(aq) + OH-(aq)
sodium carbonate or soda ash dissolves in water to give 2 sodium cations and one carbonate anion
The balanced equation for the dissolution of sodium carbonate in water produces two moles of Na+ for every mole of Na2CO3. For 0.207 mol of sodium carbonate, it yields 0.414 mol of Na+ ions.
Explanation:The balanced equation for the dissolution of sodium carbonate (Na2CO3) in water is:
Na2CO3(s) → 2Na+(aq) + CO3^2-(aq)
When 0.207 mol of sodium carbonate dissolves in water, it produces two moles of Na+ ions for every mole of sodium carbonate. Therefore, to find the number of moles of Na+ produced we multiply:
0.207 mol Na2CO3 × 2 mol Na+ / 1 mol Na2CO3 = 0.414 mol Na+
A compound analyzes to be 37.5% C, 12.5% H, and 50.0% O. The molecular mass of the compound is 93.0. What is the molecular formula of the compound? (Type your answer using the format CH4 for CH4. Keep the elements in the order given.)
Answer: The molecular formula of the compound is [tex]C_3H_{12}O_3[/tex]
Explanation:
If percentage are given then we are taking total mass is 100 grams.
So, the mass of each element is equal to the percentage given.
Mass of C = 37.5 g
Mass of H = 12.5 g
Mass of O = 50.0 g
Step 1 : convert given masses into moles.
Moles of C =[tex]\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{37.5g}{12g/mole}=3.125moles[/tex]
Moles of H = [tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{12.5g}{1g/mole}=12.5moles[/tex]
Moles of O = [tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{50.0g}{16g/mole}=3.125moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C =[tex]\frac{3.125}{3.125}=1[/tex]
For H= [tex]\frac{12.5}{3.125}=4[/tex]
For O =[tex]\frac{3.125}{3.125}=1[/tex]
The ratio of C: H: O = 1: 4: 1
Hence the empirical formula is [tex]CH_4O[/tex].
The empirical weight of [tex]CH_4O[/tex] = 1(12)+4(1)+1(16)= 32g.
The molecular weight = 93.0 g/mole
Now we have to calculate the molecular formula.
[tex]n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{93}{32}=3[/tex]
The molecular formula will be=[tex]3\times CH_4O=C_3H_{12}O_3[/tex]
The molecular formula of the compound is C3H12O3. This was determined by converting percent composition to grams, then to moles to find a simplest ratio for the empirical formula and comparing its mass with the molecular mass.
Explanation:The compound could firstly be analyzed into its empirical formula by treating the percentage composition directly as masses, which would then provide us the simplest ratio of elements. Hence for 37.5g of C we have 37.5/12.01 = ~3.1 moles of C, for 12.5g of H we have 12.5/1.01 = ~12.4 moles of H, and for 50.0g of O we have 50.0/16.00 = ~3.1 moles of O. The simplest ratio (empirical formula) will then be C3H12O3, with empirical formula mass equals to 94.1 g/mol. This mass is very close to the mentioned molecular mass of 93.0 g/mol, so our molecular formula is presumably the same as the empirical formula. Therefore, the molecular formula of the compound is C3H12O3.
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If you want to use a serial dilution to make a 1/50 dilution. The first dilution you make is a 1/5 dilution with a total volume of 1 ml. To make that dilution you would add ______of your stock solution and______ of your solvent. Then you make a 1/10 dilution and add ______ of your first dilution and ______ of your solvent.Select one:a. 0.1 ml, 0.9 ml, 0.2 ml, 0.8 mlb. 0.3 ml, 0.7 ml, 0.1 ml, 0.9 mlc. 0.2 ml, 0.8 ml, 0.1 ml, 0.9 mld. 0.2 ml, 0.8 ml, 0.3 ml, 0.7 mle. None of these answer choice
Answer:
c.
Explanation:
A serial dilution is a dilution that is made fractionated. The stock solution is diluted, then this now solution is diluted, and then successively. The final dilution is the multiplication of the steps dilutions.
The representation of the dilution is v/v (volume per volume) indicates how much of the stock solution is in the total volume of the solution. So 1/5 indicates 1 mL to 5 mL of the solution. If the final volume must be 1 mL, then the stock solution must be 0.2 mL (0.2/1 = 1/5), and the volume of the solvent is 1 mL - 0.2 = 0.8 mL.
