Answer:
\large \boxed{\text{2.0 atm}}
Explanation:
We can use Dalton's Law of Partial Pressures:
Each gas in a mixture of gases exerts its pressure separately from the other gases.
0.25 mol of O₂ exerts 0.50 atm.
If you add 0.75 mol of CO, the total amount of gas is
0.25 mol + 0.75 mol = 1.00 mol
[tex]p_{\text{total}} = \text{1.00 mol} \times \dfrac{\text{0.50 atm}}{\text{0.25 mol}}= \textbf{2.0 atm}\\\\\text{The total pressure in the flask is $\large \boxed{\textbf{2.0 atm}}$}[/tex]
The pressure of the closed flask after the addition of 0.75 moles of CO has been 2 atm.
Partial pressure can be defined as the pressure exerted by each gas in a given solution.
The total moles of gas in the container by the addition of CO has been:
Total moles = moles of oxygen + moles of CO
Total moles = 0.25 + 0.75
Total moles = 1 mol.
By using Dalton's law of partial pressure:
Total pressure = total moles [tex]\rm \times\;\dfrac{pressure\;of\;oxygen}{moles\;of\;oxygen}[/tex]
Total pressure = 1 [tex]\rm \times\;\dfrac{0.50}{0.25}[/tex]
Total pressure = 2 atm.
The pressure of the closed flask after the addition of 0.75 moles of CO has been 2 atm.
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What is the atomic mass of AlNO2?
Answer:
I am not sure, but I think this is the answer 72.987 g/mol
How many moles of H2 are needed to produce 34.8 moles of NH3?
2 i hope this helps
:)✨✨✨✨✨✨
What can be known about the salt sample that Gerry is looking at?
Answer:
That its small pointed. Pink(Himalayan salt)or white(normal salt)
Explanation:
Summa dees questions are so stupid, deys makin me salty.
Reactive oxygen species (ROS) are unstable, or reactive, compounds that result from the partial reduction of oxygen. ROS can cause damage to molecules, including membrane lipids and nucleic acids, and may be associated with some diseases. Which of these compounds are reactive oxygen species? Choose all that apply.
a. OH
b. OH-
c. O2-
d. H2O
e. H2O2
f. H-
Classify the following unbalanced chemical reaction Fe(s) + Cl2(aq) = Fe2+(aq) + Cl-(aq)
1. Acid-Base Reaction
2. Precipitation Reaction
3. Oxidation-Reduction Reaction
4. Combustion Reaction
Answer:
3. Oxidation-Reduction Reaction
Explanation:
Fe(s) + Cl2(aq) = Fe2+(aq) + Cl-(aq)
Fe(s) -2e- ----> Fe2+(aq) oxidation
Cl2(aq) + 2e- -----> 2Cl-(aq) reduction
The given unbalanced chemical reaction is the oxidation-reduction reaction. Therefore, option (3) is correct.
What is an oxidation-reduction reaction?Redox reactions can be defined as oxidation-reduction chemical reactions in which the reactants of the reaction undergo a change in their oxidation states. All the redox reactions are further broken down into two different processes: a reduction process and an oxidation process.
The oxidation and reduction reactions take place simultaneously in an Oxidation-Reduction reaction. The substance that is getting reduced in a reaction is known as the oxidizing agent, while a substance that is getting oxidized is the reducing agent.
The given chemical reaction is:
[tex]Fe(s) + Cl_2(aq) \longrightarrow Fe^{2+}(aq) + Cl^-(aq)[/tex]
The oxidation reaction for this reaction is: Fe (s) → Fe²⁺ (aq) + 2e⁻
The reduction reaction: Cl₂ (g) + 2e⁻ → 2Cl⁻ (aq)
Therefore, the given reaction between the iron and chlorine gas is the oxidation-reduction reaction or redox reaction.
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Nitroglycerin, an explosive, decomposes according to the following equation 4C3H5(NO3)3(s) → 12CO2(g) + 10H2O(g) + 6N2(g) + O2(g) Calculate the total volume of gases produced when collected at 1.45 atm, and 18.0°C from 2.70 × 102 g of nitroglycerin.
