Answer:
1703 mmHg
Explanation:
Volume and pressure are presumed to be inversely proportional. Hence a change in volume by a factor of 125/325 = 5/13 is expected to change the pressure by a factor of 13/5:
(13/5)(655 mmHg) = 1703 mmHg
3. Scientific methods may include three steps of study as listed below. Explain each step in detail with a complete content related sentence. (ref: p.12-16) a. Hypothesis b. Theory c. Scientific law
Answer:
Hypothesis is an assumption or idea about a particular topic or argument. An hypothesis should be one which is able to be tested and measurable to determine its authenticity.
A theory is an explanation of a scientific observation which has undergone series of experiments and is reproducible in any part of the world.
A law is simply a rule which gives an in depth explanation of a scientific finding. If new findings emerge the law could be changed or modified.
Draw the Lewis structure of H2O. Include any nonbonding electron pairs. Draw the molecule by placing atoms on the grid and connecting them with bonds. Include all lone pairs of electrons. - CHONSPFBrClIXMore Request Answer Part B What is the electron geometry of H2O
Answer:
Concepts and reason
Lewis structure is a structure that explains the bonding between atoms of a molecule and lone pair of electrons that is present in the molecule is called a Lewis structure.
With the help of Lewis structure the electronic geometry of a molecule can be determine.
Fundamentals
According to Lewis structure, every atom and their position in the structure of a molecule by using its chemical symbol.
Lines connecting the atoms that are bonded to them are drawn. Lone pairs are expressed by pairs of dots and are located beside the atoms.
Lewis structure of [tex]H_{2}O[/tex] is, the total number of valence electrons is eight in [tex]H_{2}O[/tex].
Consider this reaction:
2Cl2O5 —> 2Cl2 + 5O2
At a certain temperature it obeys this rate law.
rate = (2.7.M^-1•s^-1) [Cl2O5]^2
Suppose a vessel contains Cl2O5 at a concentration of 0.600M. calculate how long it takes for the concentration of Cl2O5 to decrease by 94%. you may assume no other reaction is important. round your answer to two digits
Answer:
[tex]t=9.7s[/tex]
Explanation:
Hello,
In this case, we have a second order kinetics given the second power of the concentration of chlorine (V) oxide in the rate expression, thus, the integrated equation for the concentration decay is:
[tex]\frac{1}{[Cl_2O_5]}=kt+\frac{1}{[Cl_2O_5]_0}[/tex]
Thus, the final concentration for a 94% decrease is:
[tex][Cl_2O_5]=0.600M-0.600M*0.94=0.036M[/tex]
Therefore, we compute the time for such decrease:
[tex]kt=\frac{1}{[Cl_2O_5]}-\frac{1}{[Cl_2O_5]_0}=\frac{1}{0.036M}-\frac{1}{0.60M} =26.1M^{-1}[/tex]
[tex]t=\frac{26.1M^{-1}}{k}= \frac{26.1M^{-1}}{2.7M^{-1}*s^{-1}}\\\\t=9.7s[/tex]
Regards.
Two scientists study data collected during an experiment and reach different conclusions. How would the scientific community address their disagreement?
Please
Answer: D. They would device an experiment that could test the two scientists conclusions.
Explanation:
The results of the scientific study must be verified by peer scientists or members of the scientific community to proof whether the research has been conducted produce a valid evidence.
In the given situation, the two scientists had developed different conclusion for the same experiment. This may mean either of the two may have put up an incorrect conclusion.
The scientific community may address this issue by performing the experiment. Every scientific conclusion is based upon the results of the experimental approach.
Answer:d
Explanation:
A 25.0 mLsample of an acetic acid solution is titrated with a 0.175 M NaOH solution. The equivalence point is reached when 37.5 mL of the base is added. The concentration of acetic acid in the sample was:_______,
A) 0.263
B) 0.365
C) 0.175
D) 1.83×10−4
E) 0.119
Answer:
0.263M of CH₃COOH is the concentration of the solution.
