Answer:
N2H4
Explanation:
A hydride is a binary compound of hydrogen and another element. Binary compounds contain only two atoms. We have to x-ray the hydrides of nitrogen given in the question in order to make our choice. Remember that we were told that that the hydride contains its own volume of nitrogen and twice its volume of hydrogen.
Now consider the hydride N2H4.
N2H4(g) -----> N2(g) + 2H2(g)
The volume ration of nitrogen gas to hydrogen gas in N2H4 is 1:2.
The molecular mass of the compound is;
N2H4= 2(14) + 4(1)= 28+4= 32
Since
molecular mass= 2 vapour density
Vapour density= molecular mass/2
Vapour density= 32/2
Vapour density = 16
Therefore the hydride of nitrogen referred to in the question is N2H4
A thermometer is placed in water in order to measure the water’s temperature. What would cause the liquid in the thermometer to rise? The molecules in the water move closer together. The molecules in the thermometer’s liquid spread apart. The kinetic energy of the water molecules decreases. The kinetic energy of the thermometer’s liquid molecules decreases.
Answer:
The molecules in the thermometer’s liquid spread apart
Explanation:
The molecules in the thermometer’s liquid spread apart.
What is thermometer?A thermometer is a device that measures temperature or a temperature gradient.
What causes the liquid in the thermometer to rise?The liquid (water) in thermometer exhibits convex meniscus, as a result of this meniscus, the water molecules in the thermometer will spread apart when temperature is measured.
Learn more about thermometer here: https://brainly.com/question/21720093
If 25.8 mL of an AgNO3 solution is needed to precipitate all Cl- ions in a 1570 mg of KCl (forming AgCl), what is the molarity of the AgNO3nsolution?
Answer:
M=0.816M
Explanation:
Hello,
In this case, we should consider the following reaction:
[tex]AgNO_3+KCl\rightarrow KNO_3+AgCl[/tex]
Thus, by knowing the 1:1 molar ratio of silver nitrate and potassium chloride, we can easily compute the moles of silver nitrate precipitating the 1570 mg of potassium chloride considering its molar mass of 74.5513 g/mol:
[tex]n_{AgNO_3}=1570mgKCl*\frac{1gKCl}{1000mgKCl} *\frac{1molKCl}{74.5513gKCl}*\frac{1molAgNO_3}{1molKCl} \\\\n_{AgNO_3}=0.021molAgNO_3[/tex]
Then, by using the volume of silver nitrate in liters (0.0258 L), we can directly compute the molarity:
[tex]M=\frac{0.021molAgNO_3}{0.0258L}\\ \\M=0.816M[/tex]
Regards.
How many moles of PC15 can be produced from 51.0 g of Cl2 (and excess P4)?
Express your answer to three significant figures and include the appropriate units.
LIT....ITS NOT .227 or .228!!!!
Answer:
0.287 mole of PCl5.
Explanation:
We'll begin by calculating the number of mole in 51g of Cl2. This is illustrated below:
Molar mass of Cl2 = 2 x 35.5 = 71g/mol
Mass of Cl2 = 51g
Number of mole of Cl2 =..?
Mole = Mass /Molar Mass
Number of mole of Cl2 = 51/71 = 0.718 mole
Next, we shall write the balanced equation for the reaction. This is given below:
P4 + 10Cl2 → 4PCl5
Finally, we determine the number of mole of PCl5 produced from the reaction as follow:
From the balanced equation above,
10 moles of Cl2 reacted to produce 4 moles of PCl5.
Therefore, 0.718 mole of Cl2 will react to produce = (0.718 x 4)/10 = 0.287 mole of PCl5.
Therefore, 0.287 mole of PCl5 is produced from the reaction.