Answer:
6.65dm³
Explanation:
Equation of reaction,
4C3H5(NO3)3(s) → 12CO2(g) + 10H2O(g) + 6N2(g) + O2(g)
From the equation of reaction, 4 moles of Nitroglycerin gave 29 moles of various gases.
Molar mass of nitroglycerin C₃H₅(NO₃)₃ = 908g
Since all the product of the reaction are in gaseous phase, let's assume that law of conservation of matter is held hence there's no loss in mass.
908g of C₃H₅(NO₃)₃ = 908g of products
2.70×10²g of C₃H₅(NO₃)₃ = 2.70×10²g of products
Number of moles = mass / molar mass
Molar mass of C₃H₅(NO₃)₃ = 908g/mol
Number of moles = 2.70×10² / 908
Number of moles = 0.297 moles
But 1 mole = 22.4dm³
0.297mole = x dm³
x = (0.297 × 22.4) / 1
x = 6.65dm³
The volume of gas that'll be produced when 2.70×10²g of C₃H₅(NO₃)₃ would be 6.65dm³
Suppose that while you're in the lab performing a simple distillation you encountered one of the following errors: The components within the mixture, isopropanol and dichloromethane, distilled well below their boiling point. Poor separation between isopropanol and dichloromethane was observed. The initial volume of the distillation mixture has decreased significantly, almost dry, but no distillate was collected.
Answer:
Isopropanol and dichloromethane, distilled well below their boiling point.
Explanation:
The best way to separate isopropanol and dichloromethane is the method of fractional distillation. In this method, different compounds separate from each other due to difference in boiling. The boiling point of dichloromethane is 39.6 degree Celsius which is lower than the boiling point of isopropanol which is 82.5 degree Celsius. So dichloromethane will be evaporated when the temperature reaches to 40 degree Celsius and separated from isopropanol before reaching its boiling point.
Write the limiting forms (or Canonical forms) of the following ions:
i. H3O+
, ii. CO3
2-
, iii. NO3-
Answer:
Canonical structures of a chemical specie explain its observed properties from a valence bond theory perspective.
Explanation:
Resonance is a valence bond concept introduced by Linus Pauling to explain the observed properties of certain chemical species such as bond lengths, bond angles, bond order , etc.
There are certain chemical species for which a single chemical structure does not suffice in explaining its observed properties. For instance, the bond order in CO3^2- is about 1.33. Its bond length, shows that the C-O bond present in CO3^2- is neither a pure C-O single bond nor a pure C-O double bond. Hence the structure of CO3^2- is 'somewhere in between' three contributing canonical structures as shown in the image attached to this answer. The resonance structures of NO3^- are also shown.
5. Rubbing alcohol is a commonly used disinfectant and has a cooling effect when applied to the skin. The active ingredient in rubbing alcohol is isopropanol. In drugstores, the most common concentration of rubbing alcohol sold contains 70% (vol/vol) isopropanol in water. Assuming the rubbing alcohol manufacturer uses a 100% isopropanol solution, what volume of pure isopropanol is required to produce a 200-mL bottle of rubbing alcohol
Answer:
Explanation:
70% (vol/vol) means
cotnaimns 70 %(vol/vol) 70 ml of isoprapnol is there in 100 ml of Rubbing sold alcohol.
if it is 200 ml then obvouly it has the 70*2 =140 ml of isoproanol required.
Alcohol is an organic compound that when rubbed on the skin it evaporates quickly leaving a cool effect on the skin. The reason why it evaporates is because it has loosely bound molecules and a low boiling temperature.
The volume of pure isopropanol required to produce a 200-ml bottle of rubbing alcohol is 140 ml
From the question:
Alcohol sold contains 70%(vol/vol). This means 70 ml of the solute of isopropanol can be found in 100 ml of solution.
Hence:
100ml of solution = 70ml of isopropanol
200ml of solution = ?
