Answers
We can describe a positive energetic process as any process which increases the internal energy of the system.
A positive energetic reaction or process is often referred to as being Endothermic. This means that the system which is performing the process absorbs energy. Some examples include:
Boiling an EggRoasting food over a fire (the food is the reference system)etc
Therefore, we can confirm that a positive energetic process is one in which the system in question absorbs energy, thus increasing its internal energy.
Since I could not locate the options online, I have provided a general explanation of the concept coupled with a few examples.
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Related Questions
Balance the following skeleton reaction and identify the oxidizing and reducing agents: Include the states of all reactants and products in your balanced equation. You do not need to include the states with the identities of the oxidizing and reducing agents.
NO_2(g) rightarrow NO_3^-(aq) +NO_2^- (aq) [basic]
The oxidizing agent is:______.
The reducing agent is:_______.
Answers
Answer:
a. 2NO₂ (g) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻ (aq) + H₂O (l)
b. i. NO₂⁻ is the oxidizing agent
ii. NO₃⁻ is the reducing agent.
Explanation:
a. Balance the following skeleton reaction
The reaction is
NO₂ (g) → NO₃⁻ (aq) + NO₂⁻ (aq)
The half reactions are
NO₂ (g) → NO₃⁻ (aq) (1) and
NO₂ (g) → NO₂⁻ (aq) (2)
We balance the number of oxygen atoms in equation(1) by adding one H₂O molecule to the left side.
So, NO₂ (g) + H₂O (l) → NO₃⁻ (aq)
We now add two hydrogen ions 2H⁺ on the right hand side to balance the number of hydrogen atoms
NO₂ (g) + H₂O (l) → NO₃⁻ (aq) + 2H⁺ (aq)
The charge on the left hand side is zero while the total charge on the right hand side is -1 + 2 = +1. To balance the charge on both sides, we add one electron to the right hand side.
So, NO₂ (g) + H₂O (l) → NO₃⁻ (aq) + 2H⁺ (aq) + e⁻ (4)
Since the number of atoms in equation two are balanced, we balance the charge since the charge on the left hand side is zero and that on the right hand side is -1. So, we add one electron to the left hand side.
So, NO₂ (g) + e⁻ → NO₂⁻ (aq) (5)
We now add equation (4) and (5)
So, NO₂ (g) + H₂O (l) → NO₃⁻ (aq) + 2H⁺ (aq) + e⁻ (4)
+ NO₂ (g) + e⁻ → NO₂⁻ (aq) (5)
2NO₂ (g) + H₂O (l) + e⁻ → NO₃⁻ (aq) + NO₂⁻ (aq) + 2H⁺ (aq) + e⁻ (4)
2NO₂ (g) + H₂O (l) → NO₃⁻ (aq) + NO₂⁻ (aq) + 2H⁺ (aq)
We now add two hydroxide ions to both sides of the equation.
So, 2NO₂ (g) + H₂O (l) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻ (aq) + 2H⁺ (aq) + 2OH⁻ (aq)
The hydrogen ion and the hydroxide ion become a water molecule
2NO₂ (g) + H₂O (l) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻ (aq) + 2H₂O (l)
2NO₂ (g) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻ (aq) + H₂O (l)
So, the required reaction is
2NO₂ (g) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻ (aq) + H₂O (l)
b. Identify the oxidizing agent and reducing agent
Since the oxidation number of oxygen in NO₂ is -2. Since the oxidation number of NO₂ is zero, we let x be the oxidation number of N.
So, x + 2 × (oxidation number of oxygen) = 0
x + 2(-2) = 0
x - 4 = 0
x = 4
Since the oxidation number of oxygen in NO₂⁻ is -1. Since the oxidation number of NO₂⁻ is -1, we let x be the oxidation number of N.
So, x + 2 × (oxidation number of oxygen) = 0
x + 2(-2) = -1
x - 4 = -1
x = 4 - 1
x = 3
Also, the oxidation number of oxygen in NO₃⁻ is -1. Since the oxidation number of NO₃⁻ is -1, we let x be the oxidation number of N.
So, x + 2 × (oxidation number of oxygen) = -1
x + 3(-2) = -1
x - 6 = -1
x = 6 - 1
x = 5
i. The oxidizing agent
The oxidation number of N changes from +4 in NO₂ to +3 in NO₂⁻. So, Nitrogen is reduced and thus NO₂⁻ is the oxidizing agent
ii. The reducing agent
The oxidation number of N changes from +4 in NO₂ to +5 in NO₃⁻. So, Nitrogen is oxidized and thus and NO₃⁻ is the reducing agent.
