Answer:
K.E. = 5.4362 × 10⁻¹⁹ J
Explanation:
The expression for Bohr velocity is:
[tex]v=\frac{Ze^2}{2 \epsilon_0\times n\times h}[/tex]
Applying values for hydrogen atom,
Z = 1
Mass of the electron ([tex]m_e[/tex]) is 9.1093×10⁻³¹ kg
Charge of electron (e) is 1.60217662 × 10⁻¹⁹ C
[tex]\epsilon_0[/tex] = 8.854×10⁻¹² C² N⁻¹ m⁻²
h is Plank's constant having value = 6.626×10⁻³⁴ m² kg / s
We get that:
[tex]v=\frac {2.185\times 10^6}{n}\ m/s[/tex]
Given, n = 2
So,
[tex]v=\frac {2.185\times 10^6}{2}\ m/s[/tex]
[tex]v=1.0925\times 10^6\ m/s[/tex]
Kinetic energy is:
[tex]K.E.=\frac {1}{2}\times mv^2[/tex]
So,
[tex]K.E.=\frac {1}{2}\times 9.1093\times 10^{-31}\times ({1.0925\times 10^6})^2[/tex]
K.E. = 5.4362 × 10⁻¹⁹ J
Suppose the average concentration of copper in water is measured to be 2.5 x 10" M. a. Express this concentration in mg/L. b. Express this concentration in ug/L. c. Does this concentration of copper exceed the freshwater acute criteria maximum concentration of 65 ppb?
Answer:
a) C Cu = 1588.6 mg/L
b) C Cu = 1588600 μg/L
c) this concentration exceeds the acute freshwater criteria:
∴ C Cu = 1588.6 mg/L >> 0.065 mg/L
Explanation:
∴ C Cu = 2.5 E-2 mol/L
∴Mw Cu = 63.546 g/mol
a) C Cu = 2.5 E-2 mol/L * 63.546 g/mol = 1.5886 g/L
⇒ C Cu = 1.5886 g/L * ( 1000 mg/g ) = 1588,6 mg/L
b) C Cu = 1588.6 mg/L * ( μg / 0.001 mg ) = 1588600 μg/L
c) C Cu = 65 ppb * ( ppm / 1000ppb ) = 0.065 ppm = 0.065 mg/L
∴ ppm ≡ mg/L
⇒ C Cu = 1588.6 mg/L >> 0.065 mg/L; this concentration exceeds the acute freshwater criteria
A boy on the beach holds a spherical balloon filled with air. At 7:00 am, the temperature on the beach is 35°F and the balloon has a diameter of 20 cm. Six hours later, the outside temperature is 100°F. Assume that the air is an ideal gas, that the atmospheric pressure remains constant, and that no air was lost or added to the balloon. What is the final diameter of the balloon (in cm)?
Answer:
The final diameter of the balloon is 20.84 cm.
Explanation:
To answer this problem, because both the number of moles (because no air was lost or added to the balloon) and the pressure remain constant, we can use Charles' law, or the law of volumes:
V₁T₂=V₂T₁
Where V is Volume (in L) and T is temperature (in K). The subscripts 1 and 2 refer to different states of the gas: for this problem, let's say that ₁ refers to the state of the gas at 7:00 am and that ₂ refers to the state 6 hours later.
To calculate the volume occupied by the gas we use the formula for the volume of a sphere:
[tex]V=\frac{4}{3}*\pi (\frac{d}{2})^{3}[/tex]
Thus the volume of the balloon at 7:00 am is 4188.79 cm³
Then we convert °F into K:
(35°F + 459.67) * 5/9 = 274.82 K = T₁(100°F + 459.67) * 5/9 = 310.93 K = T₂Now we use Charles' law to calculate V₂:
V₂=V₁T₂/T₁
V₂= [tex]\frac{4188.79cm^{3}*310.93K}{274.82 K}=4739.18cm^{3}[/tex]
Lastly we calculate the final diameter of the balloon:
[tex]4739.18cm^{3}=\frac{4}{3}*\pi (\frac{d}{2})^{3}\\3554.38cm^{3}=\pi (\frac{d}{2})^{3}\\1131.40cm^{3}=(\frac{d}{2})^{3}\\10.42cm=\frac{d}{2}\\20.84=d[/tex]
A researcher prepares a buffer of acetic acid and sodium acetate with a pH of 5.0.5.0. The total concentration of both components in the buffer is 250 mM,250 mM, and acetic acid has a pKapKa of 4.75.4.75. What are the concentrations of acetic acid and sodium acetate in the buffer? concentration of acetic acid: mM concentration of sodium acetate: mM How many moles of acetic acid and sodium acetate are in 2 L2 L of the buffer? amount of acetic acid: mol amount of sodium acetate: mol How many grams of acetic acid and sodium acetate are in 2 L2 L of the buffer? The molar mass of acetic acid is 60.05 g/mol;60.05 g/mol; the molar mass of sodium acetate is 82.03 g/mol.82.03 g/mol. mass of acetic acid: g mass of sodium acetate:
1.Concentration of acetic acid is 89.99 mM. Concentration of sodium acetate is 160.0 mM . 2.0.180 moles of acetic acid. 0.32 moles of sodium acetate. 3. Mass of acetic acid in 2 L solution is 10.81 g. Mass of sodium acetate in 2 L solution is 26.25 g.
