The question involves calculating the CO2 emissions from the annual production of calcium oxide in the U.S., based on the stoichiometry of the reaction converting CaCO3 to CaO, which releases CO2.
Explanation:The question is about calculating the yearly release of CO2 to the atmosphere resulting from the annual production of calcium oxide (CaO) in the United States, given that the reaction for producing CaO from calcium carbonate (CaCO3) releases CO2. Since the reaction is CaCO3(s) → CaO(s) + CO2(g), for each mole of CaCO3 decomposed, one mole of CO2 is released. Calcium carbonate (mol weight = 100.09 g/mol) decomposes to give calcium oxide (mol weight = 56.08 g/mol) and carbon dioxide (mol weight = 44.01 g/mol). Given the annual production of CaO is 8.6 × 10¹° kg, we first convert this mass to moles (using CaO's molar mass), and then calculate the corresponding moles (and mass) of CO2 released.
"Determine the equilibrium constant, Kgoal, for the reaction 4PCl5(g)⇌P4(s)+10Cl2(g), Kgoal=? by making use of the following information: P4(s)+6Cl2(g)⇌4PCl3(g), K1=2.00×1019 PCl5(g)⇌PCl3(g)+Cl2(g), K2=1.13×10−2"
Answer:
Kgoal = 8.15 X 10⁻²⁸
Explanation:
The goal reaction is
4PCl₅(g)⇌P₄(s)+10Cl₂(g) Kgoal = ?
The given reactions are
P₄(s)+6Cl₂(g)⇌4PCl₃(g), K₁ = 2.00×10¹⁹
PCl₅(g)⇌PCl₃(g)+Cl₂(g), K₂ = 1.13×10⁻²
We can obtain the goal equation by
i) multiplying the second equation with four
ii) subtracting the equation one from above equation
We know that
i) If we multiply an equation with a number the equilbirium constant increases that times (we have to raise the power of equilibrium constant by that number)
ii) if we subtract two equations the equilibrium constants are divided
Kgoal = (K₂)⁴ / K₁
Kgoal = 8.15 X 10⁻²⁸
To determine the equilibrium constant, Kgoal, for the reaction 4PCl5(g)⇌P4(s)+10Cl2(g), we can use the given equilibrium constants for the two reactions involving the same species. By multiplying the equations for the given reactions, we can obtain the equation for the desired reaction and calculate the value of Kgoal. The value of Kgoal is found to be 2.26×1017.
Explanation:To determine the equilibrium constant, Kgoal, for the reaction 4PCl5(g)↔P4(s)+10Cl2(g), we can use the given equilibrium constants for the two reactions involving the same species. We start by writing the equation for the desired reaction as the sum of the two given reactions:
P4(s)+6Cl2(g)↔4PCl3(g), with K1=2.00×1019PCl5(g)↔PCl3(g)+Cl2(g), with K2=1.13×10-2By multiplying these two equations together, we can obtain the equation for the desired reaction and calculate the value of Kgoal. Multiplying the equations gives:
4PCl5(g)+24Cl2(g)↔16PCl3(g)+4P(s)+(4)(10)Cl2(g)
Since Kgoal is the product of the equilibrium constants for the forward and reverse reactions, we can calculate it as:
Kgoal = K1 × K2 = (2.00×1019)(1.13×10-2) = 2.26×1017
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The combustion of titanium with oxygen produces titanium dioxide:
Ti (s) + O2 (g) → TiO2 (s)
When 2.060 g of titanium is combusted in a bomb calorimeter, the temperature of the calorimeter increases from 25.00 °C to 91.60 °C. In a separate experiment, the heat capacity of the calorimeter is measured to be 9.84 kJ/K. The heat of reaction for the combustion of a mole of Ti in this calorimeter is __________ kJ/mol.
Ti = 47.867 amu
O2 = 31.9988 amu
TiO2 = 79.8650 amu
Report answer in scientific notation use En rather than x 10n
Answer:
The heat of reaction for the combustion of a mole of Ti in this calorimeter is [tex]7.769\times 10^4 kJ/mol[/tex].
Explanation:
[tex]Ti (s) + O_2 (g) \rightarrow TiO_2 (s)[/tex]
Moles of titanium =[tex]\frac{2.060 g}{47.867 g/mol}=0.04303 mol[/tex]
Heat absorbed by the bomb caloriometer on combustion of 0.04303 mol of titanium be Q
The heat capacity of the bomb caloriometer =c = 9.84 kJ/K
Change in temperature of the bomb caloriometer :
=ΔT=91.60 °C-25.00 °C=66.6 °C = 339.75 K
Q = c × ΔT
[tex]Q= 9.84 kJ/K\times 339.75 K=3,343.14 kJ[/tex]
3,343.14 kJ of heat energy was released when 0.04303 moles of titanium undergone combustion.
So for 1 mol of titanium:
[tex]\frac{3,343.14 kJ}{0.04303 moles}=77,693.237 kJ/mol=7.769\times 10^4 kJ/mol[/tex]
The heat of reaction for the combustion of a mole of Ti in this calorimeter is [tex]7.769\times 10^4 kJ/mol[/tex].
A heliox deep-sea diving mixture delivers an oxygen partial pressure of 0.10 atm when the total pressure is 12.5 atm . What is the partial pressure of helium in this mixture?
Answer : The partial pressure of helium in this mixture is, 12.4 atm.
Explanation : Given,
Partial pressure of oxygen = 0.10 atm
Total partial pressure = 12.5 atm
Now we have to calculate the partial pressure of helium.
According to the Dalton's law, the total pressure of the gas is equal to the sum of the partial pressure of the mixture of gasses.
[tex]P_T=p_{He}+p_{O_2}[/tex]
where,
[tex]P_T[/tex] = total partial pressure = 12.5 atm
[tex]P_{O_2}[/tex] = partial pressure of oxygen = 0.10 atm
[tex]P_{He}[/tex] = partial pressure of hydrogen = ?
Now put all the given values is expression, we get the partial pressure of the helium gas.
[tex]12.5atm=p_{He}+0.10atm[/tex]
[tex]p_{He}=12.4atm[/tex]
Therefore, the partial pressure of helium in this mixture is, 12.4 atm.
The partial pressure of helium in a deep-sea diving mixture of heliox, given a total pressure of 12.5 atm and an oxygen partial pressure of 0.10 atm, is 12.4 atm.
Explanation:The subject of this question pertains to the principles of gas laws and partial pressures in a gas mixture, specifically applicable to scuba diving environments where deep-sea divers use a unique mixture of gases like heliox (mixture of helium and oxygen).