Determine the boiling point of a solution that contains 150.0 g of naphthalene (C10H8, molar mass = 128.16 g/mol) dissolved in 950 mL of benzene (d = 0.877 g/mL). Pure benzene has a boiling point of 80.1°C and a boiling point elevation constant of 2.53°C/m.
Answers
Answer:
Boiling T° of solution → 83.6°C
Explanation:
To solve this, we apply Elevation of boiling point, property
ΔT = Kb . m . i
As we talk about organic solute, i = 1. No ions are formed.
m = molality (moles of solute in 1kg of solvent)
We determine mass of solvent by density
D = m /V so D . V = m
950 mL . 0.877 g/mL = 833.15 g
We convert to kg → 833.15 g . 1 kg/ 1000g = 0.833 kg
Moles of solute (naphtalene): 150 g . 1 mol/ 128.16g = 1.17 mol
m = 1.17mol / 0.833 kg = 1.41 mol/kg
We replace data:
Boiling T° of solution - 80.1°C = 2.53°C/m . 1.41 m . 1
Boiling T° of solution = 2.53°C/m . 1.41 m . 1 + 80.1°C → 83.6°C
Answer:
The answer is c or 17.1 g
11. An isotope Q has 18 neutrons a mass number of 34. (a) (i) What is an isotope? An isotope is one of two or C (b) Write its electron arrangement. Mass number=34 Number of neutrons=18 Number of Protons = 34-15-16 (c) To which period and group does Q belong? Protors - Electons - Atomic number Period - Group (d) How does Q form its ion?
Answers
An isotope is an element with the same atomic number but different mass number due to differences in number of neutrons.
electron configuration is 2,8,6.
Belongs to group 6 and period group 3.
It forms an ion by accepting 2 electrons
The specific heat capacity of lead is 0.13 J/g-K. How much heat (in J) is required to raise the temperature of 15 g of lead from 22 °C to 37 °C? a. 5.8 × 10-4 J b. 0.13 J c. 29 J d. 2.0 J e. -0.13 J
Answers
Answer:
c. 29 J
Explanation:
Step 1: Given data
Specific heat capacity of Pb (c): 0.13 J/g.K (= 0.13 J/g.°C)Mass of Pb (m): 15 gInitial temperature: 22 °CFinal temperature: 37 °CStep 2: Calculate the temperature change
ΔT = 37 °C - 22 °C = 15 °C
Step 3: Calculate the heat (Q) required to raise the temperature of the lead piece
We will use the following expression.
Q = c × m × ΔT
Q = 0.13 J/g.°C × 15 g × 15 °C = 29 J
explain in details how triacylglycerol have an advantage over carbohydrates as stored fuel
Answers
Answer:
As stored fuels, triacylglycerols have two significant advantages over polysaccharides such as glycogen and starch. The carbon atoms of fatty acids are more reduced than those of sugars, and oxidation of triacylglycerols yields more than twice as much energy, gram for gram, as that of carbohydrates.
Explanation:
Three peptides were obtained from a trypsin digestion of two different polypeptides. Indicate the possible sequences from the given data.
a. Val-Gly-Arg
b. Ala-Val-Lys
c. Ala-Gly-Phe
Answers
Answer:
A) Val-Gly-Arg-Ala-Val-Lys-Ala-Gly-Phe
B) Ala-Val-Lys-Val-Gly-Arg-Ala-Gly-Phe
Explanation:
The possible sequences that could be obtained from the available dat provided are :
A) Val-Gly-Arg-Ala-Val-Lys-Ala-Gly-Phe
B) Ala-Val-Lys-Val-Gly-Arg-Ala-Gly-Phe
Trypsin is generally a catalyst for the breakdown of proteins into smaller peptides ( during the hydrolysis of peptide bonds )
GIVING BRAINLIEST
Which equations are used to calculate the velocity of a wave?
O velocity = distance ~ time
velocity = wavelength x frequency
velocity = distance/time
velocity = wavelength/frequency
velocity = distance/time
velocity = wavelength x frequency
velocity = distance ~ time
velocity = wavelength/frequency
Answers
Answer:
velocity = distance/time
velocity = wavelength × frequency
Both of these are commonly known equations to calculate velocity with different variables.
