Answer:
An organism that is completely composed D amino acids cannot survive.
Explanation:
Most of amino acids in all organism are present in L conformation.As result in all organism the enzymes are specific for L amino acids but not for D amino acids.
Bacterial cell wall contain some D amino acids such as D glutamate,D alanine but not entirely composed of D amino acids.
Balance the following redox reaction occurring in an acidic solution. The coefficient of Cr2O72−(aq) is given. Enter the coefficients (integers) into the cells before each substance below the equation. 1 Cr2O72−(aq) + ___ Ti3+(aq) + ___ H+(aq) \rightarrow→ ___ Cr3+(aq) + ___ TiO2+(aq) + ___ H2O(l)
Answer:
Cr₂O₇²⁻ (aq) + 6Ti³⁺ (aq) + 2 H⁺(aq) → 2 Cr³⁺ (aq) + 6TiO²⁺(aq) + H2O(l)
Explanation:
Cr₂O₇²⁻ (aq) + Ti³⁺ (aq) + H⁺ (aq) → Cr³⁺ (aq) + TiO²⁺ (aq) + H₂O(l)
This is the reaction, without stoichiometry.
We have to notice the oxidation number of each element.
In dicromate, Cr acts with +6 and we have Cr³⁺, so oxidation number has decreased. .- REDUCTION
Ti³⁺ acts with +3, and in TiO²⁺ acts with +4 so oxidation number has increased. .- OXIDATION
14 H⁺ + Cr₂O₇²⁻ + 6e⁻ → 2Cr³⁺ + 7H₂O
To change 6+ to 3+, Cr had to lose 3 e⁻, but as we have two Cr, it has lost 6e⁻. As we have 7 Oxygens in reactant side, we have to add water, as the same amount of oxygens atoms we have, in products side. Finally, we have to add protons in reactant side, to ballance the H.
H₂O + Ti³⁺ → TiO²⁺ + 1e⁻ + 2H⁺
Titanium has to win 1 e⁻ to change 3+ to 4+. We had to add 1 water in reactant, and 2H⁺ in products, to get all the half reaction ballanced.
Now we have to ballance the electrons, so we can cancel them.
(14 H⁺ + Cr₂O₇²⁻ + 6e⁻ → 2Cr³⁺ + 7H₂O) .1
(H₂O + Ti³⁺ → TiO²⁺ + 1e⁻ + 2H⁺) .6
Wre multiply, second half reaction .6
14 H⁺ + Cr₂O₇²⁻ + 6e⁻ → 2Cr³⁺ + 7H₂O
6H₂O + 6Ti³⁺ → 6TiO²⁺ + 6e⁻ + 12H⁺
Now we can sum, the half reactions:
14 H⁺ + Cr₂O₇²⁻ + 6e⁻ + 6H₂O + 6Ti³⁺ → 6TiO²⁺ + 6e⁻ + 12H⁺ + 2Cr³⁺ + 7H₂O
Electrons are cancelled and we can also operate with water and protons
7H₂O - 6H₂O = H₂O
14 H⁺ - 12H⁺ = 2H⁺
The final ballanced equation is:
2H⁺ + Cr₂O₇²⁻ + 6Ti³⁺ → 6TiO²⁺ + 2Cr³⁺ + H₂O
To balance the given redox reaction in an acidic solution, follow the steps: Determine oxidation numbers, identify oxidation and reduction, balance atoms and charges in each half-reaction, balance electrons, balance hydrogen and oxygen atoms, combine balanced half-reactions. Start by balancing the chromium atoms on both sides of the equation and proceed to balance the other coefficients.
Explanation:To balance the given redox reaction in an acidic solution, we need to follow these steps:
Determine the oxidation numbers of each element in the equation.Identify the element(s) that undergo oxidation and reduction.Balance the atoms and charges in each half-reaction.Balance the electrons by multiplying the half-reactions.Balance the hydrogen and oxygen atoms using H+ and H2O.Combine the balanced half-reactions to get the final balanced equation.Since the coefficient of Cr2O72−(aq) is given as 1, we can start by balancing the chromium atoms on both sides of the equation:
1 Cr2O72−(aq) + ___ Ti3+(aq) + ___ H+(aq) → ___ Cr3+(aq) + ___ TiO2+(aq) + ___ H2O(l)
By inspection, we need 2 Cr2O72− on the left side, so the equation becomes:
2 Cr2O72−(aq) + ___ Ti3+(aq) + ___ H+(aq) → ___ Cr3+(aq) + ___ TiO2+(aq) + ___ H2O(l)
The remaining coefficients can be determined by balancing the oxygen atoms, hydrogen atoms, and charges in each half-reaction, and then combining the two half-reactions to get the final balanced equation.
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Toluene, C7H8, is used in the manufacture of explosives such as TNT (trinitrotoluene). A 1.500 g sample of liquid toluene was placed in a bomb calorimeter along with excess oxygen. When the combustion of the toluene was initiated, the temperature of the calorimeter rose from 25.000 °C to 26.413 °C. The products of the combustion are CO2(g) and H2O(l), and the heat capacity of the calorimeter was 45.06 kJ °C-1. Write the balanced chemical equation for the reaction in the calorimeter. How many joules were liberated by the reaction? How many joules would be liberated under similar conditions if 1.000 mol of toluene was burned?
Answer:
There were 63.67 kJ liberated
If 1 mol of toluene was burned, there would be liberated 3906 kJ / mole
Explanation:
Step 1: Data given
Mass of the toluene sample = 1.500 grams
The temperature of the calorimeter rose from 25.000 °C to 26.413 °C
Write the balanced chemical equation for the reaction in the calorimeter.
C7H8(l) + 9O2(g) ⇒ 7CO2(g) + 4H2O(l)
Step 2: Calculate Q of the calorimeter
Q = C * Δt
Q = 45.06 kJ/ °C * ( 26.413 - 25)
Q = 63.67 kJ
Step 3: Calculate moles of toluene
Moles toluene = mass toluene / molar mass toluene
Moles toluene = 1.500 grams /92.14 g/mol
Moles toluene = 0.0163 moles
Step 4: Calculate energy per mole
energy per mole = 63.67 kJ / 0.0163 moles = 3906 kJ / mole
Final answer:
The balanced chemical equation for the combustion of toluene is 2 C7H8 + 21 O2 -> 14 CO2 + 8 H2O. The heat liberated by the reaction in the calorimeter is 63.30 kJ. If 1.000 mol of toluene is burned, 0.507 kJ of heat will be liberated.
Explanation:
The balanced chemical equation for the combustion of toluene can be written as:
2 C7H8 + 21 O2 -> 14 CO2 + 8 H2O
To calculate the heat liberated by the reaction, we can use the formula:
q = Ccal * ΔT
where q is the heat liberated, Ccal is the heat capacity of the calorimeter, and ΔT is the change in temperature.