Explanation:
The reaction of acetic acid (CH₃COOH) with NaOH is:
CH₃COOH + NaOH → CH₃COO⁻Na⁺ + H₂O
1 mole of acetic acid reacts per mole of NaOH to produce sodium acetate and water.
In the equivalence point, moles of acetic acid are equal to moles of NaOH and moles of NaOH are:
0.0375L × (0.175 moles / L) = 6.56x10⁻³ moles of NaOH = moles of CH₃COOH.
As the sample of acetic acid had a volume of 25.0mL = 0.025L:
6.56x10⁻³ moles of CH₃COOH / 0.0250L =
0.263M of CH₃COOH is the concentration of the solution
Question 5 of 20:
Select the best answer for the question.
5. Which of the following is a homonuclear diatomic molecule?
O A. NH3
O B. 2002
O C. Hz
O D. CO
Answer:
Homo nuclear molecule mean having atoms of only one element,
I cant see clearly the option B and C can you correct them , 2002? Hz?
Explanation:
Answer:
H2
Explanation:
What happens in a neutralization reaction?
a
The hydrogen (H+) ions from the base and the hydroxide (OH-) ions from the acld come together to form water.
The hydrogen (H+) ions from the acid and the hydroxide (OH) ions from the base come together to form water.
A substance's pH is increased to a value greater than 7.
A solution of a known concentration and volume is added until the reaction is completed.
Answer:
Answer is letter B
Explanation:
The first one is wrong because acids release H+, not bases.
The third one is wrong because the pH is exactly 7, not greater.
The last one is wrong because it is vague and does not fit a neutralization reaction.
An ideal gaseous reaction occurs at a constant pressure of 35.0 atm and releases 66.8 kJ of heat. Before the reaction, the volume of the system was 8.20 L. After the reaction, the volume of the system was 2.21 L. Calculate the total change in internal energy for the system. Enter your answer numerically in units of kJ.
Answer:
U = -45.557kj
Explanation:
Before we can calculate the totally internal energy change in kilojoules firstly we need to calculate W
U=q + w .
We know that
w = PΔ V
where P is the pressure of
and V is the volume
then we can calculate the work
w = 35 atm * ( 8.20L - 2.21L)
W=35atm* 5.99L
W=209.65atmJ
But 1 atm = 101.325J
then ,
w = 209.65* 101.325 J = 21242.79 J
let us convert it to Kj
But we know that 1kJ = 10^3 J .
Then w = 21.243 kJ .
Then we can now calculate the internal energy as
U = 21.243- 66.8 kJ = -45.557kj
But we know that heat was released. Theeefore, the total internal energy change was -45.557kj
The aluminum in a package containing 75 ft2 of kitchen foil weighs approximately 12 ounces. Aluminum has a density of 2.70 g/cm3 . What is the approximate thickness of the aluminum foil in millimeters?(1 ounce = 28.4g)
Answer:
18130 mm
Explanation:
Now we have a lot of unit conversions to do in order to correctly answer this question. We shall do these conversions gradually.
First we convert the weight in ounce to grams.
