Answer:
[tex]pH=10.5[/tex]
Explanation:
Hello,
In this case, the dissociation of the given weak acid is:
[tex]HA\rightleftharpoons H^++A^-[/tex]
Therefore, the law of mass action for it turns out:
[tex]Ka=\frac{[H^+][A^-]}{[HA]}[/tex]
That in terms of the change [tex]x[/tex] due to the reaction extent is:
[tex]1x10^{-23}=\frac{x*x}{0.1-x}[/tex]
Thus, by solving with the quadratic equation or solver, we obtain:
[tex]x=31.6x10^{-12}M[/tex]
Which clearly matches with the hydrogen concentration in the solution, therefore, the pH is:
[tex]pH=-log(-31.6x10^{-12})\\pH=10.5[/tex]
Regards.
Give the IUPAC name for the following compound
Answer:
3–bromo–5–chloro–4–methylhexane.
Explanation:
To name the compound given in the question, the following must be observed:
1. Locate the longest continuous carbon chain. This gives the parent name of the compound. In this case, the longest chain is carbon 6 i.e Hexane.
2. Identify the substituents attached. In this case the substituents attached are:
a. Chloro i.e Cl.
b. Bromo ie Br.
c. Methyl i.e CH3.
3. Give the substituents the lowest possible count alphabetically. Bromo comes before Chloro alphabetically, so we shall consider bromo first. Their positions are given below:
Bromo i.e Br at carbon 3
Chloro i.e Cl is at carbon 5
Methyl i.e CH3 is at carbon 4
4. Combine the above to get the name of the compound.
Therefore, the name of the compound is:
3–bromo–5–chloro–4–methylhexane.
What type of bond will be formed for atoms that have a +1 or -1 charge?
Given the information you now know, what is the effect of hyperventilation on blood pH?pH? During hyperventilation, the rapid in the blood CO2CO2 concentration shifts the equilibrium to the which the concentration of H+,H+, thereby the blood pH.
Answer:
When hypercapnia processes occur, where the concentration of carbon dioxide gas increases in the blood, the protonization of the blood increases, this means that the H + ions increase in concentration, thus generating metabolic acidosis.
This metabolic acidosis is regulated by various systems, but the respiratory system collaborates by generating hyperventilation, to increase blood oxygen pressures, decrease CO2 emissions, and indirectly decrease acidity.
Explanation:
This method of regulating the body is crucial, since the proteins in our body will not be altered if they do not happen.
The enzymes, the red globules, and many more fundamental things for life ARE PROTEINS, that in front of acidic media these modify their structure by denaturing themselves and ceasing to fulfill their functions. This is the reason why it seeks to neutralize the blood pH when it comes to an increase in CO2.
Consider the following reaction. I– 2 H2O2 (l) 2 H2O (l) + O2 (g) A solution contains 15 mL 0.1 M KI, 15 mL of DI water and 5 mL of 3% H2O2. After the decomposition of H2O2 is complete, you titrate the solution with 0.1 M AgNO3. If the catalyst, I–, is not consumed in the reaction and is completely recovered, what volume of the 0.1 M AgNO3 is required to reach the end point?
Answer:
Explanation:
The given chemical reaction is:
[tex]2H_2O_{(l)} \to^{I^-}} 2H_2O_{(l)}+O_2_{(g)}[/tex]
From above equation [tex]I^-[/tex] serves as catalyst which is not consumed by the reaction and also it is completely recovered; as a result to that , the full volume of KI will definitely react with AgNO₃.
Given that :
the volume of potassium iodide [tex]V_{KI} = 15 \ ml[/tex]
the molarity of potassium [tex]M_{KI} = 0.1 \ M[/tex]
the volume of distilled water [tex]V_W = 15 \ mL[/tex]
The volume of 3% [tex]H_2O_2 \ \ V_{H_2O_2} = 5 \ mL[/tex]
Molarity of AgNO₃ [tex]M_{AgNO_3} = 0.1 \ M[/tex]
Let take an integral look with the reaction between KI and AgNO₃; we have
[tex]KI + AgNO_3 \to KNO_3 + AgI[/tex]
At the end point; the moles of KI will definitely be equal to the moles of AgNO₃
So;
[tex]M_{KI}V_{KI}= M_{AgNO_3}V_{AgNO_3} \\ \\ V_{AgNO_3} = \dfrac{M_{KI}V_{KI}}{M_{AgNO_3}} \\ \\ \\ V_{AgNO_3} = \dfrac{ 0.1*15}{0.1}[/tex]
[tex]V_{AgNO_3} = 15 \ ml[/tex]
Thus; the volume of 0.1 M AgNO₃ needed to reach the end point is 15 mL
g The solution you created in this simulation was a 0.300M NH4Cl solution. The lab also stated that, in g/L, this concentration was 16.0 g/L. Show the calculations that prove that to be true.