Plugging in the values, we get:
q = (45.06 kJ/°C) * (26.413 °C - 25.000 °C) = 63.30 kJ
To calculate the heat liberated when 1.000 mol of toluene is burned, we can use the molar mass of toluene to convert grams to moles.
The molar mass of toluene is 92.14 g/mol, so:
1.500 g * (1 mol / 92.14 g) = 0.016 mol
Using stoichiometry, we know that for every 2 moles of toluene burned, 63.30 kJ of heat is liberated.
Therefore, for 0.016 mol of toluene:
q = (63.30 kJ / 2 mol) * 0.016 mol = 0.507 kJ
When 1.010 of sucrose (C12H22O11) undergoes combustions in a bomb calorimeter, the temperature rises from 24.92 ◦ C to 28.33 ◦ C. Find DErxn for the combustion of sucrose in kJ/mol sucrose. The heat capacity of the bomb calorimeter determined in a separate experiment is 4.90 kJ/◦ C. (Answer: -5.66 x 103 kJ/mole).
Answer:
Heat of combustion = 5.6 ×10³ kj/mol
Explanation:
Given data:
Mass of sucrose = 1.010 g
Initial temperature = 24.92 °C
Final temperature = 28.33 °C
Heat capacity of calorimeter = 4.90 KJ/°C
Heat of combustion = ?
Solution:
ΔT = 28.33 °C - 24.92 °C = 3.41 °C
Q = - c. ΔT
Q = 4.90 KJ/°C . 3.41 °C
Q = - 16.7 kj
Number of moles of sucrose :
Number of moles of sucrose = mass/ molar mass
Number of moles of sucrose = 1.010 g / 342.3 g/mol
Number of moles of sucrose = 0.003 mol
Heat of combustion:
Heat of combustion = Q/n
Heat of combustion = - 16.7 kj/0.003 mol
Heat of combustion = -5.6 ×10³ kj/mol
The combustion of sucrose in a bomb calorimeter can be calculated using the formula for heat transfer: q = C * ΔT. Given the heat capacity is 4.90 kJ/°C and the change in temperature is 3.41 °C, the energy released per gram is 16.709 kJ. By multiplying this by the molecular weight of sucrose, over 342.3 g/mol, yields approximately -5.66 x 103 kJ/mole.
Explanation:The combustion of sucrose in a bomb calorimeter can be determined through use of the formula for heat transfer, q = C x ΔT. Here, C represents the heat capacity of the calorimeter, and ΔT is the change in temperature. Given the heat capacity (C) is 4.90 kJ/°C, and the change in temperature (ΔT) is 28.33 °C - 24.92 °C = 3.41 °C, we plug these values into the equation yielding q = 4.90 kJ/°C x 3.41° C = 16.709 kJ. This is the energy released per gram. To find the energy released per mole, we need to multiply this value by the molecular weight of sucrose which is around 342.3 g/mole. This gives us an answer of -5.66 x 103 kJ/mole, and the negative sign indicates this is an exothermic reaction, meaning heat is released.
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A hot iron ball is dropped into a 200 g sample of water initially at 50 degrees Celsius.
If 8.4 kJ of heat is transferred from the ball to the water, what is the final temperature of the water? (The specific heat of water is 4.2 J/gC).a. 40 C
b. 50 C
c. 60 C
d. 70 C
Answer:
The final temperature is 60 °C
Explanation:
Step 1: Data given
Mass of the iron ball = 200 grams
Initial temperature = 50 °C
8.4 kJ of heat is transferred from the ball to the water.
The specific heat of water = 4.2 J/g°C
Step 2: Calculate the final temperature
Q = m*c*ΔT
⇒ with Q = heat is transferred from the ball to the water = 8400 J
⇒ m = the mass of the ball = 200 grams
⇒ c = the specific heat of water = 4.2 J/g°C
⇒ ΔT = T2 - T1 = T2 - 50
8400 J = 200g*4.2 J/g°C * (T2-50°C)
8400 = 840T2 - 42000
50400 = 840T2
T2 = 50800/840
T2 = 60 °C
The final temperature is 60 °C
Would you expect the nonbonding electron-pair domain in NH3 to be greater or less in size than for the corresponding one in PH3?
* greater
* lesser
Answer:
Less
Explanation:
The bond angle in NH3 is greater than the bond angle in PH3.
According to the valence shell electron pair repulsion theory, the shape of molecules depends on the number of electron pairs that surround the central atom. The electronegativity of the central atom determines the bond angle of the molecule. The greater the electronegativitty of the central atom, the greater the bond angle.
Nitrogen is more electronegative than Phosphorus hence, the bond angle in NH3 is greater than the bond angle in PH3.
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Which oxide(s) of manganese have a mass percent of the metal that is greater than 50%?MnO MnO2 Mn2O3All of these210
Answer:
All of these oxides of manganese have a mass percent of metal greater than 50%
Explanation:
Step 1: Data given
Molar mass of Mn = 54.94 g/mol
Molar mass of O = 16 g/mol
Step 2: Calculate mass % of Mn in MnO
Molar mass of MnO = 54.94 + 16 = 70.94 g/mol
% Mn = (54.94/70.94)*100 % = 77.45 %
Step 3: Calculate % of Mn in MnO2
Molar mass of MnO2 = 54.94 + 2*16 = 86.94 g/mol
% Mn = (54.94/86.94) * 100% = 63.19 %
Step 4: Calculate % of Mn in Mn2O3
Molar mass of Mn2O3 = 2*54.94 + 3*16 = 157.88 g/mol
% Mn = ((2*54.94)/157.88)*100 % = 69.6 %
All of these oxides of manganese have a mass percent of metal greater than 50%
For a tablet containing 500. mg of vitamin C, calculate how many ml of 0.095 M NaOH is required to reach the equivalence point.
Answer:
mL of NaOH required =29.9mL
Explanation:
Let us calculate the moles of vitamin C in the tablet:
The molar mass of Vitamin C is 176.14 g/mole
[tex]moles=\frac{mass}{molarmass}=\frac{500mg}{176.14}=\frac{0.5}{176.14}=0.0028[/tex]
Thus we need same number of moles of NaOH to reach the equivalence point.
For NaOH solution:
[tex]moles=MolarityXvolume=0.095Xvolume[/tex]
[tex]0.00283=0.095Xvolume[/tex]
[tex]volume=0.0299L=29.9mL[/tex]
The beta decay of cesium-137 has a half-life of 30.0 years. How many years must pass to reduce a 24 mg sample of cesium 137 to 6.0 mg?
