Answer:
The two elements are POLONIUM and RADIUM.
Explanation:
Maria Curie is a French physicist and chemist, though she was of a Polish naturals. She was the first woman to receive a Noble Price which she earned for conducting leading and head way research on radioactivity. She discovered the theory of radioactivity; also the techniques isolating radioactive isotopes. These helped her and her husband discover Polium and Radium.
Answer: She discovered polonium and radium.
Explanation: In 1896, intrigued by the physicist Henri Becquerel’s accidental discovery of radioactivity, Curie began studying uranium rays; Pierre soon joined her in her research. Two years later, the Curies discovered polonium—named after Marie’s homeland—and radium. In 1903 they shared the Nobel Prize in physics with Becquerel for their groundbreaking work on radioactivity.
In response to action potentials arriving along the transverse tubules, the sarcoplasmic reticulum releasesA) acetylcholine.B) sodium ions.C) potassium ions.D) calcium ions.E) hydrogen ions
Answer:
Calcium ions.
Explanation:
The generation of the action potential helps in the transfer of information to the different body parts. This potential occurs to the difference in membrane potential inside and outside of the cell.
The sarcoplasmic reticulum is the homologous to the endoplasmic reticulum of the cells. The sarcoplasmic reticulum contains calcium ions in it and releases the stored calcium ions on the generation of the action potential. This calcium ion is important for the action of the actin and myosin.
Thus, the correct answer is option (D).
When we think about the carbon cycle and human activities, it is important to differentiate between facts and hypotheses. Which of the following is NOT a fact, but is a hypothesis? A. The amount of carbon dioxide in the atmosphere has increased since 1950. B. Increasing atmospheric carbon dioxide will cause mean global temperature to increase by 2 degrees Celsius over the next century. C. The burning of fossil fuels contributes substantially to the ongoing rise of atmospheric CO2. D. In the past, atmospheric CO2 levels reached levels higher than those observed today.
Answer:
B. Increasing atmospheric carbon dioxide will cause mean global temperature to increase by 2 degrees Celsius over the next century.
Explanation:
By definition, a hypothesis is a tentative statement or prediction with little or no experimental test. Hypotheses are always formulated such that they can be rejected if experimental findings are against them.
Hypotheses are predictive and as such, the tone is often in future tense.
From the available options, only option B sound predictive of what might happen in the future.
Hence, the correct option is B.
The equation shows a --the breaking and forming of chemical bonds that leads to a change in the composition of matter. 2. In the equation, CO2 is a . 3. In the equation, C6H12O6 is a . 4. In O2, the type of bond that holds the two oxygen atoms together is a . 5. In H2O, the type of bond that holds one of the hydrogen atoms to the oxygen atom is a . 6. The number of oxygen atoms on the left side of the equation is the number of oxygen atoms on the right side.
A chemical equation is said to be balanced if the quantity of each type of atom in the reaction is the same on both the reactant and product sides. In a balanced chemical equation, the mass and the charge are both equal. Here the given equation is C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O.
This means that new products are formed due to the change in the chemical composition of the reactants.
Hence, the equation shows a chemical reaction - the breaking and forming of chemical bonds that leads to a change in the composition of matter.
1. The equation is C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O.
2. And, in the given equation CO₂ is a product.
3. In the equation, C₆H₁₂O₆ is a reactant.
4. In O₂, the type of bond that holds the two oxygen atoms together is a non-polar covalent bond.
5. In H₂O, the type of bond that holds one of the hydrogen atoms to the oxygen atom is a polar covalent bond.
6. The number of oxygen atoms on the left side of the equation is equal to the number of oxygen atoms on the right side.
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Precipitation reactions and ionic equations
(look at attachment)
Answer:
Net ionic equation: 2OH⁻(aq) + Fe²⁺(aq) → Fe(OH)₂(s)
Net ionic equation: 2K⁺(aq) + CH₃COO⁻ (aq) + 2Na⁺(aq) + S⁻²(aq) → NO REACTION
Net ionic equation: CO₃⁻²(aq) + Pb⁺²(aq) → PbCO₃(s)
Explanation:
a. Solutions of calcium hydroxide and Iron (II) chloride are mixed:
We identify the reactants:
Ca(OH)₂ , FeCl₂
In excess, the Fe(OH)₂ can make precipitate
Salts from chlorides with elements from group II are soluble.
The reaction is: Ca(OH)₂(aq) + FeCl₂(aq) → Fe(OH)₂(s) + CaCl₂(aq)
Ca(OH)₂(aq) + FeCl₂(aq) → Fe(OH)₂(s) + CaCl₂(aq)
We dissociate the compounds, except for the solid
Ca²⁺(aq) + 2OH⁻(aq) + Fe²⁺(aq) + 2Cl⁻(aq) → Fe(OH)₂(s) + Ca²⁺ + 2Cl⁻(aq)
Net ionic equation: 2OH⁻(aq) + Fe²⁺(aq) → Fe(OH)₂(s)
b. Solutions of potassium acetate and sodium sulfide are mixed:
The reactants are: KCH₃COOH and Na₂S
In this case there are no precipitates, because all the salts are soluble
We make the complete reaction:
2KCH₃COO (aq) + Na₂S(aq) → K₂S(aq) + 2NaCH₃COO (aq)
Net ionic equation is:
2K⁺(aq) + CH₃COO⁻ (aq) + 2Na⁺(aq) + S⁻²(aq) → 2K⁺(aq) + S⁻²(aq) + 2Na⁺(aq) + CH₃COO⁻ (aq)
c. Solutions of ammonium carbonate and lead(II) nitrate are mixed:
In this case, the reactants are: (NH₄)₂CO₃ and Pb(NO₃)₂
All salts from nitrate are soluble.