Answer:
16.0473 g/L
Explanation:
0.300 M=
0.300 mol/L x 53.491 grams/mol = 16.0473 grams/L
The concentration of the 0.300M NH₄Cl solution in g/L will be equal to 16.04 g/L.
What is the molarity?The concentration of the solution can be determined if we have the molecular formula of the compound and its molecular weight. We can easily determine the majority of a solution from the moles of solute and the volume of the solution.
The molarity of a solution can be evaluated from the number of moles of a solute per liter of a solution.
The Molarity can be determined from the formula mentioned below:
Molarity (M) = Moles of solute (n)/Solution's volume ( in L)
Given, the molarity of NH₄Cl solution = 0.300 M
We can also write it as 0.300 mol/L
It means 0.300 moles in one liter.
The molar mass of NH₄Cl = 53.5 g/mol
Then the mass of 0.300 mol of NH₄Cl = 0.300 ×53.5 = 16.04 g
Therefore, the concentration of NH₄Cl solution is 16.04g/L is equivalent to 0.300 M.
Learn more about molarity, here:
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An electrochemical cell is constructed with a zinc metal anode in contact with a 0.052 M solution of zinc nitrate and a silver cathode in contact with a 0.0042 M solution of silver(I) nitrate. What is the value of Q to use in the Nernst equation for this cell
Answer:
Q = 12.38
Explanation:
The Nernst equation is given as; Ecell = E°cell - (2.303RT/nF) log Q ;where Q is the reaction quotient.
The reaction quotient, Q in a reaction, is the product of the concentrations of the products divided by the product of the concentrations of the reactants.
In an electrochemical cell, Q is the ratio of the concentration of the electrolyte at the anode to that of the electrolyte at the cathode.
Q = [anode]/[cathode]
therefore , Q = 0.052/0.0042 = 12.38
Combustion analysis of a 13.42-g sample of the unknown organic compound (which contains only carbon, hydrogen, and oxygen) produced 39.61 g CO2 and 9.01 g H2O. The molar mass of equilin is 268.34 g/mol. Find its molecular formula.
Answer: The molecular formula for the given organic compound is [tex]C_{18}H_{20}O_2[/tex]
Explanation:
The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:
[tex]C_xH_yO_z+O_2\rightarrow CO_2+H_2O[/tex]
where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.
We are given:
Mass of [tex]CO_2=39.61g[/tex]
Mass of [tex]H_2O=9.01g[/tex]
We know that:
Molar mass of carbon dioxide = 44 g/mol
Molar mass of water = 18 g/mol
For calculating the mass of carbon:
In 44 g of carbon dioxide, 12 g of carbon is contained.
So, in 39.61 g of carbon dioxide, [tex]\frac{12}{44}\times 39.61=10.80g[/tex] of carbon will be contained.
For calculating the mass of hydrogen:
In 18 g of water, 2 g of hydrogen is contained.
So, in 9.01 g of water, [tex]\frac{2}{18}\times 9.01=1.00g[/tex] of hydrogen will be contained.
Mass of oxygen in the compound = (13.42) - (10.80 + 1.00) = 1.62 g
To formulate the empirical formula, we need to follow some steps:
Step 1: Converting the given masses into moles.Moles of Carbon = [tex]\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{10.80g}{12g/mole}=0.9moles[/tex]
Moles of Hydrogen = [tex]\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1g}{1g/mole}=1moles[/tex]
Moles of Oxygen = [tex]\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.62g}{16g/mole}=0.10moles[/tex]
Step 2: Calculating the mole ratio of the given elements.For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.10 moles.
