Answer:
Explanation:
From the given information:
The surface area of the can = (30 × 20 × 2) +(20× 15 × 2) +(30 × 15 × 2)
= 1200 + 600 + 900
= 2700 cm²
Since 1 inch² = 0.155 inch²
The surface area in inches² = 2700 × 0.155 inch²
= 418.5 inches²
The total force can be determined by using the expression:
Force = Pressure ×Area
Force = 14.7 psi × 418.5 inches²
Force = 6151.95 lbs
Yes, the gasoline can will be able to withstand the force.
Tema: Métodos de Separação de Misturas – Homogêneas e Heterogêneas;
1. Capa (0,5 ponto)
2. Índice ou Sumário (0,5 ponto)
3. Texto do trabalho
a) Introdução (1,0 ponto)
b) Objetivos (0,5 ponto)
c) Método (0,5 ponto)
d) Desenvolvimento: Fundamentação Teórica (5,0 pontos)
e) Conclusão (1,0 ponto)
4. Bibliografia (1,0 ponto)
Answer:
fjskeowkcnekvo Dee five votes come vote for dog even r
Solutions of Cu2+ turn blue litmus red because of the equilibrium: Cu(H2O)62+(aq) + H2O(l) ↔ Cu(H2O)5(OH)+(aq) + H3O+(aq) for which Ka = 1.0 x 10-8. Calculate the pH of 0.10 M Cu(NO3)2(aq).
Answer: The pH of 0.10 M [tex]Cu(NO_{3})_{2}(aq)[/tex] is 4.49.
Explanation:
Given: Initial concentration of [tex]Cu(H_{2}O)^{2+}_{6}[/tex] = 0.10 M
[tex]K_{a} = 1.0 \times 10^{-8}[/tex]
Let us assume that amount of [tex]Cu(H_{2}O)^{2+}_{6}[/tex] dissociates is x. So, ICE table for dissociation of [tex]Cu(H_{2}O)^{2+}_{6}[/tex] is as follows.
[tex]Cu(H_{2}O)^{2+}_{6} \rightleftharpoons [Cu(H_{2}O)_{5}(OH)]^{+} + H_{3}O^{+}[/tex]
Initial: 0.10 M 0 0
Change: -x +x +x
Equilibrium: (0.10 - x) M x x
As the value of [tex]K_{a}[/tex] is very small. So, it is assumed that the compound will dissociate very less. Hence, x << 0.10 M.
And, (0.10 - x) will be approximately equal to 0.10 M.
The expression for [tex]K_{a}[/tex] value is as follows.
[tex]K_{a} = \frac{[Cu(H_{2}O)^{2+}_{6}][H_{3}O^{+}]}{[Cu(H_{2}O)^{2+}_{6}]}\\1.0 \times 10^{-8} = \frac{x \times x}{0.10}\\x = 3.2 \times 10^{-5}[/tex]
Hence, [tex][H_{3}O^{+}] = 3.2 \times 10^{-5}[/tex]
Formula to calculate pH is as follows.
[tex]pH = -log [H^{+}][/tex]
Substitute the values into above formula as follows.
[tex]pH = -log [H^{+}]\\= - log (3.2 \times 10^{-5})\\= 4.49[/tex]
Thus, we can conclude that the pH of 0.10 M [tex]Cu(NO_{3})_{2}(aq)[/tex] is 4.49.
Identify the oxidation half-reaction for this reaction:
Fe(s) + 2HCl(aq) → FeCl2(aq) + H2(g)
A. Fe2+ + 2e → Fe(s)
O B. H2(g) → 2H+ + 2e
O C. Fe(s) → Fe2+ + 2e
O D. 2H+ + 2e → H2(9)
Answer:
Fe(s)->Fe2+2e-
Explanation:
A.p.e.x
The oxidation half-reaction for the given reaction is Fe(s) → Fe²⁺ + 2e⁻ Hence, Option (C) is correct
What is Oxidation reaction ?Oxidation reaction is a chemical reaction which can be described as follows ;
Addition of oxygen Removal of hydrogen Loss of ElectronAddition of electronegative atomRemoval of Electropositive elementIn the given reaction ;
Fe(s) + 2HCl(aq) → FeCl₂(aq) + H₂(g)
Fe at RHS got converted to Fe²⁺ state at LHS which shows the gain of electron by Fe with in the reaction.