Calculate the standard entropy of reaction at 298 K for the reaction Hg(liq) + Cl2(g) → HgCl2(s) The standard molar entropies of the species at that temperature are: Sºm (Hg,liq) = 76.02 J / (K mol) ; Sºm (Cl2,g) = 223.07 J / (K mol) ; Sºm (HgCl2,s) = 146.0 J / (K mol)
Answer:
−153.1 J / (K mol)
Explanation:
Calculate the standard entropy of reaction at 298 K for the reaction Hg(liq) + Cl2(g) → HgCl2(s) The standard molar entropies of the species at that temperature are: Sºm (Hg,liq) = 76.02 J / (K mol) ; Sºm (Cl2,g) = 223.07 J / (K mol) ; Sºm (HgCl2,s) = 146.0 J / (K mol)
Hg(liq) + Cl2(g) → HgCl2(s)
Given that;
The standard molar entropies of the species at that temperature are:
Sºm (Hg,liq) = 76.02 J / (K mol) ;
Sºm (Cl2,g) = 223.07 J / (K mol) ;
Sºm (HgCl2,s) = 146.0 J / (K mol)
The standard molar entropies of reaction = Sºm[products] - Sºm [ reactants]
= 146.0 J / (K mol) – [76.02 J / (K mol) +223.07 J / (K mol) ]
= -153.09 J / (K mol)
= or -153.1 J / (K mol)
Hence the answer is −153.1 J / (K mol)
What is Keq for the reaction 2HCl(9) = H2(g) + Cl2(g)?
Answer:
Keq= [(Cl2) (H2)] / (HCl)^2
Explanation:
Equilibrium Constant, Keq, is written as products/reactants.
So it's going to be Keq= [(Cl2) (H2)] / (HCl)^2
Given a K value of 0.43 for the following aqueous equilibrium, suppose sample Z is placed into water such that it’s original concentration is 0.033 M. Assume there was zero initial concentration of either A(aq) or B(aq). Once equilibrium has occurred, what will be the equilibrium concentration of Z?
2A(aq) + B(aq) <> 2Z (aq)
Answer:
[Z] = 0.00248M
Explanation:
Based in the reaction:
2A(aq) + B(aq) ⇄ 2Z (aq)
K of the reaction is defined as:
K = [Z]² / [A]²[B] = 0.43
If you add, in the first, just 0.033M of Z, concentrations in equilibrium are:
[Z] = 0.033M - 2X
[A] = 2X
[B] = X
Replacing in K equation:
0.43 = [0.033M - 2X]² / [2X]² [X]
0.43 = [0.033M - 2X]² / [2X]² [X]
0.43 = 4X² -0.132X + 0.001089 / 4X³
1.72X³ - 4X² + 0.132X - 0.001089 = 0
Solving for X:
X = 0.01526M
Replacing, concentration in equilibrium of Z is:
[Z] = 0.033M - 2*0.01526M = 0.00248M
Answer:
Less than 0.033 M
Explanation:
Hello,
In this case, given the equilibrium:
[tex]2A(aq) + B(aq) \rightleftharpoons 2Z (aq)[/tex]
Thus, the law of mass action is:
[tex]K=\frac{[Z]^2}{[A]^2[B]}[/tex]
Nevertheless, given the initial concentration of Z that is 0.033 M, we should invert the equilibrium since the reaction will move leftwards:
[tex]\frac{1}{K} =\frac{[A]^2[B]}{[Z]^2}=2.33[/tex]
Know, by introducing the change due to the reaction extent, we can write:
[tex]2.33=\frac{(2x)^2*x}{(0.033M-2x)^2}[/tex]
Which has the following solution:
[tex]x_1=2.29M\\x_2=0.0181M\\x_3= 0.0153M[/tex]
But the correct solution is [tex]x=0.0153M[/tex] since the other solutions make the equilibrium concentration of Z negative which is not possible. In such a way, its concentration at equilibrium is:
[tex][Z]_{eq}=0.033M-2*0.0153M=0.0024M[/tex]
Which is of course less than 0.033 M since the addition of a product shift the reaction leftwards in order to reestablish equilibrium (Le Chatelier's principle).
Regards.
2. Points
Which of the following is not a characteristic of a transverse mechanical
wave?
A. It travels at less than the speed of light.
B. It involves displacing the medium perpendicular to the motion of
the wave
C. It looks a little bit like a snake.
D. It is also known as a compression wave.
Answer:
D
Explanation:
Logitudinal waves also known as compression waves.
It involves displacing the medium perpendicular to the motion of the wave is not a characteristic of a transverse mechanical wave. Option B is correct.
What are transverse mechanical waves?A transverse mechanical wave is a disturbance created by it to transfer energy from one point to another. while the proposition happens the particle present within the medium get vibrates.
in a transverse wave, the particle present will vibrate up and down and are perpendicular to the wave's propagation direction. The particles shake in a directional wave in the longitudinal wave propagation.
Therefore, is not a characteristic of a transverse mechanical wave. Option B is correct. It involves displacing the medium perpendicular to the motion of the wave.