The second solution is done with a dilution of 1/10 or 1 mL of the first solution in 10 mL of the total solution. Because the solution has 1 mL, then the volume of the first solution must be 0.1 mL (0.1/1 = 1/10), and the volume of the solvent that must be added is 1 mL - 0.1 mL = 0.9 mL.
A 20.0 L container at 303 K holds a mixture of two gases with a total pressure of 5.00 atm. If there are 1.70 mol of Gas A in the mixture, how many moles of Gas B are present?
Answer:
moles B = 2.32 moles
Explanation:
In this case, we can assume that both gases are ideals, so we can use the expression for an ideal gas which is:
PV = nRT
From here, we can calculate the total moles (n) that are in the container, and then, by difference, we can calculate how much we have of gas B.
For this case, we will use R = 0.082 L atm / mol K. Solving for n:
n = PV/RT
n = 5 * 20 / 0.082 * 303
n = 4.02 moles
If we have 4.02 moles between the two gases, and we have 1.70 from gas A, then from gas B we simply have:
Total moles = moles A + moles B
moles B = Total moles - moles A
moles B = 4.02 - 1.70
moles B = 2.32 moles
We have 2.32 moles of gas B
Considering the ideal gas law, 2.32 moles of Gas B are present in the mixture.
Ideal gases are a simplification of real gases that is done to study them more easily. It is considered to be formed by point particles, do not interact with each other and move randomly. It is also considered that the molecules of an ideal gas, in themselves, do not occupy any volume.
The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:
P×V = n×R×T
where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas. The universal constant of ideal gases R has the same value for all gaseous substances.
In this case, you know:
P= 5 atmV= 20 Ln= ?R= 0.082[tex]\frac{atmL}{molK}[/tex]T= 303 KReplacing in the ideal gas law:
5 atm× 20 L= n× 0.082[tex]\frac{atmL}{molK}[/tex]× 303 K
Solving:
[tex]n=\frac{5 atmx20 L}{0.082\frac{atmL}{molK}x303 K }[/tex]
n= 4.02 moles
You have 4.02 moles between the two gases, and you have 1.70 from gas A. Then the number of moles of gas B can be calculated as:
Total moles = moles A + moles B
4.02 moles= 1.70 moles + moles B
4.02 moles - 1.70 moles= moles B
2.32 moles= moles B
Finally, 2.32 moles of Gas B are present in the mixture.
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https://brainly.com/question/4147359?referrer=searchResultsThe ship that will transport the terranauts to the core is built of what material?A. Cobalt B. Diamond C. Kryptonite D. Unobtainium
The ship meant to transport the 'terranauts' to the Earth's core could potentially be made of a hypothetical material, like 'Unobtainium', capable of enduring extreme conditions.
Explanation:The material of the ship transporting the terranauts to the core is not explicitly mentioned, but the answer would depend on the context provided by the source material (for instance, a science-fiction story or movie). In general, however, a material capable of withstanding extreme temperatures and pressures would be required for such a journey. This could potentially be a hypothetical, highly resilient element, such as Unobtainium (D).
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Using this information together with the standard enthalpies of formation of O2(g), CO2(g), and H2O(l) from Appendix C, calculate the standard enthalpy of formation of acetone.
Complete combustion of 1 mol of acetone (C3H6O) liberates 1790 kJ:
C3H6O(l)+4O2(g)?3CO2(g)+3H2O(l)?H?=?1790kJ
Answer:
ΔHacetone = - 247.5 kJ/mol
Explanation:
The enthalpy equation is as follows
ΣnΔHproducts – ΣmΔHreactants =ΔHreaction
3×ΣnΔH(CO2(g)) + 3×ΣnΔH(H2O) - ΣnΔH (C3H6O(l)) =ΔHreaction = -1790 kJ/mol
[3(-393.5) + 3(-285.8)] – ΔHacetone
= -1790 kJ/mol
(-1180.5 – 857.4)kJ/mol - ΔHacetone =
-1790 kJ/mol
-2037.9 kJ/mol - ΔHacetone
= -1790kJ/mol
-2037.9 kJ/mol + 1790kJ/mol = ΔHacetone
- 247.5 kJ/mol = ΔHacetone
ΔHacetone = - 247.5 kJ/mol
To determine the standard enthalpy of formation of acetone (C3H6O), we can use the combustion reaction equation and the standard enthalpies of formation of other compounds involved. By applying Hess's Law and the enthalpy change of the reaction, we can calculate the standard enthalpy of formation.