The amount of matter in an item is indicated by its mass, which is a fundamental physical attribute of matter. It is a scalar quantity that is typically measured in kilogrammes (kg) or grammes (g). Mass is separate from weight, which is the gravitational force exerted on an object and can change depending on the strength of the gravitational field.
1.pH = pKa + log [Sodium acetate] / [Acetic acid]
5.0 = 4.75 + log [Sodium acetate] / [Acetic acid]
log [Sodium acetate] / [Acetic acid] = 5.0 - 4.75
= 0.25
[Sodium acetate] = 1.778 x [Acetic acid]
[Sodium acetate] + [Acetic acid] = 250 mM
1.778 x [Acetic acid] + [Acetic acid] = 250 mM
2.778 x [Acetic acid] = 250 mM
[Acetic acid] = 250 mM / 2.778
= 89.99 mM
[Sodium acetate] = 1.778 x [Acetic acid]
= 1.778 x 89.99 mM
= 160.0 mM
2.Concentration of acetic acid = 89.99 mM
= 0.08999 M
Concentration of sodium acetate = 160.0 mM
= 0.16 M
2 x 0.08999 = 0.180 moles of acetic acid.
2 x 0.16 = 0.32 moles of sodium acetate.
3.Mass of acetic acid = no of moles x molar mass
= 0.180 mol x 60.05 g /mol
= 10.81 g
Mass of sodium acetate = no of moles x molar mass
= 0.32 mol x 82.03 g /mol
= 26.25 g
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Final answer:
Using the Henderson-Hasselbalch equation, the concentrations of acetic acid and sodium acetate in the buffer solution are calculated to be approximately 89 mM and 161 mM, respectively. When considering a 2 L volume of this buffer, there would be approximately 0.178 moles of acetic acid and 0.322 moles of sodium acetate, which translates to 10.69 grams of acetic acid and 26.25 grams of sodium acetate.
Explanation:
To calculate the concentrations of acetic acid and sodium acetate in the buffer, we use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
Given:
pH = 5.0pKa = 4.75Total concentration ([A-] + [HA]) = 250 mMLet's calculate [A-] and [HA] (where [A-] is the concentration of sodium acetate and [HA] is the concentration of acetic acid):
5.0 = 4.75 + log([A-]/[HA])
=> log([A-]/[HA]) = 0.25
=> [A-]/[HA] = 10^0.25
=> [A-]/[HA] ≈ 1.78
Since [A-] + [HA] = 250 mM, we can set up a system of equations:
[A-] = 1.78[HA]
[A-] + [HA] = 250 mM
Solving for [HA] and [A-] gives us:
[HA] ≈ 89 mM
[A-] ≈ 161 mM
Next, we calculate the amount of acetic acid and sodium acetate in 2 L of the buffer:
amount of acetic acid = [HA] × 2 L = 0.089 mol/L × 2 L = 0.178 mol
amount of sodium acetate = [A-] × 2 L = 0.161 mol/L × 2 L = 0.322 mol
Finally, to calculate the mass of acetic acid and sodium acetate in 2 L of the buffer, we use their molar masses:
mass of acetic acid = 0.178 mol × 60.05 g/mol = 10.69 g
mass of sodium acetate = 0.322 mol × 82.03 g/mol = 26.25 g
Calculate the molality of a 150% by mass solution of MgCl, FW-95.3 g/mol in H.O. The density of tis solution is 1.127 gim 0.0134 m 0.157 m 1.58 m 1.86 m 11.8 m igator Delete Backspace U 10 pilli
Answer: 1.86 m
Explanation:
Molality of a solution is defined as the number of moles of solute dissolved per kg of the solvent.
[tex]Molality=\frac{n\times 1000}{W_s}[/tex]
where,
n = moles of solute
[tex]W_s[/tex] = weight of solvent in g
Given : 15.0 grams of [tex]MgCl_2[/tex] is present in 100 g of solution
Moles of solute [tex]H_2O=\frac{\text {given mass}}{\text {Molar mass}}=\frac{15.0g}{95.3g/mol}=0.16[/tex]
mass of solution = 100 g
density of solution = 1.127g/ml
mass of solvent = mass of solution - mass of solute = 100 g - 15.0g = 85.0 g
[tex]Molality=\frac{0.16\times 1000}{85.0g}=1.86[/tex]
Thus molality of a 15.0% by mass solution of [tex]MgCl_2[/tex] is 1.86m.
If you mix 50mL of 0.1 M TRIS acid with 60 mL of0.2 M
TRIS base, what will be the resulting pH?
Answer: The pH of resulting solution is 8.7
Explanation:
To calculate the number of moles for given molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}[/tex]
For TRIS acid:Molarity of TRIS acid solution = 0.1 M
Volume of solution = 50 mL
Putting values in above equation, we get:
[tex]0.1M=\frac{\text{Moles of TRIS acid}\times 1000}{50mL}\\\\\text{Moles of TRIS acid}=0.005mol[/tex]
For TRIS base:Molarity of TRIS base solution = 0.2 M
Volume of solution = 60 mL
Putting values in above equation, we get:
[tex]0.2M=\frac{\text{Moles of TRIS base}\times 1000}{60mL}\\\\\text{Moles of TRIS base}=0.012mol[/tex]
Volume of solution = 50 + 60 = 110 mL = 0.11 L (Conversion factor: 1 L = 1000 mL)
To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:[tex]pH=pK_a+\log(\frac{[salt]}{[acid]})[/tex]
[tex]pH=pK_a+\log(\frac{[\text{TRIS base}]}{[\text{TRIS acid}]})[/tex]
We are given:
[tex]pK_a[/tex] = negative logarithm of acid dissociation constant of TRIS acid = 8.3
[tex][\text{TRIS acid}]=\frac{0.005}{0.11}[/tex]
[tex][\text{TRIS base}]=\frac{0.012}{0.11}[/tex]
pH = ?