Consider a galvanic (voltaic) cell that has the generic metals X and Y as electrodes. If X is more reactive than Y (that is, X more readily reacts to form a cation than Y does), classify the following descriptions by whether they apply to the X or Y electrode.
i. anode
ii. cathode
iii. electrons in the wire flow toward
iv. electrons in the wire flow away
v. cations from salt bridge flow toward
vi. anions from salt bridge flow toward
vii. gains mass
viii. loses mass
Answers
Answer:
X
anode
electrons in the wire flow away
anions from salt bridge flow toward
loses mass
Y
cathode
electrons in the wire flow toward
cations from salt bridge flow toward
gains mass
Explanation:
In a galvanic cell, oxidation occurs at the anode while reduction occurs at the cathode. The metal that is more reactive functions as the anode while the less reactive metal functions as the cathode.
Electrons leave the anode and travel via a wire to the cathode. At the anode cations give up electrons and enter into the solution.
At the cathode, cations pick up electrons and are deposited on the cathode leading to a gain in mass at the cathode.
Positive ions from the salt bridge flow towards the cathode while negative ions from the salt bridge flow towards the anode.
A 0.204 g sample of a CO3 2- antacid is dissolved with 25.0ml of 0.0981 M HCL. The hydrochloric acid that is not neutralized by the antacid is titrated to a bromophenol blue endpoint with 5.83 ml of 0.104 M NaOH. Assuming the active ingredient in the antsacid sample is CaCO3, calculate the mass of CaCO3 in the sample.
Answers
Answer:
0.0922 g
Explanation:
Number of moles of acid present = 25/1000 × 0.0981
= 0.00245 moles
Number of moles of base = 5.83/1000 × 0.104
= 0.000606 moles
Since the reaction of HCl and NaOH is 1:1
Number of moles of HCl that reacted with antacid = 0.00245 moles - 0.000606 moles
= 0.001844 moles
From the reaction;
CaCO3 + 2HCl ----> CaCl2 + H2O + CO2
1 mole of CaCO3 reacts with 2 moles of HCl
x moles of CaCO3 reacts with 0.001844 moles ofHCl
x = 1 × 0.001844/2
= 0.000922 moles
Mass of CaCO3 = 0.000922 moles × 100 g/mol
= 0.0922 g
How many grams of sodium chloride are contained in 250.0 g of a 15% NaCl solution?
Please explain and show work.
Answers
Given concentration of NaCl=15%
Means ,
In every 100g of Solution 15g of NaCl is present .
Now
Given mass=250gSo ,
[tex]\\ \Large\sf\longmapsto 250\times 15\%[/tex]
[tex]\\ \Large\sf\longmapsto 250\times \dfrac{15}{100}[/tex]
[tex]\\ \Large\sf\longmapsto 37.5g[/tex]
37.5g of NaCl present in 250g of solution.
Answer:
Given concentration of NaCl=15%
Means ,
In every 100g of Solution 15g of NaCl is present .
Now
Given mass=250g
So ,
➡250 × 15%
➡250×15/100
➡37.5g
37.5g of NaCl present in 250g of solution.
Un sistema formado por una única sustancia, ¿será siempre homogéneo? ¿Porqué? Piensa a partir de las definiciones y trata de corroborar o negar usando ejemplos concretos.
Answers
Una sustancia homogénea es una sustancia que se compone de una sola fase.
Recordemos que definimos una fase en química como "cantidad química y físicamente uniforme u homogénea de materia que se puede separar mecánicamente de una mezcla no homogénea y que puede consistir en una sola sustancia o una mezcla de sustancias" según Ecyclopedia Britiannica.
El hecho de que un sistema esté compuesto por una sola sustancia no lo hace es autóctono. A veces, un sistema puede estar compuesto por partículas sólidas de una sustancia en equilibrio con su líquido. El sistema contiene solo una sustancia pero en diferentes fases, por lo tanto, el sistema contiene una sustancia pero no es homogéneo.
Por tanto, el hecho de que un sistema contenga una sola sustancia no significa necesariamente que sea homogéneo.
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Once you have collected 40 mL of distillate, you should ________. turn off your hot plate lower your lab jack carelessly use your hand to remove the heating block turn off the hot plate and carefully lower the lab jack, making sure that no cords or hoses get caught in it
Answers
Answer:
Once you have collected 40 mL of distillate, you should ________.
turn off the hot plate and carefully lower the lab jack, making sure that no cords or hoses get caught in it.