Learn more about transverse mechanical waves, here:
https://brainly.com/question/23374194
#SPJ6
Which process is used to make lime (calcium oxide) from limestone (calcium carbonate)?
Answer:
Explanation:
Calcium oxide is fromed by the decomopostion of CaCO3 at high temperature.
CaCO3 ------> CaO +CO2
Hope this helps you
The substances nitrogen monoxide and hydrogen gas react to form nitrogen gas and water. Unbalanced equation: NO (g) + H2 (g) N2 (g) + H2O (l) In one reaction, 76.2 g of H2O is produced. What amount (in mol) of H2 was consumed? What mass (in grams) of N2 is produced?
Answer:
H2 consumed 4.22 mol
N2 produced 59.107 g
Explanation:
Balanced equation:
2NO (g) + 2H2 (g) N2 (g) + 2H2O (l)
To perform the calculations, the molecular weights of the following compounds must be known:H2O MW = 18.02 g/mol
N2 MW = 28.01 g/mol
To determine the moles of H2O produced, the following formula should be used:
[tex]MW=\frac{mass}{mol}[/tex]
The value of moles is cleared:
[tex]mol=\frac{mass}{MW} =\frac{76.2g}{18.02\frac{g}{mol} } =4.22 mol[/tex]
Now, to calculate the grams of N2 consumed, we look at the balanced equation and note that 2 moles of H2 produce 1 mole of N2. Therefore, through said observation, the amount of moles of H2 consumed can be determined.2 mol H2 ⇒ 1 mol N2
4.22 mol H2 ⇒ X
[tex]X=\frac{4.22mol*1 mol}{2 mol} =2.11 mol[/tex]
To calculate the mass of H2 consumed, the molecular weight equation is used again:
[tex]mass=MW*mol=28.013\frac{g}{mol}*2.11mol=59.107g[/tex]
(a) show that the pressure exerted by a fluid P (in pascals) is given by P= hdg, where h is the column of the fluid in metres, d is density in kg/m3, and g is the acceleration due to gravity (9.81 m/s2). (Hint: see appendix 2.). (b) The volume of an air bubble that starts at the bottom of a lake at 5.24 degree celsius increases by a factor of 6 as it rises to the surface of water where the temperature is 18.73 degree celsius and the air pressure is 0.973 atm. The density of the lake water is 1.02 g/cm3. Use the equation in (a) to determine the depth of the lake in metres.
Answer:
56.4 m
Explanation:
volume increases by factor of 6, i.e [tex]\frac{V2}{V1}[/tex] = 6
Initial temperature T1 at bottom of lake = 5.24°C = 278.24 K
Final temperature T2 at top of lake = 18.73°C = 291.73 K
NB to change temperature from °C to K we add 273
Final pressure P2 at the top of the lake = 0.973 atm
Initial pressure P1 at bottom of lake = ?
Using the equation of an ideal gas
[tex]\frac{P1V1}{T1}[/tex] = [tex]\frac{P2V2}{T2}[/tex]
P1 = [tex]\frac{P2V2T1}{V1T2}[/tex] = [tex]\frac{0.973*6*278.24}{291.73}[/tex]
P1 = 5.57 atm
5.57 atm = 5.57 x 101325 = 564380.25 Pa
Density Ρ of lake = 1.02 g/[tex]cm^{3}[/tex] = 1020 kg/[tex]m^{3}[/tex]
acceleration due to gravity g = 9.81 [tex]m/s^{2}[/tex]
Pressure at lake bottom = pgd
where d is the depth of the lake
564380.25 = 1020 x 9.81 x d
d = [tex]\frac{564380.25}{10006.2}[/tex] = 56.4 m
A stock solution will be prepared by mixing the following chemicals together:
3.0 mL of 0.00200 M KSCN
10.0 mL of 0.200 M Fe(NO3)3
17.0 mL of 0.5 M HNO3
Determine the molar concentration of Fe(NO3)3 in the stock solution.
Answer:
0.067M Fe(NO3)3
Explanation:
A stock solution is a concentrated solution that is diluted to prepare the solutions that you will use.
The volume of the stock solution is 3.0mL + 10.0mL + 17.0mL= 30.0mL.