Therefore,
The oxidation half-reaction for the given reaction is Fe(s) → Fe²⁺ + 2e⁻ Hence, Option (C) is correct
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Bond length is the distance between the centers of two bonded atoms. On the potential energy curve, the bond length is the internuclear distance between the two atoms when the potential energy of the system reaches its lowest value.
a. True
b. False
Answer:
True
Explanation:
When two atoms are at infinite distance from each other, the both atoms posses high energy.
However, as they begin to approach each other, the distance between them gradually decreases and so does their energy.
A point is eventually reached when the potential energy curve reaches its minimum value. The internuclear distance between the two atoms at this point is called the bond length of the system.
the carbon tetrachloride molecule CCI 4 is
Answer:
is a nonpolar molecule with polar bonds
DATA SHEET p 45. TRIAL 1 TRIAL 2 1. Mass of the ground pretzel 1.00 gram 1.03 g 2. Initial volume of the AgNO3 solution 0.00 mL 9.10 mL 3. Final volume of the AgNO3 solution 9.10 mL 17.25 mL 4. Volume of AgNO3 solution used 9.10 mL 8.15 mL Line 3 – Line 2 5. Volume of AgNO3 solution in liters _____ L _____ L 6. Molarity of AgNO3 solution 0.01 M 0.01 M (given) 7. Number of moles of AgNO3 ______ mol _____ mol (Line 5 × Line 6) 8. Number of mol of NaCl present in pretzel ______ mol _____ mol (Line 7) number of mol NaCl = number of mol AgNO3 9. Mass of NaCl present in the titrated sample ______ gram _____ gram (Line 8) × 58.5 g/mol
Answer:
1. 1.00 gm
2. 50 ml
3. 38.93 ml
4. 11.07 ml
5. 0.01107 L
6. 0.010 moles / L
7. 0.0001107 moles
8. 0.0001107 moles
9. 0.00647042 grams
Explanation:
Silver nitrate can react with various compounds to form different products. The weight of products may be different from the original solution introduced due to combustion reaction, as heat energy is released during the chemical process.
When 4.41g of phosphoric acid (H3PO4) react with 9.25g of barium hydroxide, water and insoluble barium phosphate form. [T/I-7] a. Write and balance the chemical equation.
Answer:
2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + 6 H₂O(l)
Explanation:
Let's consider the unbalanced equation that occurs when phosphoric acid reacts with barium hydroxide to form water and barium phosphate. This is a neutralization reaction.
H₃PO₄(aq) + Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + H₂O(l)
We will balance it using the trial and error method.
First, we will balance Ba atoms by multiplying Ba(OH)₂ by 3 and P atoms by multiplying H₃PO₄ by 2.
2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + H₂O(l)
Finally, we will get the balanced equation by multiplying H₂O by 6.
2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + 6 H₂O(l)
An electrolysis reaction is
A)
exothermic.
B)
hydrophobic.
C)
spontaneous.
D)
non-spontaneous.
Answer:
the electrolysis reaction is a non- spontaneous reaction
Explanation:
Since electrons flow from it, the anode in an electrolytic cell is positive, while the cathode is negative when electrons flow into it. The device functions like a galvanic cell in that direction. In an electrolytic cell, an external voltage is applied and that is what causes a non spontaneous reaction
Write the symbol for every chemical element that has atomic number greater than 70 and atomic mass less than 185.2
Answer:
HF...Ta... W....Lu...