Carbonate makes a precipitate when it bonds Pb.
The complete reaction is:
(NH₄)₂CO₃(aq) + Pb(NO₃)₂(aq) → PbCO₃(s) + 2NH₄NO₃(aq)
We dissociate all of the compounds, except for the solid in order to make the net ionic equation:
2NH₄⁺(aq) +CO₃⁻²(aq) + Pb⁺²(aq) +2NO₃⁻(aq) → PbCO₃(s) + 2NH₄⁺ (aq) + 2NO₃⁻(aq)
The net ionic equation is: CO₃⁻²(aq) + Pb⁺²(aq) → PbCO₃(s)
The ions that are repeated, are called spectators ions. We all cancel them.
The pressure of a gas changes from 120kPa to 50kPa. The volume changes from 45L to 40L. If the initial temperature is 81°C, what is the final temperature in Kelvin units?
Answer:
The final temperature is 131 K
Explanation:
Step 1: Data given
The initial pressure = 120 kPa = 1.18431 atm
The final pressure = 50 kPa = 0.493462 atm
The initial volume = 45 L
The final volume = 40 L
The initial temperature = 81 °C = 354 K
Step 2: Calculate the final temperature
(P1*V1)/T1 = (P2*V2)/T2
⇒with P1 = the initial pressure = 1.18431 atm
⇒with V1 = the initial volume = 45 L
⇒with T1 = the initial temperature = 354 K
⇒with P2 = the final pressure = 0.493462 atm
⇒with V2 = the final volume = 40 L
⇒with T2 = the final temperature = ?
(1.18431 * 45)/354 = (0.493462*40)/T2
0.15054788 = 19.73848/T2
T2 = 19.73848/0.15054788
T2 = 131 K
The final temperature is 131 K
Final answer:
To determine the final temperature of a gas with given initial and final pressures and volumes, one applies the combined gas law. After converting the initial Celsius temperature to Kelvin, the final temperature is calculated to be 262.33K.
Explanation:
The problem at hand involves the application of the combined gas law, which allows us to calculate changes in a gas's condition. This law is represented as (P1 * V1) / T1 = (P2 * V2) / T2, where P stands for pressure, V for volume, and T for temperature in Kelvin. Given that the pressure of a gas changes from 120kPa to 50kPa and its volume changes from 45L to 40L, with an initial temperature of 81°C (which is 354.15K), we can find the final temperature.
To solve, we start with the conversion of temperatures to Kelvin and then apply the formula. The rearranged formula to find the final temperature (T2) is T2 = ((P2 * V2) * T1) / (P1 * V1). By substituting the given values, T2 = ((50kPa * 40L) * 354.15K) / (120kPa * 45L), we get a final temperature of 262.33K.
In a research project, a scientist adds 0.1 mole of HCN, 0.1 mole of H3O , and 0.1mole of CN- to water to make a total volume of 1 L. Will this reaction proceed to a greater extent in the forward direction or in the reverse direction
Answer:
Reverse; since the value of Q is greater than the value of K
Explanation:
The equation for the reaction can be expressed as:
[tex]HCN + H_2 O[/tex] ⇄ [tex]H_3O^+ +CN^{-}[/tex] [tex]k_a = 6.2 * 10^ {-10[/tex]
[tex]Q = \frac{[H_3O^+][CN^-]}{[HCN]}[/tex]
[tex]Q = \frac{(0.1)(0.1)}{(0.1)}[/tex]
Q = 0.1
[tex]k_a = 6.2 * 10^ {-10[/tex]
Q > [tex]K_{a}[/tex]
If the value of Q is greater than [tex]K_{a}[/tex] Value; the reaction will definitely shift to reverse direction.
The overall energy involved in the formation of CsCl from Cs(s) and Cl2(g) is −443 kJ mol−1. Given the following information: heat of sublimation for Cs is +76 kJ mol−1, bond dissociation energy for 12Cl2 is +121 kJ mol−1, Ei1 for Cs is +376 kJ mol−1, and Eea for Cl(g) is 349 kJ mol−1. what is the magnitude of the lattice energy for CsCl?
Explanation:
It is given that total energy is -443 kJ/mol and formula to calculate the lattice energy is as follows.
Total energy = heat of sublimation + bond dissociation energies + ionization energy for Cs + EA of [tex]Cl^{-}[/tex] + lattice energy
-443 kJ/mol = 76 + 121 + 376 - 349 + Lattice energy
Lattice energy = (-443 - 76 -121 - 376 + 349) kJ
Lattice energy = -667 kJ
Therefore, we can conclude that -667 kJ is the magnitude of the lattice energy for CsCl.
Early life arose in an oxygen-free environment, but if any of these microbes had somehow come in contact with oxygen, the most likely effect would have been ________.