The ratio between volume of the aliquot (10.0mL) and total volume (30.0mL) is called dilution factor, that is: 30.0mL / 10.0mL = 3
That means the Fe(NO3)3 is diluted 3 times. That means the molar concentration of the stock solution is:
0.200M / 3 =
0.067M Fe(NO3)3Take a series of observations to determine if process is spontaneous. Based upon those observations, you will create an activity series, listing the metals in order of their reactivity. Second, you will construct a series of virtual galvanic cells and use those to power a stopwatch. Third, you will determine the standard reduction potential of an unknown metal; comparing its reduction potential to a standard list, you will identify the unknown. Finally, you will create a situation in which the cells are not in the standard condition and measure the cell potential; using the Nernst equation, you will determine the concentration of an unknown solution
Answer the below questions for the portion of the activity in which Sn(s) is placed in AgNO3(aq)
1. Is there a reaction? (circle the correct response) Yes / No
2. How many electrons are transferred 4 electrons
3. Write the balanced redox reaction for the combination of AgNO3(aq) and Sn(s)Sn(s) + Ag+(aq) Sn2+(aq) + Ag(s)
Answer:
Explanation:
2AgNO₃ + Sn ⇄ Sn( NO₃)₂ + 2Ag
Ag⁺/Ag = .80 V
Sn⁺²/Sn = - .14 V
Hence Ag will be reduced and Sn will be oxidised . Hence the reaction will take place . YES .
2 ) 2 electrons are transferred .
3 )
2Ag⁺ + 2e = 2Ag
Sn = Sn⁺² + 2e
---------------------------
2Ag⁺ + Sn = Sn⁺² + 2Ag .
1 Ammonia, NH3, reacts with incredibly strong bases to produce the amide ion, NH2 -. Ammonia can also react with acids to produce the ammonium ion, NH4 +. (a) Which species (amide ion, ammonia, or ammonium ion) has the largest H ¬ N ¬ H bond angle? (b) Which species has the smallest H¬N¬H bond angle?
Answer:
a) ammonium ion
b) amide ion
Explanation:
The order of decreasing bond angles of the three nitrogen species; ammonium ion, ammonia and amide ion is NH4+ >NH3> NH2-. Next we need to rationalize this order of decreasing bond angles from the valence shell electron pair repulsion (VSEPR) theory perspective.
First we must realize that all three nitrogen species contain a central sp3 hybridized carbon atom. This means that a tetrahedral geometry is ideally expected. Recall that the presence of lone pairs distorts molecular structures from the expected geometry based on VSEPR theory.
The amide ion contains two lone pairs of electrons. Remember that the presence of lone pairs causes greater repulsion than bond pairs on the outermost shell of the central atom. Hence, the amide ion has the least H-N-H bond angle of about 105°.
The ammonia molecule contains one lone pair, the repulsion caused by one lone pair is definitely bless than that caused by two lone pairs of electrons hence the bond angle of the H-N-H bond in ammonia is 107°.
The ammonium ion contains four bond pairs and no lone pair of electrons on the outermost nitrogen atom. Hence we expect a perfect tetrahedron with bond angle of 109°.
Classify the following as Arrhenius, Bronsted-Lowry, or Lewis acid-base reactions. A reaction may fit all, two, one, or none of the categories:I. Cu2+ + 4 Cl− CuCl42−II. Al(OH)3 + 3HNO3 Al3+ + 3H2O + 3 NO3−III. N2 + 3 H2 2NH3IV. CN− + H2O HCN + OH
Answer:
I. Lewis acid-base reaction
II. Arrhenius, Brønsted-Lowry, and Lewis' acid-base reaction
III. Brønsted-Lowry and Lewis'acid-base reaction
IV. Lewis acid-base reaction
Explanation:
According to Arrhenius, an acid is a substance that dissolves in water to produce H+ ions, and a base is a substance that dissolves in water to produce hydroxide (OH−) ions.
In the reaction below, AH is an avid, BOH is a base reacting together to form a salt(A-B+) and water only.
AH + BOH ---> A-B+ + H2O
According to Brønsted-Lowry definition, an acid is any substance that can donate a proton, and a base is any substance that can accept a proton.
In the reaction below, AH is an acid while B is a base, reacting together to form an acid-base conjugate pair.
AH + B <-----> BH+ + A-
According to Lewis' definition, an acid is a species that accepts an electron pair while a base donates an electron pair resulting in a coordinate covalently bonded compound, also known as an adduct. In the reaction below, A+ is an acid, B- is a base, reacting together to form product A-B.