If 1 ounce = 28.4g
12 ounces = 12×28.4 = 340.8 g
Next we convert the area of aluminum from ft2 to m2
1ft2= 0.0929 m2
75 ft2= 75 × 0.0929= 6.9675m2
Now density of aluminum= 2.70 gcm-3
Density= mass/volume
But volume= area× thickness
Density= mass/ area × thickness
Density × area × thickness= mass
Thickness= mass/ density × area
Thickness= 340.8g / 2.70gcm-3 × 6.9675m2
Thickness= 340.8/18.8
Thickness= 18.13 m
Since 1000 milimeters make 1 metre
Thickness= 18130 mm
An ideal gas sealed in a rigid 4.86-L cylinder, initially at pressure Pi=10.90 atm, is cooled until the pressure in the cylinder is Pf=1.24 atm. What is the enthalpy change for this process? ΔH =
Answer:
[tex]\Delta H=-11897J[/tex]
Explanation:
Hello,
In this case, it is widely known that for isochoric processes, the change in the enthalpy is computed by:
[tex]\Delta H=\Delta U+V\Delta P[/tex]
Whereas the change in the internal energy is computed by:
[tex]\Delta U=nCv\Delta T[/tex]
So we compute the initial and final temperatures for one mole of the ideal gas:
[tex]T_1= \frac{P_1V}{nR}=\frac{10.90atm*4.86L}{0.082*n}=\frac{646.02K }{n} \\\\T_2= \frac{P_2V}{nR}=\frac{1.24atm*4.86L}{0.082*n}=\frac{73.49K }{n}[/tex]
Next, the change in the internal energy, since the volume-constant specific heat could be assumed as ³/₂R:
[tex]\Delta U=1mol*\frac{3}{2} (8.314\frac{J}{mol*K} )*(73.49K-646.02K )=-7140J[/tex]
Then, the volume-pressure product in Joules:
[tex]V\Delta P=4.86L*\frac{1m^3}{1000L} *(1.24atm-10.90atm)*\frac{101325Pa}{1atm} \\\\V\Delta P=-4756.96J[/tex]
Finally, the change in the enthalpy for the process:
[tex]\Delta H=-7140J-4757J\\\\\Delta H=-11897J[/tex]
Best regards.
The change in enthalpy is 70.42J
Data;
Volume of the gas = 4.86LInitial Pressure = 10.90 atmFinal Pressure = 1.24 atmChange in Enthalpy = ?Change in EnthalpyThe change of enthalpy is calculated as
[tex]\delta H = \delta V + \delta nRT\\\delta n = 0\\\delta H = \delta U \\[/tex]
The volume change is negligible
The change in enthalpy here is equal to change in internal energy over ΔE
[tex]\delta H = \delta U = nCv\delta T\\\delta H = \frac{3}{2}(nR\delta T)\\\delta H = \frac{3}{2}\{\delta PV)\\ \delta H = \frac{3}{2}[(10.90-1.24)*4.86] \\\delta H = 70.42J[/tex]
The change in enthalpy is 70.42J
Learn more on change in enthalpy here;
https://brainly.com/question/14047927
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.
How many kg of gas fill a 11.6 gal gas tank
Answer:
43.964
Explanation:
i think i used a calculator so let me know if its wrong
Answer:
39.49 kg
Explanation:
:)
Suppose you were preparing 1.0 L of a bleaching solution in a volumetric flask, and it calls for 0.21 mol of NaOCl. If all you had available was a jug of bleach that contained 0.78 M NaOCl, what volume of bleach would you need to add to the volumetric flask before you added enough water to reach the 1.0 L line
Answer: brainliesss plssssssss
0.256 L
Explanation:
We should use the following formula:
concentration (1) × volume (1) = concentration (2) × volume (2)
concentration (1) = 0.82 M NaOCl
volume (1) = ?
concentration (2) = 0.21 M NaOCl
volume (2) = 1 L
volume (1) = [concentration (2) × volume (2)] / concentration (1)
volume (1) = [0.21 / 1] / 0.82 = 0.256 L
Each of the insoluble salts below are put into 0.10 M hydrobromic acid solution. Do you expect their solubility to be more, less, or about the same as in a pure water solution ?
a. Calcium sulfite
b. Calcium fluoride
c. Silver bromide
Answer:
A. Solubility of calcium sulfite increases
B. Solubility of calcium fluoride increases
C. Solubility of Silver bromide decreases
Explanation:
The solubility factor is proportional to ions' concentration. The solubility of a solution can be predicted from Le Chatelier's principle which states that if an external constraint is imposed on a system in equilibrium, the equilibrium position will shift in order to annul the effect of the external constraint. So, If the reactant's concentration increases, the equilibrium shifts to the right indicating a higher solubility of the solution and if the product's concentration increases, the equilibrium shifts to the left indicating a lesser solubility of the solution.