Answer:
the effect of oxygen on these types of microbes is it will kill them.
Explanation:
When oxygen present in the environment come in contact with anaerobe bacteria it kill them because oxygen in air act as excited oxygen singlet molecule which will react with the water present in the cell of bacteria and convert it into hydrogen peroxides and bacteria do not have any defense system from hydrogen peroxide and ultimately it kill the bacteria.
The number of substrate molecules converted to product in a given unit of time by a single enzyme molecule at saturation is referred to as the: dissociation constant maximum velocity turnover number Michaelis constant
The turnover number refers to the number of substrate molecules converted to product per unit time by a single enzyme molecule at saturation. The Michaelis constant (Km) indicates substrate concentration for half-maximal enzyme activity, and the maximum velocity (Vmax) is reached when enzyme active sites are saturated.
The number of substrate molecules converted to product in a given unit of time by a single enzyme molecule at saturation is referred to as the turnover number, also known as kcat. This measure of enzymatic activity provides a direct indication of the active site's catalytic efficiency within the enzyme's turnover rate.
In contrast, the Michaelis constant (Km) represents the substrate concentration at which the enzyme achieves half of its maximum reaction rate, or Vmax/2. This constant is used to determine the enzyme's affinity for a substrate, with a low Km indicating a high affinity, and vice versa.
When an enzyme operates in an environment with a high concentration of substrate, it will eventually reach a point where every active site is saturated with substrate -- this is the maximum velocity (Vmax) of the reaction. The Vmax is dependent on both the speed of the enzyme and the total number of enzyme molecules available.
The relationship between Vmax, Km, and substrate concentration ([S]) is described by the Michaelis-Menten equation, which is fundamental in the study of enzyme kinetics.
The Na /K pump generates a membrane potential across the plasma membrane (PM) of the cell by transporting Na and K across the PM. The action of this pump results in the cell having a slightly ____________ charge inside the cell compared to the extracellular space.
Answer:
NEGATIVE CHARGE can best fill in the gap
Explanation:
The Na /K pump functions to maintain resting potential so that the cells will be kept in a state of a low concentration of sodium ions and high levels of potassium ions within the cell.
The processes of Na - K pump illustrates active transport since it moves Na+ and K+ ions against their concentration gradient. The energy required is supplied by the breakdown of ATP (adenosine triphosphate) to ADP (adenosine diphosphate). In nerve cells the pump is used to generate gradients of both sodium and potassium ions.
How does the sodium-potassium pump contribute to the net negative charge of the interior of the cell?
The sodium-potassium pump forces out three (positive) Na+ ions for every two (positive) K+ ions it pumps in, thus the cell loses a positive charge at every cycle of the pump.
A wooden object from a prehistoric site has a carbon-14 activity of 10 cpm compared to 40 cpm for new wood. If carbon-14 has a half-life of 5730 yr, what is the age of the wood?
Answer:
The answer to your question is 11460 years
Explanation:
Data
Carbon-14 activity 10 cpm
half-life = 5730 yr
Real carbon-14 activity 40 cpm
Process
1.- Write a chart to solve this problem
Real carbon-14 40 cpm Time 0 years
After a half-life 20 cpm Time 5730 years
After a half-life 10 cpm Time 5730 years
Total time 11460 years
2.- Conclusion
The wooden object is 11460 years old.
In the laboratory 6.67 g of Sr(NO3)2 is dissolved in enough water to form 0.750 L. A 0.100 L sample is withdrawn from this stock solution and titrated with a 0.0460 M solution of Na3PO4. a. What is the concentration of the Sr(NO3)2stock solution? b. Write a balanced molecular equation for the titration reaction. c. How many milliliters of the Na3PO4 solution are required to precipitate all the Sr2+ ions in the 0.100 L sample? (MM's: Sr(NO3)2 = 211.64; Na3PO4 =163.94)Name
Answer:
The answer to your question is below
Explanation:
Data
mass of Sr(NO₃)₂ = 6.67 g Final volume = 0.750 L
Sample 0.100 L
[Na₃PO₄] = 0.046 M
a) [Sr(NO₃)₂
MM = 211.64 g
211.64 g ---------------------- 1 mol
6.67 g ---------------------- x
x = (6.67 x 1) / 211.64
x = 0.032 mol
Molarity = 0.032 / 0.75
Molarity = 0.042
b)
3Sr(NO₃)₂ + 2Na₃PO₄ ⇒ Sr₃(PO₄)₂ + 6NaNO₃
Reactants Elements Products
3 Sr 3
6 N 6
6 Na 6
2 P 2
24 O 24
c)
Calculate the moles of Sr(NO₃)₂ in 100 ml or 0.1 L
Molarity = moles / volume
Moles = Molarity x volume
Moles = 0.042 x 0.1
Moles = 0.0042
3 moles of Sr(NO₃)₂ --------------- 2 moles of Na₃PO₄
0.0042 moles of Sr(NO₃)₂ -------- x
x = (0.0042 x 2) / 3
x = 0.0028 moles of Na₃PO₄
Molarity = moles / volume
Volume = moles / Molarity
Volume = 0.0028 / 0.046
Volume = 0.060 L or 60.9 mL
A 2.00-L glass soda bottle filled only with air is tightly capped at 25°C and 728.0 mmHg. If the bottle is placed in water at 65°C, what is the pressure in the bottle?