Answer:
60.0 years must pass to reduce a 24 mg of cesium 237 to 6.0 mg
Explanation:
For radioactive decay of a radioactive nuclide-
[tex]N_{t}=N_{0}(\frac{1}{2})^{(\frac{t}{t_{\frac{1}{2}}})}[/tex]
Where, [tex]N_{t}[/tex] is amount of radioactive nuclide after "t" time , N_{0} is initial amount of radioactive nuclide and [tex]t_{\frac{1}{2}}[/tex] is half-life of radioactive nuclide
Here N_{0} = 24 mg, N_{t} = 6.0 mg and [tex]t_{\frac{1}{2}}[/tex] = 30.0 years
So, [tex]6.0mg=24mg\times (\frac{1}{2})^{(\frac{t}{30.0years})}[/tex]
or, [tex]t=60.0years[/tex]
So 60.0 years must pass to reduce a 24 mg of cesium 237 to 6.0 mg
The bond between carbon and hydrogen is one of the most important types of bonds in chemistry. The length of an H—C bond is approximately 1.1 Å. Based on this distance and differences in electronegativity, do you expect the dipole moment of an individual H—C bond to be larger or smaller than that of an H—I bond?
Answer:
The dipole moment of H-C bond will be smaller than that of an H-I bond.
Explanation:
The electronegativity of iodine is greater than that of hydrogen.As a result the iodine atom tends to attract the bond electron pair of H-I bond towards itself creating a bond dipole which does not occur in case of H-C bond as the electronegativity of carbon and hydrogen are almost same.
That"s why dipole moment of H-C bond is smaller than that of H-I bond.
Ethylene, C2H4, and tetrafluoroethylene, C2F4, are used to make the polymers polyethylene and polytetrafluoroethylene (Teflon), respectively. (a) Draw the Lewis structures for C2H4 and C2F4, and give the ideal H-C-H and F-C-F bond anges. (b) The actual H-C-H and F-C-F bond angles are 117.4∘ and 112.4∘ respectively. Explain these deviations.
Explanation:
The Lewis structures can be seen in the Figure below.
The ideal bond angle for both H-C-H and F-C-F is 120° because the have a triangular geometry.
The H-C-H has 117.4° angles due to the repulsion caused by the electrons of the double bond and the electrons of the H-C bond. This repulsion brings closer the H and decreases the angle.
On the other hand, in the F-C-F the same phenomenom happens but given that the F have lone pairs of electrons the repulsion is bigger, the F get closer and the angle is smaller than the H-C-H.
Final answer:
The Lewis structure for tetrafluoroethylene (C2F4) shows each carbon atom double-bonded to the other and single-bonded to two fluorine atoms, with an sp2 hybridization. The ideal bond angles for sp2 hybridized carbon are 120 degrees, but the actual angles differ due to variations in electronegativity and substituent size.
Explanation:
Polymerization and Lewis Structures
To draw the Lewis structure for the monomer tetrafluoroethylene (C2F4) used to make the polymer polytetrafluoroethylene (Teflon), we start by considering that carbon and fluorine both follow the octet rule, meaning that they both seek to be surrounded by eight electrons to reach a stable configuration. In C2F4, each carbon atom is double-bonded to the other carbon atom and single-bonded to two fluorine atoms. The hybridization of carbon in C2F4 is sp2 because it has three sigma bonds (one with the other carbon and two with fluorine atoms) and one pi bond (part of the double bond with the other carbon).
The ideal bond angles for an sp2 hybridized atom are 120 degrees. However, in the case of C2H4 and C2F4, the actual bond angles are 117.4 degrees for H-C-H and 112.4 degrees for F-C-F respectively. These deviations from the ideal angles can be attributed to the differences in electronegativity and size of the substituents, causing variations in electron cloud repulsions and thus slightly adjusting the bond angles.
All of the following increase from left to right across the second period of the periodic table (Na - Ar) EXCEPT
effective nuclear charge
number of core electrons
number of valence electrons
atomic number
Answer: effective nuclear charge
Explanation: This is because electrons are added to tell same shell at about the same distance from the nucleus
Copper(II) sulfate pentahydrate, CuSO4 ·5 H2O, (molar mass 250 g/mol) can be dehydrated by repeated heating in a crucible. Which value is closest to the percentage mass of water lost from the total mass of salt in the crucible when the crucible undergoes repetitive heatings until a constant mass is reached?36%26%13%25%
Answer:
The water lost is 36% of the total mass of the hydrate
Explanation:
Step 1: Data given
Molar mass of CuSO4*5H2O = 250 g/mol
Molar mass of CuSO4 = 160 g/mol
Step 2: Calculate mass of water lost
Mass of water lost = 250 - 160 = 90 grams
Step 3: Calculate % water
% water = (mass water / total mass of hydrate)*100 %
% water = (90 grams / 250 grams )*100% = 36 %
We can control this by the following equation
The hydrate has 5 moles of H2O
5*18. = 90 grams
(90/250)*100% = 36%
(160/250)*100% = 64 %
The water lost is 36% of the total mass of the hydrate
BNCT relies on the initial targeting of tumor cells by an appropriate chemical compound tagged with 10 5B, which preferentially concentrates in tumor cells. During the irradiation of the tumor site by neutrons (10n) the 10 5B absorbs a low-energy neutron (10n), and it ejects an energetic short-range alpha particle (42α or 42He) and lithium ion along with gamma radiation (γ). This radiation deposits most of its energy within the cell containing the original 10 5B atom. Therefore, if a higher concentration of 10 5B exists in tumor cells relative to other normal tissues, a concomitantly higher dose will be delivered to the tumor cells during neutron irradiation. What is the nuclear reaction that takes place in the tumor cell?
In Boron Neutron Capture Therapy (BNCT), the nuclear reaction involves boron-10 absorbing a neutron, leading to the emission of an alpha particle, lithium ion, and gamma radiation. This reaction selectively deposits high-energy particles within tumor cells, maximizing damage to cancerous tissue while sparing normal cells. The main equation is [tex]\(^{10}_{5}B + ^{1}_{0}n \rightarrow ^{11}_{5}B^* \rightarrow ^{4}_{2}\alpha + ^{7}_{3}Li + \gamma\)[/tex].
Boron Neutron Capture Therapy (BNCT) is a cancer treatment that involves the use of boron-10 (10B) compounds targeted to tumor cells. The key nuclear reaction in BNCT occurs when boron-10 absorbs a low-energy neutron (10n). This results in the formation of an excited boron-11 (11B) nucleus, which promptly undergoes a nuclear reaction. The boron-11 nucleus decays into an alpha particle (42α or 42He) and a lithium-7 ion, releasing gamma radiation (γ) in the process.
The nuclear reaction can be represented as follows:
[tex]\[ ^{10}_{5}B + ^{1}_{0}n \rightarrow ^{11}_{5}B^* \rightarrow ^{4}_{2}\alpha + ^{7}_{3}Li + \gamma \][/tex]
In this equation, [tex]\(^{10}_{5}B\)[/tex] represents boron-10, [tex]\(^{1}_{0}n\)[/tex] represents a neutron, [tex]\(^{11}_{5}B^*\)[/tex] is the excited boron-11 nucleus, [tex]\(^{4}_{2}\alpha\)[/tex] is the alpha particle, [tex]\(^{7}_{3}Li\)[/tex] is the lithium-7 ion, and [tex]\(\gamma\)[/tex] is the gamma radiation.