Explanation:
Distillate is the product obtained from the process of distillation. Distillation is the separation of components of a liquid mixture based on different boiling points. Distillation can be used to purify alcohol, for desalination, refining of crude oil, and for obtaining liquefied gases. A lab jack is an essential tool that supports and lifts hotplates, glassware, baths, and other small lab equipment requiring stable surfaces at a specific height.
A decomposition of a sample of diphosphorus trioxide forms 1.29 g phosphorus to every 1.00 g oxygen. The decomposition of a sample of diphosphorus pentoxide forms 0.775 g phosphorus to every 1.00 g oxygen.
Required:
How many grams of P205 are formed when 5.89 g of P react with excess oxgen?
Answers
Answer:
There is 13.48 grams of P2O5 formed
Explanation:
Step 1: Data given
A decomposition of a sample of diphosphorus trioxide forms 1.29 g phosphorus to every 1.00 g oxygen.
Mass of P = 5.89 grams
Molar mass of O2 = 32.0 g/mol
atomic mass of P = 30.97 g/mol
molar mass of P2O5 = 141.94 g/mol
Step 2: The balanced equation
4P(s)+5O2(g)⇔ 2P2O5(s)
Step 3: Calculate moles of P
Moles P = Mass P / atomic mass P
Moles P = 5.89 grams / 30.97 g/mol
Moles P = 0.190 moles
Step 4: Calculate moles of P2O5
For 4 moles P we need 5 moles O2 to produce 2 moles P2O5
For 0.190 moles of P we'll have 0.190/2 = 0.095 moles P2O5
Step 5: Calculate mass of P2O5
Mass P2O5 = moles P2O5 * molar mass P2O5
Mass P2O5 = 0.095 moles * 141.94 g/mol
Mass P2O5 = 13.48 grams
There is 13.48 grams of P2O5 formed
Which of the following have only a -C-O-C- functional group?
Answers
Answer:
B) ethers
Explanation:
The functional group of an organic compound defines its specificity. The functional group is responsible for the chemical behavior of an organic compound. For example, alkenes are known to have a carbon-carbon double bond (C=C) functional group.
Likewise, organic compounds known as ETHERS are known to possess an ethoxy functional group i.e. oxygen atom bonded to two alkyl groups (R- OR; where R is an alkyl group). Members of ether functional group includes dimethyl ether (CH3-O-CH3), diethyl ether (C2H5-O-C2H5).
who much the velocity of a body when it travels 600m in 5 min
Answers
Answer:
2 m/s
Explanation:
Applying the formulae of velocity,
V = d/t............. Equation 1
Where V = Velocity of the body, d = distance, t = time
From the question,
Given: d = 600 m, t = 5 minutes = (5×60) = 300 seconds.
Substitute these values into equation 1
V = 600/300
V = 2 m/s.
Hence the velocity of the body when it travels is 2 m/s
At 445oC, Kc for the following reaction is 0.020. 2 HI(g) <--> H2 (g) + I2 (g) A mixture of H2, I2, and HI in a vessel at 445oC has the following concentrations: [HI] = 1.5 M, [H2] = 2.50 M and [I2] = 0.05 M. Which one of the following statements concerning the reaction quotient, Qc, is TRUE for the above system?
a. Qc = Kc; the system is at equilibrium.
b. Qc is less than Kc; more H2 and I2 will be produced.
c. Qc is less than Kc; more HI will be produced.
d. Qc is greater than Kc; more HI will be produced.
Answers
Explanation:
The given balanced chemical equation is:
[tex]2 HI(g) <--> H_2 (g) + I_2 (g)[/tex]
The value of Kc at 445oC is 0.020.
[HI]=1.5M
[H2]=2.50M
[I2]=0.05M
The value of Qc(reaction quotient ) is calculated as shown below:
Qc has the same expression as the equilibrium constant.
[tex]Qc=\frac{[H_2][I_2]}{[HI]^2} \\Qc=(2.50Mx0.05M)/(1.5M)^2\\Qc=0.055[/tex]
Qc>Kc,
Hence, the backward reaction is favored and the formation of Hi is favored.
Among the given options, the correct answer is option d. Qc is greater than Kc; more HI will be produced.
công thức của định lý pytago
Answers
The sum of the squares of two sides of a right angle is equal to the square of the hypotenuse
How many mL of 0.200M KI would contain 0.0500 moles of KI?
Please explain and show work.