Answer:
The pressure in the bottle is 826 mmHg
Explanation:
In this case it is assumed that the volume of the soda bottle does not change, so it remains constant with a value of 2.00 L. Then it is possible to apply the Gay Lussac law.
This law indicates that when there is a constant volume, as the temperature increases, the gas pressure increases. And when the temperature decreases, gas pressure decreases. That is, the gas pressure is directly proportional to its temperature.
Gay-Lussac's law can be expressed mathematically as follows:
[tex]\frac{P}{T}=k[/tex]
Where P = pressure, T = temperature and K = Constant
Having a gas that is at a pressure P1 and a temperature T1, as the temperature varies to a new T2 value, then the pressure will change to P2. It is then fulfilled:
[tex]\frac{P1}{T1} =\frac{P2}{T2}[/tex]
Remember that the temperature must be in degrees Kelvin (° K) and that 0 ° C is 273.15 ° K
In this case you know:
P1= 728 mmHgT1= 25°C+273.15°K= 298.15°KP2= ?T2= 65°C+273.15°K= 338.15°KReplacing:
[tex]\frac{728 mmHg}{298.15K} =\frac{P2}{338.15K}[/tex]
Resolving you get:
[tex]\frac{728 mmHg}{298.15K} *338.15K=P2[/tex]
P2=825.67 mmHg≅826 mmHg
The pressure in the bottle is 826 mmHg
The pressure of the bottle is 825.7 mmHg
The parameters given in the question are
Pressure 1= 728 mmHg
Temperature 1= 25°C
= 25+273
T1= 298K
Temperature 2= 65°C
= 65 +273
= 338K
P1/T1= P2/T2
728/298= P2/338
Cross multiply
298 × P2= 728 × 338
298 × P2= 246,064
P2= 246,064/298
P2= 825.7
Hence the pressure in the bottle is 825.7 mmHg
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An 8.10-g sample of was placed in an evacuated container, where it decomposed at 590°C according to the following reaction: At equilibrium the total pressure and the density of the gaseous mixture were 1.83 atm and 1.57 g/L respectively. Calculate for this reaction.
Correct question:
An 8.10-g sample of SO3 was placed in an evacuated container, where it decomposed at 590°C according to the following reaction:
SO3(g) <-----> SO2(g) + 1/2 O2 (g)
At equilibrium the total pressure and the density of the gaseous mixture were 1.83 atm and 1.57 g/L respectively. Calculate Kp for this reaction
Answer:
Kp for this reaction is 0.149atm
Explanation:
Given details
The state reaction
SO3(g) <-----> SO2(g) + 1/2 O2 (g)
Density = 1.57 g/L
Temperature = 590°C = 863K
The given mass of SO3 is 8.10g
The molar mass of SO3 is
S + 3O = {(32) + 3(16)} = 80g/mol
Numbers of mole =
Given mass/molar mass = 8.10/80
Numbers of mole of SO3 = 0.1013mol
From density = mass/volume
Volume V = 8.10/1.56 = 5.2L
Initial pressure from PV = nRT
R = Universal gas constant
R = 0.0821 atm/K/mol
P = (0.1013*0.0821*863)/5.2
P = 1.38 atm
At equilibrium
moles SO3 = 0.10 - X
moles SO2 = X
moles O2 = X/2
moles total = (0.10 - X) + X + X/2
Total mole = 0.10 + X/2
Ptot = (0.10 + X/2)*0.0821*863/5.2 = 1.83
(0.10 + X/2)* 70.8523 = 9.516
X/2 = 0.1343 -0.10 = 0.0343
X = 0.0686
At equilibrium
moles SO3 = 0.10 - X = 0.10 - 0.0686 = 0.0314
moles SO2 = X = 0.0686
moles O2 = X/2 = 0.0343
moles total = 0.10 + X/2 = 0.10 + 0.0343 = 0.1343
P(SO3) = Ptot*X(SO3) = 1.83*0.0314/0.1343 = 0.428atm
P(SO2) = 1.83*0.0686/0.1343 = 0.935atm
P(O2) = 1.83*0.0343/0.1343 = 0.467atm
Kp for this reaction is
Kp = [P(SO2)*P(O2)^1/2]/P(SO3)
Kp = {0.935*(0.467)^0.5}/0.428
Kp = 0.149atm
The combustion of glucose is represented by the following balanced equation: C6H12O6+6 O2→6 H2O+6 CO2. The reaction uses 1 gram of both C6H12O6 and O2. What is the percent yield if 0.45 g of H2O is produced? a 0.558% b 100% c 0.31% d 80%
Answer : The correct option is, (d) 80 %
Solution : Given,
Mass of [tex]C_6H_{12}O_6[/tex] = 1 g
Mass of [tex]O_2[/tex] = 1 g
Molar mass of [tex]C_6H_{12}O_6[/tex] = 180 g/mole
Molar mass of [tex]O_2[/tex] = 32 g/mole
Molar mass of [tex]H_2O[/tex] = 18 g/mole
First we have to calculate the moles of [tex]C_6H_{12}O_6[/tex] and [tex]O_2[/tex].