Putting values in above equation, we get:
[tex]pH=8.3+\log(\frac{0.012/0.11}{0.005/0.11})\\\\pH=8.7[/tex]
Hence, the pH of resulting solution is 8.7
Which of the following does not describe a measurement standard? Group of answer choices Measurement standards avoid ambiguity. A standard need not agree with a previously defined size. Confusion is eliminated when the correct measurement is applied. Measurement standards must be unchanging.
Answer:
Confusion is eliminated when the correct measurement is applied
Explanation:
A measurement standard is often described as the reference to which other measurements are judged or based on. Most times, measurement standard are used to determine how accurate an experimental measurement is.
A measurement standard may not eliminate confusion when the correct measurement is applied. If the scientist does not agree with the measurement standard being used, then problem will set in.
The statement 'A standard need not agree with a previously defined size' does not describe a measurement standard because consistency with previously defined sizes is crucial for accuracy. Measurement standards ensure comparability of data across various domains, driving advancements in both science and technology.
Explanation:The statement that does not describe a measurement standard is 'A standard need not agree with a previously defined size.' Measurement standards must indeed be consistent with previously defined sizes to ensure accuracy and eliminate confusion. Measurement standards are crucial to avoid ambiguity and to make sure that experimental data from different labs can be compared accurately. By having unchanging standards, like for the size of a meter, we ensure that a unit of measure remains constant over time, which is indispensable for consistency in scientific experiments and technology development.
Furthermore, the correct measurement standards are essential for the growth of technology as they allow for precise and reliable data. The relationship between science and technology is symbiotic; they drive each other's advancement. As technology improves, new measurement techniques become possible, leading to more detailed observations and more accurate scientific experiments, which in turn can further advance technology.
Convert 59.9 m^2 to square centimeters. cm^2
Answer:
59.9 m² = 599000 cm²
Explanation:
According to the International System of Units, square metre (m²) is the SI unit of area. 1 m² is defined as the area of the square having sides 1 m long.
Some other units of area are square millimetres (mm²), square kilometres (km²), square centimetres (cm²).
Now, converting square metre (m²) to square centimetres (cm²)
Since, 1 m² = 10⁴ cm²
Therefore, 59.9 m² = 59.9 × 10⁴ cm² = 599000 cm².
Therefore, 59.9 m² = 599000 cm²
Rearrange the following equation and solve for x when the hydrogen ion concentration is 0.01 M. pH=4+logx (Recall that pH=−log[H+]. Use the hints if you do not obtain one of the answers.) View Available Hint(s) Rearrange the following equation and solve for when the hydrogen ion concentration is 0.01 . (Recall that . Use the hints if you do not obtain one of the answers.) x=2 x=0.40 x=0.1 x=1×10−2
To solve the equation pH = 4 + log(x) and find the value of x when the hydrogen ion concentration is 0.01 M, rearrange the equation by subtracting 4 from both sides. Substitute the value of pH with its equivalent -log(0.01) and simplify the equation. The value of x is 0.0001.
Explanation:To solve the equation pH = 4 + log(x) and find the value of x when the hydrogen ion concentration is 0.01 M, we need to rearrange the equation. Subtracting 4 from both sides, we get log(x) = pH - 4. Using the fact that pH = -log[H+], we substitute the value of pH with its equivalent -log(0.01). Simplifying the equation, we have log(x) = -log(0.01) - 4. Taking the antilog (10^x) of both sides, we get x = 10^(-log(0.01) - 4). Evaluating this expression gives us x = 0.0001.
Air at 293 K and 750 mm Hg pressure has a relative humidity of 80%. What is its percent humidity? The vapour pressure of water at 293 K is 17.5 mm Hg. (a) 80.38 (b) 80 (c) 79.62 (d) 78.51
Answer : The correct option is, (c) 79.62
Explanation :
The formula used for percent humidity is:
[tex]\text{Percent humidity}=\text{Relative humidity}\times \frac{p-p^o_v}{p-p_v}\times 100[/tex] ..........(1)
The formula used for relative humidity is:
[tex]\text{Relative humidity}=\frac{p_v}{p^o_v}[/tex] ...........(2)
where,
[tex]p_v[/tex] = partial pressure of water vapor
[tex]p^o_v[/tex] = vapor pressure of water
p = total pressure
First we have to calculate the partial pressure of water vapor by using equation 2.
Given:
[tex]p=750mmHg[/tex]
[tex]p^o_v=17.5mmHg[/tex]
Relative humidity = 80 % = 0.80
Now put all the given values in equation 2, we get:
[tex]0.80=\frac{p_v}{17.5mmHg}[/tex]
[tex]p_v=14mmHg[/tex]
Now we have to calculate the percent humidity by using equation 1.
[tex]\text{Percent humidity}=0.80\times \frac{750-17.5}{750-14}\times 100[/tex]
[tex]\text{Percent humidity}=79.62\%[/tex]
Therefore, the percent humidity is 79.62 %
A candidate material for a turbine blade application oxidizes by diffusion of metal atoms through the oxide to the metal surface, where metal and oxygen react to form the oxide. After 10 hours at 550°C, an oxide layer 8 um has formed. What will the thickness be after 100 hours. (hint: turbine blade can be considered as a cylinder)
Answer:
The thickness of the oxide layer will 80 μm after 100 hours.