The key aspect of BNCT is that the alpha particle and lithium ion released during this nuclear reaction have high linear energy transfer (LET) and are ejected within a very short range. As a result, the majority of the deposited energy occurs within the tumor cell containing the original boron-10 atom. This selective deposition of high-energy particles within the tumor cells aims to maximize the damage to cancerous tissue while minimizing harm to surrounding normal tissues. The success of BNCT relies on achieving a higher concentration of boron-10 in tumor cells compared to normal tissues, ensuring an effective and targeted treatment approach.
Classify the possible combinations of signs for a reaction's ∆H and ∆S values by the resulting spontaneity.
A) Spontaneous as written at all temperatures.
B) Spontaneous in reverse at all temperatures.
C) Spontaneous as written above a certain temperature.
D) Spontaneous as written, below a certain temperature.
1) ∆H = + // ∆S = +
2) ∆H = - // ∆S = +
3) ∆H = - // ∆S = -
4) ∆H = + // ∆S = -
Answer:
Correct combinations are
For option A it is 2
For option B it is 4
For option D it is 3
For option C it is 1
Explanation:
For checking the spontaneity of a reaction, we have to check the sign of ΔG using the below formula
ΔG = ΔH - T×ΔSwhere
ΔG is the change in Gibbs free energy
ΔH is the change in enthalpy
T is the temperature
ΔS is the change in entropy
For spontaneous reactions, ΔG must be less than zero and for non-spontaneous reactions ΔG must be greater than zero but for an equilibrium reaction ΔG must be equal to zeroSo in case of 1 as ΔH and ΔS are positive if the temperature is above a certain value then ΔG will be less than zero
So in case of 2 as ΔH is negative and ΔS is positive then ΔG will always be less than zero at all temperatures
So in case of 3 as ΔH and ΔS are negative if the temperature is below a certain value then ΔG will be less than zero
So in case of 4 as ΔH is positive and ΔS is negative then ΔG will always be greater than zero but in reverse direction as ΔG is less than zero therefore in reverse direction the reaction will be spontaneous at all temperatures
Copper may be used for electroplating with a half reaction of Cu2+ + 2e- ---> Cu. If a current of 10.0A is applied to a Cu2+ solution for 60.0 min, the mass of copper plated out can be calculated as
A.) 10.0 x 3600 x (1/96500) x (2.00/1) x 63.5
B.) 10.0 x 3600 x (1/96500) x (1/2.00) x 63.5
C.) 96500 x (1/10.0) x (1/2.00) x (1/3600) x 63.5
D.) 96500 x (1/10.0) x (1/2.00) x (1/3600) x (1/63.5)
Answer:
The correct option is: B.) 10.0 x 3600 x (1/96500) x (1/2.00) x 63.5
Explanation:
Given reduction half-reaction: Cu²⁺ + 2e⁻ → Cu
Given: Electrons transferred: n = 2, Current: I = 10.0 A,
Time: t = 60.0 min = 60 × 60 sec = 3600 sec (∵ 1 min = 60 sec)
Since, Electric charge (Q) = current × time
∴ Q = I × t
⇒ Q = 10.0 A × 3600 sec = (10.0 × 3600) C
Since, one faraday charge is equal to the charge of one mole electrons.
∴ One mole electron = 1 Faraday (F) = 96,500 coulombs (C)
⇒ 1 C = 1 ÷ 96,500 mole electron
∴ (10.0 × 3600) C = [(10.0 × 3600) × (1 ÷ 96,500)] mole electron
Now since 2 mole electrons reduces 1 mole Cu²⁺ to Cu.
So, 1 mole electrons reduces (1/2) mole Cu²⁺
Therefore, moles of Cu²⁺ reduced by [(10.0 × 3600) × (1 ÷ 96,500)] mole electrons = [(10.0 × 3600) × (1 ÷ 96,500)] mole electron × (1 mole Cu²⁺ ÷ 2 mole electron)
= [(10.0 × 3600) × (1 ÷ 96,500) × (1 ÷ 2)] moles of Cu²⁺
As number of moles = mass taken ÷ molar mass
⇒ mass of copper plated = number of moles × molar mass
As the molar mass of copper = 63.5 g/ mol
∴ mass of copper plated = [(10.0 × 3600) × (1 ÷ 96,500) × (1 ÷ 2)] moles × 63.5 g/mol = [(10.0 × 3600) × (1 ÷ 96,500) × (1 ÷ 2) moles × 63.5] g
Therefore, the mass of copper plated = [(10.0 × 3600) × (1 ÷ 96,500) × (1 ÷ 2) moles × 63.5] g
If 60.0 grams of carbonic acid are sealed in a 2.00 L soda bottle at room temperature (298 K) and decompose completely via the equation below, what would be the final pressure of carbon dioxide assuming it had the full 2.00 L in which to expand? H₂CO₃(aq) → H₂O(l) + CO₂(g)
Answer:
The final pressure of the carbon dioxide gas will 11.84 atm.
Explanation:
Moles of carbonic acid = [tex]\frac{60.0 g}{62 g/mol}=0.9677 mol[/tex]
[tex]H_2CO_3(aq)\rightarrow H_2O(l) + CO_2(g)[/tex]
According to reaction, 1 mol of carbonic acid gives 1 mole of carbon dioxide gas.
Then 0.9677 moles of carbonic acid will give :
[tex]\frac{1}{1}\times 0.9677 mol=0.09677 mol[/tex] of carbon dioxide
Moles of carbon dioxide gas = n = 0.09677 mol
Volume of soda bottle = [tex]V_1=2 L[/tex]
Pressure of the carbon dioxide gas = P
Temperature of the carbon dioxide gas = T = 298 K
[tex]PV=nRT[/tex] (ideal gas law)
[tex]P=\frac{nRT}{V}=\frac{0.9677 mol\times 0.0821 atm L/mol K\times 298 K}{2 L}=11.84 atm[/tex]
The final pressure of the carbon dioxide gas will 11.84 atm.
By using the ideal gas law equation PV = nRT, calculating the number of moles of CO₂ formed from 60.0 grams of decomposed carbonic acid, and inputting the given values for the volume (V) and temperature (T), we calculated that the final pressure of the CO₂ in the soda bottle would be approximately 1.18 atm.
Explanation:To calculate the final pressure of carbon dioxide in the soda bottle we must utilize the ideal gas law equation, which is PV = nRT. Where P stands for pressure, V is volume, n is the number of moles of gas, R is the gas constant and T is the temperature.