A+ + B- ------> A-B
Considering the above definitions;
I. Cu²+ + 4 Cl− ---> CuCl4²− is a Lewis acid-base reaction because it involves electron sharing only.
II. Al(OH)3 + 3HNO3 ---> Al3+ + 3H2O + 3 NO3− is an Arrhenius, Brønsted-Lowry, and a Lewis acid-base reaction because it involves protons, electrons and hydroxide ions.
III. N2 + 3 H2 ---> 2NH3 is a Lewis acid-base reaction because it involves sharing of electrons only.
IV. CN− + H2O ---> HCN + OH is both a Lewis and Brønsted-Lowry acid-base reaction because both protons and electrons sharing is involved.
In a Bronsted-Lowry acid-base reaction reaction, an acid donates protons which is accepted by the base.
The following are useful definitions of acids and bases;
An Arrhenius acid produces hydrogen ion as its only positive ion in solution while an Arrhenius base produces hydroxide ion as its only negative ion in solution.A Bronsted-Lowry acid donates hydrogen ions while a Bronsted-Lowry base accepts hydrogen ionsA Lewis acid accepts lone pairs of electrons while a Lewis base donates lone pairs of electrons.Based on these, we can now classify the reactions accordingly;
Cu^2+ + 4Cl− ------>[CuCl4[^2− Lewis acid-base reaction Al(OH)3 + 3HNO3 -----> Al^3+ + 3H2O + 3NO3^− Arrhenius acid-base reactionN2 + 3H2 ----> 2NH3 NoneCN− + H2O ------> HCN + OH^- Bronsted-Lowry acid-base reactionLearn more: https://brainly.com/question/9352088
The mass of an object with 500 J of kinetic energy moving with a velocity of 5 m/s is kg.
Answer:
[tex]m=20kg[/tex]
Explanation:
Hello,
In this case, we define the kinnetic energy as:
[tex]K=\frac{1}{2} m*v^2[/tex]
Thus, for finding the mass we simply solve for it on the previous equation given the kinetic energy and the velocity:
[tex]m=\frac{2*K}{v^2}=\frac{500kg*\frac{m^2}{s^2} }{(5\frac{m}{s})^2} =\frac{500kg*\frac{m^2}{s^2} }{25\frac{m^2}{s^2}}\\\\m=20kg[/tex]
Best regards.
Answer:
The answer is 40 kg
Explanation:
You will this formula below:
m=[tex]\frac{2*\\KE}{v^{2} }[/tex]
Now we know our formula, now we plug in the given numbers:
m=[tex]\frac{2(500J)}{(5m/s)^2}[/tex]
Simplify and we get:
m=40 kg
I hope this was helpful.
Which table represents a relation that is not function?
Please
Answer:
1
Explanation:
Any relation with a repeated input value is not a function.
Table 1 has the input value 2 listed twice, so does not represent a function.
If you prepare a solution by adding sufficient amount of solute so that after heating and cooling the solution there is a visible amount of solid solute left in the bottom of the beaker, the solution would be considered ________.
Answer:
saturated
Explanation:
Given the equation 2KCIO3(s)=2KCI(s) + 3O2(g). A 3.00-g sample of KCIO3 is decomposed and the oxygen at 24 degrees C and 0.982 atm is collected. What volume of oxygen gas will be collected assuming 100% yield?
Answer:
0.912 L or 912 mL
Explanation:
M(KClO3) = 122.55 g/mol
3.00 g KClO3 * 1 mol/122.55 g = 3.00/122.55 mol =0.02449 mol
2KCIO3(s)=2KCI(s) + 3O2(g)
from reaction 2 mol 3 mol
given 0.02449 mol x
x = 0.02449*3/2 =0.03673 mol O2
T = 24 + 273.15 = 297.15 K
PV = nRT
V= nRT/P = (0.03673 mol*0.082057 L*atm/K*mol*297.15 K)/0.982 atm =
= 0.912 L or 912 mL
The equilibrium constant for the reaction NO2(g)+NO3(g)→N2O5(g) is 2.1x10-20 , therefore: a. At equilibrium, the concentration of products and reactants is about the same. b. At equilibrium, the concentration of products is greater than the reactants. c. At equilibrium, the concentration of reactants is greater than the products
Answer: c. At equilibrium, the concentration of reactants is greater than the products
Explanation:
Equilibrium constant for a reaction is the ratio of concentration of products to the concentration of reactants each raised to the power its stoichiometric coefficients.