[tex]\text{ Moles of }C_6H_{12}O_6=\frac{\text{ Mass of }C_6H_{12}O_6}{\text{ Molar mass of }C_6H_{12}O_6}=\frac{1g}{180g/mole}=0.00555moles[/tex]
[tex]\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{1g}{32g/mole}=0.0312moles[/tex]
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
[tex]C_6H_{12}O_6+6O_2\rightarrow 6H_2O+6CO_2[/tex]
From the balanced reaction we conclude that
As, 6 mole of [tex]O_2[/tex] react with 1 mole of [tex]C_6H_{12}O_6[/tex]
So, 0.0312 moles of [tex]O_2[/tex] react with [tex]\frac{0.0312}{6}=0.0052[/tex] moles of [tex]C_6H_{12}O_6[/tex]
From this we conclude that, [tex]C_6H_{12}O_6[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]O_2[/tex] is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of [tex]H_2O[/tex]
From the reaction, we conclude that
As, 6 mole of [tex]O_2[/tex] react to give 6 mole of [tex]H_2O[/tex]
So, 0.0312 mole of [tex]O_2[/tex] react to give 0.0312 mole of [tex]H_2O[/tex]
Now we have to calculate the mass of [tex]H_2O[/tex]
[tex]\text{ Mass of }H_2O=\text{ Moles of }H_2O\times \text{ Molar mass of }H_2O[/tex]
[tex]\text{ Mass of }H_2O=(0.0312moles)\times (18g/mole)=0.562g[/tex]
Theoretical yield of [tex]H_2O[/tex] = 0.562 g
Experimental yield of [tex]H_2O[/tex] = 0.45 g
Now we have to calculate the percent yield of the reaction.
[tex]\% \text{ yield of reaction}=\frac{\text{ Experimental yield of }H_2O}{\text{ Theoretical yield of }H_2O}\times 100[/tex]
[tex]\% \text{ yield of reaction}=\frac{0.45g}{0.562g}\times 100=80\%[/tex]
Therefore, the percent yield of reaction is, 80 %
Answer:
The yield would D. 80%!
Explanation:
Since 1 gram of O2 only produces 0.56 g of H2O, whereas 1 g of C6H12O6 produces 0.60 g of H2O, the O2 is the limiting reagent.
The volume of a sample of chlorine gas is 8.00 liters at 45.0 degrees Celsius and 0.966 atm. How many moles of chlorine are present in the sample? R = 0.0821 (atm)(L)(mol-1)(K-1)
Answer:
Number of moles of Cl₂ = 0.3 mol
Explanation:
Given data:
Number of moles of Cl₂ = ?
Pressure = 0.966 atm
Volume = 8.00 L
Temperature = 45°C
Solution:
The given problem will be solve by using general gas equation, which is,
PV = nRT
R = general gas constant (0.0821 atm.L/mol.K)
Now we will convert the °C into K.
Temperature = 45+ 273 = 318 K
Now we will put the values in formula.
n = PV/RT
n = 0.966 atm × 8.00 L / 318 K ×0.0821 atm.L/mol.K
n = 7.728/26.1078 /mol
n = 0.3 mol
What mass (g) of barium iodide is contained in 188 ml of a barium iodide solution that has an iodide ion concentration of 0.532m 19.6 39.1 19,600 39,100 276
The mass of barium iodide in 188 ml of a 0.532M solution is calculated to be approximately 39.11 grams using concepts of molarity, moles, and molar mass.
Explanation:The question is asking to find out the mass of barium iodide in a solution, given a certain volume and a concentration. The question is a standard calculation involving molarity and volume. Molarity (M) is defined as the number of moles of solute per liter of solution. Therefore, you can calculate the number of moles of solute first using the formula M = n/V, where n is the number of moles and V is the volume in liters. We can convert 188 mL to liters (0.188 L) and we know the molarity is 0.532M, so the number of moles in the solution is 0.532M * 0.188 L = 0.10 moles of iodide ions. To find the mass of barium iodide, we multiply this number by the molar mass of barium iodide, which is roughly 391.136 g/mol (from barium's molar mass of 137.327 g/mol and iodine's molar mass of 253.809 g/mol). Thus, the mass of barium iodide is 0.10 moles * 391.136 g/mol = 39.11 g.
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Whch compounds is most reactive toward nucleophilic acyl substitution?
Answer:
There are five main types of acyl derivatives. Acid halides are the most reactive towards nucleophiles, followed by anhydrides, esters, and amides. Carboxylate ions are essentially unreactive towards nucleophilic substitution, since they possess no leaving group
A metal object with mass of 23.2 g 23.2 g is heated to 97.0 °C 97.0 °C and then transferred to an insulated container containing 90.0 g 90.0 g of water at 20.5 °C. 20.5 °C. The water temperature rises and the temperature of the metal object falls until they both reach the same final temperature of 22.6 °C. 22.6 °C. What is the specific heat of this metal object? Assume that all the heat lost by the metal object is absorbed by the water.