Cross Multiply
200 ml x 70 ml / 100 ml
= 140 ml
Therefore, the volume of pure isopropanol required to produce a 200-ml bottle of rubbing alcohol is 140 ml
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Which of the following is a chemical property of iron? It
Answer:
is capable of combining with oxygen to form iron oxide
g Provide the complete balanced chemical equation for each reaction. Include the phases (s, l, g, or aq) for each substance. If there is no reaction, write NR. Also, provide the type of reaction (combination, decomposition, combustion, single replacement, double replacement, or neutralization). Gaseous methane (CH4) reacts with gaseous oxygen.
Answer:
CH4 (g) + O2 (g) → CO2 (g) + H2O (g)
This is a combustion reaction.
Explanation:
The combination of methane (CH4) and oxygen (O2) yields carbon dioxide (CO2) and water (H2O).
CO2 is typically a gas, and water, in this case, is in a gas form because it evaporated.
The reaction is combustion because the methane reacts with the oxygen to produce carbon dioxide and water. Combustion reactions must involve O2 as one reactant.
ilicon has three naturally occurring isotopes (Si-28, Si-29, and Si-30). Masses and natural abundances for two isotopes are listed here. Isotope Mass (amu) Abundance (%) Si-28 27.9769 92.2 Si-29 28.9765 4.67 Si-30 ? ? Part A Find the natural abundance of Si-30 . Express your answer using two significant figures.
Answer:
Part A
The natural abundance of Si-30 = 3.1
Part B
The mass of Si-30 = 29.9551 amu
Explanation:
Part A
The sum of the natural abundances = 100
Si-28 has a mass of 27.9769 amu and a natural abundance of 92.2%.
Si-29 has a mass of 28.9765 amu and a natural abundance of 4.67%
Si-30's mass and natural abundance are unknown.
Natural abundance for Si-30 = 100% - 92.2% - 4.67% = 3.13% = 3.1% to 2 s.f.
Part B
The total atomic mass of an element is an addition combination of the mass and natural abundances of all the isotopes of that element.
Molar mass of Silicon normally = 28.0855 amu
Let the mass of Si-30 be m
28.0855 = (27.9769×0.922) + (28.9765×0.0467) + (m×0.0313)
28.0855 = 27.14790435 + 0.0313m
0.0313m = 28.0855 - 27.14790435 = 0.93759565
m = (0.93759565/0.0313) = 29.9551325879 amu = 29.9551 amu
Hope this Helps!!
Part A: The natural abundance of Si-30 = 3.1
Lets solve the question:
The sum of the natural abundances = 100 Si-28 has a mass of 27.9769 amu and a natural abundance of 92.2%. Si-29 has a mass of 28.9765 amu and a natural abundance of 4.67% Si-30's mass and natural abundance are unknown.Natural abundance (NA) refers to the abundance of isotopes of a chemical element as naturally found on a planet.
Natural abundance for Si-30 = [tex]100\% - 92.2\% - 4.67\% = 3.13\% = 3.1\%[/tex] in significant figures.
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What is the systematic name of the following compound?
Mn3(PO4)2
The polyatomic ion phosphate has the formula PO
Answer:
Manganese(II) phosphate | Mn3(PO4)2 - PubChem
Answer:
Magnese(ll) posphate M23 (p042) Molecular weight.
That is what the leters stand for!
IF THIS HELPED AND IF YOU DON'T MIND CAN YOU PLEASE MARK ME BRAINLIEST?How would I find the quantity of heat absorbed or released when 2.0g of LiOH is dissolved in 100g of H₂0 when the enthalpy of the solution is -23.6KJ/mol?
Answer:
1.97kJ of energy are released.
Explanation:
The dissolution of LiOH in water is:
LiOH(s) → Li⁺(aq) + OH⁻(aq) ΔH = -23.6kJ
That means, when 1 mole of LiOH is dissolved, there are released (Because of the - in the enthalpy) 23.6kJ
2.0 g of LiOH (Molar mass: 23.95g/mol) are:
2.0g LiOH × (1 mol / 23.95g) =0.0835 moles of LiOH.
As 1 mole of LiOH release 23.6kJ, 0.0835moles release:
0.0835moles × (-23.6kJ / 1mole) = 1.97kJ of energy are released
What is the rate of a reaction if the value of kis 0.1, [A] is 1 M, and [B] is 2 M?