For the reaction:
[tex]NO_2(g)+NO_3(g)\rightleftharpoons N_2O_5(g)[/tex]
Equilibrium constant is given as:
[tex]K_{eq}=\frac{[N_2O_5]}{[NO_2]\times [NO_3]}[/tex]
[tex]2.1\times 10^{-20}=\frac{[N_2O_5]}{[NO_2]\times [NO_3]}[/tex]
When
a) K > 1, the concentration of products is greater than the concentration of reactants
b) K < 1, the concentration of reactants is greater than the concentration of products
c) K= 1, the reaction is at equilibrium, the concentration of reactants is equal to the concentration of products
Thus as [tex]K_{eq}[/tex] is [tex]2.1\times 10^{-20}[/tex] which is less than 1,
the concentration of reactants is greater than the concentration of products
Constructive interference occurs when the compression of one wave meets
up with the compression of a second wave.
A. True
B. False
Answer:
True
Explanation:
Ammonia will decompose into nitrogen and hydrogen at high temperature. An industrial chemist studying this reaction fills a tank with of ammonia gas, and when the mixture has come to equilibrium measures the amount of nitrogen gas to be 13. mol. Calculate the concentration equilibrium constant for the decomposition of ammonia at the final temperature of the mixture.
Complete Question
The complete question is shown on the first uploaded image
Answer:
The concentration equilibrium constant is [tex]K_c = 14.39[/tex]
Explanation:
The chemical equation for this decomposition of ammonia is
[tex]2 NH_3[/tex] ↔ [tex]N_2 + 3 H_2[/tex]
The initial concentration of ammonia is mathematically represented a
[tex][NH_3] = \frac{n_1}{V_1} = \frac{29}{75}[/tex]
[tex][NH_3] = 0.387 \ M[/tex]
The initial concentration of nitrogen gas is mathematically represented a
[tex][N_2] = \frac{n_2}{V_2}[/tex]
[tex][N_2] = 0.173 \ M[/tex]
So looking at the equation
Initially (Before reaction)
[tex]NH_3 = 0.387 \ M[/tex]
[tex]N_2 = 0 \ M[/tex]
[tex]H_2 = 0 \ M[/tex]
During reaction(this is gotten from the reaction equation )
[tex]NH_3 = -2 x[/tex](this implies that it losses two moles of concentration )
[tex]N_2 = + x[/tex] (this implies that it gains 1 moles)
[tex]H_2 = +3 x[/tex](this implies that it gains 3 moles)
Note : x denotes concentration
At equilibrium
[tex]NH_3 = 0.387 -2x[/tex]
[tex]N_2 = x[/tex]
[tex]H_2 = 3 x[/tex]
Now since
[tex][NH_3] = 0.387 \ M[/tex]
[tex]x= 0.387 \ M[/tex]
[tex]H_2 = 3 * 0.173[/tex]
[tex]H_2 = 0.519 \ M[/tex]
[tex]NH_3 = 0.387 -2(0.173)[/tex]
[tex]NH_3 = 0.041 \ M[/tex]
Now the equilibrium constant is
[tex]K_c = \frac{[N_2][H_2]^3}{[NH_3]^2}[/tex]
substituting values
[tex]K_c = \frac{(0.173) (0.519)^3}{(0.041)^2}[/tex]
[tex]K_c = 14.39[/tex]
if 196L of air at 1 atm is compressed to 2600mL,what is the new temperature?
Answer:
Around 3.62 degrees kelvin
Explanation:
Assuming this is at STP:
The first step is to convert 2600mL to liters. There are 1000 milliliters in a liter, meaning that this is equal to 2.6L.
Now:
[tex]\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\\\\\dfrac{196(1)}{273}=\dfrac{2.6(1)}{T_2} \\\\\\T_2\approx 3.62K[/tex]
Hope this helps!
Calculate the osmotic pressure of a solution prepared by dissolving 65.0 g of Na2SO4 in enough water to make 500 mL of solution at 20°C. (Assume no ion pairing – in other words, assume that the electrolyte completely dissociates into its constituent ions.)
Answer:
66.0 atm
Explanation:
We can calculate the osmotic pressure (π) using the following expression.
[tex]\pi = i \times M \times R \times T[/tex]
where,
i: van 't Hoff indexM: molarityR: ideal gas constantT: absolute temperatureStep 1: Calculate i
Sodium sulfate completely dissociates according to the following equation.
Na₂SO₄ ⇒ 2 Na⁺ + SO₄²⁻
Since it produces 3 ions, i = 3.