Case 1. Calcium sulfite
The dissociation reaction of CaSO3 is given below:
CaSO3 ----> Ca²+ + SO3²-
SO3²- is the conjugate base of the weak acid, H2SO3. Therefore, on the addition of hydrobromic acid, some of the sulfite ion is removed from the solution by the following reaction;
H+ + SO3²- ---> HSO3-
This shifts the equilibrium to the right, more dissociation, thereby resulting in more solubility of the solute.
Case 2. Calcium fluoride
The dissociation reaction of calcium fluoride (CaF2) is shown below.
CaF2 ----> Ca²+ + 2F-
Fluoride ion (F-) is a strong conjugate base of the weak acid. Therefore, some of fluoride ions is removed by the addition of hydrobromic acid as shown below:
H+ + F- ---->. HF
Hence, the concentration of fluoride ions reduces, shifting equilibrium in the forward direction. Therefore, the solubility will be more than in pure water solution.
Case 3: Silver bromide
The dissociation reaction of AgBr is as follows:
AgBr ----> Ag+ + Br-
The addition of HBr will increase the concentration of bromide ions. Hence, equilibrium will shift in backward direction resulting in a lesser solubility than in water.
The solubility of calcium sulfite and calcium fluoride is greater in 0.10 M hydrobromic acid solution than in pure water while the solubility of silver bromide is lesser in 0.10 M hydrobromic acid solution than in pure water.
Common ion effect refers to the decrease in solubility of a substance in a solution that contains another solute with which it has a common ion. If a substance is dissolved in a solution that contains a solute with which it has a common ion, the solubility of the substance in that solution is less than its solubility in pure water.
Considering the substances given, the solubility of calcium sulfite and calcium fluoride in 0.10 M hydrobromic acid solution is more than their solubility in pure water the equilibrium position is shifted in the forward direction.
However, solubility of silver bromide in 0.10 M hydrobromic acid solution is less than its solubility in pure water due to common ion effect.
Learn more: https://brainly.com/question/6505878
A researcher placed 25.0 g of silver chloride, AgCl, in sunlight and allowed the substance to decompose completely to form silver, Ag, with the release of chlorine gas, Cl2. The gas was collected in a container during the decomposition. The researcher determined that the mass of the silver formed was 18.8 g, and the mass of the chlorine gas formed was 6.2 g. The equation for the reaction is:
Answer:
A. The law of definite proportions states that all pure samples of a particular chemical compound contain the same elements combines in the same proportion by mass.
B. The law of conservation of mass states that during ordinary chemical reactions, matter can neither be created or destroyed.
Note: The full question is as follows;
A researcher placed 25.0 g of silver chloride, AgCl, in sunlight and allowed the substance to decompose completely to form silver, Ag, with the release of chlorine gas, Cl2. The gas was collected in a container during the decomposition. The researcher determined that the mass of the silver formed was 18.8 g, and the mass of the chlorine gas formed was 6.2 g. The equation for the reaction is:
2AgCl ----> 2Ag + Cl2
a. State the law of definite proportions. Then use the researcher's data to confirm the law of definite proportions. Show your calculations.
b. State the law of conservation of matter. Then use the researcher's data to confirm the conservation of matter. Show your calculations.
Explanation:
A. Mass of silver obtained from AgCl = 18.8g.
Percentage mass of silver in the chloride = (18.8/25.0) * 100 = 75.2 %
Mass of chlorine obtained from AgCl = 6.2
Percentage mass of chlorine = (6.2/25) * 100 = 24.8 %
In one mole of AgCl with a molar mass of 143.3 g/mol; mass of silver = 107.8, mass of Cl = 35.5
Percentage mass of Ag = (107.8/143.3) * 100 = 75.2%
Percentage mass of Cl = (35.5/143.3) * 100 = 24.8%
Since the percentages by mass of Ag and AgCl obtained from the sample is the same to that obtained from a mole of AgCl, the law of definite proportions which states that all pure samples of a particular chemical compound contain the same elements combined in the same proportion by mass is verified.
B. Mass of reactant; AgCl sample = 25.0
Mass of products; At = 18.8 g; Cl = 6.2 g
Sum of products masses = 18.8 + 6.2 = 25.0 g
Therefore mass of reactant = mass of products.