First, we need to determine the number 'n' of moles of CO₂ formed. Knowing that 60.0 grams of carbonic acid decompose completely into water and CO₂, and knowing that the molar mass of carbonic acid H₂CO₃ is approximately 62.03 g/mol, we can find that the number of moles (n) is equal to the mass (60g) divided by the molar mass (62.03 g/mol), so n equals about 0.967 moles.
Because it specifies that we should assume CO₂ to 'have the full 2.00 L in which to expand', the volume V in the ideal gas law equation is 2.00 L. Similarly, the temperature T is given as 298 K. We will now plug these values into the ideal gas law equation, using 0.0821 as the value for R (the ideal gas constant in units of L*atm/(mol*K)). This gives us P = (nRT) / V, so P is about 1.18 atm, which is the final pressure of CO₂.
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For which of these processes is the value of ΔH expected to be negative?I. The temperature (of water) increases when calcium chloride dissolves in waterII. Steam condenses to liquid waterIII. Water freezesIV. Dry ice sublimesA.) IV onlyB.) I, II, and IIIC.) I onlyD.) II and III only
Answer: B.) I, II, and III
Explanation:
Exothermic reactions are defined as the reactions in which energy of the product is lesser than the energy of the reactants. The total energy is released in the form of heat and [tex]\Delta H[/tex] for the reaction comes out to be negative.
Endothermic reactions are defined as the reactions in which energy of the product is greater than the energy of the reactants. The total energy is absorbed in the form of heat and [tex]\Delta H[/tex] for the reaction comes out to be positive.
I) The temperature (of water) increases when calcium chloride dissolves in water : Thus the reaction is exothermic and [tex]\Delta H[/tex] for the reaction comes out to be negative.
II) Steam condenses to liquid water : The energy is released when bonds are formed when it coverts from gas to liquid and thus [tex]\Delta H[/tex] for the reaction comes out to be negative.
III) Water freezes : The energy is released when bonds are formed to get converted from liquid to solid and thus [tex]\Delta H[/tex] for the reaction comes out to be negative.
IV) Dry ice sublimes : The energy is absorbed when bonds are broken to get converted from solid to gas and thus [tex]\Delta H[/tex] for the reaction comes out to be positive.
A 8.00 L tank at 26.9 C is filled with 5.53 g of dinitrogen difluoride gas and 17.3 g of sulfur hexafluoride gas. You can assume both gases behave as ideal gases under these conditions. Calculate the mole fraction and partial pressure of each gas, and the total pressure in the tank
To calculate the mole fraction and partial pressure of each gas, we calculate the number of moles of each gas using the given masses and molar masses. The mole fraction is then determined by dividing the moles of each gas by the total moles. The partial pressure of each gas and the total pressure in the tank are found using the mole fraction and the given total pressure.
Explanation:To calculate the mole fraction and partial pressure of each gas, we first need to calculate the number of moles of each gas. For dinitrogen difluoride (N2F2), we divide the given mass by the molar mass of N2F2 to get the number of moles.
Moles of N2F2 = 5.53 g / (28.01 g/mol) = 0.1977 mol
Similarly, for sulfur hexafluoride (SF6), the moles are calculated as follows:
Moles of SF6 = 17.3 g / (146.06 g/mol) = 0.1185 mol
The mole fraction of each gas is then calculated by dividing the moles of the gas by the total moles:
Mole fraction of N2F2 = 0.1977 mol / (0.1977 mol + 0.1185 mol) = 0.6258
Mole fraction of SF6 = 0.1185 mol / (0.1977 mol + 0.1185 mol) = 0.3742
To calculate the partial pressure of each gas, we use the formula:
Partial pressure = Mole fraction * Total pressure
Given that the total pressure is 26.9 C, we substitute the values to find:
Partial pressure of N2F2 = 0.6258 * 26.9 C = 16.82 C
Partial pressure of SF6 = 0.3742 * 26.9 C = 10.08 C
Finally, the total pressure in the tank is the sum of the partial pressures:
Total pressure = Partial pressure of N2F2 + Partial pressure of SF6
Total pressure = 16.82 C + 10.08 C = 26.9 C
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The mole fraction of N₂F₂ is 0.4141 with a partial pressure of 0.256 atm, while the mole fraction of SF₆ is 0.5859 with a partial pressure of 0.361 atm. The total pressure in the tank is 0.617 atm.
To solve this problem, we need to use the Ideal Gas Law and the concept of partial pressures.
Step 1: Calculate Moles of Each Gas
First, we calculate the number of moles for each gas:
Molar mass of dinitrogen difluoride N₂F₂ = 2(14) + 2(19) = 66 g/molMoles of N₂F₂ = 5.53 g / 66 g/mol = 0.0838 molMolar mass of sulfur hexafluoride SF₆ = 32 + 6(19) = 146 g/molMoles of SF₆ = 17.3 g / 146 g/mol = 0.1185 molStep 2: Total Moles in the Tank
Total moles = 0.0838 mol + 0.1185 mol = 0.2023 molStep 3: Calculate the Mole Fraction
Mole fraction of N₂F₂ = 0.0838 mol / 0.2023 mol = 0.4141Mole fraction of SF₆ = 0.1185 mol / 0.2023 mol = 0.5859Step 4: Calculate Partial Pressures
We use the Ideal Gas Law: PV = nRT
R (Ideal Gas Constant) = 0.0821 atm·L/mol·KTemperature (T) = 26.9°C = 300.05 K (T = 26.9 + 273.15)First, calculate the total pressure:
P = (nRT) / V = (0.2023 mol × 0.0821 atm·L/mol·K × 300.05 K) / 8.00 L = 0.617 atmThen, calculate the partial pressures:
Partial pressure of N₂F₂ = 0.4141 × 0.617 atm = 0.256 atmPartial pressure of SF₆ = 0.5859 × 0.617 atm = 0.361 atmSummary
In summary, the mole fraction of dinitrogen difluoride is 0.4141, and its partial pressure is 0.256 atm. The mole fraction of sulfur hexafluoride is 0.5859, and its partial pressure is 0.361 atm. The total pressure in the tank is 0.617 atm.In part D of the experiment, a student begins with a beaker containing 10.0 mL of 0.010 M HCl solution and adds different volumes of 0.010 M NaOH. (a) How many millimoles of acid are initially in the beaker? (If in doubt, look back at your calculation from experiment 9).
Answer:
There are initially 0.1 millimoles of acid (HCl) in the beaker.
Explanation:
Step 1: Data given
volume of 0.010 M HCl = 10.0 mL = 0.01 L
Step 2: Calculate the number of moles HCl
Number of moles = Molarity HCl * volume of solution
Number of moles HCl = 0.010 M * 0.01 L = 0.0001 moles HCl = 0.1 *10^-3 mol = 0.1 mmol HCl
There are initially 0.1 millimoles of acid (HCl) in the beaker.