Si enfriamos mercurio de 100C. Calcular la cantidad de calor que se debe restar sabiendo que la masa de mercurio es de 1800gr
Answer:
I do not speak Spanish.
Explanation:
How many atoms are in .45 moles of P4010
Answer:
5×6.02×1023
Explanation:
there are 5×6.02×1023 molecules of p4010 in 5mole. there are four P atoms in a single molecule of p4010
2.50 L of 0.700 M phosphoric acid reacts with 5.25 moles of sodium hydroxide. How many moles of hydrogen ions will completely neutralize the moles of hydroxide ions present in this amount of sodium hydroxide? a) 0.583 b) 1.75 c) 3.00 d) 15.75 e) 5.25
Answer:
5.25 moles of protons. Option e
Explanation:
Reaction between phosphoric acid and sodium hydroxide is neutralization.
We can also say, we have an acid base equilibrium right here:
H₃PO₄ + 3NaOH → Na₃PO₄ + 3H₂O
Initially we have 5.25 moles of base.
We have data from the acid, to state its moles:
M = mol/L, so mol = M . L
mol = 1.75 moles of acid
If we think in the acid we know:
H₃PO₄ → 3H⁺ + PO₄⁻³
We know that 1 mol of acid can give 3 moles of protons (hydrogen ions)
If we have 1.75 moles of acid, we may have
(1.75 . 3) /1 = 5.25 moles of protons
These moles will be neutralized by the 5.25 moles of base
H₃O⁺ + OH⁻ ⇄ 2H₂O Kw
In a titration of a weak acid and a strong base, we have a basic pH
A molecule of acetone and a molecule of propyl aldehyde are both made from 3 carbon atoms, 6 hydrogen atoms, and 1 oxygen atom. The molecules differ in their arrangement of atoms. How do formulas for the two compounds compare? Both compounds have the same molecular formula, but have unique structural formulas. Both compounds have unique molecular formulas and structural formulas. Both compounds have the same structural formula, but have unique molecular formulas.
Explanation:
The structures of both acetone and propanal are shown below:
In the formula of propanal there is -CHO functional group at the end.
In acetone -CO- group is present in the middle that is on the second carbon.
The molecular formula is C3H6O.
Both have same molecular formula but different structural formulas.
C. A sample may contain any or all of the following ions Hg2 2, Ba 2, and Al 3. 1) No precipitate forms when an aqueous solution of NaCl was added to the sample solution. 2) No precipitate forms when an aqueous solution of Na2SO4 was added to the sample. 3) A precipitate forms when the sample solution was made basic with NaOH. Which ion or ions were present. Write the net ionic equation(s) for the the reaction (s).
Answer:
Al^3+
Explanation:
Solubility rules tell us what substances are soluble in water. Since NaCl was added and no precipitate was observed, the mercury II ion is absent.
Addition of Na2SO4 does not form a precipitate meaning that Ba^2+ is absent.
If a precipitate is formed when NaOH is added, the the ionic reaction is as follows;
Al^3+(aq) + 3 OH^-(aq) ------> Al(OH)3(s)
An endothermic reaction will start when the required
energy is received from the environment or solution.
AH
activation
thermal
kinetic
Answer:
A: ΔH
Explanation:
Endothermic reactions are this that occur as a result of absorption of heat energy from the surroundings by the reactants to form new products.
Thus, we can say it is one with an increase in enthalpy (ΔH) of the system.
Thus, option A is correct.
9. How can you separate sugar from a sugar solution contained in a glass without taste? Explain
Answer:
See explanation
Explanation:
Sugar is a polar crystalline substance. The sugar crystal is capable of dissolving in water since it is polar.
When sugar dissolves in water, a sugar solution is formed. If I want to separate the sugar from the water in the solution, I have to boil the solution to a very high temperature.
When I do that, the water in the sugar solution is driven off and the pure sugar crystal is left behind.
When butane reacts with Br2 in the presence of Cl2, both brominated and chlorinated products are obtained. Under such conditions, the usual selectivity of bromination is not observed. In other words, the ratio of 2-bromobutane to 1-bromobutane is very similar to the ratio of 2-chlorobutane to 1-chlorobutane. Can you offer and explanation as to why we do not observe the normal selectivity expected for bromination
Answer:
Bromine radical formation is carried out in the presence of Br₂ and Cl₂ causing the normal selectivity not to be observed ( this causes the difference in activation energy to be reduced )
Explanation:
Why the normal selectivity expected for bromination is not observed
On the basis of selectivity and applying the Arrhenius equation the greater the difference between the activation energies the more the selectivity.
as seen in the formation of primary and secondary radicals in the Bromine radical formation. this difference is caused mainly by the propagation step ( exothermic ) . But the main reason why the the usual selectivity of bromination is not observed is because it Bromine radical formation is carried out in the presence of Br₂ and Cl₂ ( this causes the difference in activation energy to be reduced )
Calculate percent yield when you start with 0.78 grams of camphor and end with 0.23 grams of iso/borneol. The molecular weight of camphor is 152.23 g/mol, and the molecular weight of iso/borneol is 154.25 g/mol.
Answer:
29.1%
Explanation:
First we convert 0.78 g of camphor to moles, using its molar mass:
0.78 g ÷ 152.23 g/mol = 0.00512 mol camphorThen we convert 0.23 g of isoborneol to moles, using its molar mass:
0.23 g ÷ 154.25 g/mol = 0.00149 mol isoborneolFinally we calculate the percent yield:
0.00149 mol / 0.00512 mol * 100% = 29.1%0.50 g of hydrogen chloride (HCl) is dissolved in water to make 4.0 L of solution. What is the pH of the resulting hydrochloric acid solution
Explanation:
Given the mass of HCl is ---- 0.50 g
The volume of solution is --- 4.0 L
To determine the pH of the resulting solution, follow the below-shown procedure:
1. Calculate the number of moles of HCl given by using the formula:
[tex]number of moles of a substance=\frac{given mass of the substance}{its molecular mass}[/tex]
2. Calculate the molarity of HCl.
3. Calculate pH of the solution using the formula:
[tex]pH=-log[H^+][/tex]
Since HCl is a strong acid, it undergoes complete ionization when dissolved in water.
[tex]HCl(aq)->H^+(aq)+Cl^-(aq)[/tex]
Thus, [tex][HCl]=[H^+][/tex]
Calculation:
1. Number of moles of HCl given:
[tex]number of moles of a substance=\frac{given mass of the substance}{its molecular mass}\\=0.50g/36.5g/mol\\=0.0137mol[/tex]
2. Concentration of HCl:
[tex]Molarity of HCl=\frac{number of moles of HCl}{its molar mass}\\=\frac{0.0137 mol}{4.0 L} \\= 0.003425 M[/tex]
3. pH of the solution:
[tex]pH=-log[H^+]\\=-log(0.003425)\\=2.47[/tex]
Hence, pH of the given solution is 2.47.
Which of the following are examples of physical properties of ethanol? Select all that apply.
The boiling point is 78.37°C
It is a clear, colorless liquid
It is flammable
It is a liquid at room temperature
A gas mixture, with a total pressure of 300. torr, consists of equal masses of Ne (atomic weight 20.) and Ar (atomic weight 40.). What is the partial pressure of Ar, in torr
Answer:
The partial pressure will be "100 torr".