Explanation:The standard enthalpy of formation of acetone (C3H6O) can be calculated using the given equation for the combustion of acetone and the standard enthalpies of formation of O2(g), CO2(g), and H2O(l). For the combustion reaction:
C3H6O(l) + 4O2(g) → 3CO2(g) + 3H2O(l)
The enthalpy change or ΔH of the reaction is -1790 kJ. Using Hess's Law and the enthalpy change, we can determine the standard enthalpy of formation of acetone.
We can set up an equation using the standard enthalpy of formation values:
ΔH = Σ(ΔHf(products)) - Σ(ΔHf(reactants))
By rearranging the equation, we can solve for the standard enthalpy of formation of acetone.
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If a particular ore contains 55.1 % calcium phosphate, what minimum mass of the ore must be processed to
obtain 1.00 kg of phosphorus?
Express your answer with the appropriate units.
Answer:
6049.69 g or 6.04969 Kg of ore must be processed to get 1 Kg of phosphorous.
Explanation:
Given
An ore has 51% Calcium phosphate
To find
The minimum mass of ore to be processed to get 1.00 Kg of phosphorous
First find the mass of phosphorous in 1 mole = molar mass of Calcium phosphate, Ca₃(PO₄)₃.
Molar mass of Ca₃(PO₄)₃ is 310 g
Molar mass of P is 31 g
1 mole of Ca₃(PO₄)₃ has 3 atoms of phosphorous
i.e 310 g of Ca₃(PO₄)₃ has 3× 31g of P
= 93 g P
93 g P wil be present in 310 g of Ca₃(PO₄)₃
1 Kg or 1000 g of P will be in (1000÷93) ×310
=3333.33 g of Ca₃(PO₄)₃
but the ore has only 55.1% Ca₃(PO₄)₃
i.e
100 g of Ca₃(PO₄)₃ will have 55.1g Ca₃(PO₄)₃
we need 3333.33g of Ca₃(PO₄)₃
100 g of ore will have 55.1g Ca₃(PO₄)₃
3333.33g of Ca₃(PO₄)₃ will be present in
(3333.33÷ 55.1) ×100
= 6049.60 g of the ore
So 6049.69 g or 6.04969 Kg of ore must be processed to get 1 Kg of phosphorous.
Use the images to explain why carbon forms a bond with four chlorine atoms.
Explanation:
The Lewis dot diagram shows how electrons participate in a bond with Carbon and Chlorine. This is shown by the sticks and the 2 paired electrons near the carbon atom which represent the bonds. These electrons form these bonds because they form octets when they are bonded which most molecules and compounds follow
Hoped this helped, 2Trash4U
Answer: Chlorine and Carbon exhibits covalent bonding which involves the sharing if electrons. Carbon is less electronegative than Chlorine so it is considered as the central atom on this compound. Carbon shares it 4 valence electrons on Cl to attain octet. Now Chlorine is more electronegative so it will receive the lone pairs based from the Lewis structure.
Since C os in group 4 and Cl in group 7 the number of electrons will be:
C = 4
Cl= 7 x 4
# of e- = 32 - 8 ( the number of bonds)
Total number of e-. = 26 e-
Distribute the electrons to the most electronegative atom which is Cl have 6 e- each to attain octet.
Explanation:
For action potential generation and propagation, are there any other cation channels that could substitute for the voltage-gated sodium channels if the sodium channels were blocked?
Final answer:
Other cation channels, such as voltage-gated calcium and potassium channels, can substitute for the function of blocked voltage-gated sodium channels in action potential generation and propagation.
Explanation:
If the voltage-gated sodium channels are blocked, other cation channels can substitute for their function in action potential generation and propagation. One such cation channel is the voltage-gated calcium channel. Although calcium channels primarily play a role in synaptic transmission, they can also contribute to the depolarization phase of the action potential in the absence of functional sodium channels.
Another cation channel that can substitute for voltage-gated sodium channels is the voltage-gated potassium channel. While potassium channels primarily repolarize the membrane during the action potential, they can also contribute to the depolarization phase in the absence of sodium channels.