The initial beaker contains 0.1 millimoles of HCl. The process of adding NaOH to HCl is called titration, which helps measure the concentration of a solution and involves achieving an equivalence point where all reactant molecules have reacted.
Explanation:The question is asking about the quantity of acid initially present in a beaker before the addition of sodium hydroxide. The millimoles of acid can be calculated using the formula: Molarity (M) = Moles/Liter. For this experiment, you have an initial volume of 10.0 mL (or 0.010 L) of a 0.010 M HCl solution. Therefore, you can calculate the millimoles of acid in the beaker by multiplying the molarity by the volume in liters: 0.010 moles/L * 0.010 L = 0.0001 moles or 0.1 millimoles of HCl.
Titration, mentioned in the references, is a technique used to find the concentration of a solution, where a reaction is created that can easily be observed when the amount of reactant in a solution is exactly sufficient to react with another compound. In your experiment, NaOH is used to titrate HCl. The point when all the molecules of one reactant have reacted with another is called the equivalence point, which you can determine from a titration curve.
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Select all of the properties of ionic bonds. (select all that apply)
Intermediate bond strength
Formed from nonmetals only
Electrons are transferred
Formed from metals and nonmetals
Formed from metals only
Weakest bond
Strongest bond
Electrons are lost from metal atoms
Formed from cations in a sea of electrons
Electrons are shared
Formed from cations and anions
Answer:from the given for ionic are;
Electrons are transferred
Strongest bond
Electrons are lost from metal atoms
Formed from cations and anions
Metallic bond are;
Formed from cations in a sea of electrons
Intermediate bond strength
Formed from metals only
For covalent are;
Weakest bond
Electrons are shared
Formed from metals and nonmetal
Formed from metals and nonmetals
The properties of ionic bonds are it is the strongest bond, formed by metal atom, and in between cation and anion due to transfer of electrons.
What is ionic bond?Ionic bond is one of the strongest bond in chemistry and it is formed by the high attraction force between cations and anions.
In the ionic bond, cation atom is a metal atom and that was formed by the transfer of electron.Metallic bonds are the bonds which are having intermediate bond strength and formed by metal atoms only. In this bond cations are formed in a sea of electrons.Covalent bond is the weakest bond formed by metals and non metals by the sharing of electrons.So, properties of ionic bonds are strongest bond, formed by cations & anions through electron transfer.
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The net equation for the oxidative reactions of the pentose phosphate pathway is glucose − 6 − phosphate + 2 NADP + + H 2 O ⟶ ribulose − 5 − phosphate + 2 NADPH + CO 2 + 2 H + Select true statements about the pentose phosphate pathway. Glucose is a precursor of the pentose phosphate pathway. Carbon atoms from the pentose sugar products may enter the glycolytic pathway. Pentose phosphate pathway activity is high in rapidly dividing cells. Glucose‑6‑phosphate is reduced to ribulose‑5‑phosphate in this series of reactions.
Final answer:
The pentose phosphate pathway generates NADPH and key metabolites for cell function.
Explanation:
The pentose phosphate pathway (PPP) is a metabolic pathway that runs parallel to glycolysis, generating NADPH and pentoses, as well as ribose 5-phosphate. It plays a crucial role in providing crucial components for nucleotide synthesis and redox regulation in cells.
True statements about the pentose phosphate pathway include glucose being a precursor, carbon atoms from pentose sugars entering glycolytic pathway, and the pathway's high activity in rapidly dividing cells. The conversion of glucose-6-phosphate to ribulose-5-phosphate is a key step in the series of reactions.
A 2.50 g sample of powdered zinc is added to 100.0 mL of a 2.00 M aqueous solution of hydrobromic acid in a calorimeter. The total heat capacity of the calorimeter and solution is 448 J/K. The observed increase in temperature is 21.1 K at a constant pressure of one bar. Calculate the standard enthalpy of reaction using these data. Zn ( s ) + 2 HBr ( aq ) ⟶ ZnBr 2 ( aq ) + H 2 ( g )
Answer: The standard enthalpy of the reaction is -248.78 kJ/mol
Explanation:
The heat released by the reaction is absorbed by the calorimeter and the solution.
The chemical equation used to calculate the heat released follows:
[tex]q=c\times \Delta T[/tex]
where,
c = heat capacity of calorimeter = 448 J/K
[tex]\Delta T[/tex] = change in temperature = 21.1 K
Putting values in above equation, we get:
[tex]q=448J/K\times 21.1K=9452.8J[/tex]
Sign convention of heat:
When heat is absorbed, the sign of heat is taken to be positive and when heat is released, the sign of heat is taken to be negative.
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass of zinc = 2.50 g
Molar mass of zinc = 65.4 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of zinc}=\frac{2.50g}{65.4g/mol}=0.038mol[/tex]
To calculate the standard enthalpy of the reaction, we use the equation:
[tex]\Delta H^o_{rxn}=\frac{q}{n}[/tex]
where,
[tex]q[/tex] = amount of heat released = -9452.8 J
n = number of moles = 0.038 moles
[tex]\Delta H^o_{rxn}[/tex] = standard enthalpy of the reaction
Putting values in above equation, we get:
[tex]\Delta H^o_{rxn}=\frac{-9452.8J}{0.038mol}=-248757.9J/mol=-248.78kJ/mol[/tex]
Conversion factor used: 1 kJ = 1000 J
Hence, the standard enthalpy of the reaction is -248.78 kJ/mol
A chemistry student needs 65.0g heptane for an experiment. He has available 20.0g of a 38.1% w/w solution of heptane in chloroform.
Calculate the mass of solution the student should use. If there's not enough solution, press the "No solution" button.
Round your answer to 3 significant digits.
Answer : The correct answer is "No solution".
Explanation :
As we are given that 38.1 % w/w solution of heptane in chloroform. That means 38.1 gram of heptane present in 100 gram of solution.
Now we have to determine the mass of solution for 65.0 grams of hepatne.
As, 38.1 grams of heptane present in 100 grams of solution
So, 65.0 grams of heptane present in [tex]\frac{65.0}{38.1}\times 100=171[/tex] grams of solution
Since, there is only 20.0 grams of solution available. That means, there is not enough solution.
Thus, the correct answer is "No solution".
Final answer:
The student would need 170.6g of the 38.1% w/w heptane solution to obtain 65.0g of heptane. However, the student only has 20.0g of solution, which is not enough to provide the required amount of heptane, so the student does not have a viable solution.
Explanation:
The student needs to calculate the mass of solution to use to obtain 65.0g of heptane from a 38.1% w/w solution of heptane in chloroform. The weight/weight percentage (w/w%) indicates that for every 100g of solution, there are 38.1g of heptane. To find the mass of the solution needed to obtain 65.0g of heptane, the following calculation can be performed:
m = (desired mass of heptane) / (percentage of heptane in solution as a decimal)
m = 65.0g / 0.381
m = 170.6g
The student would need to use 170.6g of the 38.1% w/w heptane solution to obtain the required amount of pure heptane. However, since the student only has 20.0g of solution, there is not enough to obtain 65.0g of heptane without obtaining more solution. Therefore, the student cannot proceed with the experiment as is and must either get more solution or adjust the experiment.