This is in accordance with the law of conservation of mass which states that during ordinary chemical reactions, matter is neither created nor destroyed.
When 25ml of sulphuric acid, was titrated with 0.0820 mol/L sodium hydroxide solution the end point was detected (with phenolphthalein) at 22.5ml . Calculate the concentration of sulphuric
acid in mol/L.
Answer:
the concentration of sulphuric acid is 14mol/l
The boiling point of diethyl ether, CH3CH2OCH2CH3, is 34.500 °C at 1 atmosphere. Kb(diethyl ether) = 2.02 °C/m In a laboratory experiment, students synthesized a new compound and found that when 14.94 grams of the compound were dissolved in 279.5 grams of diethyl ether, the solution began to boil at 35.100 °C. The compound was also found to be nonvolatile and a non-electrolyte. What is the molecular weight they determined for this compound ?
Answer:
The correct answer is 179.94 g/mol.
Explanation:
Based on the given question, the boiling point of diethyl ether us 34.500 degree C at 1 atm pressure. The boiling point of the solution given is 35.100 degree C. The Kb of diethyl ether given is 2.02 degree C/m. The weight of the compound given is 14.94 grams, the weight of the solvent (diethyl ether) is 279.5 grams.
The molecular weight of the compound can be determined by using the formula,
deltaTb = Kb * molality
Tb-To = Kb * molality
Tb-To = Kb*wt/mol.wt*1000/w (solvent)
35.100 - 34.500 = 2.02 * 14.94 / mol. wt * 1000 g / 279.5 g
0.6 = 2.02 * 53.45/ mol.wt
mol. wt = 2.02*53.45/0.6
mol. wt = 179.94 g/mol
Hence, the molecular weight of the compound is 179.94 gram per mol.
what are mineralocorticoids
Explanation:
it is used to describe those action of adrenal corticosteroids that produce sodium
What is the name of this molecule? (will give BRAINLIEST)
A straight chain of four carbons. There is a triple bond between the second and third carbons when counting from left to right or right to left.
Answer:
2 - Butyne
Explanation:
The name of the molecule with a carbon atoms arranged in a straight chain with a triple bond between the second and third carbons is 2 - Butyne.
2- Butyne is an alkyne with structural formula given below. Some of the properties of Butyne include it is a produced artificially, it is volatile and colorless in nature.
Hence, the given molecules described is 2 - Butyne.
Identify the state(s) of matter that each property describes.
Answer:solid,liquid,gas,plasma
Explanation:
This question seems incomplete. I believe the full question is as followed:
Identify the state(s) of matter that each property describes.
1.) takes the shape of its container:
O gas
O liquid
O solid
2.) fills all available space:
O gas
O liquid
O solid
3.) maintains its shape:
O gas
O liquid
O solid
4.) can be poured:
O gas
O liquid
O solid
5.) is compressible:
O gas
O liquid
O solid
6.) has a fixed volume:
O gas
O liquid
O solid
The answers to the 1st are gas and liquid.
The answer to the 2nd is gas.
The answer to the 3rd is solid.
The answer to the 4th is liquid.
The answer to the 5th is gas.
The answers to the 6th are liquid and solid.
A salt solution was found to contain 1.50 g of salt dissolved in 50 mL of water. On evaporation, the recovered salt weighed 1.47 g. What percent of salt was recovered?
A) 20.4%
B) 107%
C) 98%
D) 20.0%
Answer:
C = 98%
Explanation:
Hello,
To determine the percentage of salt recovered, we'll divide the mass of the salt recovered over by the original mass of the salt.