Explanation:
Thickness of oxide layer in 10 hours= 8 μm
Thickness of oxide layer in 1 hour = [tex]\frac{8 \mu m}{10 hour}[/tex]
Thickness of oxide layer in 100 hour :
[tex]\frac{8 \mu m}{10 hour}\times 100=80 \mu m[/tex]
The thickness of the oxide layer will 80 μm after 100 hours.
Match with the correct definition.
A.
Each electron state can hold no more than two electrons, which must have opposite spins.
B.
Electron sharing between two adjacent atoms such that each atom assumes a stable electron configuration.
C.
Electrostatic attraction between oppositely charged ions.
D.
The positively charged ion cores are shielded from one another, and also "glued" together by the sea of valence electrons.
Ionic bonding: A, B, C, or D?
Covalent Bonding: A,B,C or, D?
Metallic bonding: A, B, C, or D?
Pauli exclusion principle: A, B C, or D?
Answer:
Ionic bonding: C
Covalent bonding: B
Metallic bonding: D
Pauli exclusion principle: A
Explanation:
All the electrons in 1 atom are characterized by a series of 4 numbers, known as quantum numbers. These numbers (n, l, ml, ms) describe the state of each electron (in which level, sublevel, orbital it is and its spin). For 2 electrons to coexist in the same atom they must differ in at least of these numbers. If they occupy the same level, sublevel and orbital, then they must have different (and opposite) spins. This is known as Pauli exclusion principle.
Also, to gain stability atoms can gain, lose or share electrons. In doing so they form bonds. There are 3 kinds of bonds:
Ionic bonding: these are formed between metals and nonmetals. Metals tend to lose electrons and form cations (positive ions) and nonmetals tend to gain electrons and form anions (negative ions). Cations and anions attract each other due to electrostatic forces between oppositely charged ions.Covalent bonding: these are formed between nonmetals, which share pairs of electrons so as to reach the electron configuration of the closest noble gas (the most stable electron configuration).Metallic bonding: valence electrons are loose in metals, so they move together as a "sea of electrons", acting as "glue" of the remaining positive cores (electrons that are negative charges serve to attract the positive charges of the cores).Part B Compute 1240.64/64. Round the answer appropriately Express your answer numerically using the proper number of significant figures. View Available Hint(s) Submit
Answer:
19
Explanation:
1240.64/64
This is a simple division that can easily be evaluated using any form of calculator.
Here, the main challenge is in expressing the answer in the proper number of significant figure.
The approach here is to express the answer to the which is least precise. This is usually the number with the lower significant values.
Here, 64 has two significant figures and it is of a lower rank. Our answer should be expressed this way:
[tex]\frac{1240.64}{64}[/tex] = 19.385 = 19
Combustion of 8.9541 grams of C4H10 will yield (Example answer: 2.345) Do not enter units. grams of CO2. (Assume 100% yield)
Answer: The mass of carbon dioxide produced is 27.1 grams.
Explanation:
Combustion reaction is defined as the reaction in which a hydrocarbon reacts with oxygen gas to produce carbon dioxide gas and water molecule.
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
Given mass of butane = 8.9541 g
Molar mass of butane = 58.12 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of butane}=\frac{8.9541g}{58.12g/mol}=0.154mol[/tex]
The chemical equation for the combustion of butane follows:
[tex]2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_2O[/tex]
By Stoichiometry of the reaction:
2 moles of butane produces 8 moles of carbon dioxide
So, 0.154 moles of butane will produce = [tex]\frac{8}{2}\times 0.154=0.616mol[/tex] of carbon dioxide
Now, calculating the mass of carbon dioxide by using equation 1, we get:
Moles of carbon dioxide = 0.616 moles
Mass of carbon dioxide = 44 g/mol
Putting values in equation 1, we get:
[tex]0.616mol=\frac{\text{Mass of carbon dioxide}}{44g/mol}\\\\\text{Mass of carbon dioxide}=27.1g[/tex]
Hence, the mass of carbon dioxide produced is 27.1 grams.
Define "Homoglycan" and "Heteroglycan" and give an example of each
Answer:
Polysaccharides, also known as glycans, are the polymers of carbohydrate. These polymers are composed of the monosaccharide monomeric units that are joined together by glycosidic bonds.
A polysaccharide molecule composed of the same type of monosaccharide units is known as homopolysaccharide or homoglycan.
Example: Cellulose, glycogen, and cellulose.
Whereas, a polysaccharide molecule composed of more than type of monosaccharide is known as heteropolysaccharides or heteroglycans.
Examples: Hyaluronic acid and heparin
Will chromium metal dissolve in 10 M hydrochloric acid? a) yes b) no c) impossible to tell
Answer:
The correct option is: a) yes
Explanation:
Chromium is a chemical element that belongs to the group 6 of the periodic table and a member of the d-block. It is a hard and brittle transition metal having atomic number 24.
The hydrated chlorides of chromium can be obtained by dissolving the chromium metal in sulphuric acid or hydrochloric acid.