When the total pressure is 12.5 atm and the oxygen has a partial pressure of 0.10 atm, the partial pressure of helium in the mixture can be found via subtraction. Hence, the partial pressure of helium in this mixture can be found by subtracting the partial pressure of oxygen from the total pressure. That is, 12.5 atm - 0.10 atm = 12.4 atm. In other words, helium, as part of this heliox mixture, offers a partial pressure of 12.4 atm.
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A weather balloon is infated to a volume of 26.5 L at a pressure of 745 mmHg and a temperature of 23.3 °C. The balloon rises in the atmosphere to an altitude where the pressure is 360. mmHg and the temperature is -13.7°C Assuming the balloon can freely expand, calculate the volume of the balloon at this altitude.
Answer: 48 Liters.
Explanation:
Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.
The combined gas equation is,
[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]
where,
[tex]P_1[/tex] = initial pressure of gas = 745 mmHg
[tex]P_2[/tex] = final pressure of gas =360 mmHg
[tex]V_1[/tex] = initial volume of gas = 26.5 L
[tex]V_2[/tex] = final volume of gas = ?
[tex]T_1[/tex] = initial temperature of gas = [tex]23.3^oC=(273+23.3)K=296.3K[/tex]
[tex]T_2[/tex] = final temperature of gas = [tex]-13.7^oC=273+(-13.7)K=259.3K[/tex]
Now put all the given values in the above equation, we get the final pressure of gas.
[tex]\frac{745mmHg\times 26.5L}{296.3K}=\frac{360mmhg\times V_2}{259.3K}[/tex]
[tex]V_2=48L[/tex]
Therefore, the volume of the gas will be 48 Liters.
Combine the two half-reactions that give the spontaneous cell reaction with the smallest E∘. Fe2+(aq)+2e−→Fe(s) E∘=−0.45V I2(s)+2e−→2I−(aq) E∘=0.54V Cu2+(aq)+2e−→Cu(s) E∘=0.34V
Answer: The spontaneous cell reaction having smallest [tex]E^o[/tex] is [tex]I_2+Cu\rightarrow Cu^{2+}+2I^-[/tex]
Explanation:
We are given:
[tex]E^o_{(Fe^{2+}/Fe)}=-0.45V\\E^o_{(I_2/I^-)}=0.54V\\E^o_{(Cu^{2+}/Cu)}=0.34V[/tex]
The substance having highest positive [tex]E^o[/tex] potential will always get reduced and will undergo reduction reaction. Here, iodine will always undergo reduction reaction, then copper and then iron.
The equation used to calculate electrode potential of the cell is:
[tex]E^o_{cell}=E^o_{oxidation}+E^o_{reduction}[/tex]
The combination of the cell reactions follows:
Case 1:Here, iodine is getting reduced and iron is getting oxidized.
The cell equation follows:
[tex]I_2(s)+Fe(s)\rightarrow Fe^{2+}(aq.)+2I^-(aq.)[/tex]
Oxidation half reaction: [tex]Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-[/tex] [tex]E^o_{oxidation}=0.45V[/tex]
Reduction half reaction: [tex]I_2(s)+2e^-\rightarrow 2I_-(aq.)[/tex] [tex]E^o_{reduction}=0.54V[/tex]
[tex]E^o_{cell}=0.45+0.54=0.99V[/tex]
Thus, this cell will not give the spontaneous cell reaction with smallest [tex]E^o_{cell}[/tex]
Case 2:Here, iodine is getting reduced and copper is getting oxidized.
The cell equation follows:
[tex]I_2(s)+Cu(s)\rightarrow Cu^{2+}(aq.)+2I^-(aq.)[/tex]
Oxidation half reaction: [tex]Cu(s)\rightarrow Cu^{2+}(aq.)+2e^-[/tex] [tex]E^o_{oxidation}=-0.34V[/tex]
Reduction half reaction: [tex]I_2(s)+2e^-\rightarrow 2I_-(aq.)[/tex] [tex]E^o_{reduction}=0.54V[/tex]
[tex]E^o_{cell}=-0.34+0.54=0.20V[/tex]
Thus, this cell will give the spontaneous cell reaction with smallest [tex]E^o_{cell}[/tex]
Case 3:Here, copper is getting reduced and iron is getting oxidized.
The cell equation follows:
[tex]Cu^{2+}(aq.)+Fe(s)\rightarrow Fe^{2+}(aq.)+Cu(s)[/tex]
Oxidation half reaction: [tex]Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-[/tex] [tex]E^o_{oxidation}=0.45V[/tex]
Reduction half reaction: [tex]Cu^{2+}(aq.)+2e^-\rightarrow Cu(s)[/tex] [tex]E^o_{reduction}=0.34V[/tex]
[tex]E^o_{cell}=0.45+0.34=0.79V[/tex]
Thus, this cell will not give the spontaneous cell reaction with smallest [tex]E^o_{cell}[/tex]
Hence, the spontaneous cell reaction having smallest [tex]E^o[/tex] is [tex]I_2+Cu\rightarrow Cu^{2+}+2I^-[/tex]
A chemist classifying five elements determines the following: Element Property S: Sulfur Powdery yellow solid Ag: Silver Shiny solid at room temperature that is also a good conductor of heat and electricity Si: Silicon Semiconductor Ne: Neon Inert gas at room temperature Hg: Mercury Silvery liquid that is used in thermometers and fluorescent lights Use these properties to classify the substances as metals, metalloids, or nonmetals.
Hey there!:
* S,and Ne are non-metals.
* Si is a metalloid.Generally metalloids acts as semiconductor.
* Ag and Hg are metals because metals exhibits lustre property, melleable and ductility properties.
Hope this helps!
Explanation:
Metals are the substances which lose electrons to attain stability and hence they form cations.
Metals are good conductors of heat and electricity and generally they are solid at room temperature. Metals have a shiny surface and they are also malleable and ductile.
For example, Ag: Silver Shiny solid at room temperature that is also a good conductor of heat and electricity. Hence, silver is a metal.
Non-metals are the substances which gain electrons to gain stability and hence they form anions.
Non-metals are bad conductors of heat and electricity. They are brittle and non-shiny in nature.
For example, S: Sulfur Powdery yellow solid is a non-metal.
Metalloids are the substances which show properties of both metals and non-metals.
Metalloids are moderately able to conduct heat and electricity. Generally metalloids are solid at room temperature. They are also able to react with other molecules.
For example, Si: Silicon Semiconductor is a metalloid.
Hence, the given substances are classified as follows.
S: Sulfur Powdery yellow solid - non-metalAg: Silver Shiny solid at room temperature that is also a good conductor of heat and electricity - metalSi: Silicon Semiconductor - metalloidNe: Neon Inert gas at room temperature - non-metalHg: Mercury Silvery liquid that is used in thermometers and fluorescent lights - metalOn a cool, rainy day, the barometric pressure is 746 mmHg. Calculate the barometric pressure in centimeters of water (cmH2O) (d of Hg = 13.5 g/mL; d of H2O = 1.00 g/mL).