Answer:
The specific heat of the object [tex]C_{obj}[/tex] = 0.457 [tex]\frac{KJ}{kg K}[/tex]
Explanation:
Mass of the object [tex]m_{obj}[/tex] = 23.2 gm
Initial temperature [tex]T_{obj}[/tex] = 97 ° c
Mass of the water [tex]m_{w}[/tex] = 90 gm
Initial temperature of water [tex]T_{w}[/tex] = 20.5 ° c
Final temperature of both water & object [tex]T_{f}[/tex] = 22.6 ° c
It is given that heat lost by the object = heat gain by the water
⇒ [tex]m_{obj}[/tex] [tex]C_{obj}[/tex] ( [tex]T_{obj}[/tex] - [tex]T_{f}[/tex] ) = [tex]m_{w}[/tex] [tex]C_{w}[/tex] ( [tex]T_{f}[/tex] - [tex]T_{w}[/tex])
Put all the values in above formula we get
⇒ 23.2 × [tex]C_{obj}[/tex] ( 97 - 22.6 ) = 90 × 4.18 × ( 22.6 - 20.5 )
⇒ [tex]C_{obj}[/tex] = 0.457 [tex]\frac{KJ}{kg K}[/tex]
This is the specific heat of the object.
Calculate the pH after 0.018 mole of HCl is added to 1.00 L of each of the four solutions. (Assume that all solutions are at 25°C.) (a) 0.129 M acetic acid (HC2H3O2, Ka = 1.8 ✕ 10−5) (b) 0.129 M sodium acetate (NaC2H3O2) (c) pure H2O (d) 0.129 M HC2H3O2 and 0.129 M NaC2H3O2
Explanation:
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Regarding the formula Al203 which of the following is accurate
Answer:
D. The subscript 2 indicates that two atoms of aluminum are present in the substance.
Explanation:
The answer choices are:
A. The coefficient 3 indicates that there are a total of three atoms of oxygen present in the substance. B. The subscript 2 indicates that two atoms of oxygen are present in the substance. C. The chemical symbol Al indicates that oxygen is present in the substance. D. The subscript 2 indicates that two atoms of aluminum are present in the substance.Solution
The correct formula is:
[tex]Al_2O_3[/tex]
What you must know to answer this question is that the chemical formulas indicate the number of atoms of each element in the formula by placing a subscript to the right of the chemical symbol that represents the atom.
Thus, we can deal with each statement:
A. The coefficient 3 indicates that there are a total of three atoms of oxygen present in the substance.
Incorrect. 3 is not a coefficient but a subscript. Thus this option is wrong.
B. The subscript 2 indicates that two atoms of oxygen are present in the substance.
Incorrect. The subscript 2 is to the right of tha aluminum symbol, thus it does not represent the number of atoms of oxygen.
C. The chemical symbol Al indicates that oxygen is present in the substance.
Incorrect. The chemical symbol Al indicates that aluminum is present in the substance. Thus, this is wrong.
D. The subscript 2 indicates that two atoms of aluminum are present in the substance.
Correct. The subscript 2 is to the right of the symbol Al, which is the chemical symbol for aluminum. Thus, this indicates that there are two atoms of aluminum in the substance.
How many hours will it take for the concentration of methyl isonitrile to drop to 14.0 %% of its initial value?
This is an incomplete question, here is a complete question.
The rearrangement of methyl isonitrile (CH₃NC) to acetonitrile (CH₃NC) is a first-order reaction and has a rate constant of 5.11 × 10⁻⁵ s⁻¹ at 472 K. If the initial concentration of CH₃NC is 3.00 × 10⁻² M :
How many hours will it take for the concentration of methyl isonitrile to drop to 14.0 % of its initial value?
Answer : The time taken will be, 10.7 hours
Explanation :
Expression for rate law for first order kinetics is given by:
[tex]t=\frac{2.303}{k}\log\frac{a}{a-x}[/tex]
where,
k = rate constant = [tex]5.11\times 10^{-5}s^{-1}[/tex]
t = time passed by the sample = ?
a = let initial amount of the reactant = 100
a - x = amount left after decay process = 14 % of 100 = 14
Now put all the given values in above equation, we get
[tex]t=\frac{2.303}{5.11\times 10^{-5}}\log\frac{100}{14}[/tex]
[tex]t=38482.72s=\frac{38482.72}{3600}=10.7hr[/tex]
Therefore, the time taken will be, 10.7 hours
The decay series for 238 92U is represented in Figure 19.1. Write the balanced nuclear equation for each of the following radioactive decays. a. Alpha-particle production by 226 88 Ra b. Beta-particle production by 214 82 Pb
Answer: The decay process of the radioisotopes are written below.
Explanation:
For the given options:
Option a:Alpha decay is defined as the process in which the nucleus of an atom disintegrates into two particles. The first one which is the alpha particle consists of two protons and two neutrons, also known as helium nucleus. The second particle is the daughter nuclei which is the original nucleus minus the alpha particle released.
[tex]_Z^A\textrm{X}\rightarrow _{Z-2}^{A-4}\textrm{Y}+_2^4\alpha[/tex]
The chemical equation for the alpha decay process of Ra-226 isotope follows:
[tex]_{88}^{226}\textrm{Ra}\rightarrow _{86}^{222}\textrm{Rn}+_2^4\alpha[/tex]
Option b:Beta decay is defined as the process in which the neutrons get converted into an electron and a proton. The released electron is known as the beta particle. In this process, the atomic number of the daughter nuclei gets increased by a factor of 1 but the mass number remains the same.