Answers
Answer:
250ml
Explanation:
call it V
V*0.2=0.05 (moles)
so V=0.05/0.2 = 0.25l = 250ml
We know
[tex]\boxed{\Large{\sf Molarity=\dfrac{No\:of\:moles\:of\:solute}{Volume\:of\:solution\:in\;\ell}}}[/tex]
[tex]\\ \Large\sf\longmapsto Volume\:of\:KI=\dfrac{0.05}{0.2}[/tex]
[tex]\\ \Large\sf\longmapsto Volume\:of\:KI=0.25L[/tex]
[tex]\\ \Large\sf\longmapsto Volume\:of\:KI=250mL[/tex]
Give the change in condition to go from a gas to a solid. Group of answer choices cool or increase pressure cool or reduce pressure increase heat or reduce pressure increase heat or increase pressure none of the above
Answers
Answer:
cool or increase pressure
Explanation:
For a gas to form solid. There must be reduced heat and pressure. The deposition of gas into solid occurs through the removal of thermal energy. The air looses thermal energy and changes into solid.Determine the mmol of both starting materials (factoring in that formic acid is not pure, but rather 88% weight/volume, or 88g/100 ml), showing your work. Determine the limiting reagent in this synthesis. Lastly, calculate the theoretical yield of benzimidazole that you could expect to form.
Answers
Solution :
Molecular Molar Mass Volume Density Mass Moles nmoles
formula (g/mol) (mL) (g/mL) (g)
[tex]$C_6H_8N_2$[/tex] 108.14 0.108 0.001 1
HCOOH 46.02 0.064 1.22 0.07808 0.0017 1.7
mmoles of o-phenylenediamine = 1 mmoles
mmoles of formic acid = 1.7 [tex]\approx[/tex] 2 mmoles
From the reaction of o-phenylenediamine and formic acid, we see,
1 mmole of o-phenylenediamine reacts with 1 mmole of formic acid.
But here, 2 mmoles of the formic acid , this means that the formic acid is an excess reagent and the o-phenylenediamine is the limiting reagent here.
The amount of product depends on the limiting reagent that is o-phenylenediamine. So, 1mmole of o-phenylenediamine will give 1mmole of product.
molar mass of Benzimidazole = [tex]118.14[/tex] g/mol
mmoles of Benzimidazole formed = [tex]1[/tex] mmol
Mass of benzimidazole formed = molar mass x [tex]\frac{nmoles}{1000}[/tex]
[tex]$=\frac{118.14 \times 1}{1000}$[/tex]
= 0.11814 g
So the theoretical yield of Benzimidazole is = 0.118 g = 118mg
In the reaction HCI + NH4OH --> NH4CI+H2O, which compound has an element ratio of 1:4:1?
H2O
NH4Cl
HCI
ΝΗ4ΟΗ
Answers
The compound in this reaction which is having the elemental ratio of 1:4:1 is NH₄Cl where nitrogen and chlorine are of one mole each with 4 hydrogens.
What is elemental ratio?Elemental ratio of a compound is the ratio of number of atoms of each elements in that compound. The elemental ratio can be determined from the molecular formula of compounds.
The given reaction is a double displacement reaction. Here, the Cl group is replaced to the ammonia and OH group is replaced to the water. Thus, two species is replaced in the reaction.
In NH₄Cl, there are one nitrogen, 4 hydrogens and one chlorine atom. Therefore, the elemental ratio of the compound is 1:4:1. The elemental ratio of water is 2:1 and HCl is 1:1 and that in NH₄OH is 1:5:1. Hence, option b is correct.
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Given the equation: 2C6H10(l) 17 O2(g) ---> 12 CO2(g) 10 H2O(g) MM( g/mol): 82 32 44 18 If 115 g of C6H10 reacts with 199 g of O2 and 49 g of H2O are formed, what is the percent yield of the reaction
Answers
Answer:
74%
Explanation:
Step 1: Write the balanced equation
2 C₆H₁₀(l) + 17 O₂(g) ⇒ 12 CO₂(g) + 10 H₂O(g)
Step 2: Determine the limiting reactant
The theoretical mass ratio (TMR) of C₆H₁₀ to O₂ is 164:544 = 0.301:1.
The experimental mass ratio (EMR) of C₆H₁₀ to O₂ is 115:199 = 0.578:1.
Since EMR > TMR, the limiting reactant is O₂.