Mass of salt recovered = 1.47g
Initial mass of salt = 1.50g
Percentage of salt recovered = (mass recovered/ initial mass of salt) × 100
Percentage of salt recovered = (1.47 / 1.50) × 100
Percentage of salt recovered = 0.98 × 100
Percentage of salt recovered = 98%
The percentage of salt recovered is equal to 98%
Calculate the number of moles of magnesium, chlorine, and oxygen atoms in 4.90 moles of magnesium perchlorate, Mg(ClO4)2.]\
Answer:
4.90 moles of [tex]Mg(ClO_4)_2[/tex] will produce (9.8) moles of [tex]Cl^{-}[/tex] ,
(4.90) moles of [tex]Mg^{2+}[/tex] and
(39.2) moles of [tex]O^{2-}[/tex]
Explanation:
From the question we are told that
The number of moles of is [tex]n = 4.90 \ mols[/tex]
The formation reaction of [tex]Mg(ClO_4)_2[/tex] is
[tex]Mg^{2+} + 2 Cl^{-} + 8O^{2+} \to Mg(ClO_4)_2[/tex]
From the reaction we see that
1 mole of [tex]Mg(ClO_4)_2[/tex] is formed by 2 moles of [tex]Cl^{-}[/tex] 1 mole of [tex]Mg^{2+}[/tex] and 4 [tex]O^{2-}[/tex]
This implies that
4.90 moles of [tex]Mg(ClO_4)_2[/tex] will produce (2 * 4.90) moles of [tex]Cl^{-}[/tex] ,
(1 * 4.90) moles of [tex]Mg^{2+}[/tex] and
(8 * 4.90) moles of [tex]O^{2-}[/tex]
So
4.90 moles of [tex]Mg(ClO_4)_2[/tex] will produce (9.8) moles of [tex]Cl^{-}[/tex] ,
(4.90) moles of [tex]Mg^{2+}[/tex] and
(39.2) moles of [tex]O^{2-}[/tex]
Answer:
- [tex]n_{Mg}=4.90molMg[/tex]
- [tex]n_{Cl}=9.6molCl[/tex]
- [tex]n_{O}=38.4molO[/tex]
Explanation:
Hello,
In this case, for the given 4.90 moles of magnesium perchlorate, we can compute the moles of each atom by identifying the subscript each atom has in the molecule as shown below:
- Moles of magnesium atoms: here, one mole of magnesium perchlorate has only one mole of magnesium atom (subscript is one), this the moles of magnesium atoms are also 4.90 moles.
- Moles of chlorine atoms: here, one mole of magnesium perchlorate has two moles of chlorine atoms as it has a two out of the parenthesis enclosing the perchlorate anion, thus, we have:
[tex]n_{Cl}=4.80molMg(ClO_4)_2*\frac{2molCl}{1molMg(ClO_4)_2} =9.6molCl[/tex]
- Moles of oxygen atoms: here, one mole of magnesium perchlorate has eight moles of oxygen atoms as it has a four in the oxygen subscript and a two out of the parenthesis enclosing the perchlorate anion, thus, we have:
[tex]n_{O}=4.80molMg(ClO_4)_2*\frac{8molO}{1molMg(ClO_4)_2} =38.4molO[/tex]
Best regards.
What is the equilibrium constant for the following reaction:HCO2H(aq) + CN–(aq) HCO2–(aq) + HCN(aq)Does the reaction favor the formation of reactants or products? The acid dissociation constant, Ka, for HCO2H is 1.8 x 10–4and the acid dissociation constant for HCN is 4.0 x 10–10.(A) K = 1.00. The reaction favors neither the formation of reactants nor products.(B) K = 2.2 x 10–6. The reaction favors the formation of products.(C) K = 2.2 x 10–6. The reaction favors the formation of reactants.(D) K = 4.5 x 105. The reaction favors the formation of products.(E) K = 4.5 x 105. The reaction favors the formation of reactants.
Answer:
(D) K = 4.5 x 10⁵. The reaction favors the formation of products
Explanation:
HCOOH + CN⁻ ⇆ HCOO⁻ + HCN
K = [HCOO⁻] [ HCN ] / [ HCOOH] [ CN⁻]
HCOOH ⇄ H ⁺ + COO⁻
K₁ = [ H⁺] [ COO⁻ ] / [HCOOH ]
HCN ⇆ H⁺ + CN⁻
K₂ = [ H⁺] [ CN⁻] / [ HCN ]
K₁ / K₂
= [ H⁺] [ COO⁻ ] / [HCOOH ] X [ HCN ] / [ H⁺] [ CN⁻]
= [ COO⁻ ][ HCN ] / [HCOOH ] [ CN⁻]
= K
K = K₁ / K₂
= 1.8 x 10⁻⁴ / 4 x 10⁻¹⁰
= 4.5 x 10⁵
So equilibrium constant of the reaction
HCOOH + CN⁻ ⇆ HCOO⁻ + HCN
is very high . Hence reaction favours the formation of product.
option (D) is correct.