[tex]_Z^A\textrm{X}\rightarrow _{Z+1}^{A}\textrm{Y}+_{-1}^0\beta[/tex]
The chemical equation for the alpha decay process of Pb-214 isotope follows:
[tex]_{82}^{214}\textrm{Pb}\rightarrow _{83}^{214}\textrm{Bi}+_{-1}^0\beta[/tex]
Hence, the decay process of the radioisotopes are written above.
The combustion of ethane ( C 2 H 6 ) produces carbon dioxide and steam. 2 C 2 H 6 ( g ) + 7 O 2 ( g ) ⟶ 4 CO 2 ( g ) + 6 H 2 O ( g ) How many moles of CO 2 are produced when 5.30 mol of ethane is burned in an excess of oxygen?
Answer:
10.6 moles of CO₂ are produced in this combustion
Explanation:
The combustion reaction is:
2C₂H₆ (g) + 7O₂ (g) ⟶ 4CO₂ (g) + 6H₂O (g)
We assume the ethane as the limiting reactant because the excersise states that the O₂ is in excess.
We make a rule of three:
2 moles of ethane can produce 4 moles of CO₂
Therefore 5.30 moles of ethane will produce (5.3 . 4) /2 = 10.6 moles
Which compounds will be soluble in water?
Answer:
salt
Explanation:
Choose the chemical equation that is correctly balanced. 2Ca(s) + Cl2(g) → CaCl2(s) 4Mg(s) + O2(g) → 2MgO(s) Li(s) + Cl2(g) → 2LiCl(s) C(s) + O2(g) → CO2(g)
Answer:
Last option C(s) + O2(g) → CO2(g)
Explanation:
The reactions are:
2Ca(s) + Cl2(g) → CaCl2(s)
4Mg(s) + O2(g) → 2MgO(s)
Li(s) + Cl2(g) → 2LiCl(s)
C(s) + O2(g) → CO2(g)
Let's count the atoms (check out the stoichiometry):
1. We have 2 Ca in reactant side and 2Cl, in product side we have 1 Ca and 2 Cl. UNBALANCED
2. We have 4 Mg in reactant side and 2 O. In product side we have 2 Mg and 2 O. UNBALANCED
3. In reactant side we have 1 Li and 2 Cl. Then, in product side we have 2Li and 2Cl. UNBALANCED
C(s) + O2(g) → CO2(g)
1 C and 2 O ⇒ 1 C and 2 O Correctly balanced
The chemical equation [tex]\( \text{C(s)} + \text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} \)[/tex] is correctly balanced. The correct option is (D).
To determine which chemical equation is correctly balanced, we need to ensure that the number of atoms for each element is the same on both sides of the equation. Let’s examine each option:
A) [tex]\( 2\text{Ca(s)} + \text{Cl}_2\text{(g)} \rightarrow \text{CaCl}_2\text{(s)} \)[/tex]
- Reactants: 2 Ca and 2 Cl
- Products: 1 Ca and 2 Cl
In this equation, the calcium (Ca) atoms are not balanced (2 Ca atoms on the left vs. 1 Ca atom on the right).
This equation is not balanced.
B) [tex]\( 4\text{Mg(s)} + \text{O}_2\text{(g)} \rightarrow 2\text{MgO(s)} \)[/tex]
- Reactants: 4 Mg and 2 O
- Products: 2 Mg and 2 O
In this equation, the magnesium (Mg) atoms are not balanced (4 Mg atoms on the left vs. 2 Mg atoms on the right).
This equation is not balanced.
C) [tex]\( \text{Li(s)} + \text{Cl}_2\text{(g)} \rightarrow 2\text{LiCl(s)} \)[/tex]
- Reactants: 1 Li and 2 Cl
- Products: 2 Li and 2 Cl
In this equation, the lithium (Li) atoms are not balanced (1 Li atom on the left vs. 2 Li atoms on the right).
This equation is not balanced.
D) [tex]\( \text{C(s)} + \text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} \)[/tex]
- Reactants: 1 C and 2 O
- Products: 1 C and 2 O
In this equation, both carbon (C) and oxygen (O) atoms are balanced.
After analyzing the chemical equations, the only equation that is correctly balanced is:
[tex]\*\*D) \( \text{C(s)} + \text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} \)\*\*[/tex]
To verify that the equation is balanced:
Carbon [tex](C)[/tex]:
- Reactants: 1 atom
- Products: 1 atom
Oxygen [tex](O)[/tex]:
- Reactants: 2 atoms
- Products: 2 atoms
Since the number of atoms for each element is the same on both sides of the equation, it confirms that option D is the correct and balanced chemical equation.
The complete question is:
Choose the chemical equation that is correctly balanced.
A) [tex]2Ca(s) + Cl_2(g) \rightarrow CaCl_2(s)[/tex]
B) [tex]4Mg(s) + O_2(g) \rightarrow 2MgO(s)[/tex]
C) [tex]Li(s) + Cl_2(g) \rightarrow 2LiCl(s)[/tex]
D) [tex]C(s) + O_2(g) \rightarrow CO_2(g)[/tex]
The molarity of an aqueous solution of nacl is defined as the
Answer: Number of moles of NaCl per liter of solution
Explanation: Molarity can be defined as the number of moles
of solute per liter of solution.