Rate = K[A]2[B]2
A. 1.6 (mol/L)/s
B. 0.8 (mol/L)/S
C. 0.2 (mol/L)/S
D. 0.4 (mol/L)/S
Answer:
D. 0.4 (mol/L)/S
Explanation:
You simply have to plug in the given values into the rate law.
Rate = k[A][B]
Rate = (0.1)(1)²(2)²
Rate = (0.1)(1)²(4)²
Rate = 0.4
Indicate whether the following represents a Chemical or Physical change: Gasoline burns in air
Answer:
Chemical
Explanation:
Whenever "burns" is being used, most of the time it is a chemical change
Answer:
chemical change
Explanation:
it's a chemical change because when gas burn it produces water and carbon dioxide
Which of these tasks would a geologist be most likely to perform?
A. Determining the species of a recently collected specimen
O B. Hypothesizing how pieces of ancient pottery were used
O C. Creating a new kind of material using polymers
O D. Determining the best method to extract underground natural gas
SUBMIT
Answer:
Explanation:
O B. Hypothesizing how pieces of ancient pottery were used
A chemist working as a safety inspector finds an unmarked bottle in a lab cabinet. A note on the door of the cabinet says the cabinet is used to store bottles of diethylamine, tetrahydrofuran, chloroform, ethanolamine, and acetone. First, from her collection of Material Safety Data Sheets (MSOS), the chemist finds the following information:
liquid density
diethylamine 1.1 gcm-3
tetrahydrofuran 0.7 9gcm-3
chloroform 0.71 gcm-3
ethanolamine 0.89 gcm-3
acetone 1.6 gcm-3
Next, the chemist measures the volume of the unknown liquid as 0.767 L and the mass of the unknown liquid as 682 g.
1. Calculate the density of the liquid.
2. Given the data above, is it possible to identify the liquid?
3. If it is possible to identify the liquid, do so.
a. dimethyl sulfoxide.
b. acetone.
c. diethylamine.
d. tetrahydrofuran .
e. carbon tetrachloride
Answer:
1. density = 0.89 g/cm3
2. Yes is possible to identify the liquid
3. ethanolamine
Explanation:
Data:
mass = 682 g
volume = 0.767 L = 767 mL or cm3
1.
To calculate the density of the liquid it is necessary to know that the density formula is:
[tex]density=\frac{mass(g)}{volume(cm^{3}) }[/tex]
The data obtained is replaced in the formula:
[tex]density=\frac{682g)}{767(cm^{3}) }=0.89\frac{g}{cm^{3} }[/tex]
2.
With the given data it is possible to identify the liquid, this because the density value is a basic property of each liquid.
3.
It is possible to determine what liquid it is, since when comparing the value obtained with those reported in the collection of Material Safety Data Sheets (MSOS), the value that agrees is that of ethanolamine.
Magnesium and nitrogen react in a combination reaction to produce magnesium nitride:
3 Mg + N2 → Mg3N2
In a particular experiment, a 8.33-g sample of N2 reacts completely. The mass of Mg consumed is ________ g.
Answer:
21.7 g
Explanation:
Step 1: Write the balanced equation
3 Mg + N₂ → Mg₃N₂
Step 2: Calculate the moles corresponding to 8.33 g of nitrogen
The molar mass of N₂ is 28.01 g/mol.
[tex]8.33 g \times \frac{1mol}{28.01g} =0.297mol[/tex]
Step 3: Calculate the moles of magnesium that reacts with 0.297 moles of nitrogen
The molar ratio of Mg to N₂ is 3:1. The reacting moles of Mg are 3/1 × 0.297 mol = 0.891 mol
Step 4: Calculate the mass corresponding to 0.891 moles of magnesium
The molar mass of Mg is 24.31 g/mol.