What is the pH of 0.30 M ethanolamine, HOCH2CH2NH2, (Kb = 3.2 x 10−5)?
Answer:
pH= 11.49
Explanation:
Ethanolamine is an organic chemical compound of the formula; HOCH2CH2NH2. Ethanolamine, HOCH2CH2NH2 is a weak base.
From the question, the parameters given are; the concentration of ethanolamine which is = 0.30M, pH value= ??, pOH value= ??, kb=3.2 ×10^-5
Using the formula below;
[OH^-]=√(kb×molarity)----------------------------------------------------------------------------------------------------------(1)
[OH^-] =√(3.2×10^-5 × 0.30M)
[OH^-]= √(9.6×10^-6)
[OH^-]=3.0984×10^-3
pOH= -log[OH^-]
pOH= -log 3.1×10^-3
pOH= 3-log 3.1
pH= 14-pOH
pH= 14-(3-log3.1)
pH= 11+log 3.1
pH= 11+ 0.4914
pH= 11.49
The pH of the ethanolamine solution has been 11.5.
The pH of the solution has been defined as the concentration of hydrogen ion in the solution.
The pH has been the difference of pOH from 14.
Computation for the pH of ethanolamineThe ethanolamine has been a weak base. The OH concentration of ethanolamine has been given as:
[tex]\rm OH^-=\sqrt{Kb\;\times\;Molarity}[/tex]
The Kb of sample has been given as [tex]3.2\;\times\;10^-^5[/tex]
The molarity of sample has been, 0.3 M.
Substituting the values for hydroxide ion concentration:
[tex]\rm OH^-=\sqrt{3.2\;\times\;10^-^5\;\times\;0.3}\\OH^-=3.098\;\times\;10^-^3[/tex]
The pOH has been the negative logarithmic values of hydroxide ion concentration. The pOH has been given as:
[tex]\rm pOH=-log\;OH^-\\pOH=-log\;3.098\;\times\;10^-^3\\pOH=2.5[/tex]
The pOH of the sample has been 2.5.
The pH of ethanolamine sample has been given as:
[tex]\rm pH=14-pOH\\pH=14-2.5\\pH=11.5[/tex]
The pH of the ethanolamine solution has been 11.5.
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Draw all of the monochlorination products that you might obtain from the free-radical chlorination of ethylcyclopropane.You do not have to consider stereochemistry.Draw one structure per sketcher. Add additional sketchers using the dropdown menu in the bottom right corner.Separate multiple products using the + sign from the dropdown menu.
Answer:
-) (2-chloroethyl)cyclopropane
-) (1-chloroethyl)cyclopropane
-) 1-chloro-2-ethylcyclopropane
-) 1-chloro-1-ethylcyclopropane
Explanation:
The monochlorination are products in which Cl is added only 1 time. Now, in the ethylcyclopropane we will have 5 carbons, so we will have 5 options for monochlorination (see figure 1). But, for carbons 2 and 3 we will have equivalent structures, therefore at the end, we will have only 4 possible structures. In carbon 5, carbon 4, carbon 3 and carbon 1 (see figure 2).
Final answer:
Monochlorination of ethylcyclopropane can produce several isomers, based on the position where the chlorine atom is added - either on the cyclopropane ring or the ethyl group attached to it.
Explanation:
The subject of the question is the monochlorination products from the free-radical chlorination of ethylcyclopropane. During the free-radical chlorination, chlorine radicals can react with the different types of hydrogens in ethylcyclopropane - those on the cyclopropane ring as well as those on the ethyl group. Chlorination at the ring will yield two isomers due to two different types of hydrogens (geminal and vicinal), while chlorination on the ethyl group could occur at either one of the methylene hydrogens or the methyl group hydrogens. When considering the ethyl group, monochlorination would give a 1-chloroethylcyclopropane or a 2-chloroethylcyclopropane, depending on which hydrogen is substituted. With regard to the cyclopropane ring, substitution could lead to two different chlorocyclopropanes due to the different positions of hydrogen on the ring.
Glycolysis produces NADH. However, NADH cannot cross the inner mitochondrial membrane to be used in the electron transport chain (respiratory chain). The malate-aspartate shuttle in some animal cells transfers electrons from cytosolic NADH to the matrix. A diagram of the malate-aspartate shuttle is given below. You can use it as a reference to answer parts (a) and (b)
a) Which compound accepts electrons from NADH, producing a compound that can pass through the inner membrane?
A. glutamate
B. aspartate
C. α- ketoglutarate
C. oxaloacetate
b) Which compound transfers reducing equivalents (electrons) into the mitochondrial matrix? O
A. malate
B. α-ketoglutarate
C. oxaloacetate
D. glutamate
E. aspartate
Answer:
a) Oxaloacetate.
b) Malate.
Explanation:
The malate-aspartate shuttle works in the mitochondria of the liver, kidney, and heart.
The cytosolic NADH reducing equivalents transfer to the cytosolic oxalacetate, producing malate, by the cytosolic malate dehydrogenase. This malate crosses the inner mitochondrial membrane throughout the malate-α-ketoglutarate transport system.
Inside the matrix, the reducing equivalents pass from the malate to the NAD⁺, forming NADH, by the action of the mitochondrial malate dehydrogenase. This formed NADH passes electrons directly to the respiratory chain.
Aqueous hydrochloric acid reacts with solid sodium hydroxide to produce aqueous sodium chloride and liquid water . What is the theoretical yield of sodium chloride formed from the reaction of of hydrochloric acid and of sodium hydroxide? Round your answer to significant figures.
The theoretical yield of sodium chloride formed from the reaction of 3.3 g of hydrochloric acid and 6.7 g of sodium hydroxide is 5.3 g
How to obtain the theoretical yield of sodium chloride?