When 1.550 gg of liquid hexane (C6H14)(C6H14) undergoes combustion in a bomb calorimeter, the temperature rises from 25.87 ∘C∘C to 38.13 ∘C∘C. Find ΔErxnΔErxn for the reaction in kJ/molkJ/mol hexane. The heat capacity of the bomb calorimeter, determined in a separate experiment, is 5.73 kJ/∘CkJ/∘C.
Answer:
ΔErxn[tex]= -3.90*10^3KJ[/tex]
Explanation:
Given from the question
T1 = 25.87∘C
T2= 38.13∘C.
C= 5.73Kj/C
CHECK THE ATTACHMENT FOR DETAILED EXPLATION
What occurs when potassium reacts with bromine to form potassium bromide?
1) Electrons are shared and the bonding is ionic.
2) Electrons are shared and the bonding is covalent.
3) Electrons are transferred and the bonding is ionic
4) Electrons are transferred and the bonding is covalent.
Consider the following system at equilibrium: P(aq)+Q(aq)⇌3R(aq) Classify each of the following actions by whether it causes a leftward shift, a rightward shift, or no shift in the direction of the net reaction. Drag the appropriate items to their respective bins.
Items:1) Increase [P]2) Increase [Q]3) Increase [R]4) Decrease [P]5) Decrease [Q]6) Decrease [R]7) Triple [P] and reduce [Q] to one third8) Triple both [Q] and [R]
Explanation:
P(aq)+Q(aq)⇌3R(aq)
This problem involves applying LeChatelier's principle.
LeChatelier's principle states that whenever a system in equilibrium is disturbed, the equilibrium position would change in order to annul that change.
1) Increase [P]
This would cause the equilibrium position to shift to the right. This is because more reactions have been added, to annul that change more products have to be formed.
2) Increase [Q]
This would cause the equilibrium position to shift to the right. This is because more reactions have been added, to annul that change more products have to be formed.
3) Increase [R]
This would cause the equlibrium position to shift to the left. This is because more products have been formed, to annul that change more reactants have to be formed.
4) Decrease [P]
This would cause the equlibrium position to shift to the left. This is because there are now less reactants, to annul that change more reactants have to be formed.
5) Decrease [Q]
This would cause the equilibrium position to shift to the left. This is because there are now less reactants, to annul that change more reactants have to be formed.
6) Decrease [R]
This would cause the equilibrium position to shift to the right. This is because there are now less products, to annul that change more products have to be formed.
7) Triple [P] and reduce [Q] to one third
No shift in the direction of the net reaction because both changes cancels each other.
8) Triple both [Q] and [R]
No shift in the direction of the net reaction because both changes cancels each other.
Why are sediments carried by wind deposited in a sorted manner? ( that is the largest particles at the bottom and the smallest particles on top)
Answer:
Explanation:
The larger sediment particles are having more weright(mass), hence fall quickly (early) and smaller particles with low mass are carried by wind for longer time and falls slowly , hence you observe sorted kind of things. Hope this helps you to understadn this phenomenon.
consider an exceptionally weak acid, HA, with Ka= 1 x 10-20. you make 0.1M solution of the salt NA. what is the pH.