Step 2: Calculate M
We can calculate the molarity of Na₂SO₄ using the following expression.
[tex]M = \frac{mass\ of\ solute }{molar\ mass\ of\ solute\ \times liters\ of\ solution} = \frac{65.0g}{142.04g/mol \times 0.500L} =0.915M[/tex]
Step 3: Calculate T
We will use the following expression.
K = °C + 273.15
K = 20°C + 273.15 = 293 K
Step 4: Calculate π
[tex]\pi = 3 \times 0.915M \times \frac{0.08206atm.L}{mol.K} \times 293K =66.0 atm[/tex]
1. There are how many mol of oxygen in 3.5 mol of caffeine.
Answer:
7 mol
Explanation:
Caffeine molecular formula C8H10N4O2. It has 2 atoms of oxygen.
C8H10N4O2 - 2O
1 mol 2 mol
3.5 mol x mol
x = 3.5*2/1 = 7 mol
Enter your answer in the provided box. To make use of an ionic hydrate for storing solar energy, you place 409.0 kg of sodium sulfate decahydrate on your house roof. Assuming complete reaction and 100% efficiency of heat transfer, how much heat (in kJ) is released to your house at night
Answer:
409.0 kg of sodium sulfate decahydrate will produce 4.49×10⁵ kJ
of heat energy.
Explanation:
CHECK THE COMPLETE QUESTION BELOW
To make use of an ionic hydrate for storing solar energy, you place 409.0 kg of sodium sulfate decahydrate on your house roof. Assuming complete reaction and 100% efficiency of heat transfer, how much heat (in kJ) is released to your house at night? Note that sodium sulfate decahydrate will transfer 354 kJ/mol
EXPLANATION
Here we were asked to calculate the amount of heat will be generated by 409.0 kg of sodium sulfate decahydrate at night assuming there Isa complete reaction and 100% efficiency of heat transfer in the process
The molecular weight of sodium sulfate decahydrate (H₂₀Na₂O₁₄S) is needed here, so it must be firstly calculated.
The molecular weight of sodium sulfate decahydrate (H₂₀Na₂O₁₄S)
( 1*20) + (22.98*2) + (16*14)+ (32*14)= 322.186 g/mol.
Thus 409.0 kg of H₂₀Na₂O₁₄S will have a value which is equivalent to = (409000g)/(322.186 g/mol.)
=1269.453mol of H₂₀Na₂O₁₄S.
But it was stated in the the question that per mole of H₂₀Na₂O₁₄S will transfer 354 kJ heat.
Therefore, 1269.453mol will transfer 1269.453× 354 kJ = 4.49×10⁵ kJ of heat.
Hence, 409.0 kg of sodium sulfate decahydrate will produce
4.49×10⁵ kJ of heat energy.
In general,for a gas at a constant volume?
Answer:
The pressure of a gas is directly proportional to its Kelvin temperature if the volume is kept constant. At constant volume and temperature, the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of the component gases.
Explanation:
When nitrogen dioxide (NO2) gas from car exhaust combines with water in the air, it forms nitrogen oxide and nitric acid (HNO3), which causes acid rain, and nitrogen oxide. Balanced eqjation:
(NO); 3NO2(g) + H20(l) --> 2HNO3(aq) + NO(g).
A) How many molecules of NO2 are needed to react with 0.250 mol of H2O?
B) How many grams of HNO3 are produced when 60.0 g of NO2 completly reacts?
C) How many grams of HNO3 can be produced if 225 g of NO2 is mixed with 55.2 g of H2O?
Answer:
A. 0.75 moles NO2 are required
B. 82.2 gnof HNO3 are produced
C. 205.3 g of HNO3 are produced
Explanation:
Check attachment below for explanation and calculations
A chemistry student weighs out of an unknown solid compound and adds it to of distilled water at . After minutes of stirring, only some of the has dissolved. The student drains off the solution, then washes, dries and weighs the that did not dissolve. It weighs 0.570 kg.
Required:
a. Using the information above, can you calculate the solubility of X?
b. If so, calculate it. Remember to use the correct significant digits and units. .
Complete Question
A chemistry student weighs out 0.950 kg of an unknown solid compound and adds it to 2.00 L of distilled water at . After minutes of stirring, only some of the has dissolved. The student drains off the solution, then washes, dries and weighs the that did not dissolve. It weighs 0.570 kg.