Step 3: Calculate the theoretical yield of H₂O
The theoretical mass ratio of O₂ to H₂O 544:180.
199 g O₂ × 180 g H₂O/544 g O₂ = 65.8 g H₂O
Step 4: Calculate the percent yield of H₂O
%yield = (experimental yield/theoretical yield) × 100%
%yield = (49 g/65.8 g) × 100% = 74%
Answer:
Percentage yield of H₂O = 74.24%
Explanation:
The balanced equation for the reaction is given below:
2C₆H₁₀ + 17O₂ —> 12CO₂ + 10H₂O
Next, we shall determine the masses of C₆H₁₀ and O₂ that reacted and the mass of H₂O produced from the balanced equation. This is can be obtained as follow:
Molar mass of C₆H₁₀ = 82 g/mol
Mass of C₆H₁₀ from the balanced equation = 2 × 82 = 164 g
Molar mass of O₂ = 32 g/mol
Mass of O₂ from the balanced equation = 17 × 32 = 544 g
Molar mass of H₂O = 18 g/mol
Mass of H₂O from the balanced equation = 10 × 18 = 180 g
SUMMARY:
From the balanced equation above,
164 g of C₆H₁₀ reacted with 544 g of O₂ to produce 180 g of H₂O.
Next, we shall determine the limiting reactant. This can be obtained as follow:
From the balanced equation above,
164 g of C₆H₁₀ reacted with 544 g of O₂.
Therefore, 115 g of C₆H₁₀ will react to produce = (115 × 544)/164 = 381 g of O₂.
From the calculations made above, we can see that a higher mass (i.e 381 g) of O₂ than what was given (i.e 199 g) is needed to react with 115 g of C₆H₁₀.
Therefore, O₂ is the limiting reactant and C₆H₁₀ is the excess reactant.
Next, we shall determine the theoretical yield of H₂O. This can be obtained by using the limiting reactant as shown below:
From the balanced equation above,
544 g of O₂ reacted to produce 180 g of H₂O.
Therefore, 199 g of O₂ will react to produce = (199 × 180)/544 = 66 g of H₂O.
Thus, the theoretical yield of H₂O is 66 g.
Finally, we shall determine the percentage yield. This can be obtained as follow:
Actual yield of H₂O = 49 g
Theoretical yield of H₂O = 66 g
Percentage yield of H₂O =?
Percentage yield = Actual yield /Theoretical yield × 100
Percentage yield of H₂O = 49/66 × 100
Percentage yield of H₂O = 74.24%
When selling on the street, dealers may not know the purity of the ketamine they have, and thus users do not know exactly how much ketamine they are receiving. It is unlikely that the ketamine is pure, or even that different batches of ketamine have the same purity. Assume the drug the user typically buys is only 25% ketamine, and therefore, the user actually dissolved 0.250 g ketamine in 1/4 cup of water to make the solution instead of 1 g in the previous question. 1 cup = 236.5 mL What volume of this ketamine solution would the 65.0 kg user have to inject to experience a high at 0.400 mg/kg? volume: mL What volume of this ketamine solution would the user have to inject to become unconscious at 2.00 mg/kg? of use contact us help What volume of this ketamine solution would the user have to inject to become unconscious at 2.00 mg/kg?
Answers
Answer:
a. 6.15 mL b. 30.73 mL
Explanation:
a. What volume of this ketamine solution would the 65.0 kg user have to inject to experience a high at 0.400 mg/kg?
Since we have 0.250 g of ketamine in 1/4 cup of water and 1 cup of water equals 236.5 mL, we need to find the concentration of ketamine we have.
So concentration of ketamine C = mass of ketamine, m/volume of water, V
m = 0.250 g and V = 1/4 cup = 1/4 × 236.5 mL = 59.125 mL
So, C = m/V = 0.250 g/59.125 mL = 0.00423 g/mL = 4.23 mg/mL
Since the user has a mass of 65 kg and requires a high at 0.400 mg/kg, the mass of ketamine for this high is M = 65 kg × 0.400 mg/kg = 26 mg
Since mass, M = concentration ,C × volume, V
M = CV
V = M/C
The volume of ketamine required for the 0.400 mg/kg high is
V = 26 mg/4.23 mg/mL
V = 6.15 mL
b. What volume of this ketamine solution would the user have to inject to become unconscious at 2.00 mg/kg?