Answer:
[tex]pH=10.5[/tex]
Explanation:
Hello,
In this case, the dissociation of the given weak acid is:
[tex]HA\rightleftharpoons H^++A^-[/tex]
Therefore, the law of mass action for it turns out:
[tex]Ka=\frac{[H^+][A^-]}{[HA]}[/tex]
That in terms of the change [tex]x[/tex] due to the reaction extent is:
[tex]1x10^{-23}=\frac{x*x}{0.1-x}[/tex]
Thus, by solving with the quadratic equation or solver, we obtain:
[tex]x=31.6x10^{-12}M[/tex]
Which clearly matches with the hydrogen concentration in the solution, therefore, the pH is:
[tex]pH=-log(-31.6x10^{-12})\\pH=10.5[/tex]
Regards.
The following data show the rate constant of a reaction measured at several different temperatures. Temperature (K) Rate Constant (1/s) 310 0.194 320 0.554 330 1.48 340 3.74 350 8.97 Part APart complete Use an Arrhenius plot to determine the activation barrier for the reaction. Express your answer using three significant figures. Ea
Complete Question
The complete question is shown on the first uploaded image
Answer:
Part A
activation barrier for the reaction [tex]E_a = 84 .0 \ KJ/mol[/tex]
Part B
The frequency plot is [tex]A = 2.4*10^{13} s^{-1}[/tex]
Explanation:
From the question we are told that
at [tex]T_1 = 300 \ K[/tex] [tex]k_1 = 5.70 *10^{-2}[/tex]
and at [tex]T_2 = 310 \ K[/tex] [tex]k_2 = 0.169[/tex]
The Arrhenius plot is mathematically represented as
[tex]ln [\frac{k_2}{k_1} ] = \frac{E_a}{R} [\frac{1}{T_1} - \frac{1}{T_2} ][/tex]
Where [tex]E_a[/tex] is the activation barrier for the reaction
R is the gas constant with a value of [tex]R = 8.314*10^{-3} KJ/mol \cdot K[/tex]
Substituting values
[tex]ln [\frac{0.169}{6*10^-2{}} ] = \frac{E_a}{8.314*10^{-3}} [\frac{1}{300} - \frac{1}{310} ][/tex]
=> [tex]E_a = 84 .0 \ KJ/mol[/tex]
The Arrhenius plot can also be mathematically represented as
[tex]k = A * e^{-\frac{E_a}{RT} }[/tex]
Here we can use any value of k from the data table with there corresponding temperature let take [tex]k_2 \ and \ T_2[/tex]
So substituting values
[tex]0.169 = A e ^{- \frac{84.0}{8.314*10^{-3} * 310} }[/tex]
=> [tex]A = 2.4*10^{13} s^{-1}[/tex]
A galvanic cell at a temperature of 25.0°C is powered by the following redox reaction:
2MnO4^-(aq)+16H+(aq)+5Pb(s)-->2Mn^2+(aq)+8H2O(l)+5Pb^2+(aq)
Suppose the cell is prepared with 1.87 M MnO−4 and 1.37 M H+ in one half-cell and 3.23 M Mn+2 and 6.62 M Pb+2 in the other. Calculate the cell voltage under these conditions. Round your answer to 3 significant digits.
Answer:
1.63 V
Explanation:
Let us state the reaction equation again for the purpose of clarity;
2MnO4^-(aq)+16H+(aq)+5Pb(s)-->2Mn^2+(aq)+8H2O(l)+5Pb^2+(aq)
The reduction potentials for the two half reaction equations are;
MnO 4 - (aq) + 8H + (aq) + 5e - → Mn2+(aq) + 4H2O(l) Eo=1.51 V
Pb2+(aq) + 2e - → Pb(s) Eo= -0.13 V
E°cell = E°red – E°Ox
E°cell = 1.51 - (-0.13)
E°cell = 1.51 + 0.13
E°cell = 1.64 V
But Q= [Mn^2+]^2 [Pb^2+]^5/[MnO4^-]^2 [H^+]^16
Q= [3.23]^2 [6.62]^5/[1.87]^2 [1.37]^16
Q= 10.43 × 12714.22/3.4969 × 154
Q= 132609.3/538.5226
Q= 246.25
From Nernst equation
E= E° - 0.0592/n log Q
Where n=10
E= 1.64- 0.0592/10 log 246.25
E= 1.64-0.0142
E= 1.63 V