It is important to note that these cation channels may not fully replicate the function of voltage-gated sodium channels, and the exact impact on action potential generation and propagation can vary depending on the specific circumstances.
Can you determine the average atomic mass for an element if you know only the atomic number and mass number of an atom? If not, what other information would you need?
Answer: To determine the average atomic mass, we require abundance of every isotope in the nature.
Explanation:
Average atomic mass is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.
Formula used to calculate average atomic mass follows:
[tex]\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i[/tex]
For Example: There are 3 isotopes of silicon, Si-28, Si-29 and Si-30.
We know that:
Average atomic mass of silicon = 28.086 amu
In order to calculate the average atomic mass of an element, we need the atomic mass of every isotope and also the abundance of every isotope in the nature.
Percentage abundance of Si-28 isotope = 92.2 %
Percentage abundance of Si-29 isotope = 4.7 %
Percentage abundance of Si-30 isotope = 3.1 %
Hence, to determine the average atomic mass, we require abundance of every isotope in the nature.
Predict the mass of oxygen required to react with 14 g of nitrogen to make N2O5 if 16 g of oxygen reacts with 14 g of nitrogen to make N2O2.
The mass of oxygen reacted with 14 g nitrogen has been 16 grams.
The formation of dinitrogen dioxide has been mediated with the reaction of oxygen with nitrogen.
Mass of Oxygen requiredThe mass of oxygen required for the reaction has been given with the balanced chemical equation as:
[tex]\rm N_2\;+\;O_2\;\rightarrow\;N_2O_2[/tex]
Thus, for the reaction with 1 mole of oxygen, 1 mole of nitrogen has been required.
The mass of 1 mole oxygen has been 32 g.
The mass of 1 mole nitrogen has been 28 g.
Thus, for the reaction with 14 g nitrogen, the mass of oxygen required has been:
[tex]\rm 28\g\;N_2=32\;g\;O_2\\14\;g\;N_2=\dfrac{32}{28}\;\times\;14\g\;O_2\\14\;g\;N_2=165\;g\;O_2[/tex]
The mass of oxygen reacted with 14 g nitrogen has been 16 grams.
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The mass of oxygen required to react with 14 g of nitrogen to make [tex]N_2O_5[/tex] is 80 g.
The balanced chemical equation for the formation of [tex]N_2O_5[/tex] from nitrogen [tex](N_2)[/tex] and oxygen [tex](O_2)[/tex] is:
[tex]\[ N_2 + 5O_2 \rightarrow 2N_2O_5 \][/tex]
The balanced chemical equation for the formation of [tex]N_2O_2[/tex] from nitrogen [tex](N_2)[/tex] and oxygen [tex](O_2)[/tex] is:
[tex]\[ N_2 + 2O_2 \rightarrow 2N_2O_2 \][/tex]
From the second equation, we can see that 16 g of oxygen reacts with 14 g of nitrogen to form [tex]N_2O_2[/tex]. This gives us the molar ratio of oxygen to nitrogen for [tex]N_2O_2[/tex], which is 2:1 by mass.
Now, we need to find out the molar ratio of oxygen to nitrogen for [tex]N_2O_5[/tex]. From the first equation, the molar ratio of oxygen to nitrogen is 5:1.
Since the mass of nitrogen (14 g) is the same in both reactions, we can directly compare the ratios of oxygen used in both reactions.
For [tex]N_2O_2[/tex], we have 16 g of oxygen for every 14 g of nitrogen, and for [tex]N_2O_5[/tex], we need 5 times as much oxygen for the same amount of nitrogen.
Therefore, the mass of oxygen required for [tex]N_2O_5[/tex] is:
[tex]\[ \frac{5}{2} \times 16 \text{ g of oxygen} = 40 \text{ g of oxygen} \][/tex]
Using the molar masses of nitrogen (14 g/mol for [tex]N_2[/tex]) and oxygen (32 g/mol for [tex]O_2[/tex]), we can calculate the amount of oxygen needed:
[tex]\[ \frac{14 \text{ g of N2}}{28 \text{ g/mol of N2}} \times \frac{5 \text{ mol of O2}}{1 \text{ mol of N2}} \times 32 \text{ g/mol of O2} = 80 \text{ g of O2} \][/tex]
arrange the following solutions in order of decreasing freezing point: 0.10m Na3PO4; 0.35m NaCl; 0.20m MgCl2; 0.15m C6H12O6; 0.15m CH3COOH.