Explanation:
Given:
[tex]P_{Ar} = 300 \ torr[/tex]
By assuming Ar and Ne having 50 gm each, we get
mol of Ne = [tex]\frac{50}{20}[/tex]
= [tex]2.5 \ mol[/tex]
mol of Ar = [tex]\frac{50}{40}[/tex]
= [tex]1.25 \ mol[/tex]
now,
[tex]n_T= mol.A_r+mol.N_e[/tex]
[tex]=1.25+2.5[/tex]
[tex]=3.75[/tex]
then,
[tex]X_{Ar}=\frac{n_{Ar}}{n_T}[/tex]
[tex]=\frac{1.25}{3.75}[/tex]
[tex]=0.33[/tex]
hence,
The partial pressure of Ar will be:
⇒ [tex]P_{Ar} = P_T\times X_{AT}[/tex]
By substituting the values, we get
[tex]=300\times 0.33[/tex]
[tex]=100 \ torr[/tex]
The partial pressure of Ar in the mixture is 99.9 torr
Let the mass of both gas be 10 g
Next, we shall determine mole of each gas.
For Ne:Mass = 10 g
Molar mass of Ne = 20 g/mol
Mole of Ne =?Mole = mass / molar mass
Mole of Ne = 10 / 20
Mole of Ne = 0.5 mole For Ar:Mass = 10 g
Molar mass of Ar = 40 g/mol
Mole of Ar =?Mole = mass / molar mass
Mole of Ar = 10 / 40
Mole of Ar = 0.25 moleNext, we shall determine the mole fraction of Ar
Mole of Ne = 0.5 mole
Mole of Ar = 0.25 mole
Total mole = 0.5 + 0.25 = 0.75 mole
Mole fraction of Ar =?[tex]mole \: fraction \: = \frac{mole}{total \: mole} \\ \\ mole \: fraction \: of \:Ar = \frac{0.25}{0.75} \\ \\ mole \: fraction \: of \:Ar = 0.333 \\ \\ [/tex]
Finally, we shall determine the partial pressure of Ar
Mole fraction of Ar = 0.333
Total pressure = 300 torr
Partial pressure of Ar =?Partial pressure = mole fraction × total pressure
Partial pressure of Ar = 0.333 × 300
Partial pressure of Ar = 99.9 torrLearn more on partial pressure: https://brainly.com/question/15577259
5. A beam of photons with a minimum energy of 222 kJ/mol can eject electrons from a potassium surface. Estimate the range of wavelengths of light that can be used to cause this phenomenon. Show your calculations with units of measure (dimensional analysis) and briefly explain your reasoning.
Answer: The range of wavelengths of light that can be used to cause given phenomenon is [tex]8.953 \times 10^{21} m[/tex].
Explanation:
Given: 222 kJ/mol (1 kJ = 1000 J) = 222000 J
Formula used is as follows.
[tex]E = \frac{hc}{\lambda}[/tex]
where,
E = energy
h = Planck's constant = [tex]6.625 \times 10^{-25} Js[/tex]
c = speed of light = [tex]3 \times 10^{8} m/s[/tex]
Substitute the values into above formula as follows.
[tex]E = \frac{hc}{\lambda}\\222000 J = \frac{6.625 \times 10^{-34}Js \times 3 \times 10^{8} m/s}{\lambda}\\\lambda = 8.953 \times 10^{21} m[/tex]
Thus, we can conclude that the range of wavelengths of light that can be used to cause given phenomenon is [tex]8.953 \times 10^{21} m[/tex].
Balance the following chemical equation.
CCl4 -> ___ C+ ___ Cl2
Answer:
Explanation:
CCl4 => C + 2Cl2
When perchloric acid (HClO4) reacts with tetraphosphorus decaoxide, phosphoric acid and dichlorine heptaoxide are produced.
a. Trei
b. False
Answer:
я не знаю ответа :(
Explanation:
How many grams of sodium nitrate (NaNO3) are needed to
prepare 100 grams of a 15.0 % by mass sodium nitrate
solution?
Answer:
15.0 g
Explanation:
15.0% =0.150
100.0 g × 0.150= 15.0g
Sodium nitrate is "an inorganic compound with the formula of NaNO₃.
What is an inorganic compound?Inorganic compound is "a chemical compound that lacks carbon–hydrogen bonds".