Name the following ternary acids. a. H2 SO4 e. HMnO4 i. HNO2 m. HClO4 b. H2 SO3 f. H2 CrO4 j. H2 CO3 n. HClO3 c. H2 S2 O3 g. H3 BO3 k. H2 C2 O4 o. HClO2 d. H3 PO4 h. HNO3 l. CH3 COOH p. HClO
Answer:
H2 SO4 (Sulfuric acid) - HMnO4 (Permanganic acid) - HNO2 (Nitrous acid) HClO4 (Perchloric acid) - H2 SO3 (Sulphurous acid) - H2CrO4 (Chromic acid) H2CO3 (Carbonic acid) - HClO3 (Chloric acid) - H3BO3 (boric acid) - HClO2 (Chlorous acid) H3PO4 (Phosphoric acid) HNO3 (Nitric acid) - HClO (Hypochlorous acid) - CH3 COOH (acetic acid) - H2S2O3 (Thiosulfuric acid)- H2C2O4 (oxalic acid)
Explanation:
To name the ternary acid, they have to obbey the following formula
Hx - Non metal - Oy
where the oxidation state in H and O are +1 y -2 respectively.
When the oxidation state of the central Non metal is odd, the atomicity of H is 1 and in the O is deduced in such a way that the sum of oxidation states is 0.
When the oxidation state of the central No metal is even, the atomicity of H is 2 and in the O is deduced in such a way that the sum of oxidation states is 0.
IV V VI VII
- - 1 Hypo ……. Ous
- 3 4 3 … ous
4 5 6 5 … ic
7 Per …… Ous
The oxalic acid is an organic compound of two carboxyl groups, so it is also called ethanedioc acid; It is actually known as oxalic by some plants of the genus oxalis. (prefix et (2 carbons))
The acetic acid that comprises a carboxylic group and a methyl group is also an organic compound. It is popularly known as acetic acid but it is actually called methylcarboxylic acid or having two carbons, ethanoic acid.
List in order the 9 most electronegative elements of the periodic table
Answer:
F (3.98)> O (3.44)> Cl (3.16)> N (3.04)> Kr (3.00)> Br (2.96)> I (2.66)> Xe (2.6)> S(2.58)
Explanation:
Electronegativity is described as the tendency or ability of an element to attract electron density or bond pair of electrons towards itself. It is a chemical property of an element.
In the periodic table, the least electronegative element is caesium (0.79) and the most electronegative element is fluorine (3.98).
List of 9 most electronegative elements are:
F (3.98)> O (3.44)> Cl (3.16)> N (3.04)> Kr (3.00)> Br (2.96)> I (2.66)> Xe (2.6)> S(2.58).
Use the Ref An aqueous solution of chromium(II) acetate has a concentration of 0.260 molal. The percent by mass of chromium(II) acetate in the solution is Submit Answer Retry Entire Group 9 more group attempts remaining
Answer:
The percent by mass of chromium(II) acetate in the solution is 4.42%.
Explanation:
Molality of the chromium(II) acetate solution = 0.260 m = 0.260 mol/kg
This means that in 1 kg of solution 0.260 moles of chromium(II) acetate are present.
1000 g = 1 kg
So, in 1000 grams of solution 0.260 moles of chromium(II) acetate are present.
Then in 100 grams of solution = [tex]\frac{0.260 mol}{1000}\times 100=0.0260 mol[/tex]
Mass of 0.0260 moles chromium(II) acetate:
= 0.0260 mol × 170 g/mol = 4.42 g
[tex](w/w)\%=\frac{\text{mass of solute in 100 gram solution}}{100}\times 100[/tex]
[tex]=\frac{4.42 g}{100 g}\times 100=4.42\%[/tex]
The percent by mass of chromium(II) acetate in the solution is 4.42%.
You have 15.42g of C2H6. How many moles of H2O can be made?
Answer: The moles of water produced are 1.54 moles.
Explanation:
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 ethane = 15.42 g
Molar mass of ethane = 30.07 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of ethane}=\frac{15.42g}{30.07g/mol}=0.513mol[/tex]
The chemical equation for the combustion of ethane follows:
[tex]2C_2H_6+5O_2\rightarrow 4CO2+6H_2O[/tex]
By Stoichiometry of the reaction:
2 moles of ethane produces 6 moles of water
So, 0.513 moles of ethane will produce = [tex]\frac{6}{2}\times 0.513=1.54mol[/tex] of water
Hence, the moles of water produced are 1.54 moles.
In 1983, a pilot miscalculated the fuel requirement for Air Canada Fight 143 from Montreal to Edmonton. Halfway through the flight, the plane ran out of fuel because of an English-metric conversion error. Suppose that based on the distance and conditions, the flight would require at least 2.2 x 104 kg of fuel. The plane started off having 5, 756 liters in the tank when the pilot arrived. If there are 803 g of jet fuel in 1 liter, determine how much fuel (in liters) would be needed for a successful trip.
Answer:
27,397.23 L would be needed for a successful trip.
Explanation:
The problem gives us the density (ρ) of the fuel, by telling us that there are 803 g of fuel in 1 L, in which case:
ρ=[tex]\frac{mass}{Volume}=\frac{803g}{1L} =803\frac{g}{L}[/tex]
The required mass of fuel is 2.2 * 10⁴ kg, we can convert this value into g:
2.2 * 10⁴ kg * [tex]\frac{1000g}{1kg}[/tex] = 2.2 * 10⁷ g
We calculate the required volume (V), using the mass and density:
[tex]803 g/L = \frac{2.2*10^{7}g }{V} \\V=\frac{2.2*10^{7}g }{803g/L}\\ V=27397.26 L[/tex]
Thus 27,397.23 L would be needed for a successful trip.