Answer:
0.15 m C6H12O6 > 0.15 m CH3COOH > 0.10 m Na3PO4 > 0.20 m MgCl2 > 0.35 m NaCl
Explanation:
Step 1:
0.10m Na3PO4
Na3PO4 → 3Na+ + PO4^-3-
⇒ van't Hoff factor is 3 + 1 = 4
0.10 m * 4 = 0.40
0.35m NaCl
NaCl → Na+ + Cl-
⇒ van't Hoff factor = 1+1 = 2
0.35 * 2 = 0.70
0.20m MgCl2
MgCl2 → Mg^2+ + 2Cl-
⇒ Van't Hoff factor = 1+2 = 3
0.20 * 3 = 0.60
0.15m C6H12O6
⇒ for non-ionic compounds in solution, like glucose (C6H12O6) , the van't Hoff factor is 1. They do not dissociate in water.
0.15 * 1 = 0.15
0.15m CH3COOH.
CH3CO2H ⇄ CH3CO2− + H+
⇒ Van't hoff factor ≈ 1<x<2
0.15 * 2 = 0.30
Larger the concentration of the concentration of the particles, smaller the freezing point.
0.15 m C6H12O6 > 0.15 m CH3COOH > 0.10 m Na3PO4 > 0.20 m MgCl2 > 0.35 m NaCl
The order of decreasing freezing point is -
0.15m C6H12O6 < 0.15m CH3COOH < 0.10 m Na3PO4 < 0.20 m MgCl2 < 0.35 m NaCl,
The Freezing Point DepressionIt is calculated as:
dT = i Kf m
m = The concentration values in molalities
Kf = the cryoscopic constant for water (-1.86, the same for all solutions)
i = the Van’t Hoff factor, which is the number of ion particles per individual molecule/formula of solute
Solution:
So you have to arrange in increasing order of the product i·m
0.15 m C6H12O6
im = 1 x 0.15 = 0.15 (undissociated, i=1)
0.15 m CH3COOH
im =1.1 x 0.15 = 0.17 (partially dissociated) ( 1<i<2)
0.10 m Na3PO4
im = 4 x 0.1 = 0.4 (i=4)
0.20 m MgCl2
im = 3 x 0.2 = 0.6 (i=3)
0.35 m NaCl,
im = 2 x 0.35 = 0.7 (i=2)
Thus, the order of decreasing freezing point is -
0.15m C6H12O6 < 0.15m CH3COOH < 0.10 m Na3PO4 < 0.20 m MgCl2 < 0.35 m NaCl,
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__________ bonds form between the oxygen and hydrogen within water molecules, whereas __________ bonds form between different water molecules.
Answer:
A. Covalent bond
B. Hydrogen bond
Explanation:
Covalent bonds form within a water molecule between the oxygen and hydrogen atoms, while hydrogen bonds form between different water molecules due to electrical attraction.
Explanation:In a water molecule, covalent bonds form between the oxygen and hydrogen atoms. This occurs because they share electrons to fulfill the Octet Rule for stable electron configurations. On the other hand, when different water molecules interact with each other, a form of weak connection known as hydrogen bonds are formed. This happens as the negatively charged oxygen end of one water molecule is attracted to the positively charged hydrogen end of another water molecule. This type of bonding results in some unique properties of water, such as its ability to act as a solvent for many substances.
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Nitrogen (N) is much more electronegative than hydrogen (H). Which of the following statements is correct about the atoms in ammonia (NH3)?
A) Each hydrogen atom has a partial positive charge.
B) The nitrogen atom has a strong positive charge.
C) Each hydrogen atom has a slight negative charge.
D) The nitrogen atom has a partial positive charge.
E) There are covalent bonds between the hydrogen atoms.
In the molecule ammonia (NH3), each hydrogen atom has a partial positive charge and the nitrogen atom has a partial negative charge. There are covalent bonds between the hydrogen atoms.
Explanation:In the molecule ammonia (NH3), each hydrogen atom has a partial positive charge (A) because nitrogen is more electronegative than hydrogen. The nitrogen atom, on the other hand, has a partial negative charge (D) due to the uneven distribution of electrons. The presence of covalent bonds between nitrogen and hydrogen atoms (E) results in the formation of ammonia.
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