Answer:
100.71 cm H2O
Explanation:
In a fluid column barometer, the height of the fluid column is proportional to the pressure. The pressure is by definition:
[tex]P=\frac{F}{A}[/tex], where F is a force and A is the area.
In a column barometer the force is given by the weight of the fluid:
[tex]F=m*g[/tex], and the mass may be expressed as [tex]m=p*V[/tex], where [tex]p[/tex] is the density and V is the volume.
Replacing this in the pressure definition:
[tex]P=\frac{pVg}{A}[/tex]
In a constant cross section area column, the volume may be calculated as:
[tex]V=A*h[/tex] where A is the area and h the height. Replacing this in the previous equation:
[tex]P=\frac{pAhg}{A}=pgh[/tex]
Different columns may be over the same pressure, so:
[tex]P_{w}=P_{Hg}\\ p_{w}h_{w}g=p_{Hg}h_{Hg}g\\[/tex]
Dividing each part for gravity constant:
[tex]p_{w}h_{w}=p_{Hg}h_{Hg}[/tex]
And isolating hw:
[tex]h_{w}=\frac{p_{Hg}h_{Hg}}{p_{w} } \\h_{w}=\frac{13.5*746}{1.00}=10071[/tex] mm
It is equal to 1007,1 cm.
Thischapteropenswithadiscussionofocean acidiication. a. Seawater contains many salts, including sodium chloride. Write its chemical formula. b. Sodium chloride is soluble in water. What chemical process takes place when solid sodium chloride dissolves? Hint: See Section 5.8
Answer:
a. NaCl
b. Dissociation (the compound breaking down into ions)
Identify the Brønsted-Lowry acid, the Brønsted-Lowry base, the conjugate acid, and the conjugate base in each reaction: (a) (CH3)3N(aq)+H2O(l)⇌(CH3)3NH+(aq)+OH−(aq) (b) HNO3(aq)+H2O(l)⇌H3O+(aq)+NO3−(aq) Drag the appropriate items to their respective bins.
Answer:
Explanation:
Writing the equation properly:
(CH₃)₃N + H₂O ⇄ (CH₃)₃NH⁺ + OH⁻
HNO₃ + H₂O ⇄ H₂O + NO₃⁻
The bronsted-lowry theory defines an acid as a proton donor and a base as a proton acceptor.
In a bronsted-lowry acid-base reaction, the original acid gives up its proton and becomes a conjugate base. Also, the original base accepts a proton and becomes a conjugate acid. For every acid, there is a conjugate base and for every base there is a conjugate acid.
What differentiates an acid from its conjugate base is a proton. The difference between a base and its conjugate acid is a proton.
(CH₃)₃N + H₂O ⇄ (CH₃)₃NH⁺ + OH⁻
Bronsted-lowry base acid conjugate acid conjugate base
HNO₃ + H₂O ⇄ H₃O⁺ + NO₃⁻
Bronsted-lowry acid base conjugate acid conjugate base
In the reaction (a), (CH3)3N is the Brønsted-Lowry base, H2O is the Brønsted-Lowry acid, (CH3)3NH+ is the conjugate acid, and OH- is the conjugate base. In reaction (b), HNO3 is the Brønsted-Lowry acid, H2O is the Brønsted-Lowry base, H3O+ is the conjugate acid, and NO3- is the conjugate base.
Explanation:In the reaction (a) (CH3)3N(aq) + H2O(l) <-> (CH3)3NH+(aq) + OH-(aq), (CH3)3N is the Brønsted-Lowry base, H2O is the Brønsted-Lowry acid, (CH3)3NH+ is the conjugate acid, and OH- is the conjugate base.
In the reaction (b) HNO3(aq) + H2O(l) <-> H3O+(aq) + NO3-(aq), HNO3 is the Brønsted-Lowry acid, H2O is the Brønsted-Lowry base, H3O+ is the conjugate acid, and NO3- is the conjugate base.
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Sphingolipids are particularly important in these structures: a. Membranes b. Lipoproteins c. The brain and nervous system. d. Membranes, the brain and the nervous system. e. All of these are correct
Answer:
The correct option is d. Membranes, the brain and the nervous system.
Explanation:
The sphingolipids are a class of lipids which contain a sphingosine amino alcohol.
The main function of sphingolipid is the protection of the cells against harmful environment by forming a chemically resistant and mechanically stable membrane.
Sphingolipids are present in abundance in the brain and in the central nervous system where they are the important constituents of the plasma membranes. They are also important for the proper functioning, development and maintenance of the nervous system.
Therefore, Sphingolipids are particularly present in the membranes, the brain and the nervous system.
Sodium carbonate reacts with nitric acid to form sodium nitrate, carbon dioxide, and water according to the balanced reaction below. How many grams of sodium carbonate are required to completely react with 25.0 mL of 0.155 M nitric acid?
Answer: The mass of sodium carbonate reacted is 0.205 g.
Explanation:
To calculate the moles of a solute, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]
We are given:
Volume of nitric acid = 25 mL = 0.025 L (Conversion factor: 1 L = 1000 mL)
Molarity of the solution = 0.155 moles/ L
Putting values in above equation, we get:
[tex]0.155mol/L=\frac{\text{Moles of nitric acid}}{0.025L}\\\\\text{Moles of nitric acid}=0.003875mol[/tex]
For the given chemical reaction:
[tex]Na_2CO_3+2HNO_3\rightarrow H_2CO_3+2NaNO_3[/tex]
By Stoichiometry of the reaction:
If 2 moles of nitric acid reacts with 1 mole of sodium carbonate.
So, 0.003875 moles of nitric acid will react with [tex]\frac{1}{2}\times 0.003875=0.0019375moles[/tex] of sodium carbonate.
To calculate the mass of sodium carbonate, we use the equation:
Molar mass of sodium carbonate = 105.98 g/mol
Moles of sodium carbonate = 0.0019375 moles
Putting values in above equation, we get:
[tex]0.0019375mol=\frac{\text{Mass of sodium carbonate}}{105.98g/mol}\\\\\text{Mass of sodium carbonate}=0.205g[/tex]
Hence, the mass of sodium carbonate reacted is 0.205 g.
To find out how many grams of sodium carbonate are required to completely react with 25.0 mL of 0.155 M nitric acid, we used stoichiometry. First, we calculated the number of moles of nitric acid, then used the balanced equation to find out the moles of sodium carbonate, which is equal to the moles of nitric acid. Finally, we multiplied this by the molar mass of sodium carbonate to get the weight in grams, which is 0.411 grams.