Since the concentration of ketamine is C = 4.23 mg/mL, and Since the user has a mass of 65 kg and requires an injection of 2.00 mg/kg to be unconscious, the mass of ketamine required to be unconscious is M' = 65 kg × 2.00 mg/kg = 130 mg
Since mass, M' = concentration ,C × volume, V
M' = CV
V = M/C
The volume of ketamine required for the 2.00 mg/kg unconscious injection is
V = 130 mg/4.23 mg/mL
V = 30.73 mL
A reaction vessel is charged with phosphorus pentachloride, which partially decomposes to phosphorus trichloride and molecular chlorine according to the following reaction:
PCl5(g)â PCl3(g)+Cl2(g)
When the system comes to equilibrium at 250.0°C, the equilibrium partial pressures are: PPCl5 = 0.688 atm and PPCl3 = PCl2 = 0.870 atm.
Required:
What is the value of Kp at this temperature?
Answers
Answer:
Kp = 1.10.
Explanation:
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In this case, according to the given information about the chemical reaction at equilibrium, it turns out possible for us to find the partial pressures-based equilibrium expression for the decomposition of phosphorous pentachloride by applying the law of mass action whereas the pressure of products is divided by that of the reactants as shown below:
[tex]Kp=\frac{p_{PCl_3}p_{Cl_2}}{p_{PCl_5}}[/tex]
Now, we plug in the given pressures to obtain:
[tex]Kp=\frac{0.870}{0.688} \\\\Kp=1.10[/tex]
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g 32.53 g of a solid is heated to 100.oC and added to 50.0 g of water in a coffee cup calorimeter and the contents are allowed to sit until they finally have the same temperature. The water temperature changes from 25.36 oZ to 34.4 oC. What is the specific heat capacity (in J/goC) of the solid
Answers
Answer:
0.886 J/g.°C
Explanation:
Step 1: Calculate the heat absorbed by the water
We will use the following expression
Q = c × m × ΔT
where,
Q: heatc: specific heat capacitym: massΔT: change in the temperatureQ(water) = c(water) × m(water) × ΔT(water)
Q(water) = 4.184 J/g.°C × 50.0 g × (34.4 °C - 25.36 °C) = 1.89 × 10³ J
According to the law of conservation of energy, the sum of the energy lost by the solid and the energy absorbed by the water is zero.
Q(water) + Q(solid) = 0
Q(solid) = -Q(water) = -1.89 × 10³ J
Step 2: Calculate the specific heat capacity of the solid
We will use the following expression.
Q(solid) = c(solid) × m(solid) × ΔT(solid)
c(solid) = Q(solid) / m(solid) × ΔT(solid)
c(solid) = (-1.89 × 10³ J) / 32.53 g × (34.4 °C - 100. °C) = 0.886 J/g.°C
There are _______ alkanes with molecular formula C10H22
a. 74
b. 75
c. 76
d. 77
Answers
Ammonia reacts with oxygen to produce nitrogen monoxide and water:
4 NH3(g) + 5 O2(g) ---> 4 NO(g) + 6 H2O(g)
Which of the following are stoichiometric amounts of the two reactants?
a) 1.0 g, 1.25 g
b) 0.75 mol, 0.9375 mol
Answers
Answer:
b) 0.75 mol, 0.9375 mol
Explanation:
According to this question, ammonia reacts with oxygen to produce nitrogen monoxide and water. The chemical equation is as follows:
4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)
Based on this balanced equation, 4 moles of ammonia (NH3) reacts with 5 moles of oxygen (O2).
A stoichiometric amount of the two reactants (NH3 and O2) must represent the ratio 4:5.
Given the provided options;
0.75 mol of ammonia (NH3) will react with (0.75 × 5/4) = 0.935 mol of O2 for them to be in stoichiometry.
N.B: 1 mol of NH3 will react with 1.25mol of O2 and not 1g, 1.25g.
Na Sa Bant HCL -> 50g Hao pt Soy
Answers
North America and south africa
A solution has a OH- concentration of 7.7x10-3. What is the pH of this solution?
Answers
Answer:
11.9 pH
Explanation:
First, we need to find pOH
To find that, we use the formula -log[OH]
-log[7.7x10^-3] = 2.11351
To find the pH, we'll use this formula: 14 = pH + pOH
14 = pH + 2.11351
Subtract boths sides by 2.11351
14 = pH + 2.11351
-2.11351 -2.11351
pH = 11.88649
write the balanced equation for
[B]⁴[C][D]/[A]²