15% = 0.15
100.0 g × 0.15= 15g
Hence, 15g of Sodium nitrate are needed to prepare 100 gms of a 15% by mass sodium nitrate.
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Chemical reactions can exhibit different rate constants at differing: Select the correct answer below: initial concentrations volumes of container temperatures none of the above
Explanation:
Chemical reactions can exhibit different rate constants at differing:
i)initial concentrations
ii)volumes of container
iii) temperatures
iv)none of the above.
The rate constant of a reaction is constant at a particular temperature.
It is not depending on the initial concentration of the reactants. It varies with temperature.
Thus, among the given options the correct answer is Temperatures.
Chemical reactions can exhibit different rate constant at different temperatures. Hence, the correct option is temperature.
What is a rate constant?
The rate constant is the proportionality constant in the equation that expresses the relationship between the rate of a chemical reaction and the concentrations of the reacting substances.
Chemical reactions proceed at vastly different speeds depending on the nature of the reacting substances, the type of chemical transformation, the temperature, etc.
For a given reaction, the speed of the reaction will vary with the temperature, the pressure, and the amounts of reactants present.
The rate constant goes on increasing as the temperature goes up, but the rate of increase falls off quite rapidly at higher temperatures.
On the other hand, the volume of the container, initial concentration does not affect the rate constant.
Therefore, Chemical reactions can exhibit different rate constant at different temperatures. Hence, the correct option is temperature.
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Which statement is true with respect to standard reduction potentials?
SRP values that are greater than zero always represent a reduction reaction.
SRP values that are less than zero always represent a reduction reaction.
Half-reactions with SRP values greater than zero are spontaneous.
Half-reactions with SRP values greater than zero are nonspontaneous.
Answer:
C). Half-reactions with SRP values greater than zero are spontaneous.
Explanation:
SRPs or Standard Reduction Potentials are characterized as the ability of a probable distinction among the anode and cathode of a usual/standard cell. It aims to examine the capacity of chemicals to reduce themselves.
The third statement asserts a true claim regarding the SRPs(Standard Reduction Potentials) that the 'half-reactions which take place with the SRP possesses the values higher than zero and they are unconstrained.' The other statements are incorrect as they either show the estimation of SRPs more than 0 or display them as being restricted. Thus, option C is the correct answer.
PLEASE HELP ASAP MOLES TO MOLECULES
Answer:
4.77mol is the correct answer
During the postabsorptive state, metabolism adjusts to a catabolic state.
a. True
b. False
Answer:
The postabsorptive state (also called the fasting state) occurs when the food is already digested and absorbed, and it usually occurs overnight, when you sleep (if you skip meals for some days, you will enter in this state).
The catabolic state is the metabolic breakdown of molecules into simpler ones, releasing energy (heat) and utilizable resources.
Now, when you are in a postabsorptive state, the glucose levels start to drop, then the body starts to depend on the glycogen stores, which are catabolized into glucose, this is defined as the start of the postabsorptive state.
So yes, as the postabsorptive states, catabolic processes start to happen, so the statement is true.
What is the correct order for the reactions that produce the following transformation. a. (1) H2/Lindlar (2) CH3CO2OH b. (1) H2/Lindlar (2) O3, Zn, HCl c. (1) H2/Pd (2) CH3CO2OH d. (1) Na, NH3 (2) CH3CO2OH
Answer:
Explanation:
Can you provide a picture? I can outline the reactions though. a) will make a Z double bond from a triple bond and then peroxyacid can do epoxidation. b) will make the Z double bond then ozonolysis to double bond will create to aldehyde compounds. c) is essentially useless unless there is a ketone or aldehyde in the compound already since H2/Pd will fully reduce the alkyne (which I am assuming is present) and so the peroxyacid can't do epoxidation and can only do baeyer villiger oxidation, and d) reduces the alkyne to an E alkene and then do epoxidation to give an epoxide (with trans steroechemistry)