Tris is a weak base. The basic form of tris can be written tris0 and the protonated (conjugate acid) form can be written tris. The pKa of tris is 8.1. a. What is the ratio of the basic and the acidic forms of tris at pH 6.1? b. What is the useful buffering range of tris?
Answer:
a) [Tris0] : [Tris] = 1 : 100
b) Range = 7.1 to 9.1
Explanation:
a) Calculation of ratio of the basic and the acidic forms of tris
pH of a buffer is calculate using Henderson-Hasselbalch equation
[tex]pH = pKa+log\frac{Salt}{Acid}[/tex]
Conjugate acid of Tris dissociated as
[tex]Tris \leftrightharpoons Tris0 + H^+[/tex]
For tris,
Salt or Basic form = tris0
Acid or Acidic form = Tris
pKa = 8.1
pH = 6.1
[tex]pH = pKa+log\frac{Tris0}{Tris}[/tex]
[tex]6.1 = 8.1+log\frac{Tris0}{Tris}[/tex]
[tex]log\frac{Tris0}{Tris} = -2[/tex]
[tex]\frac{Tris0}{Tris} = antilog (-2)[/tex]
[tex]\frac{Tris0}{Tris} = 10^{-2}[/tex]
[Tris0] : [Tris] = 1 : 100
b) Range of Tris
Range of any buffer is:
From (pKa -1) to (pKa+1)
So, range of Tris is:
From (8.1 - 1) to (8.1 +1)
or from 7.1 to 9.1
The oxidation of the sugar glucose, C6H12O6, is described by the following equation. C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(l) ΔH = −2802.5 kJ/mol The metabolism of glucose gives the same products, although the glucose reacts with oxygen in a series of steps in the body.
(a) How much heat in kilojoules can be produced by the metabolism of 18.1 g of glucose?
(b) How many Calories can be produced by the metabolism of 18.1 g of glucose?
Answer:
(a) 282 kJ
(b) 67.4 Calories
Explanation:
(a) The molar enthalpy, ΔH = −2802.5 kJ/mol, means that the heat produced by the reaction is 2802.5 kJ per mol of glucose.
We can multiply the enthalpy by the number of moles of glucose to get the heat produced by the metabolism. Grams of glucose will be converted to moles using the molar mass of glucose (180.156 g/mol):
(18.1 g)(mol/180.156g)(2802.5 kJ/mol) = 282 kJ
(b) Using the result we obtained above, kJ will be converted to Calories using the conversion factor of 4.184J = 1 cal. Calorie with a capital C is the same as a kilocalorie.
(282 kJ)(1 cal/4.184J) = 67.4 kcal = 67.4 Calories
Answer:
For a: The amount of heat produced for given amount of glucose is -283.05 kJ
For b: The amount of heat produced for given amount of glucose is -67648.9 Cal
Explanation:
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 glucose = 18.1 g
Molar mass of glucose = 180.16 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of glucose}=\frac{18.1g}{180.16g/mol}=0.101mol[/tex]
The given chemical reaction follows:
[tex]C_6H_{12}O_6(s)+6O_2(g)\rightarrow 6CO_2(g)+6H_2O(l);\Delta H=-2802.5kJ/mol[/tex]
For a:By Stoichiometry of the reaction:
When 1 mole of glucose is reacted, the amount of heat released is 2802.5 kJ
So, when 0.101 moles of glucose is reacted, the amount of heat released is [tex]\frac{2802.5}{1}\times 0.101=283.05kJ[/tex]
Hence, the amount of heat produced for given amount of glucose is -283.05 kJ
For b:To convert the heat produced in kilo joules to calories, we use the conversion factor:
1 kJ = 239 Cal
So, [tex]-283.05kJ\times (\frac{239Cal}{1kJ})=-67648.9Cal[/tex]
Hence, the amount of heat produced for given amount of glucose is -67648.9 Cal
We put 50 g of ice of 0 °C in a beaker containing 450 g of beer of 20 °C, mix, and wait for heat exchange within the beaker. What will be the final temperature?
Answer : The final temperature will be, 292 K
Explanation :
In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.
[tex]q_1=-q_2[/tex]
[tex]m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)[/tex]
where,
[tex]c_1[/tex] = specific heat of ice = [tex]2.05J/g.K[/tex]
[tex]c_2[/tex] = specific heat of beer = [tex]4.2J/g.K[/tex]
[tex]m_1[/tex] = mass of ice = 50 g
[tex]m_2[/tex] = mass of beer = 450 g
[tex]T_f[/tex] = final temperature = ?
[tex]T_1[/tex] = initial temperature of ice = [tex]0^oC=273+0=273K[/tex]
[tex]T_2[/tex] = initial temperature of beer = [tex]20^oC=273+20=293K[/tex]
Now put all the given values in the above formula, we get:
[tex]50g\times 2.05J/g.K\times (T_f-273)K=-450g\times 4.2J/g.K\times (T_f-293)K[/tex]
[tex]T_f=291.971K\approx 292K[/tex]
Therefore, the final temperature will be, 292 K
Just as the depletion of stratospheric ozone threatens life on Earth today, its accumulation was one of the crucial processes that allowed life to develop in prehistoric times: 3O2(g) → 2O3(g) At a given instant, the reaction rate in terms of [O2] is 2.17 × 10−5 mol/L·s. What is it in terms of [O3]? Enter your answer in scientific notation.