We shall begin by determining the limiting reactant for the reaction of 3.3 g of hydrochloric acid and 6.7 g of sodium hydroxide. This is shown below:
HCl + NaOH -> NaCl + H₂O
Molar mass of HCl = 36.5 g/molMass of HCl from the balanced equation = 1 × 36.5 = 36.5 g Molar mass of NaOH = 40 g/molMass of NaOH from the balanced equation = 1 × 40 = 40 gFrom the balanced equation above,
36.5 g of HCl reacted with 40 g of NaOH
Therefore,
3.3 g of HCl will react with = (3.3 × 40) / 36.5 = 3.6 g of NaOH
Now, we can see that only 3.6 g of NaOH reacted with 3.3 g of HCl
Thus, the limiting reactant is HCl
Finally, we shall calculate the theoretical yield of sodium chloride. Details below:
HCl + NaOH -> NaCl + H₂O
Molar mass of HCl = 36.5 g/molMass of HCl from the balanced equation = 1 × 36.5 = 36.5 g Molar mass of NaCl = 58.5 g/molMass of NaOH from the balanced equation = 1 × 58.5 = 58.5 gTheoretical yield of sodium chloride =?From the balanced equation above,
36.5 g of HCl reacted to produce 58.5 of NaCl
Therefore,
3.3 g of HCl will react to produce = [tex]\frac{3.3\ \times\ 58.5}{36.5}[/tex] = 5.3 g of NaCl
In conclusion, theoretical yield of sodium chloride is 5.3 g
Complete question:
Aqueous hydrochloric acid reacts with solid sodium hydroxide to produce aqueous sodium chloride and liquid water.
What is the theoretical yield of sodium chloride formed from the reaction of 3.3 g of hydrochloric acid and 6.7 g of sodium hydroxide.? Round your answer to significant figure
Which of the following does the internal energy of an ideal gas depend upon?
a. The internal energy depends upon its temperature.
b. The internal energy depends upon its volume.
c. The internal energy depends upon its pressure.
Answer:
The correct option is: a. The internal energy depends upon its temperature.
Explanation:
Ideal gas is a hypothetical gas that obeys the ideal gas law. The equation for the ideal gas law:
P·V=n·R·T
Here, V- volume of gas, P - total pressure of gas, n- total mass or number of moles of gas, T - absolute temperature of gas and R- the gas constant
Also, according to the Joule's second law, the internal energy (U) of the given amount of ideal gas depends on the absolute temperature (T) of the gas only, by the equation:
[tex]U = c_{V}nRT[/tex]
Here, [tex]c_{V}[/tex] is the specific heat capacity at constant volume
The internal energy of an ideal gas depends on its temperature and is independent of its volume and pressure. For non-ideal systems, internal energy can be influenced by volume and pressure but only through their effect on temperature.
The internal energy of an ideal gas depends primarily on its temperature. This is because, for an ideal gas, the internal energy is a function of state and is determined by the kinetic energy of its particles, which is directly related to temperature. The volume and pressure of an ideal gas do not directly determine its internal energy, although changes in these properties can indirectly affect temperature and thus internal energy. Option A
In the case of an ideal gas, the internal energy does not change with volume or pressure shifts alone; it is the temperature that directly impacts internal energy. However, for other types of systems or under non-ideal conditions, both volume and pressure can have more complex relationships with internal energy.
Water is boiling in an aluminum pan placed on an electrical element on a stovetop. The pan has a bottom that is 0.800 cm thick and 14.0 cm in diameter. Aluminum thermal conductivity is 220 W/(m oC). In every second, 1.2 g boiling water evaporates. The latent heat of vaporization for water is 2.26 × 106 J/kg. (a) How much heat is transfered to the boiling water in one second?
Explanation:
The given data is as follows.
Thickness = 0.8 cm = [tex]0.8 \times 10^{-3} m[/tex] (As 1 m = 1000 cm)
Diameter = 14.0 cm, Conductivity = [tex]220 W/(m^{o}C)[/tex]
mass = 1.2 g, L = [tex]2.26 \times 10^{6} J/kg[/tex]
Now, we will calculate the heat of vaporization for water as follows.
Q = mL
Q = [tex]0.8 \times 10^{-3} \times 2.26 \times 10^{6}[/tex]
Q = 1808 J
So, rate of heat transfer in 1 sec will be as follows.
[tex]\frac{Q}{t}[/tex] = 1808 J/s
Thus, we can conclude that 1808 J/s heat is transfered to the boiling water in one second.
Final answer:
To find out how much heat is transferred to the boiling water in one second, we multiply the mass of water that evaporates per second (converted to kilograms) by the latent heat of vaporization for water, resulting in 2712 joules.
Explanation:
To determine how much heat is transferred to the boiling water in one second, we can use the latent heat of vaporization formula, which states that the heat required for a phase change is equal to the mass of the substance times the latent heat of vaporization (Q = m x L).
In this case, 1.2 grams of water evaporates every second, and the latent heat of vaporization for water is 2.26 × 106 J/kg. First, we need to convert the mass from grams to kilograms by dividing by 1000, which gives us 0.0012 kg. Then we multiply the mass by the latent heat to find the heat transfer:
Q = 0.0012 kg × 2.26 × 106 J/kg = 2712 J
Therefore, 2712 joules of heat are transferred to the boiling water in one second.
Calculate the pH of the solution formed when 45.0 mL of 0.100M NaOH solution is added to 50.0 mL of 0.100M CH3COOH (Ka for acetic acid = 1.8 x10-5 ).
Answer:
pH of soltion will be 5.69
Explanation:
The pH of the solution will be due to excessive acid left and the salt formed. Thus, it will form a buffer solution.
The pH of buffer solution is calculated from Henderson Hassalbalch's equation, which is:
[tex]pH=pKa+log(\frac{[salt]}{[acid]} )[/tex]
[tex]pKa=-log(Ka)[/tex]
[tex]pKa=-log(1.8X10^{-5})=4.74[/tex]
The moles of acid taken :
[tex]moles=molarityXvolume=0.1X50=5mmol[/tex]
The moles of base taken:
[tex]moles=molarityXvolume=0.1X45=4.5mmol[/tex]
The moles of acid left after reaction :
[tex]5-4.5=05mmol[/tex]
The moles of salt formed = 4.5mmol
Putting values in equation
[tex]pH=pKa+log(\frac{[salt]}{[acid]} )=4.74+log(\frac{4.5}{0.5})=5.69[/tex]
The study of chemicals and bonds is called chemistry. There are two types of elements are there and these rare metals and nonmetals.
The correct answer is 5.59.
What is PH?pH, historically denoting "potential of hydrogen" is a scale used to specify the acidity or basicity of an aqueous solution. Acidic solutions are measured to have lower pH values than basic or alkaline solutionsThe pH of the solution will be due to excessive acid left and the salt formed. Thus, it will form a buffer solution.
The pH of buffer solution is calculated from Henderson Hassalbalch's equation, which is:
[tex]pH= pka+log\frac{salt}{acid}[/tex]
[tex]pka= -log(ka)[/tex]
[tex]pka =-log(1.8*10^{-5}[/tex]
The moles of acid are taken as:-
[tex]moles = M*V[/tex]
[tex]0.5*50=5[/tex]
The moles of the base are taken as:-
[tex]moles = M*V[/tex]
[tex]0.1*45=4.5[/tex]
moles of acid left is 0.5
Place all the values to the equation:-
[tex]pH=4.74+log\frac{4.5}{0.5} \\=5.69[/tex]
Hence, the correct answer is 5.69.
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