Therefore the molarity of an aqueous solution of NaCl is thus defined as the number of moles of NaCl per liter of solution.
when the temperature of a gas changes, it's volume decreases from 12 cm3 to 7 cm3 if the final temperature is measured to be 18°C what is the initial temperature in Kelvin units?
pls help :)
Answer:
The initial temperature is 499 K
Explanation:
Step 1: Data given
initial volume = 12 cm3 = 12 mL
Final volume = 7 cm3 = 7mL
The final temperature = 18 °C = 291 K
Step 2: Calculate the initial temperature
V1/T1 = V2/T2
⇒with V1 = the initial volume = 0.012 L
⇒with T1 = the initial volume = ?
⇒with V2 = the final volume 0.007 L
⇒with T2 = The final temperature = 291 K
0.012 / T1 = 0.007 / 291
0.012/T1 = 2.4055*10^-5
T1 = 0.012/2.4055*10^-5
T1 = 499 K
The initial temperature is 499 K
The______of water molecules and the hydrogen bonds between water molecules explain most of water's life-supporting properties.The_____of water molecules to each other helps transport water from the roots to the leaves in plants.
Polarity
Cohesion
Explanation:
One molecule of water joins with four other water molecule by hydrogen bonds. In a water molecule, one end has positive charge and the other end has negative charge. This difference in charge creates polarity in the molecule.
The cohesion of water molecules helps transport water from roots to the leaves in plants. Plants absorb water from the soil by osmosis. They absorb mineral ions by active transport, against the concentration gradient. Root hair cells are adapted for taking up water and mineral ions by having a large surface area to increase the rate of absorption.
Answer: Cohesion
Explanation:
Cohesion is a measure of how well molecules stick to each other. Water molecules is a typical example of Cohesion.
Hydrogen bonds are formed between each water molecule, enabling them stick to each other strongly. Due to this sticky nature of water molecules, they form droplets on surfaces (e.g dew drops) and a dome like-shape when filling a container just before it overflows.
Cohesion also produces Surface Tension; a phenomenon that makes it possible for light objects like needle to float when placed gently and insects to walk on water.
Also aided by capillary action (the movement of a liquid across the surface of a solid caused by adhesion between the two), Cohesion makes it possible for water to be taken as single huge molecule from Xylem in the roots of plant to its leaves.
A solution of HCl gas dissolved in water (sold commercially as "muriatic acid," a solution used to clean masonry surfaces) has 20.22 g of HCl per 100.0 g of solution, and its density is 1.10 g/mL.
What is its molarity?
What is its mole fraction?
Answer:
[HCl] = 6.09 M
Xm HCl = 0.11
Explanation:
Let's analyse data:
20.22 g of solute / 100 g of solution
Solution's density = 1.10 g/mL
As we have the mass of solution and its density we determine solution's volume to stablish [M]
Density = Mass / volume → 1.10 g/mL = 100 g / Volume
100 g / 1.10g/mL = 90.9 mL
Let's convert the volume to L → 90.9 mL . 1L/ 1000mL = 0.0909L
We convert the mass of solute to moles → 20.22 g . 1mol/ 36.45g =
0.554 moles
[M] = Molarity (moles of solute /1L of solution) = 0.554 mol/0.0909L = 6.09M
Mole fraction (Xm) = Moles of solute / Total moles
Total moles = Moles of solute + Moles of solvent
Mass of solvent = Mass of solution - Mass of solute
Mass of solvent = 100 g - 20.22 g = 79.78g
We convert the mass to moles → 79.78 g / 18g/mol = 4.43 moles
Total moles = 4.43 moles + 0.554moles = 4.984 moles
Xm = 0.554 / 4.984 = 0.11
Final answer:
To find the molarity and mole fraction of the HCl solution, we calculate the molarity as 6.09 M and the mole fraction of HCl as 0.111 based on the given mass of HCl, solution density, and total solution mass.
Explanation:
To determine the molarity and mole fraction of a solution of HCl gas dissolved in water, we first need to calculate the mass of the solution and then convert the mass of HCl to moles. Given that the solution has 20.22 g of HCl per 100.0 g of solution, and its density is 1.10 g/mL, we can calculate the molarity and mole fraction as follows:
Calculating Molarity
Convert the mass of HCl to moles using the molar mass of HCl (36.46 g/mol):
20.22 g HCl * (1 mol HCl / 36.46 g) = 0.554 moles HCl. To find the volume of the solution, use the density and total mass:
100.0 g solution * (1 mL / 1.10 g) = 90.91 mL. Convert this to liters: 90.91 mL = 0.09091 L. Calculate molarity (M):
M = moles of solute / liters of solution = 0.554 moles / 0.09091 L = 6.09 M.
Calculating Mole Fraction
The mole fraction of HCl (XHCl) requires the number of moles of water:
(79.78 g water / 18.015 g/mol) = 4.43 moles of water. Calculate mole fraction of HCl:
XHCl = moles of HCl / (moles of HCl + moles of water) = 0.554 / (0.554 + 4.43) = 0.111.