[tex]0.891 mol \times \frac{24.31g}{mol} = 21.7 g[/tex]
Answer:
[tex]m_{Mg}=21.7 g Mg[/tex]
Explanation:
Hello,
In this case, considering the given reaction, we are able to compute the mass of magnesium that is consumed by considering its molar mass (24.31 g/mol), the molar mass of diatomic nitrogen (28.02 g/mol), the initial mass of nitrogen (8.33 g) and the 3:1 molar ratio of magnesium to nitrogen in the reaction.
Hence we compute it by applying the shown below stoichiometric procedure:
[tex]m_{Mg}=8.33 gN_2*\frac{1molN_2}{28.02gN_2} *\frac{3molMg}{1molN_2} *\frac{24.31gMg}{1molMg} \\\\m_{Mg}=21.7 g Mg[/tex]
Regards.
According to the following reaction, what amount of Al2S3 remains when 20.00 g of Al2S3 and 2.00 g of H2O are reacted? A few of the molar masses are as follows: Al2S3 = 150.17 g/mol, H2O = 18.02 g/mol.
Answer:
Mass of Al2S3 remaining is 17.212 g
Explanation:
Equation of the reaction is given below:
Al2S3 + 6H2O -----> 2Al(OH)3 + 3H2S
From the balanced equation above
6 mole of H20 reacts with 1 mole of Al2S3
i.e. 6 * 18.02 g of H2O reacts with 1 * 150.71 g of Al2S3
= 108.12 g of H2O reacts with 150.71 g of Al2S3
Therefore 2.0 g of water will react with 2.0 * (150.71/108.12) g of Al2S3
= 2.788 g of Al2S3
Mass of Al2S3 remaining = 20.0 g - 2.788 g = 17.212 g
According to the properly balanced chemical equation, the amount (mass) of [tex]AL_2S_3[/tex] that remains after the chemical reaction is 17.22 grams.
Given the following data:
Mass of [tex]AL_2S_3[/tex] = 20.00 gramsMass of [tex]H_2O[/tex] = 2.00 gramsMolar mass of [tex]AL_2S_3[/tex] = 150.17 g/molMolar mass of [tex]H_2O[/tex] = 18.02 g/mol.To calculate the amount (mass) of [tex]AL_2S_3[/tex] that remains after the chemical reaction:
First of all, we would write a properly balanced chemical equation for this chemical reaction.
[tex]Al_2S_3 + 6H_2O ---> 2Al(OH)_3 + 3H_2S[/tex]
By stoichiometry:
1 mole of [tex]AL_2S_3[/tex] reacts with 6 moles of [tex]H_2O[/tex]
Next, we would calculate the mass of each compound.
For [tex]AL_2S_3[/tex]:
[tex]Mass = Number\;of\;moles \times molar\;mass\\\\Mass = 1 \times 150.17[/tex]
Mass = 150.17 grams
For [tex]H_2O[/tex]:
[tex]Mass = Number\;of\;moles \times molar\;mass\\\\Mass = 6 \times 18.02[/tex]
Mass = 108.12 grams
108.12 grams of [tex]H_2O[/tex] = 150.17 grams of [tex]AL_2S_3[/tex]
2.00 grams of [tex]H_2O[/tex] = X grams of [tex]AL_2S_3[/tex]
Cross-multiplying, we have:
[tex]108.12 \times X = 150.17 \times 2\\\\108.12X = 300.34\\\\X = \frac{300.34}{108.12}[/tex]
X = 2.78 grams of [tex]AL_2S_3[/tex]
Remaining mass = [tex]20.00 - 2.78[/tex]
Remaining mass = 17.22 grams of [tex]AL_2S_3[/tex]
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When vinylcyclohexane is treated with in dichloromethane, the major product is (2-bromo ethylidene)cyclohexane . Account for the formation of this product by drawing the structure of the most stable radical intermediate. Include all valence lone pairs in your answer. Include all valence radical electrons in your answer.
Answer:
Explanation:
Vinylcyclohexane is an example of a cyclic hydrocarbon where the vinyl group (-CH=CH₂ ) attaches itself to an end of a cyclohexane in ring form thereby giving rise to a vinylcyclohexane. The vinyl group are ethylene with a reduction in one hydrogen atom given them the name vinyl.