Explanation:The subject of this question is related to a chemical reaction between sodium carbonate and nitric acid. In order to determine the amount of sodium carbonate required to completely react with the given amount of nitric acid, we need to use stoichiometry.
Here are the steps:
First, we need to find the number of moles of the nitric acid. As Molarity(M) = moles/volume in liter, then moles of nitric acid = volume(L) * Molarity = 25.0 mL*0.155M = 0.003875 moles, with 25.0 mL converted to 0.025 L. According to the balanced equation, one mole of nitric acid reacts with one mole of sodium carbonate. This means, the moles of nitric acid is equal to the moles of the sodium carbonate. Therefore, sodium carbonate required is also 0.003875 moles. Lastly, convert the moles to grams. The molar mass of sodium carbonate (Na2CO3) is about 105.99 g/mol. Therefore, the weight of sodium carbonate required = moles * molar mass = 0.003875 moles * 105.99 g/mol = 0.411 g
So, we need 0.411 grams of sodium carbonate to completely react with 25.0 mL of 0.155 M nitric acid.
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The pH of saliva is normally in the range of 6.4 to 7.0. However, when a person is ill, the person's saliva becomes more acidic. (a) When Marco is sick, he tests the pH of his saliva and finds that it is 5.3. What is the hydrogen ion concentration of his saliva
Answer:
Concentration of hydrogen ion, [tex][H^+]=5.0118*10^{-6} M[/tex]
Explanation:
pH is defined as the negative logarithm of hydrogen ion's concentration.
The lower the value of pH, the higher the acidic the solution is.
The formula for pH can be written as:
[tex]pH=-log[H^+][/tex]
Given,
pH of the saliva of Marco = 5.3
To calculate: Hydrogen ion concentration in the saliva
Thus, applying in the formula as:
[tex]pH=-log[H^+][/tex]
[tex]5.3=-log[H^+][/tex]
So,
[tex]log[H^+]=-5.3[/tex]
[tex][H^+]=10^{(-5.3)}[/tex]
[tex][H^+]=5.0118*10^{-6} M[/tex]
The hydrogen ion concentration of Marco's saliva is approximately 5.01 x 10^-6 M.
Explanation:The pH scale ranges from 0 to 14, where anything below 7 is acidic and above 7 is alkaline. The hydrogen ion (H+) concentration of a solution can be determined using the pH value. A change of one unit on the pH scale represents a ten-fold change in the concentration of H+ ions. Given that Marco's saliva has a pH of 5.3, we can calculate the hydrogen ion concentration as follows:
Convert the pH value to a H+ concentration by taking the inverse logarithm of the pH value. In this case, H+ concentration = 10⁻⁵(-pH).Substitute the pH value (5.3) into the equation to find the H+ concentration: H+ concentration = 10∧5.3⇒10∧6≈10⁻⁶Calculate the value using a calculator or scientific notation. The H+ concentration is approximately 5.01 x 10⁻⁶M.Therefore, the hydrogen ion concentration of Marco's saliva is approximately 5.01 x 10⁻⁶M.
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What is the composition of Gilsonite?
Answer:
Roughly C100 H140 N3 O
Explanation:
Gilsonite is a bituminous product that resembles shiny black obsidian.
It contains more than 100 elements.
Its mass composition varies but is approximately 84 % C, 10 % H, 3 % N, and 1 % O.
Its empirical formula is roughly C100 H140 N3 O.
Consider the hypothetical reaction A(g)??2B(g). A flask is charged with 0.73atm of pure A, after which it is allowed to reach equilibrium at 0 ?C. At equilibrium the partial pressure of A is 0.37atm .
A: What is the total pressure in the flask at equilibrium?
B:What is the value of Kp?
Answer:
1.08 atm is the total pressure in the flask at equilibrium.
The value of the [tex]K_p=1.44[/tex].
Explanation:
Partial pressure of A at equilibrium = [tex]p_a=0.37 atm[/tex]
Partial pressure of B at equilibrium = [tex]p_b=2x[/tex]
A(g) ⇄ 2B(g)
at t=0 0.73 atm 0
At equilibrium (0.73- x) 2x
[tex]p_a=0.37 atm=(0.73- x)[/tex]
x = 0.36 atm
[tex]p_b=2x=2\times 036 atm=0.72 atm[/tex]
Total pressure in the flask at equilibrium = P
[tex]P=p_a+p_b=0.36 atm+0.72 atm = 1.08 atm[/tex]
The expression of equilibrium constant will be given as:
[tex]K_p=\frac{p_{b}^2}{p_a}=\frac{(0.72 atm)^2}{0.36 atm}=1.44[/tex]
The partial pressure is the pressure exerted by individual gases in the volume it is present. 1.08 atm is the total pressure and 1.44 is the equilibrium constant.
What are total pressure and equilibrium constant?The total static and velocity pressure of a system is called total pressure. While the equilibrium constant is the proportion of the partial pressure of products and reactants.
Given,
The partial pressure of A [tex](p_{a})[/tex] = 0.37 atmThe partial pressure of B [tex](p_{b})[/tex] = 2xFrom the reaction,
[tex]\begin{aligned}p_{a} &= 0.37 \rm\; atm \\\\\\&= (0.73 - x)\\\\\rm x &= 0.36 \;\rm atm\end{aligned}[/tex]
Solving the partial pressure of B:
[tex]\begin{aligned}p_{b} &= 2\rm x\\\\&= 2 \times 0.36\\\\&= 0.72\;\rm atm\end{aligned}[/tex]
Total pressure (P) in the flask will be:
[tex]\begin{aligned} \rm P &= p_{a}+ p_{b}\\\\&=0.36 + 0.72\\\\&= 1.08 \;\rm atm\end{aligned}[/tex]
Also, the equilibrium constant will be calculated as:
[tex]\begin{aligned} K_{p} &= \dfrac{p_{b}^{2}}{p_{a}}\\\\&=\dfrac{(0.72)^{2}}{0.3.6}\\\\&=1.44 \end{aligned}[/tex]
Therefore, A. 1.08 atm is the total pressure and B. 1.44 is the equilibrium constant.
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___________ minerals are dark-colored because they contain _____________. a. Mafic; iron and/or magnesium b. Felsic; iron and/or magnesium c. Mafic; calcium, sodium, or potassium d. Felsic; calcium, sodium, or potassium
Answer:
A. Mafic; iron and/or magnesium
Explanation:
Let's find the answer by naming some minerals and their chemistry.
Mafic minerals are dark-colored whereas felsic minerals are light-colored, thats way mafic rocks are dark-colored because they are mainly composed by mafic minerals and the other way around for felsic rocks.