Required:
a. Using the information above, can you calculate the solubility of X?
b. If so, calculate it. Remember to use the correct significant digits and units. .
Answer:
a
Yes the solubility of X can be calculated this is because the solubility of a substance dissolved in a solution is the amount of that substance that is needed to saturate 1 unit volume of the solvent solution at that given temperature.
And from our question we see that substance X saturated the solvent and there is still remained undissolved substance X
b
The solubility of X is [tex]S = 190 g /L[/tex]
Explanation:
From the question we are told that
The initial mass of the unknown solid is [tex]m_i =0. 950 \ kg[/tex]
The mass of the undissolved substance is [tex]m_u = 0.570 \ kg[/tex]
The volume of the solution is [tex]V =2.00\ L[/tex]
Yes the solubility of X can be calculated this is because the solubility of a substance dissolved in a solution is the amount of that substance that is needed to saturate 1 unit volume of the solvent solution at that given temperature.
And from our question we see that substance X saturated the solvent and there is still remained undissolved substance X
The mass of the substance that dissolved ([tex]m_d[/tex] ) is mathematically represented as
[tex]m_d = m_i - m_u[/tex]
[tex]m_d = 0.95 - 0.570[/tex]
[tex]m_d = 0.38 \ kg = 0.38 *1000 = 380 g[/tex]
The solubility of this substance (X) is mathematically represented as
[tex]S = \frac{m_d}{V}[/tex]
substituting values
[tex]S = \frac{ 380}{2}[/tex]
[tex]S = 190 g /L[/tex]
What is the conjugate acid in the following equation hbr + H2O yields h30 positive + BR negative
Answer:
HBr + H2O = H3O+ + Br-
So our conjugate acid is the H3O+ to H2O
Explanation:
A conjugate acid of a base results when the base accepts a proton.
Consider ammonia reacting with water to form an equilibrium with ammonium ions and hydroxide ions:
NH3 (aq) + H2O (l) ⇌ NH4+ (aq) + OH- (aq)
Ammonium, NH4+, acts as a conjugate acid to ammonia, NH3.
Which of the following aqueous solutions are good buffer systems? . 0.24 M hydrochloric acid + 0.23 M sodium chloride 0.28 M ammonia + 0.35 M ammonium nitrate 0.16 M barium hydroxide + 0.28 M barium bromide 0.15 M nitrous acid + 0.14 M potassium nitrite 0.35 M calcium nitrate + 0.21 M calcium iodide
Answer: 0.28 M ammonia + 0.35 M ammonium nitrate and 0.15 M nitrous acid + 0.14 M potassium nitrite
Explanation:
Buffer solution is the solution which resists the change in the magnitude of the pH when small additions of either acid or base is added.
Acidic Buffer solutions consist of weak acid and its conjugate base usually mixed in relatively equal and large quantities.
Basic Buffer solutions consist of weak base and its conjugate acid usually mixed in relatively equal and large quantities.
Thus 0.28 M ammonia + 0.35 M ammonium nitrate ( weak base + conjugate acid) and 0.15 M nitrous acid + 0.14 M potassium nitrite (weak acid + conjugate base) are good buffer systems
The aqueous solutions that are good buffer systems are:
0.28 M ammonia + 0.35 M ammonium nitrate. 0.15 M nitrous acid + 0.14 M potassium nitrite.We want to determine which of the given solutions would make a good buffer.
What is a buffer?A buffer is a solution used to resist abrupt changes in pH when an acid or a base is added.
What kinds of buffers exist?Acidic buffer: formed by a weak acid and its conjugate base.Basic buffer: formed by a weak base and its conjugate acid.Which of the following aqueous solutions are good buffer systems?
0.24 M hydrochloric acid + 0.23 M sodium chloride. No, since HCl is a strong acid.0.28 M ammonia + 0.35 M ammonium nitrate. Yes, it would be a good basic buffer.0.16 M barium hydroxide + 0.28 M barium bromide. No, since Ba(OH)₂ is a strong base. 0.15 M nitrous acid + 0.14 M potassium nitrite. Yes, it would be a good acidic buffer.0.35 M calcium nitrate + 0.21 M calcium iodide. No, since no acids nor bases are present.The aqueous solutions that are good buffer systems are:
0.28 M ammonia + 0.35 M ammonium nitrate. 0.15 M nitrous acid + 0.14 M potassium nitrite.Learn more about buffers here: brainly.com/question/24188850