For Carbon = [tex]\frac{0.9}{0.10}=9[/tex]
For Hydrogen = [tex]\frac{1}{0.10}=10[/tex]
For Oxygen = [tex]\frac{0.10}{0.10}=1[/tex]
Step 3: Taking the mole ratio as their subscripts.The ratio of C : H : O = 9 : 10 : 1
Hence, the empirical formula for the given compound is [tex]C_9H_{10}O[/tex]
For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.
The equation used to calculate the valency is :
[tex]n=\frac{\text{Molecular mass}}{\text{Empirical mass}}[/tex]
We are given:
Mass of molecular formula = 268.34 g/mol
Mass of empirical formula = 134 g/mol
Putting values in above equation, we get:
[tex]n=\frac{268.34g/mol}{134g/mol}=2[/tex]
Multiplying this valency by the subscript of every element of empirical formula, we get:
[tex]C_{(9\times 2)}H_{(10\times 2)}O_{(1\times 2)}=C_{18}H_{20}O_2[/tex]
Thus, the molecular formula for the given organic compound is [tex]C_{18}H_{20}O_2[/tex].
An ethylene glycol solution contains 21.4 g of ethylene glycol (C2H6O2) in 97.6 mL of water.
Required:
a. Determine the freezing point of the solution. Express you answer in degrees Celsius. (Assume a density of 1.00 g/mL for water.)
b. Compute the boiling point of the solution. (Assume a density of 1.00 g/mL for water.)
Answer:
a. TFinal = -6.57°C
b. Tfinal = 101.80°C
Explanation:
When a solute is added to a solvent producing an ideal solution, the freezing point of the solution decreases with regard to pure solvent. Also, boiling point increases with regard to pure solvent.
The formulas are:
Freezing point:
ΔT = Kf×m×i
Where Kf is freezeing point depression constant of water (1.86°C/m), m is molality of solution and i is van't Hoff factor (1 for ethylene glycol).
Boiling point:
ΔT = Kb×m×i
Where K is freezeing point depression constant of water (0.51°C/m), m is molality of solution and i is van't Hoff factor (1 for ethylene glycol).
Moles of 21.4g of ethylene glycol (Molar mass: 62.07g/mol) are:
21.4g C₂H₆O₂ ₓ (1mol / 62.07g) = 0.345 moles
And kg of 97.6mL of water = 97.6g are 0.0976kg. Molality of the solution is:
0.345mol / 0.0976kg = 3.5325m
Replacing in the formulas:
a. Freezing point:
ΔT = 1.86C/m×3.5325m×1
ΔT = 6.57°C
0°C - Tfinal = 6.57°C
TFinal = -6.57°Cb. Boiling point:
ΔT = 0.51°C/m×3.5325m×1
ΔT = 1.80°C
Tfinal - 100°C = 1.80°C
Tfinal = 101.80°C
If you have 101 g of hydrogen gas (H2) and excess amount of nitrogen gas (N2), how many grams of ammonia gas (NH3) can you make?
Answer:
572. 3 g of NH3
Explanation:
Equation of the reaction: 3H2 + N2 ----> 2NH3
From the equation of reaction, 3 moles of H2 reacts with 1 mole of N2 to produce 2 moles of NH3.
Since N2 is in excess in the given reaction, H2 is the limiting reactant.
Molar mass of H2 = 2 g/mol
Molar mass of NH3 = 17 g/mol
Therefore 3 * 2 g of H2 reacts to produce 2 * 17 g of NH3
6 g of H2 produces 34 g of NH3
101 g of H2 will produce (34 * 101)/6 g of NH3 = 572.3 g of NH3
Therefore, 572.3 g of NH3 are produced
Answer:
572.33g of NH3.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
N2 + 3H2 —> 2NH3
Next, we shall determine the mass of the H2 that reacted and the mass of NH3 produced from the balanced equation. This is illustrated below:
Molar Mass of H2 = 2x1 = 2g/mol
Mass of H2 from the balanced equation = 3 x 2 = 6g
Molar Mass of NH3 = 14 + (3x1) = 17g/mol
Mass of NH3 from the balanced equation = 2 x 17 = 34g.