Answer:
Rate in terms of formation of [tex]O_{3}[/tex] is [tex]1.45\times 10^{-5}mol/L.s[/tex]
Explanation:
According to law of mass action for this reaction: Rate = [tex]-\frac{1}{3}\frac{\Delta [O_{2}]}{\Delta t}=\frac{1}{2}\frac{\Delta [O_{3}]}{\Delta t}[/tex][tex]-\frac{\Delta [O_{2}]}{\Delta t}[/tex] represents rate of disappearance of [tex]O_{2}[/tex] and [tex]\frac{\Delta [O_{3}]}{\Delta t}[/tex] represents rate of formation of [tex]O_{3}[/tex]Here, [tex]-\frac{\Delta [O_{2}]}{\Delta t}=2.17\times 10^{-5}mol/L.s[/tex]So, [tex]\frac{\Delta [O_{3}]}{\Delta t}=\frac{2}{3}\times -\frac{\Delta [O_{2}]}{\Delta t}=\frac{2}{3}\times (2.17\times 10^{-5}mol/L.s)=1.45\times 10^{-5}mol/L.s[/tex]Hence rate in terms of formation of [tex]O_{3}[/tex] is [tex]1.45\times 10^{-5}mol/L.s[/tex]The rate of formation is the time taken by the reaction to yield the product by the chemical change in the reactants. The rate of the formation of the ozone is [tex]1.45 \times 10^{-5} \;\rm mol/Ls[/tex].
What is the law of mass action?The law of mass action states that the rate of the reaction is proportional to the product of the reactant masses.
Rate according to the law of mass action:
[tex]\rm -\dfrac{1}{3}\dfrac {\Delta[O_{2}]}{\Delta t} = \rm \dfrac{1}{2}\dfrac{\Delta [O_{3}]}{\Delta t}[/tex]
Here,
Rate of disappearance of oxygen [tex](\rm -\dfrac {\Delta[O_{2}]}{\Delta t} ) = 2.17 \times 10^{-5} \;\rm mol/Ls[/tex] Rate of formation of ozone =[tex]\rm \dfrac{\Delta [O_{3}]}{\Delta t}[/tex]Substituting values in the above equation:
[tex]\begin{aligned}\rm \dfrac{\Delta [O_{3}]}{\Delta t} &= \dfrac{2}{3}\times - \rm \dfrac{\Delta [O_{2}]}{\Delta t}\\\\&= \dfrac{2}{3} \times (2.17 \times 10^{-5})\\\\&= 1.45 \times 10^{-5}\;\rm mol/Ls\end{aligned}[/tex]
Therefore, the rate of formation in the terms of ozone is [tex]1.45 \times 10^{-5} \;\rm mol/Ls.[/tex]
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What is the approximate buffering range of formic acid? a. You have a formic acid/potassium formate solution that has a pH of 4.75. What is the ratio of the formate to the formic acid? b.
Answer:
a. The buffering range is between 2.74 and 4.74.
b. The ratio of the formate to the formic acid is 10.23.
Explanation:
a. For every buffer solution, the optimal effective range is pH = pKa ± 1. Outside this range, the buffer does not work properly.
For the formic acid, the pKa is 3.74, thus the optimal range is between 2.74 and 4.74.
b. The Henderson-Hasselbalch equation is a chemical expression used to calculate the pH of a buffer knowing the ratio of the acid to base, or to calculate the ratio knowing the pH. The expression is:
[tex]pH = pKa + Log \frac{[A^{-}]}{[HA]}[/tex]
where [A^{-}] is the concentration of the conjugate base and [HA] is the concentration of the acid.
For a formic acid/potassium formate solution that has a pH of 4.75 and pka of 3.74:
[tex]pH - pKa =4.75 - 3.74 = 1.01 = Log \frac{[A^{-}]}{[HA]}[/tex]
[tex]\frac{[A^{-}]}{[HA]} = 10^{1.01} = 10.23[/tex]
Since density depends on the mass and volume of an object, we need both of these values combined in the correct way to solve for density. Keep in mind that we may not be given the mass and volume directly, or the values may not be given in the units we want and we need to convert to the correct units before solving for density.
When converting between metric units, use the prefixes to help you determine the magnitude of a value. The prefix k indicates kilo, 1000. Therefore 1000 g = 1 kg.
A 11.0 kg iron weightlifting plate has a volume of 1400 cm3 . What is the density of the iron plate in g/cm3?
Express your answer in grams per cubic centimeter to three significant figures.
Answer:
Density=7.86g/cm³
Explanation:
To solve this exercise, we only need to convert units. To do so, we can use a three-step process:
Identify the units that the problem gives us.Identify the units that we must report in the answer.Use the appropiate conversion factor in order to convert the units in 1.) to the units in 2.)In this case, the units we're given are kg and cm³, and the units we must convert them into are g and cm³. Keeping in mind the conversion factor of 1000 g = 1 kg, we convert the mass of the plate:
[tex]11.0kg*\frac{1000g}{1kg}=11000 g[/tex]
Then we divide the mass by the volume, that is already in cubic centimeters:
[tex]Density=\frac{11000g}{1400cm^{3}}=7.86\frac{g}{cm^{3} }[/tex]
A 76.00 pound flask of mercury costs $145.00. The density of mercury is 13.534 g/cm3. Find the price of one cubic inch of mercury by calculating intermediate values. What is the price of one pound of mercury?
The price of one cubic inch of mercury is approximately $0.9308, and the price of one pound of mercury is approximately $1.9079.