SOo, when vinylcyclohexane is treated with NBS ( i.e N-Bromosuccinimide a chemical reagent used in organic reactions) ; the bromine in the NBS reacts with the cyclohexane thereby giving rise to a allyl radical first. The allyl radical is resonance stabilized radical with an unpaired electron on the allylic carbon . As a result of stabilization ; a more stable substituted cycloalkene is formed as an intermediate .
This stable substituted cycloalkene intermediate then finally react with a bromine ion to give a major product known as ; (2-bromo ethylidene)cyclohexane.
The diagram emphasizing more on the above explanation can be seen in the attached image below
What did John Dalton publish?
Answer:
An early theory describing properties of atoms.
Explanation:
Apex
Two samples of the same rainwater are tested using two indicators at an environrnental lab. The first indicator, Methyl Orange, reveals a distinct yellow color when added to the sample. The second indicator, Litmus, turns red when placed in contact with the water sample.
Required:
a. Identify a possible pH value for the rainwater.
b. Explain, in terms of hydronium ions and hydroxide ion concentrations, the pH value of the rainwater.
Answer:
A. The pH value of rainwater is acidic about 4.4
B. The molar concentrations of the Hydronium ions are more than that of the hydroxide ions. That is why the rainwater is acidic with a pH of less than 7
Explanation:
A. Methyl orange is an acid indicator that is used to detect acidic solutions which have pH values that fall within the range of about 4.4 to 7. The distinct yellow colour change that was shown by the methyl orange as it was added to the water shows that the pH value is acidic, with a value above 4.4. (it has to be like this before methyl orange changes to yellow colour)
B. The Hydronium ( H30+) ion concentrations and the hydroxide (OH-) ion concentrations are used to measure the pH values of substances.
We can tell that the Hydronium ( H30+) ion concentrations are more than the hydroxide (OH-) ion concentrations in the sample of rainwater tested. This can be detected from the colour change that both the methyl orange and the litmus paper gave. The indicators showed that the rainwater solution was indeed acidic. Hence, the pH value will be less than 7, but greater than 4.4.
At 298K, the equilibrium constant for the following reaction is 4.20×10-7: H2CO3(aq) + H2O H3O+(aq) + HCO3-(aq) The equilibrium constant for a second reaction is 4.80×10-11: HCO3-(aq) + H2O H3O+(aq) + CO32-(aq) Use this information to determine the equilibrium constant for the reaction: H2CO3(aq) + 2H2O 2H3O+(aq) + CO32-(aq)
Answer:
The correct answer is 2.016 x 10⁻¹⁷
Explanation:
We have the following chemical reactions and their equilibrium constants (K):
(1) H₂CO₃(aq) + H₂O ⇒ H₃O⁺(aq) + HCO₃⁻(aq) K₁= 4.20×10⁻⁷
(2) HCO₃⁻(aq) + H₂O ⇒ H₃O⁺(aq) + CO₃²⁻(aq) K₂= 4.80×10⁻¹¹
And we have to obtain K for the following reaction:
H₂CO₃(aq) + 2H₂O ⇒ 2H₃O⁺(aq) + CO₃²⁻(aq)
If we add equations (1) and (2) we obtain the the desired equation. Remember that when we add chemical equations, the global equilibrium constant is the product of the constants.
H₂CO₃(aq) + H₂O ⇒ H₃O⁺(aq) + HCO₃⁻(aq) K₁= 4.20×10⁻⁷
+
HCO₃⁻(aq) + H₂O ⇒ H₃O⁺(aq) + CO₃²⁻(aq) K₂= 4.80×10⁻¹¹
-------------------------------------------------------------
H₂CO₃(aq) + 2H₂O ⇒ 2H₃O⁺(aq) + CO₃²⁻(aq) K= K₁ x K₂
K = K₁ x K₂ = (4.20×10⁻⁷) x (4.80×10⁻¹¹) = 2.016 x 10⁻¹⁷
Which statement BEST describes how a golf club does "work" on a golf ball?
(A) When the club hits the ball the club transfers all of its kinetic energy to the ball.
(B) All of the kinetic energy from the club is transferred to the ball as they both move through the air.
(C)
Some of the kinetic energy from the golf club is transferred to the ball and some transforms into sound
and heat, but the total energy remains the same.
(D) The golf club loses kinetic energy when it hits the ball and the ball gains kinetic energy from the air as it
travels
Answer:
C
Explanation:
It looks pretty reasonable to me
According to the ideal gas law, what happens to the volume of a gas when the
temperature doubles (all else held constant)?
A. The volume stays constant.
B. The volume doubles.
OOO
C. It cannot be determined
D. The volume is halved
According to the ideal gas law, when the temperature of a gas doubles, its volume doubles as well (Option B).
What does the ideal law state?The ideal gas law relates the pressure, volume, number of moles and temperature of an ideal gas.
Let's consider the equation of the ideal gas law.
P . V = n . R .T
V = n . R . T / P
As we can see, there is a direct relationship between the volume and the temperature. Thus, if the temperature doubles, the volume will double as well.
According to the ideal gas law, when the temperature of a gas doubles, its volume doubles as well (Option B).
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One of the many remarkable enzymes in the human body is carbonic anhydrase, which catalyzes the interconversion of carbon dioxide and water with bicarbonate ion and protons. If it were not for this enzyme, the body could not rid itself rapidly enough of the CO2 accumulated by cell metabolism. The enzyme catalyzes the dehydration (release to air) of up to 107 CO2 molecules per second. Which components of this description correspond to the terms enzyme, substrate, and turnover number?
Answer:
Enzyme is carbonic anhydrase
Substrate is [tex]CO_2[/tex]
Turnover number is [tex]10^{7}[/tex]
Explanation:
An enzyme is used by a living organism as a catalyst to perform a specific biochemical reaction.
A substrate is a molecule upon which an enzyme acts.
Turnover number refers to the number of substrate molecules transformed by a single enzyme molecule per minute. Here, the enzyme is the rate-limiting factor.
Here,
Enzyme is carbonic anhydrase
Substrate is [tex]CO_2[/tex]
Turnover number is [tex]10^{7}[/tex]
The core of the pyruvate dehydrogenase complex is made up of eight catalytic________that make up the_______component.
a. monomers; E1b. dimers; E2c. dimers; E3d. trimers; E2
Answer:
(D.)
The core of the pyruvate dehydrogenase complex is made up of eight catalytic trimers that make up the E2 component.
Explanation:
Eight trimers assemble as a hollow truncated cube, which forms the core of the multi-enzyme complex, known as the E2 complex in human pyruvate dehydrogenase complex.
An excess of sodium carbonate, Na2CO3, in solution is added to a solution containing 15.71 g CaCl2. After performing the experiment, 13.19 g of calcium carbonate, CaCO3, is produced. Calculate the percent yield of this reaction
Answer:
93.15 %
Explanation:
We have to start with the chemical reaction:
[tex]CaCl_2~+~Na_2CO_3~->~CaCO_3~+~NaCl[/tex]
Now, we can balance the reaction:
[tex]CaCl_2~+~Na_2CO_3~->~CaCO_3~+~2NaCl[/tex]
Our initial data are the 15.71 g of [tex]CaCl_2[/tex], so we have to do the following steps:
1) Convert from grams to moles of [tex]CaCl_2[/tex] using the molar mass (110.98 g/mol).
2) Convert from moles of [tex]CaCl_2[/tex] to moles of [tex]CaCO_3[/tex] using the molar ratio. ( 1 mol [tex]CaCl_2[/tex]= 1 mol of [tex]CaCO_3[/tex]).
3) Convert from moles of [tex]CaCO_3[/tex] to grams of [tex]CaCO_3[/tex] using the molar mass. (100 g/mol).
[tex]15.71~g~CaCl_2\frac{1~mol~CaCl_2}{110.98~g~CaCl_2}\frac{1~mol~CaCO_3}{1~mol~CaCl_2}\frac{100~g~CaCO_3}{1~mol~CaCO_3}=14.16~g~CaCO_3[/tex]
Finally, we can calculate the yield percent:
[tex]%~=~\frac{13.19~g~CaCO_3}{14.16~g~CaCO_3}*100=93.15~%[/tex]
I hope it helps!
The percentage yield obtained when excess sodium carbonate, Na₂CO₃, is added to a solution containing 15.71 g CaCl₂ is 93.2%
We'll begin by writing the balanced equation for the reaction. This is given below: [tex]Na_{2}CO_{3} + CaCl_{2} - > CaCO_{3} + 2NaCl[/tex]Molar mass of CaCl₂ = 40 + (35.5×2) = 111 g/mol
Mass of CaCl₂ from the balanced equation = 1 × 111 = 111 g
Molar mass of CaCO₃ = 40 + 12 + (16×3) = 100 g/mol
Mass of CaCO₃ from the balanced equation = 1 × 100 = 100 g
SUMMARY
From the balanced equation above,
111 g of CaCl₂ reacted to produce 100 g of CaCO₃
Next, we shall determine the theoretical yield of of CaCO₃. This can be obtained as follow:From the balanced equation above,
111 g of CaCl₂ reacted to produce 100 g of CaCO₃.
Therefore,
15.71 g of CaCl₂ will react to produce = [tex]\frac{15.71 * 100}{111} \\\\[/tex] = 14.15 g of CaCO₃.
Thus, the theoretical yield of of CaCO₃ is 14.15 g
Finally, we shall determine the percentage yield. This can be obtained as follow:Actual yield of CaCO₃ = 13.19 g
Theoretical yield of CaCO₃ = 14.15 g
Percentage yield =?[tex]Percentage yield = \frac{Actual}{Theoretical} * 100\\\\= \frac{13.19}{14.15} * 100\\\\[/tex]
= 93.2%Therefore, the percentage yield of the reaction is 93.2%
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Asbestosis is a lung disease caused by inhaling asbestos fibers. The US Department of Health and Human Services considers a particular form of asbestos to be a carcinogen. The composition of this form of asbestos is 26.31% Mg, 20.20% Is, 1.45% H and the rest of the mass is due to oxygen. The molar mass of the compound is 277 g/mol. What is the molecular formula for the carcinogenic form of asbestos
Answer: The molecular formula for the carcinogenic form of asbestos [tex]Mg_3Si_2H_4O_9[/tex]
Explanation:
a) If percentage are given then we are taking total mass is 100 grams.
So, the mass of each element is equal to the percentage given.
Mass of Mg = 26.31 g
Mass of Si= 20.20 g
Mass of H= 1.45 g
Mass of O= (100-(26.31+ 20.20+ 1.45)) = 52.04 g
Step 1 : convert given masses into moles
Moles of Mg=[tex]\frac{\text{ given mass of Mg}}{\text{ molar mass of Mg}}= \frac{26.31g}{24g/mole}=1.10moles[/tex]
Moles of Si=[tex]\frac{\text{ given mass of Si}}{\text{ molar mass of Si}}= \frac{20.20g}{28g/mole}=0.72moles[/tex]
Moles of H=[tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{1.45g}{1g/mole}=1.45moles[/tex]
Moles of O=[tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{52.04g}{16g/mole}=3.25moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For Mg = [tex]\frac{1.10}{0.72}=1.5[/tex]
For Si =[tex]\frac{0.72}{0.72}=1[/tex]
For H=[tex]\frac{1.45}{0.72}=2[/tex]
For O =[tex]\frac{3.25}{0.72}=4.5[/tex]
The ratio of Mg : Si: H: O = 1.5 : 1 : 2 : 4.5
Converting them into whole numbers :
The ratio of Mg : Si: H: O = 3 : 2 : 4 : 9
Hence the empirical formula is [tex]Mg_3Si_2H_4O_9[/tex]
Empirical mass =[tex]3\times 24+2\times 28+4\times 1+9\times 16=276g[/tex]
Molecular mass = 277 g
[tex]n= \frac{\text {Molecular mass}}{\text {Empirical mass}}=\frac{277}{276}=1[/tex]
Thus molecular formula =[tex]1\times Mg_3Si_2H_4O_9=Mg_3Si_2H_4O_9[/tex]