From the balanced equation above,
6g of H2 reacted to produce 34g of NH3.
Finally, we can determine the mass of ammonia (NH3) produced by reacting 101g of H2 as follow:
From the balanced equation above,
6g of H2 reacted to produce 34g of NH3.
Therefore, 101g of H2 will react to produce = ( 101 x 34) / 6 = 572.33g of NH3.
Therefore, 572.33g of NH3 is produced from the reaction.
What is a good title for this chart?
Answer:
pH of the acid
Explanation:
I WILL GIVE BRAINLIEST
What are the number of protons, neutrons, and electrons in 19 F
9?
Answer:
This isotope of fluorine has 9 protons, 9 electrons and 10 neutrons.
Explanation:
The three‑dimensional structure of a generic molecule is given. Identify the axial and equatorial atoms in the three‑dimensional structure. What is the shape of this molecule?
Answer:
Explanation:
CHECK THE ATTACHMENT FOR THE COMPLETE QUESTION AND THE DETAILED EXPLANATION
NOTE:
Equatorial atoms are referred to atoms that are attached to carbons in the cyclohexane ring which is found at the equator of the ring.
Axial atoms are atoms that exist in a bond which is parallel to the axis of the ring in cyclohexane
For some hypothetical metal the equilibrium number of vacancies at 750°C is 2.8 × 1024 m−3. If the density and atomic weight of this metal are 5.60 g/cm3 and 65.6 g/mol, respectively, calculate the fraction of vacancies for this metal at 750°C.
Answer:
The correct answer is 5.447 × 10⁻⁵ vacancies per atom.
Explanation:
Based on the given question, the at 750 degree C the number of vacancies or Nv is 2.8 × 10²⁴ m⁻³. The density of the metal is 5.60 g/cm³ or 5.60 × 10⁶ g/m³. The atomic weight of the metal given is 65.6 gram per mole. In order to determine the fraction of vacancies, the formula to be used is,
Fv = Nv/N------ (i)
Here Nv is the number of vacancies and N is the number of atomic sites per unit volume. To find N, the formula to be used is,
N = NA×P/A, here NA is the Avogadro's number, which is equivalent to 6.022 × 10²³ atoms per mol, P is the density and A is the atomic weight. Now putting the values we get,
N = 6.022 × 10²³ atoms/mol × 5.60 × 10⁶ g/m³ / 65.6 g/mol
N = 5.14073 × 10²⁸ atoms/m³
Now putting the values of Nv and N in the equation (i) we get,
Fv = 2.8 × 10²⁴ m⁻³ / 5.14073 × 10²⁸ atoms/m^3
Fv = 5.44669 × 10⁻⁵ vacancies per atom or 5.447 × 10⁻⁵ vacancies/atom.
iron oxide + oxygen equals to ?
Answer:
It's ferric oxide Fe2O3
Explanation:
I don't say u must have to mark my ans as brainliest but if it has really helped u plz don't forget to thank me plz...
Acetonitrile, CH3CN, is a polar organic solvent that dissolves many solutes, including many salts. The density of a 1.80 M acetonitrile solution of LiBr is 0.826 g/mL. Calculate the concentration of the solution in units of (a) molality; (b) mole fraction of LiBr; (c) mass percentage of CH3CN.
Answer:
(a) [tex]m=2.69m[/tex]
(b) [tex]x_{LiBr}=0.099[/tex]
(c) [tex]\% LiBr=18.9\%[/tex]
Explanation:
Hello,
In this case, given the molality in mol/L, we can compute the required units of concentration assuming a 1-L solution of acetonitrile and lithium bromide that has 1.80 moles of lithium bromide:
(a) For the molality, we first compute the grams of lithium bromide in 1.80 moles by using its molar mass:
[tex]m_{LiBr}=1.80mol*\frac{86.845 g}{1mol}=156.32g[/tex]
Next, we compute the mass of the solution:
[tex]m_{solution}=1L*0.826\frac{g}{mL}*\frac{1000mL}{1L}=826g[/tex]
Then, the mass of the solvent (acetonitrile) in kg:
[tex]m_{solvent}=(826g-156.32g)*\frac{1kg}{1000g}=0.670kg[/tex]
Finally, the molality:
[tex]m=\frac{1.80mol}{0.670kg} \\\\m=2.69m[/tex]
(b) For the mole fraction, we first compute the moles of solvent (acetonitrile):
[tex]n_{solvent}=669.68g*\frac{1mol}{41.05 g} =16.31mol[/tex]
Then, the mole fraction of lithium bromide:
[tex]x_{LiBr}=\frac{1.80mol}{1.80mol+16.31mol}\\ \\x_{LiBr}=0.099[/tex]
(c) Finally, the mass percentage with the previously computed masses:
[tex]\% LiBr=\frac{156.32g}{826g}*100\%\\ \\\% LiBr=18.9\%[/tex]
Regards.