But remember that mafic minerals as amphiboles, pyroxenes or biotites, involve in their chemical structure iron and/or magnesium. Although calcium and sodium can be incorporated in amphiboles and clinopyroxenes, they are not involved in orthopyroxenes and biotites. On the other hand, although potassium is involved in biotite and in some extent in amphiboles, this element is not involved in pyroxenes.
So in conclusion, mafic minerals are usually dark-colored because they involve iron and/or magnesium in their chemical structures.
Calcium carbide (CaC2) reacts with water to produce acetylene (C2H2): CaC2 (s) 2H2O (g) → Ca(OH)2 (s) C2H2 (g) Production of 6.5 g of C2H2 requires consumption of ________ g of H2O.
Calcium carbide (CaC₂) reacts with water to produce acetylene (C₂H₂): CaC₂ (s) + 2 H₂O (g) → Ca (OH)₂ (s) + C₂H₂ (g) Production of 6.5 g of C₂H₂ (MW = 26.036 g/mol) requires consumption of 2.3 gm. Hence, option B is the correct option.
The balanced chemical equation given is:
CaC₂ (s) + 2 H₂O (g) → Ca (OH)₂ (s) +C₂H₂ (g)
The molar ratio between CaC₂ and C₂H₂ is 1:1, and the molar ratio between H₂O and C₂H₂ is 2:1.
To determine the mass of H₂O required to produce 6.5 g of C₂H₂, one can use the following steps:
The number of moles of C₂H₂:
Moles of C₂H₂ = Mass / Molar mass = 6.5 g / 26.036 g/mol
The molar ratio between H₂O and C₂H₂ is 2:1, the number of moles of H₂O required is half of the moles of C₂H₂:
Moles of H₂O = Moles of C₂H₂ / 2
The mass of H₂O using its moles and molar mass:
Mass of H₂O = Moles of H₂O × Molar mass of H₂O
1. Moles of C₂H₂ = 6.5 g / 26.036 g/mol ≈ 0.2494 mol
2. Moles of H₂O = 0.2494 mol / 2 ≈ 0.1247 mol
3. Mass of H₂O = 0.1247 mol × 18.016 g/mol ≈ 2.244 g
So, the consumption of H₂O required to produce 6.5 g of C₂H₂ is approximately 2.244 g.
The closest option to this value is 2.3 g, so the correct answer is:
B. 2.3
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In the first order reaction A → products, the initial concentration of A is 0.1 108M, and 44 s later, 0.0554M. What is the initial rate of this reaction? (Initial rate-k[A]). (t %-0.693/k)
Answer : The initial rate of the reaction is, [tex]1.739\times 10^{-3}s^{-1}[/tex]
Explanation :
First we have to calculate the rate constant of the reaction.
Expression for rate law for first order kinetics is given by :
[tex]k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}[/tex]
where,
k = rate constant = ?
t = time taken for the process = 44 s
[tex][A_o][/tex] = initial amount or concentration of the reactant = 0.1108 M
[tex][A][/tex] = amount or concentration left time 44 s = 0.0554 M
Now put all the given values in above equation, we get:
[tex]k=\frac{2.303}{44}\log\frac{0.1108}{0.0554}[/tex]
[tex]k=0.0157[/tex]
Now we have to calculate the initial rate of the reaction.
Initial rate = K [A]
At t = 0, [tex][A]=[A_o][/tex]
Initial rate = 0.0157 × 0.1108 = [tex]1.739\times 10^{-3}s^{-1}[/tex]
Therefore, the initial rate of the reaction is, [tex]1.739\times 10^{-3}s^{-1}[/tex]
10. In an experiment in a general chemistry laboratory, a student collected a sample of a gas over water. The volume of the gas was 265 mL at a pressure of 753 torr and a temperature of 27 °C. The mass of the gas was 0.472 g. What was the molar mass of the gas?
To find the molar mass of a gas collected over water, apply the ideal gas law to determine the moles of gas. Then, divide the mass of the gas by the number of moles.
Explanation:Calculating the Molar Mass of a GasTo determine the molar mass of a gas from an experiment in a general chemistry laboratory, you first need to use the ideal gas law, which is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant and T is the temperature in Kelvin.
The student collected a gas with a volume of 265 mL, which is equivalent to 0.265 L, at a pressure of 753 torr. Since 1 atm is equal to 760 torr, the pressure in atmospheres is 753 torr / 760 torr/atm = 0.9911 atm. The temperature must be converted from Celsius to Kelvin; thus, 27 °C is equal to 300.15 K (27 + 273.15 = 300.15 K).
To solve for n (the number of moles), you rearrange the ideal gas law to n = PV / RT. With the previously mentioned values and the gas constant R as 0.0821 L·atm/K·mol, n = (0.9911 atm × 0.265 L) / (0.0821 L·atm/K·mol × 300.15 K). After calculation, the number of moles of gas n is found.
Once n is calculated, the molar mass (M) can be found using the formula M = mass of gas (g) / number of moles (mol). Therefore, with the mass of the gas being 0.472 g, we calculate M = 0.472 g / n moles. By plugging in the value of n from our previous calculation, we can determine the molar mass.
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Final answer:
To calculate the molar mass of a gas, the ideal gas law is used with the given pressure, volume, and temperature. The molar mass is found to be approximately 44.22 g/mol after doing the conversions and calculations.
Explanation:
The student's question asks about determining the molar mass of a collected gas sample. To find out the molar mass of the gas, we use the ideal gas law formula (PV = nRT), where P is pressure, V is volume, n is number of moles, R is the gas constant, and T is temperature. Here's the step-by-step solution:
Convert the given pressure from torr to atm (753 torr / 760 torr/atm).
Convert the volume from mL to L (265 mL / 1000 mL/L).
Convert the temperature from Celsius to Kelvin (27 °C + 273.15).
Solve the ideal gas law equation for the number of moles (n).
Use the mass of the gas and the number of moles calculated to find the molar mass (molar mass = mass/n).
The detailed calculations show:
Solve for n: PV = nRT → n = PV / RT = (0.991 atm × 0.265 L) / (0.0821 L·atm/(mol·K) × 300.15 K) ≈ 0.01067 mol.
Finally, determine the molar mass: molar mass = mass/n = 0.472 g / 0.01067 mol ≈ 44.22 g/mol.
Hence, the molar mass of the gas is approximately 44.22 g/mol.
. Marble is a rock that is composed primarily of the carbonate minerals calcite (CaCO3) and dolomite (Ca,Mg(CO3)2). In Exercise 2, you saw the way carbonate minerals react to a drop of acetic acid. Based on this observation, why do you think people who care for monuments and sculptures made of marble are concerned about acid rain?
Answer:
Explanation:
People who use marble to carve monuments and sculptures are always concerned because it can easily chemically weather.