Using the above data, we must perform several calculations to determine the cost of one cubic inch of mercury and one pound of mercury. Here's a step-by-step guide to do it:
Given:
Flask of mercury mass = 76.00 poundsCost of flask of mercury = $145.00Density of mercury = 13.534 g/cm³We will convert the mass of the flask of mercury to grams
1 pound (lb) = 453.59237 grams (g)
Mass of flask of mercury in grams = 76.00 lb * 453.59237 g/lb
Then, calculate the volume of mercury in cm³
Density of mercury = Mass / Volume
Volume of mercury in cm³ = Mass of flask of mercury in grams / Density of mercury
Then, we will calculate the volume of mercury in cubic inches
1 cm³ = 0.0610237 in³ (approximately)
Volume of mercury in cubic inches = Volume of mercury in cm³ * 0.0610237
Calculate the price of one cubic inch of mercury
Price of one cubic inch of mercury = Cost of flask of mercury / Volume of mercury in cubic inches
We will calculate the price of one pound of mercury
Price of one pound of mercury = Cost of flask of mercury / Mass of flask of mercury in pounds
We will perform the calculations:
Mass of flask of mercury in grams = 76.00 lb * 453.59237 g/lb = 34473.53 g
Volume of mercury in cm³ = 34473.53 g / 13.534 g/cm³ = 2549.915 cm³
Volume of mercury in cubic inches = 2549.915 cm³ * 0.0610237 in³/cm³ = 155.61 in³
Price of one cubic inch of mercury = $145.00 / 155.61 in³ ≈ $0.9308/in³
Price of one pound of mercury = $145.00 / 76.00 lb ≈ $1.9079/lb
Therefore, the price of one cubic inch of mercury is approximately $0.9308, and the price of one pound of mercury is approximately $1.9079.
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The price of one pound of mercury can be calculated by dividing the total cost of $145.00 by the total weight, which is 76 pounds, resulting in a price of $1.9079 per pound.
Explanation:To find the price per pound of mercury, we first need to find the total mass of mercury in pounds. Given that a 76.00 pound flask of mercury costs $145.00, the price per pound can be calculated directly by dividing the total cost by the total weight in pounds.
Price per pound of mercury = Total cost / Total weight in pounds
Price per pound of mercury = $145.00 / 76.00 pounds
Price per pound of mercury = $1.9079 (rounded to four decimal places)
Note that we do not need the density of mercury for this specific calculation since the total weight and cost are directly given.
You are performing a chemical reaction in a test tube. The test tube gets colder as the reaction takes place. This chemical reaction is... O a.a double replacement reaction. O b.endothermic Oc.a single replacement reaction. O d.a decomposition reaction. e. exothermic
Answer: Option (b) is the correct answer.
Explanation:
A chemical reaction in which heat energy is absorbed by the reactant molecules is known as an endothermic reaction.
Therefore, upon completion of this reaction the container in which reaction is carried out becomes colder.
A chemical reaction in which heat energy is released by the reactant molecules is known as an exothermic reaction.
Therefore, upon completion or during this type of reaction the container in which reaction is carried out becomes hot.
Thus, we can conclude that when the test tube gets colder as the reaction takes place then it means this chemical reaction is endothermic reaction.
calculate pressure (at STP) in 10.0L vessel after reaction of 1.0L hydrochloric acid (concentration 35% and density 1.28 g/cub.cm.) with 1.0kg of limestone (calcium carbonate with 4.0% inert impurities). Yield of reaction 97%
Answer:
The pressure in the vessel is 13,3 atm.
Explanation:
The global reaction is:
2 HCl (aq)+ CaCO₃ (s) → CaCl₂(aq)+ H₂O(l)+ CO₂(g)
The increase in the pressure of the vessel after the reaction is by formation of a gas (CO₂). So we have to find the produced moles of this gas and apply the gas ideal law to find the pressure.
We have to find the limit reactant, to do so, we have to calculate the moles of each reactant in the reaction, the one that have the less moles will be the limit reactant:
HCl:
1,0L × (35/100) × (1000 cm³/1L) × (1,28 g/ 1cm³) × (1mol HCl/ 36,46 g) ÷ 2mol
(Concentration) (L to cm³) (cm³ to g) (g to mol) (moles of reaction)
moles of HCl= 6,14 mol
CaCo₃:
1,0 kg × (96/100) × (1000 g/1kg) × (1 mol/100,09g)
(Limestone) (CaCo₃ in limestone) (kg to g) (g to mol)
moles of CaCo₃= 9,59 mol
So, reactant limit is HCl
This reaction have a yield of 97%. So, the CO₂ moles are:
6,14 mol × 97÷ = 5,96 mol CO₂
The ideal gas formula to obtain pressure is:
P = nRT/V
Where: n = 5,96mol; R= 0,082 atm×L/mol×K; T = 273,15 (until STP conditions) and V= 10,0 L
Replacing this values in the equation the pressure is
P = 13,3 atm
I hope it helps!
Of the two type of bonds (pi and sigma), which is capable of rotation around that bond and which is not? Why?
Answer:
The sigma bonds are capable of rotating, the pi bonds not.
Explanation:
Sigma bonds are the strongest type of bonds, there are related to the overlapping in the atomic orbitals, and they can rotate. In the pi bound (that is a double bond), there are electrons moving on the molecule and it is not permitted the rotation on this type of bonds. Of the sigma bonds are capable of rotating while the pi bonds not.