The decomposition of hydrogen peroxide, H2O2, has been used to provide thrust in the control jets of various space vehicles. Determine how much heat (in kJ) is produced by the decomposition of 1.71 mol of H2O2 under standard conditions.
Answer:
[tex]Q=-361.56kJ[/tex]
Explanation:
Hello,
In this case, the decomposition of hydrogen peroxide is given by:
[tex]2H_2O_2\rightarrow 2H_2O+O_2[/tex]
Which occurs in gaseous phase, therefore the enthalpy of reaction is:
[tex]\Delta _rH=2\Delta _fH_{H_2O}-2\Delta _fH_{H_2O_2}[/tex]
Oxygen is not included as it is a pure element. The enthalpies of formation for both hydrogen peroxide and water are -136.11 and -241.83 kJ/mol respectively, so we compute the enthalpy of reaction:
[tex]\Delta _rH=2(-241.83kJ/mol)-2(-136.11kJ/mol)=-211.44kJ/mol[/tex]
Then, the total heat that is released for 1.71 mol of hydrogen peroxide is:
[tex]Q=n*\Delta _rH=1.71mol*-211.44kJ/mol\\\\Q=-361.56kJ[/tex]
Whose sign means a released heat.
Regards.
Liquid hexane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 4.3 g of hexane is mixed with 7.14 g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Round your answer to significant digits.
Answer:
We can produce 6.20 grams of CO2
Explanation:
Step 1: Data given
Mass of hexane = 4.3 grams
Molar mass of hexane = 86.18 g/mol
Mass of oxygen = 7.14 grams
Molar mass of oxygen = 32.0 g/mol
Step 2: The balanced equation
2C6H14 + 19O2 → 12CO2 + 14H2O
Step 3: Calculate moles
Moles = mass / molar mass
Moles hexane = 4.3 grams / 86.18 g/mol
Moles hexane = 0.0499 moles
Moles oxygen = 7.14 grams / 32.0 g/mol
Moles oxygen = 0.2231 moles
Step 4: Calculate the limiting reactant
For 2 moles hexane we need 19 moles O2 to produce 12 moles CO2 and 14 moles H2O
Oxygen is the limiting reactant. It will completely be consumed ( 0.2231 moles). Hexane is in excess. There will react 2/19 * 0.2231 = 0.02348 moles
There will be porduced 12/19 * 0.2231 = 0.1409 moles CO2
Step 5: Calculate mass CO2
Mass CO2 = moles CO2 * molar mass CO2
Mass CO2 = 0.1409 moles * 44.01 g/mol
Mass CO2 = 6.20 grams
We can produce 6.20 grams of CO2
Isomer such as acetic acid and methyl formate have
Answer:
C
Explanation:
This is the definition of an isomer.