Marble is a metamorphic rock derieved from limestone. The minerals that makes up limestone are calcite and dolomite. When a metamorphic transformation occurs, the rock is subjected to intense temperature and pressure. Marble is made of calcite and dolomite minerals.
Rain water dissolves carbon dioxide to form weak carbonic acid. This weak carbonic acid can easily dissolve marble thereby defacing the monument and the sculpture.
Acid rain can cause damage to marble monuments and sculptures because it reacts with the carbonate minerals in marble, leading to their dissolution and degradation over time.
Explanation:People who care for monuments and sculptures made of marble are concerned about acid rain because marble is primarily composed of carbonate minerals such as calcite (CaCO3) and dolomite (Ca,Mg(CO3)2). When acid rain falls on marble, the weak acid reacts with the carbonate minerals, causing them to dissolve. This can lead to the degradation and erosion of the marble over time.
The reaction between acid rain and marble can be represented by the following equation: CaCO3(s) + H₂SO4 (aq) → CaSO4(s) + H₂O(1) + CO2(g). The reaction produces calcium sulfate, water, and carbon dioxide gas, which contribute to the dissolution of the marble.
Therefore, the concern for acid rain arises from the fact that it can cause significant damage to marble monuments and sculptures, affecting their aesthetic value and structural integrity.
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Noble gas compounds like KrF, XeCl, and XeBr are used in excimer lasers. Draw an approximate molecular orbital diagram appropriate for these molecules (they will be quite similar). Give the ground-state electron configuration of KrF and predict whether the cationic analog (KrF) is likely to have a stronger bond.
Answer:
Here's what I get.
Explanation:
The MO diagrams of KrBr, XeCl, and XeBr are shown below.
They are similar, except for the numbering of the valence shell orbitals.
Also, I have drawn the s and p orbitals at the same energy levels for both atoms in the compounds. That is obviously not the case.
However, the MO diagrams are approximately correct.
The ground state electron configuration of KrF is
[tex](1\sigma_{g})^{2}\, (1\sigma_{u}^{*})^{2} \, (2\sigma_{g})^{2} \, (2\sigma_{u}^{*})^{2} \, (3\sigma_{g})^{2} \, (1\pi_{u})^{4} \, (1\pi_{g}^{*})^{4} \, (3\sigma_{g}^{*})^{1}[/tex]
KrF⁺ will have one less electron than KrF.
You remove the antibonding electron from the highest energy orbital, so the bond order increases.
The KrF bond will be stronger.
Final answer:
The molecular orbital diagram for noble gas compounds like KrF can be explained by the ground-state electron configuration and the expected bond strength of its cationic form.
Explanation:
Noble Gas Compounds Molecular Orbital Diagram:
The ground-state electron configuration of KrF is [Kr] 5s² 5p⁶. The cationic analog of KrF is likely to have a stronger bond due to the removal of an electron, leading to a decrease in repulsion between the nuclei and shared electrons.
Comparing Bond Strength in KrF vs. KrF⁺:
In KrF, the valence electrons fill the σ bonding and σ non-bonding orbitals, and the π bonding orbitals are empty.
When KrF loses an electron to become KrF⁺, it removes an electron from the σ non-bonding orbital (lone pair).
Losing this electron strengthens the bond because it removes electron density that opposes the bonding interaction in the σ bonding orbital. This is similar to losing a lone pair in other molecules.
Therefore, KrF⁺ is likely to have a stronger bond than KrF.
The concentration of water vapor in a sample of air that has a partial pressure of water of 0.91 torr and a total pressure of air of 735 torr is ________ ppm. The concentration of water vapor in a sample of air that has a partial pressure of water of 0.91 torr and a total pressure of air of 735 torr is ________ ppm. 0.81 0.12 8.1 ⋅ 10−4 1.2 1.2 ⋅ 103
Answer : The concentration of water vapor in a sample of air is, [tex]1.2\times 10^3ppm[/tex]
Explanation : Given,
Partial pressure of water = 0.91 torr
The total pressure of air = 735 torr
Parts per million (ppm) : It is defined as the mass of a solute present in one million [tex](10^6)[/tex] parts by the mass of the solution.
Now we have to calculate the concentration of water vapor in a sample of air.
[tex]\text{Concentration of water vapor}=\frac{\text{Partial pressure of water}}{\text{Total pressure of air}}\times 10^6[/tex]
Now put all the given values in this formula, we get concentration of water vapor in a sample of air.
[tex]\text{Concentration of water vapor}=\frac{0.91torr}{735torr}\times 10^6=2.1\times 10^3ppm[/tex]
Therefore, the concentration of water vapor in a sample of air is, [tex]1.2\times 10^3ppm[/tex]
The concentration of water vapor in a sample of air that has a partial pressure of water of 0.91 torr and a total pressure of air of 735 torr can be calculated using Dalton's law of partial pressures. The concentration comes out to be approximately 1200 ppm.
Explanation:To calculate the concentration of water vapor in a sample of air in parts per million (ppm), we first need to understand the relationship between partial pressures and total pressure in a mixture of gases. According to Dalton's law of partial pressures, the total pressure of a gas mixture is equal to the sum of the partial pressures of its components. Here, the partial pressure of water vapor is 0.91 torr and the total pressure of air is 735 torr.
To find the concentration of water vapor, divide the partial pressure of the water vapor by the total pressure of the air, and then multiply by 1,000,000 to convert the value to ppm. So, (0.91 torr / 735 torr) x 1,000,000 = 1238.7755 ppm. This value needs to be rounded off to the appropriate number of significant digits which in this case is two, so the concentration of water vapor in the air would be 1200 ppm in this scenario.
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What determines chemical bonding characteristics?
The configuration of the outermost orbital
The configuration of the nucleus
The atomic weight of an atom
The type of material
The number of neutrons
Answer: Option (a) is the correct answer.
Explanation:
A chemical bonding in a substance occurs when there is exchange of electrons between the atoms. These electrons are present in the outermost orbital are actually away from the nucleus of the atom.
Due to which there is no force of attraction between nucleus or valence electrons. Hence, it is easy for the outermost orbital to lose electron.
Whereas nucleus of an atom contains neutrons and protons. And, changes in the nucleus of an atom leads to nuclear reactions.
Therefore, we can conclude that the configuration of the outermost orbital determines chemical bonding characteristics.
Consider the reaction: 3 A + 5 B --> 4 C. The rate of the reaction is found to be .7511 M/s. What is the rate of change of B? Be sure to consider whether the substance is disappearing or appearing in your answer.
Answer:
-3.7555 M/s is the rate of change of B.
Explanation:
3A + 5B → 4C
Given that rate of the reaction ,R= 0.7511 M/s
Rate of the reaction is defined as change in concentration of any one of the reactant or product with respect to time.
[tex]R=\frac{-1}{3}\frac{dA}{dt}=\frac{-1}{5}\frac{dB}{dt}=\frac{1}{4}\frac{dC}{dt}[/tex]
[tex]R=0.7511 M/s=\frac{-1}{5}\frac{dB}{dt}[/tex]
[tex]\frac{dB}{dt}=5\times 0.7511 M/s=-3.7555 M/s[/tex]
The negative sign indicates the concentration of reactant B is decreasing with progress in time. This mean reactant B is disappearing.
-3.7555 M/s is the rate of change of B.
Which isomer would you expect to undergo E2 elimination faster, trans-1-bromo-4-tert-butylcyclohexane or cis-1-bromo-4-tert-butylcyclohexane? Draw each molecule in its more stable chair conformation, and explain your answer.
Answer: Cis-1-bromo-4-tert-butylcyclohexane would undergo faster elimination reaction.
Explanation:
The two primary requirements for an E-2 elimination reaction are:
1.There must be availability of β-hydrogens that is presence of hydrogen on the carbon next to the leaving group.
2.The hydrogen and leaving group must have a anti-periplanar position .
Any substrate which would follow the above two requirements can give elimination reactions.
For the structure of trans-1-bromo-4-tert-butylcyclohexane and cis-1-bromo-4-tert-butylcyclohexane to be stable it must have the tert-butyl group in the equatorial position as it is a bulky group and at equatorial position it would not repel other groups. If it is kept on the axial position it would undergo 1,3-diaxial interaction and would destabilize the system and that structure would be unstable.
Kindly find the structures of trans-1-bromo-4-tert-butylcyclohexane and cis-1-bromo-4-tert-butylcyclohexane in attachment.
The cis- 1-bromo-4-tert-butylcyclohexane has the leaving group and β hydrogens in anti-periplanar position so they can give the E2 elimination reactions easily.
The trans-1-bromo-4-tert-butylcyclohexane does not have the leaving group and βhydrogen in anti periplanar position so they would not give elimination reaction easily.
so only the cis-1-bromo-4-tert butyl cyclohexane would give elimination reaction.
Trans-1-bromo-4-tert-butylcyclohexane is expected to undergo E2 elimination faster than cis-1-bromo-4-tert-butylcyclohexane due to less steric hindrance.
Explanation:In determining the rate of E2 elimination, the trans-1-bromo-4-tert-butylcyclohexane would undergo E2 elimination faster than the cis-1-bromo-4-tert-butylcyclohexane. This is due to the larger degree of steric hindrance in the case of the cis isomer.
In trans-1-bromo-4-tert-butylcyclohexane, the bromine is at the equatorial position while the tert-butyl group is axial. It forms a structure that allows the compound to experience less steric hindrance with bromine in a more favorable position for leaving.
In comparison, cis-1-bromo-4-tert-butylcyclohexane has a bromine and tert-butyl group both at equatorial positions. This causes steric hindrance, and in turn, slows down the E2 elimination rate. Despite the more stable conformation, the bromine is not well-oriented for a leaving group in E2 elimination.
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(T) (F) The Great Decoupling means that well-being and wealth can increase while the use of resources and production of pollution can decrease.
Two systems with heat capacities 19.9 J mol-1 K-1 and 28.2 ] mol 1 K-1 respectively interact thermally and come to an equilibrium temperature of 300C. If the initial temperature of system 1 was 450C, what was the initial temperature of system 2 in °C? You may assume that the total energy of the combined systems remains constant
Answer : The initial temperature of system 2 is, [tex]19.415^oC[/tex]
Explanation :
In this problem we assumed that the total energy of the combined systems remains constant.
[tex]-q_1=q_2[/tex]
[tex]m\times c_1\times (T_f-T_1)=-m\times c_2\times (T_f-T_2)[/tex]
The mass remains same.
where,
[tex]C_1[/tex] = heat capacity of system 1 = 19.9 J/mole.K
[tex]C_2[/tex] = heat capacity of system 2 = 28.2 J/mole.K
[tex]T_f[/tex] = final temperature of system = [tex]30^oC=273+30=303K[/tex]
[tex]T_1[/tex] = initial temperature of system 1 = [tex]45^oC=273+45=318K[/tex]
[tex]T_2[/tex] = initial temperature of system 2 = ?
Now put all the given values in the above formula, we get
[tex]-19.9J/mole.K\times (303-318)K=28.2J/mole.K\times (303-T_2)K[/tex]
[tex]T_2=292.415K[/tex]
[tex]T_2=292.415-273=19.415^oC[/tex]
Therefore, the initial temperature of system 2 is, [tex]19.415^oC[/tex]
3. A student titrated 25.00 mL of a solution of formic acid with sodium hydroxide of known concentration. The student plotted pH verses volume (mL) of NaOH added and found that the titration required 26.66 mL of NaOH to reach equivalence point. Describe how you would use the students’ titration curve to determine the concentration of the formic acid solution.
Answer:
1. Multiply the equivalence point volume by the NaOH concentration.
2. Divide that result by the volume taken from the formic acid solution for titration.
Explanation:
The reaction between NaOH and formic acid is one to one, it means that a NaOH mol reacts for each mole of formic acid in the solution. So the number of moles who react in the NaOH titrator is the same number of moles that were in the volume taken of the formic acid solution for titration. So, the number of formic acid moles can be calculated as:
[tex]n_{CHOOH}=V_{NaOH}*C_{NaOH}[/tex]
Then, divide that number of moles by the volume taken from the formic acid solution:
[tex]C_{CHOOH}=\frac{V_{NaOH}*C_{NaOH}}{V_{CHOOH}}[/tex]
The concentrations must be in Molar units (Mol/Liter) and the volume in liters, so:
[tex]C_{CHOOH}=\frac{0.02666*C_{NaOH}}{V_{CHOOH}}[/tex]
The concentration of the formic acid solution can be determined using the titration curve provided by the student.
To determine the concentration of the formic acid solution using the student's titration curve, follow these steps:
1. Identify the equivalence point on the titration curve. This is the point where the pH changes rapidly and corresponds to the addition of 26.66 mL of NaOH.
2. At the equivalence point, the moles of NaOH added are equal to the moles of formic acid in the initial solution.
3. Calculate the moles of NaOH added using the concentration of the NaOH solution[tex](C_NaOH)[/tex] and the volume of NaOH added [tex](V_NaOH)[/tex] at the equivalence point:
[tex]\[ \text{moles of NaOH} = C_{\text{NaOH}} \times V_{\text{NaOH}} \][/tex]
4. Since the stoichiometry of the reaction between formic acid (HCOOH) and NaOH is 1:1, the moles of NaOH added will be equal to the moles of formic acid (HCOOH) in the initial solution.
5. Calculate the concentration of the formic acid solution using the moles of formic acid and the initial volume of the formic acid solution [tex](V_HCOOH):[/tex]
[tex]\[ \text{Concentration of formic acid} = \frac{\text{moles of HCOOH}}{V_{\text{HCOOH}}} \][/tex]
6. The concentration of formic acid can be expressed as:
[tex]\[ C_{\text{HCOOH}} = \frac{C_{\text{NaOH}} \times V_{\text{NaOH}}}{V_{\text{HCOOH}}} \][/tex]
7. Plug in the known values to find the concentration of formic acid:
[tex]\[ C_{\text{HCOOH}} = \frac{C_{\text{NaOH}} \times 26.66 \text{ mL}}{25.00 \text{ mL}} \][/tex]
8. If the concentration of NaOH is not given, it can be determined from the pH of the solution before the equivalence point where the pH starts to increase sharply. At this point, the concentration of [tex]OH^-[/tex] ions can be calculated, and since[tex][OH^-] = \(\frac{K_w}{[\text{H}_3\text{O}^+]}\),[/tex] the concentration of NaOH can be found.
9. Once the concentration of NaOH is known, use it in the equation from step 7 to find the concentration of formic acid.
Consider these two cases. Case 1: An electron jumps from energy level 4 to energy level 2 in an atom. Case 2: An electron jumps from energy level 4 to energy level 1 in an atom. For case 1, what happens when an electron jumps from energy level 4 to energy level 2 in an atom?
When an electron jumps from higher energy level to a lower energy level it radiates or gives out energy in the form of radiation.
What is an energy level?
Electrons present in an atom revolve in different orbits which are stationary states and are also called as energy levels. The energy levels are numbered as integers which are also called as principal quantum numbers.
Energy of the stationary state is given as E= -R[tex]_h[/tex] 1/n² where R[tex]_h[/tex] is the Rydberg's constant. When an electron is excited, and it moves from lower to higher energy levels there is absorption of energy, while when it moves from higher energy level to lower energy level it radiates or gives out energy in the form of radiation.
They can also be defined as the distances between electron and nucleus of an atom . Electrons present in K energy level have least energy .Energy level diagrams are studied to understand nature of bonding , placement of electrons in orbits and and elemental behavior under certain conditions.
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Be sure to answer all parts and enter your answers in scientific notation. Write the 1 g equivalents using the following prefixes: (a) mega−, (b) kilo−, (c) deci−, (d) centi−, (e) milli−, (f) micro−, (g) nano−, (h) pico−. (a) 1 g = × 10 megagrams (b) 1 g = × 10 kilograms (c) 1 g = × 10 decigrams (d) 1 g = × 10 centigrams (e) 1 g = × 10 milligrams (f) 1 g = × 10 micrograms (g) 1 g = × 10 nanograms (h) 1 g = × 10 picograms
Explanation:
Gram is unit which is used to expressed mass of a substance. Many other units can be used to express the mass of a substance. All the units are inter changeable.
Conversion factors used to convert the units are:
(a) Gram to mega gram
[tex]1g=1\times 10^{-6}Mg[/tex]
(b) Gram to kilo gram
[tex]1g=1\times 10^{-3}kg[/tex]
(c) Gram to deci gram
[tex]1g=1\times 10^{1}1dg[/tex]
(d) Gram to centi gram
[tex]1g=1\times 10^{2}cg[/tex]
(e) Gram to mili gram
[tex]1g=1\times 10^{3}mg[/tex]
(f) Gram to micro gram
[tex]1g=1\times 10^{6}\mu g[/tex]
(g) Gram to nano gram
[tex]1g=1\times 10^{9}ng[/tex]
(h) Gram to pico gram
[tex]1g=1\times 10^{12}pg[/tex]
Hence, the conversion factors for gram equivalents are given above.
To convert 1 gram to different metric prefixes in scientific notation, we use the powers of ten specific to each prefix, ensuring the numeric value is between 1 and 1000. The result is megagrams (1e-6), kilograms (1e-3), decigrams (1e1), centigrams (1e2), milligrams (1e3), micrograms (1e6), nanograms (1e9), and picograms (1e12).
Explanation:The task involves converting the mass of 1 gram into various metric prefixes and expressing them in scientific notation, where the numeric value is greater than one but less than 1000. The prefixes include mega, kilo, deci, centi, milli, micro, nano, and pico. Below are the conversions using the appropriate powers of ten to express 1 gram in the specified units:
(a) 1 g = 1 × 10-6 megagrams (Mg)(b) 1 g = 1 × 10-3 kilograms (kg)(c) 1 g = 1 × 101 decigrams (dg)(d) 1 g = 1 × 102 centigrams (cg)(e) 1 g = 1 × 103 milligrams (mg)(f) 1 g = 1 × 106 micrograms (μg)(g) 1 g = 1 × 109 nanograms (ng)(h) 1 g = 1 × 1012 picograms (pg)When 1.0 g of fructose, C6H12O6(s), a sugar commonly found in fruits, is burned in oxygen in a bomb calorimeter, the temperature of the calorimeter increases by 1.58 °C. If the heat capacity of the calorimeter and its contents is 9.90 kJ/°C, what is q for this combustion?
Answer : The heat of combustion is, 15642 J
Solution :
Formula used :
[tex]q=m\times c\times \Delta T[/tex]
where,
[tex]q[/tex] = heat of combustion = ?
[tex]m[/tex] = mass of fructose = 1.0 g
[tex]c[/tex] = heat capacity of the calorimteter = [tex]9.90KJ/g^oC=9900J/g^oC[/tex]
conversion used : 1 KJ = 1000 J
[tex]\Delta T[/tex] = change in temperature = [tex]1.58^oC[/tex]
Now put all the given values in the above formula, we get
[tex]q=1.0g\times 9900J/g^oC\times 1.58^oC[/tex]
[tex]q=15642J[/tex]
Therefore, the heat of combustion is, 15642 J
Answer:
q combustion = -15.6 kJ (exothermic)
Explanation:
Bomb calorimeter questions are interesting because we are usually given the heat capacity of the bomb calorimeter not the specific heat capacity (which includes grams in the unit). Therefore we don't actually need to include the mass of the fructose as long as we know how much the temperature of the calorimeter changed (ΔT) and the heat capacity of the calorimeter.
We know that:
-q (combustion) = q (calorimeter) and we have enough information to calculate q (calorimeter):
q (calorimeter) = (Heat Capacity)(Change in Temp.)
⇒ q(cal) = (9.90 kJ/°C)(1.58 °C) = 15.642 kJ
= 15.6 kJ (3 sig figs)
Since -q (combustion) = q (calorimeter), then:
q (combustion) = -15.6 kJ (negative sign simply means heat released)