The solubility of cadmium oxalate, , in 0.150 M ammonia is mol/L. What is the oxalate ion concentration in the saturated solution? If the solubility product constant for cadmium oxalate is , what must be the cadmium ion concentration in the solution? Now, calculate the formation constant for the complex ion
Answer:
[Cd²⁺] = 2.459x10⁻⁶M
Kf = 9.96x10⁶
Explanation:
Solubility of CdC₂O₄ is 6.1x10⁻³M and ksp is 1.5x10⁻⁸
The ksp of CdC₂O₄ is:
CdC₂O₄(s) ⇄ Cd²⁺(aq) + C₂O₄²⁻(aq)
ksp = [Cd²⁺] [C₂O₄²⁻] = 1.5x10⁻⁸
As solubility is 6.1x10⁻³M, concentration of C₂O₄²⁻ ions is 6.1x10⁻³M. Replacing:
[Cd²⁺] = 1.5x10⁻⁸ / [6.1x10⁻³M]
[Cd²⁺] = 2.459x10⁻⁶MAll Cd²⁺ in solution is 6.1x10⁻³M and exist as Cd²⁺ and as Cd(NH₃)₄²⁺. That means concentration of Cd(NH₃)₄²⁺ is:
[Cd(NH₃)₄²⁺] + [Cd²⁺] = 6.1x10⁻³M
[Cd(NH₃)₄²⁺] = 6.1x10⁻³M - 2.459x10⁻⁶M = 6.098x10⁻³M
[Cd(NH₃)₄²⁺] = 6.098x10⁻³MIn the same way, the whole concentration of NH₃ in solution is 0.150M, as you have 4ₓ6.098x10⁻³M = 0.024M of NH₃ producing the complex, the concentration of the free NH₃ is:
[0.150M] = [NH₃] + 0.024M
0.1256M = [NH₃]The equilibrium of the complex formation is:
Cd²⁺ + 4 NH₃ → Cd(NH₃)₄²⁺
The kf, formation constant, is defined as:
Kf = [Cd(NH₃)₄²⁺] / [Cd²⁺] [NH₃]⁴
Replacing:
Kf = [6.098x10⁻³M] / [2.459x10⁻⁶M] [0.1256M]⁴
Kf = 9.96x10⁶When you turn on the air conditioner during a hot summer day the cooler air will sink to the floor, while warmer air rises to the
ceiling
Which type of heat transfer is this an example of?
(A) conduction
(B) convection
(C) radiation
(D)
kinetic
what is the name of the liquid in the clinical thermometer
Answer:I suppose it is mercury...
Explanation:
I don't say u must have to mark my ans as brainliest but if it has really helped u plz don't forget to thnk me...
A mixture of compounds containing diethylamine, phenol, ammonia, and acetic acid is separated using liquid-liquid extraction as follows: Step 1: Concentrated HCl is added followed by draining the aqueous layer. Step 2: Dilute NaOH is added to the organic layer followed by draining the aqueous layer. Step 3: Concentrated NaOH is added to the organic layer followed by draining the aqueous layer. Which compound would you expect to be extracted into the aqueous layer after the addition of dilute HCl, step 1? Group of answer choices
Complete Question
The complete question is shown on the first uploaded image
Answer:
The correct option is ammonia
Explanation:
The mixture contains two base compound which are
ammonia,
and diethylamine
Now the addition of HCl which is a strong acid in step 1 will cause the protonation of the two base compound , which makes the soluble hence resulting in them being extracted to the aqueous layer as represented in below
[tex]NH_3 + HCl\to NH_4 ^{+} + Cl^-[/tex]
and
[tex](CH 3CH 2) 2NH + HCl \to (CH 3CH 2) 2NH_2^{+} + Cl[/tex]
Trans-4-hexen-3-ol can be synthesized starting from acetaldehyde. One of the key reagents is ethyl grignard.
1. Synthesize ethyl grignard from acetaldehyde in the steps below using the reagents provided.
2. Synthesize (trans)-4-hexen-3-ol from acetaldehyde.
find the given attachment
Heat is added to a 1.0-kg block of ice at OC. Determine if the process is
endothermic or exothermic. Explain your answer. *
Answer:
endothermic
Explanation:
Heat is added to make the process possible.
Which of the following is not an example of a mechanical wave?
A. Fans doing "The Wave" at a sporting event.
B. Sound waves coming out of the radio.
C. Water waves at hie beach.
D. Sunshine.
Answer:
Option D
Explanation:
A mechanical wave is a wave of energy that can travel long distances and could go through characteristics of matter such as solids, liquids, and gases. Mechanical waves can also travel through vacuums. A good example of a mechanical wave would be sound, sound is a wave spread through a object and can go through different types of matter. Which is why your answer is option D "sunshine." Light cannot go through a vacuum while sounds, and water can.
Hope this helps.
The mechanical wave example does not include the sunshine
What is mechanical waves ?It is the wave of energy that can travel long distances and considered the characteristics of matter like solids, liquids, and gases. It can also travel via vacuums. The Light cannot go via a vacuum while sounds, and water can go.
Learn more about sound here:https://brainly.com/question/16750970
If a gas occupies 12.60 liters at a pressure of 1.50 atm, what will its pressure at a volume of 2.50 liters?
Answer:
7.56 atm
Explanation:
Boyle's law states that the pressure and volume of a gas are proportional to each other
The formular for Boyle's law is
P1V1=P2V2
According to the question above, the values given are
P1=1.50 atm
P2= ?
V1=12.60 litres
V2= 2.50 litres
Let us make P2 the subject of formular
P2= P1V1/V2
P2= 1.50×12.60/2.50
P2= 18.9/2.50
P2= 7.56 atm
Hence when the volume of a gas is 2.50 litres then it's pressure is 7.56 atm
Glycine, C2H5O2N, is important for biological energy. The combustion reaction of glycine is given by the equation 4C2H5O2N(s) + 9O2(g) → 8CO2(g) + 10H2O(l) + 2N2(g) ΔH°rxn = –3857 kJ/mol Given that ΔH°f[CO2(g)] = –393.5 kJ/mol and ΔH°f[H2O(l)] = –285.8 kJ/mol, calculate the enthalpy of formation of glycine.
Answer:
ΔH°f C₂H₅O₂N(s) = -537.2kJ
Explanation:
Based on the reaction:
4 C₂H₅O₂N(s) + 9O₂(g) → 8CO₂(g) + 10H₂O(l) + 2N₂(g)
ΔHrxn = ΔH°f products - ΔH°f reactants.
As:
ΔH°fO₂(g) = 0
ΔH°fCO₂(g) = -393.5kJ/mol
ΔH°fH₂O(l) = -285.8kJ/mol
ΔH°fN₂(g) = 0
The ΔHrxn is:
ΔHrxn = (8×-393.5kJ/mol + 10×-285.8kJ/mol) - (4×ΔH°fC₂H₅O₂N(s)) = -3857kJ/mol
-6006kJ/mol - (4×ΔH°fC₂H₅O₂N(s)) = -3857kJ/mol
-4×ΔH°fC₂H₅O₂N(s) = 2149kJ/mol
ΔH°fC₂H₅O₂N(s) = 2149kJ/mol / -4
ΔH°f C₂H₅O₂N(s) = -537.2kJConsider 10.0 g of helium gas (He) in a rigid steel container. If you add 10.0 g of neon gas (Ne) to this container, which of the following best describes what happens? (Assume the temperature is constant.)
a) The pressure in the container doubles.
b) The pressure in the container more than doubles.
c) The volume of the container doubles.
d) The volume of the container more than doubles.
e) The pressure in the container increases but does not double.
Answer: (e) The pressure in the container increases but does not double.
Explanation:
To solve this, we need to first remember our gas law, Boyle's law states that the pressure and volume of a gas have an inverse relationship. That is, If volume increases, then pressure decreases and vice versa, when temperature is held constant. Therefore, increasing the volume in this case does not double the pressure owning to out gas law, but an increase in pressure would be noticed if temperature is constant
The pressure in the container increases but does not double.
At constant temperature and volume, the pressure of a given mass of gas is directly proportional to the number of moles of gas present.
Number of moles of He = 10 g/4 g/mol = 2.5 moles
Number of moles of Ne = 10 g/20 g/mol = 0.5 moles
We can see that the number of moles only increases by 1/5 of its initial value therefore, the pressure in the container increases but does not double.
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How do forces between particles in liquids compare to forces in tho other states of matter?
Answer:I hope it will be beneficial for you
Force of attraction between the particles of solid is very strong the particles of solid are held together by strong inter molecular forces leading to the formation of a rigid structure
Force of attraction between the particles of the liquid is weak as compare to solids there particles are far away from each other and have the property to move easily.
Force of attraction between the particles of gases is very weak than the two states hence the particles of gases are highly compressible having week intermolecular interaction between them and have indefinite shape and volume
Answer:
Forces between particles in Liquids are closely packed compared to other states of matter like the liquid and gaseous state of matter.
Explanation: