Answer:
272.33 J/Kg°C
Explanation:
Data obtained from the question include the following:
Weight of metal = 30 N
Heat used (Q) = 1.25×10⁴ J
Change in temperature (ΔT) = 15.0 °C.
Specific heat capacity (C) =..?
Next, we shall determine the mass of the metal.
The mass of the metal can be obtained as follow:
Weight (W) = mass (m) x acceleration due to gravity (g)
W = mg
Weight of metal = 30 N
Acceleration due to gravity = 9.8 m/s²
Mass (m) =..?
W = mg
30 = m x 9.8
Divide both side by 9.8
m = 30/9.8
m = 3.06 Kg
Finally, we shall determine the specific heat capacity of the metal as show below:
Heat used (Q) = 1.25×10⁴ J
Change in temperature (ΔT) = 15.0 °C.
Mass (m) = 3.06 Kg
Specific heat capacity (C) =..?
Q = mCΔT
1.25×10⁴ = 3.06 x C x 15
Divide both side by 3.06 x 15
C = (1.25×10⁴) / (3.06 x 15)
C = 272.33 J/Kg°C
Therefore, the specific heat capacity of metal is 272.33 J/Kg°C.
. Calculate the final Celsius temperature of sulfur dioxide gas if 50.0 mL of the gas at 20 C and 0.450 atm is heated until the pressure is 0.750 atm. Assume that the volume remains constant.
Answer:
The final temperature of sulfur dioxide gas is 215.43 C
Explanation:
Gay Lussac's Law establishes the relationship between the temperature and the pressure of a gas when the volume is constant. This law says that if the temperature increases the pressure increases, while if the temperature decreases the pressure decreases. In other words, the pressure and temperature are directly proportional quantities.
Mathematically, the Gay-Lussac law states that, when a gas undergoes a transformation at constant volume, the quotient of the pressure exerted by the temperature of the gas remains constant:
[tex]\frac{P}{T}=k[/tex]
Assuming you have a gas that is at a pressure P1 and at a temperature T1 at the beginning of the experiment, by varying the temperature to a new value T2, then the pressure will change to P2, and it will be true:
[tex]\frac{P1}{T1} =\frac{P2}{T2}[/tex]
The reference temperature is the absolute temperature (in degrees Kelvin)
In this case:
P1= 0.450 atmT1= 20 C= 293.15 K (being 0 C= 273.15 K)P2=0.750 atmT2= ?Replacing:
[tex]\frac{0.450atm}{293.15 K} =\frac{0.750 atm}{T2}[/tex]
Solving:
[tex]T2 =\frac{0.750 atm}{\frac{0.450atm}{293.15 K} }[/tex]
[tex]T2=\frac{0.750 atm}{0.450 atm} *293.15K[/tex]
T2=488.58 K
Being 273.15 K= 0 C, then 488.58 K= 215.43 C
The final temperature of sulfur dioxide gas is 215.43 C
The reaction, 2 NO(g) + O2(g) → 2 NO2(g), was found to be first order in each of the two reactants and second order overall. The rate law is therefore
Answer:
[tex]r=k[NO][O_2][/tex]
Explanation:
Hello,
In this case, rate laws allows us to compute how fast a chemical reaction is carried out by means of the change in the concentration of the species affecting the rate. In such a way, since the statement says that the reaction was found to be first order to both nitrogen monoxide and oxygen, it means that their concentrations are powered to first power by separated. It also implies that the overall order is second-order since the specific orders are added (powers properties). Therefore, the rate law is:
[tex]r=k[NO][O_2][/tex]
Whereas k is the rate constant and we find the concentration of the reactants to the first power each one.
Best regards.
Which of the following is the correct equation for the reaction below?
A. CH (g) + O2 (g) CO (g) + H2O (g)
B. CH (g) + 2O (g) CO (g) + 2HO (g)
C. CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g)
D. CH4 (g) + 2O2 (g) CO2 (g) + H2O (g)
Answer:
(c) is the correct answer
CH4(g)+2O2(g)⟶CO2(g)+2H2O(l)
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Draw the structure of beta-D-idose in its pyranose form.
Answer:
See figure 1
Explanation:
In this case, we can start with the linear structure of D-Idose. Then, if we have a "D" configuration the "OH" in the last chiral (carbon 5) will be in the right. This carbon will attack carbon 1 and we will produce a cyclic structure with 6 members (pyranose). Additionally, we have to keep in mind that we want the "beta" structure. So, the "OH" on carbon 1 must point up (red arrow). Finally, we will have a cyclic structure with 6 atoms and the "OH" on carbon 1 pointing up.
See figure 1
I hope it helps!
2) Os foguetes são utilizados para levar pessoas ao espaço (os astronautas), mas principalmente cargas como, por exemplo, os satélites artificiais, os telescópios espaciais, levar sondas a outros planetas etc. Escreva V(verdadeiro) ou F (falso) em cada afirmação.
( ) Foguetes só levam astronautas ao espaço.
( ) Satélites artificiais servem para ajudar na previsão do clima.
( ) Satélites artificiais "fotografam" o planeta para descobrir queimadas ilegais.
( ) Satélites artificiais permitem vermos jogos ao vivo até do Japão.
( ) Foguetes são movidos com pólvora e dinamite.
Answer:
F, V, V , V, F
Explanation:
1 - "Os foguetes são utilizados para levar pessoas ao espaço (os astronautas), mas principalmente cargas como, por exemplo, os satélites artificiais, os telescópios espaciais, levar sondas a outros planetas etc".
2 - Tipo Meteorologia: utilizados para monitorar o tempo e o clima no planeta Terra, por exemplo, os da série Meteosat.
3 - ...
4 - ...
5 - Usam combustivel solido, liquido, hibridos (solido e liquido), iônica:
Solido:
São sistemas simples que unem os dois propelentes envolvidos em uma massa sólida que, quando inflamada, não para de queimar até o esgotamento completo.
Liquido:
São muito mais complexos e envolvem o bombeamento de quantidades imensas de propelentes para as câmaras de combustão dos motores.
Hibridos:
O propelente sólido – normalmente o combustível – é distribuído ao longo do tanque de maneira homogênea. O propelente líquido ou gasoso "normalmente o oxidante" fica armazenado em tanques.
Podem ser desligados depois de sofrerem ignição, além de permitirem um controle de queima relativamente preciso.
Iônica:
Usando eletricidade (captada por painéis solares ou gerada por reatores atômicos) para ionizar átomos (normalmente gases nobres, como xenônio), e expulsá-los em velocidades altíssimas.
It takes 242. kJ/mol to break a chlorine-chlorine single bond. Calculate the maximum wavelength of light for which a chlorine-chlorine single bond could be broken by absorbing a single photon. Round your answer to 3 significant digits. single by absorbing a significant digit.
Answer:
495nm
Explanation:
The energy of a photon could be obtained by using:
E = hc / λ
Where E is energy of a photon, h is Planck's constant (6.626x10⁻³⁴Js), c is speed of the light (3x10⁸ms⁻¹) and λ is wavelength.
The energy to break 1 mole of Cl-Cl bonds is 242kJ = 242000J. The energy yo break a single bond is:
242000J/mol ₓ (1mol / 6.022x10²³bonds) = 4.0186x10⁻¹⁹J/bond.
Replacing in the equation:
E = hc / λ
4.0186x10⁻¹⁹J = 3x10⁸ms⁻¹ₓ6.626x10⁻³⁴Js / λ
λ = 4.946x10⁻⁷m
Is maximum wavelength of light that could break a Cl-Cl bond.
Usually, wavelength is given in nm (1x10⁻⁹m / 1nm). The wavelength in nm is:
4.946x10⁻⁷m ₓ (1nm / 1x10⁻⁹m) =
495nmWhich of the following solutions would be least acidic? Assume all of the acids are the same concentration and at 25°C. The acid is followed by its Ka.
a) Hydrofluoric acid, 3.5. 10-4
b) Hydrocyanic acid, 4.9. 10-10
c) Nitrous acid, 4.6. 10-4
d) Unable to be determined by Ka
Answer:
Option (b) Hydrocyanic acid, 4.9×10^-10
Explanation:
Data obtained from the question include:
Ka of Hydrofluoric acid = 3.5×10^-4
Ka of Hydrocyanic acid = 4.9×10^-10
Ka of Nitrous acid = 4.6×10^-4
To know which acid is least acidic, we shall determine the the pKa value for each acid.
This is illustrated below:
For Hydrofluoric acid
Ka = 3.5×10^-4
pKa =..?
pKa = –Log Ka
pKa = –Log 3.5×10^-4
pKa = 3.5
For Hydrocyanic acid
Ka = 4.9×10^-10
pKa =..?
pKa = –Log Ka
pKa = –Log 4.9×10^-10
pKa = 9.3
For Nitrous acid
Ka = 4.6×10^-4
pKa =..?
pKa = –Log Ka
pKa = –Log 4.6×10^-4
pKa = 3.3
Summary:
Acid >>>>>>>>>>>>> Ka >>>>>>>> pKa
Hydrofluoric acid >> 3.5×10^-4 >> 3.5
Hydrocyanic acid >> 4.9×10^-10 > 9.3
Nitrous acid >>>>>>> 4.6×10^-4 >> 3.3
NB: The smaller the pKa value, the more acidic the compound is and the larger the pKa value, the less acidic the compound will be.
From the above calculations, Hydrocyanic acid has the highest pKa value.
Therefore, Hydrocyanic acid is the least acidic compound
An aqueous solution is made by dissolving 29.4 grams of aluminum acetate in 433 grams of water. The molality of aluminum acetate in the solution is
Answer:
0.333 m
Explanation:
Molality (m) is moles of solute over kilograms of solvent.
Convert grams of the solute (aluminum acetate) to moles.
(29.4 g)/(204.11 g/mol) = 0.144 mol
Convert grams of the solvent (water) to kilograms.
433 g = 0.433 kg
Divide the solute by the solvent.
(0.144 mol)/(0.433 kg) = 0.333 m
The molality of the solution is 0.333 m.
Answer to the best of your ability please
Answer:
The answer to your question is given below.
Explanation:
To draw the structure of 2–methyl–1–butanamine, we following must be observed:
1. The functional group of the compound is amine –NH2.
2. The functional group is located at carbon 1.
3. The longest continuous carbon chain is carbon 4 i.e butane. Since the functional group is amine, the –e at the end of the butane is replaced with
–amine, making the name to be butanamine.
4. Methyl, CH3 is located at carbon 2.
5. Combine the above to get the structure of 2–methyl–1–butaamine.
Please see attached photo for the structure of 2–methyl–1–butanamine
1. Suppose 1.00 g of NaOH is used to prepare 250 mL of an NaOH solution. Compare the expected molarity of this solution to the actual average molarity you measured in the standardization. What do you notice? 2. Do you think the results would have been more accurate if a different type of acid or base were used in the standardization? Why, or why not? 3. There are many different primary standards that could be used in a standardization titration. What are the criteria for a primary standard?
Answer:
See explanation
Explanation:
The calculated concentration of the sodium hydroxide is;
Number of moles= mass/molar mass = 1g/40gmol-1 = 0.025 moles
Concentration= number of moles/volume= 0.025×1000/250 = 0.1 M
This calculated concentration will be different from the molarity of NaOH obtained by standardization with acid. The result will not be more accurate if a different acid is used for the standardization this is because sodium hydroxide is deliquescent and absorbs moisture thereby leading to inaccuracy in the calculated molarity.
Any substance that must be used as a primary standard must not absorb moisture, it must be stable and it must be a substance in its pure form.
The following reaction is part of the electron transport chain. Complete the reaction and identify which species is reduced. The abbreviation Q represents coenzyme Q. Use the appropriate abbreviation for the product.
FADH2+Q→
The reactant that is reduced is: _____
Answer:
[tex]FADH_2+Q --> FAD + QH_2[/tex]
The reactant that is reduced is Q.
Explanation:
The complete equation for the reaction is such that:
[tex]FADH_2+Q --> FAD + QH_2[/tex]
Two molecules of H atom is lost from [tex]FADH_2[/tex] and the H atoms are gained by the coenzyme Q. Consequently, [tex]FADH_2[/tex] becomes FAD while Q becomes [tex]QH_2[/tex].
From the definition of oxidation as loss of hydrogen and reduction as the addition of hydrogen, it can be concluded that the FADH2 that lost hydrogen is a reactant that is oxidized while the coenzyme Q that gained hydrogen is a reactant that is reduced in the reaction.
The free energy obtained from the oxidation (reaction with oxygen) of glucose (C6H12O6) to form carbon dioxide and water can be used to re-form ATP by driving the above reaction in reverse. Calculate the standard free energy change for the oxidation of glucose.
Answer:
The correct answer is -2878 kJ/mol.
Explanation:
The reaction that takes place at the time of the oxidation of glucose is,
C₆H₁₂O₆ (s) + 6O₂ (g) ⇒ 6CO₂ (g) + 6H₂O (l)
The standard free energy change for the oxidation of glucose can be determined by using the formula,
ΔG°rxn = ∑nΔG°f (products) - ∑nΔG°f (reactants)
The ΔG°f for glucose is -910.56 kJ/mol, for oxygen is 0 kJ/mol, for H2O -237.14 kJ/mol and for CO2 is -394.39 kJ/mol.
Therefore, ΔG°rxn = 6 (-237.14) + 6 (-394.39) - (-910.56)
ΔG°rxn = -2878 kJ/mol
Calculate the volume in liters of a M mercury(II) iodide solution that contains of mercury(II) iodide . Round your answer to significant digits.
Answer:
41L
Explanation:
Of a 4.8x10⁻⁵M mercury (II) iodide that contains 900mg of mercury (II) iodide. 2 significant digits
Molarity, M, is an unit of concentration in chemistry defined as the ratio between moles of solute (Mercury (II) iodide in this case) per Liter of solution.
A 4.8x10⁻⁵M solution contains 4.8x10⁻⁵ moles of solute per liter.
Now, 900mg = 0.900g of mercury (II) iodide (Molar mass: 454.4g/mol) are:
0.900g × (1mol / 454.4g) = 1.98x10⁻³moles of HgI₂
If in 1L there are 4.8x10⁻⁵ moles of HgI₂, There are 1.98x10⁻³moles of HgI₂ in:
1.98x10⁻³moles of HgI₂ ₓ (1L / 4.8x10⁻⁵moles) =
41LWhat is the balanced oxidation half-reaction for the following reaction? Cu2+(aq) + Fe(s) → Cu(s) + Fe2+(aq) Question 8 options: There is no reaction Cu2+(aq) + 2e– → Cu(s) Fe2+(aq) + 2e– → Fe(s) Cu(s) → Cu2+(aq) + 2e– Fe(s) → Fe2+(aq) + 2e–
Answer:
Fe(s) → Fe2+(aq) + 2e-
Explanation:
Cu2+(aq) + Fe(s) → Cu(s) + Fe2+(aq)
Oxidation is the loss of electrons. When the oxidation number of an element increases, that means there is a loss of electrons and that element is being oxidized.
The oxidation half equation in this reaction is;
Fe(s) → Fe2+(aq)
The loss of electron is represented in the product side and is given by;
Fe(s) → Fe2+(aq) + 2e-
Scoring Scheme: 3-3-2-1 Use one of your experimentally determined values of k, the activation energy you determined, and the Arrhenius equation to calculate the value of the rate constant at 25 °C. Alternatively, you can simply extrapolate the straight line plot of ln(k) vs. 1/T in your notebook to 1/298 , read off the value of ln(k), and determine the value of k. Please put your answer in scientific notation. k (25 °C) = my slope is: -16538 my k values are: 0.00057 0.0017 0.00525 0.0238 0.1386 my activation energy is: 137.5 and my temperatures are: 45, 55, 65, 75, and 80 in celsius. the order of temp and k values correspond to each other.
Answer:
K = 2.7x10⁻⁵ at 25ºC
Explanation:
A way to write Arrhenius equation is:
ln K = - Ea/R × (1/T) + lnA
If you graph ln K as Y and 1/T as X (Absolute temperature in K), the equation you will obtain is:
Y = -13815X +35.817
R² = 0.9927
(Taking the last k point as 0.0386) (ln 0.0386), 0.1386 has no sense)
Your slope is -13815
-13815K = - Ea/R
-13815K×8.314J/molK = 114858J/mol = Ea
And your intercept =
lnA = 35.817
A = 3.59x10¹⁵
Now, you want to know rate constant at 25ºC = 298.15K. Replacing in the equation (Where Y is ln (activation energy) and X is 1/T):
Y = -13815X +35.817
Y = -13815(1/298.15K) +35.817
Y = -10.5187
lnK = -10.5187
K = 2.7x10⁻⁵ at 25ºCchromium (iii) or chromium(ii) are frequently used to apply chrome finish to sink fixtures such as faucets. if 45.2 Amps flow through a solution of chromium (iii) for 2 hours, how many grams of chromium can be deposited on a fixture A...175.35g B...58.45g c...0.016g d....0.974g
Answer:
58.45g is the answer
Explanation:
took the test
The mass of chromium that can be deposited is equal to 58.45 g. Therefore, option B is correct.
What is electric current?For a steady flow of charge through a conductor, the current can be determined with the following equation:
[tex]{\displaystyle I={Q \over t}[/tex]
where Q is the electric charge while time t. If Q and t are measured in coulombs (C) and seconds then I will be in amperes.
Electric charge flows by electrons, from lower potential to higher electrical potential. Any stream of charged objects can constitute an electric current.
Given, the amount of electric current flowing through the solution:
I = 45.2 A
The time for which the current flows, t = 2 hrs = 2 × 60 ×60 = 7200 sec
The charge flowing through the solution, Q = I × t
Q = 45.2 × 7200
Q = 325440 C
The number of moles of electrons in 325440 C charge = 325440/96500 = 3.37 mol
We know Cr³⁺ + 3e⁻ → Cr (s)
3 moles of electrons deposit of chromium = 1 mol
3.37 mol of electrons deposit of chromium = 3.37/3 = 1.12 mol
The mass of chromium in 1.12 mol = 1.12 × 52 = 58.45 g
Learn more about electric current, here:
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If a boy (m = 50kg) at rest on skates is pushed by another boy who exerts a force of 200 N on him and if the first boy's final velocity is 8 m/s, what was the contact time? t= s
Answer:
t = 2 seconds
Explanation:
It is given that,
Mass of a boy, m = 50 kg
Initial speed of boy, u = 0
Final speed of boy, v = 8 m/s
Force exerting by another boy, F = 200 N
Let t is the time of contact. The force acting on an object is given by :
F = ma
a is acceleration
So,
[tex]F=\dfrac{m(v-u)}{t}\\\\t=\dfrac{m(v-u)}{F}\\\\t=\dfrac{50\times 8}{200}\\\\t=2\ s[/tex]
So, the contact time is 2 seconds.
Answer:
t=2 s
Explanation:
Sodium pentothal is a short-acting barbiturate derivative used as a general anesthetic and known in popular culture as truth serum. It is synthesized like other barbiturates, but uses thiourea, (H2N)2C=S, in place of urea. The mechanism involves the following steps:
1. Ethoxide ion deprotonates malonic ester, forming enolate anion 1;
2. Enolate anion 1 acts as a nucleophile in an SN2 reaction with ethyl bromide, forming alkylated intermediate 2;
3. Ethoxide ion deprotonates alkylated intermediate 2, forming enolate anion 3;
4. Enolate anion 3 acts as a nucleophile in an SN2 reaction with 2-bromopentane, forming alkylated intermediate 4;
5. Alkylated intermediate 4 reacts with thiourea to form tetrahedral intermediate 5;
6. Tetrahedral intermediate 5 collapses, expelling ethoxide ion and forming intermediate 6;
7. Intermediate 6 reacts with sodium hydroxide to form sodium pentothal.
Draw the mechanism out on a separate sheet of paper and then draw the structure of enolate anion 1. You do not have to consider stereochemistry. You do not have to explicitly draw H atoms. Do not include lone pairs in your answer. They will not be considered in the grading.
Answer:
Figure 1
Explanation:
In this case, we have to ester with a "malonic synthesis" in which we have to add a strong (ethoxide) to produce an enolate ion that would form a new C-C bond with an alkyl halide (ethyl bromide and bromo pentane). Then a "nucleophilic acyl substitution reaction" takes place to add thiourea, in this step two ethanol groups are eliminated to form a cyclic structure. Finally, an "elimination reaction" happen by the addition of sodium hydroxide generating a double bond and a negative charge in the sulfur atom that is neutralized with the positive charge of sodium.
See figure 1 for the total mechanism
I hope it helps!
What is the percent yield for a chemical reaction if the actual yield is 36 g and the theorical yield is 45 g.
Answer:
⇒ Percent yield = 80 %
Explanation:
Given:
Actual yield = 36 g
Theoretical yield = 45 g
Find:
Percent yield
Computation:
⇒ Percent yield = [Actual yield / Theoretical yield] 100%
⇒ Percent yield = [36 / 45] 100%
⇒ Percent yield =[0.8] 100%
⇒ Percent yield = 80 %
Assume that a compound is a cyclic, planar, completely conjugated ring. Which number of p electrons would make it aromatic?
a) 0 p electrons
b) 2 p electrons
c) 3 p electrons
d) 4 p electrons
e) 32 p electrons
Answer:
option b is correct
2 p electron makes aromatic
Explanation:
An aromatic compound which is cyclic, planar and has a complete conjugate ring must have (4n + 2)pi electrons(Huckel's rule)
Huckel's Standard (4n+2 rule): For a compound to be an aromatic, a particle must have a specific number of pi (electrons with pi bonds, or lone pairs inside p orbitals) inside a shut loop of parallel, adjoining p orbitals. The pi electron tally is characterized by the arrangement of numbers created from 4n+2 where n = zero or any positive whole number (i..e, n = 0, 1, 2, and so forth.). The most widely recognized case in six pi electrons (n = 1) which is found for example in benzene, pyrrole, furan, and pyridine.
where n is the number of pi electrons
where n = 0
(4n +2) pi electrons = 2pi electrons
attached is an example of aromatic which is cyclic, planar and a complete conjugate ring
Answer:
2 p electrons.
Explanation:
For any compound to be considered an aromatic compound it must be cyclic,flat, conjugated and it must obey Huckel's rule that states an aromatic compound must have 4n + 2 pi electrons in it's p orbitals for it to be an aromatic compound.
n can represent an integer from 2, 6,10, 14,........
The lone pair is actually in a pure 2p orbital perpendicular to the ring, which means they count as π electrons.
Content attribution
QUESTION 2 • 1 POINT
Which anion would bond with K+ in a 1: 1 ratio to form a neutral ionic compound?
The given question is incomplete. The complete question is :
Which anion would bond with K+ in a 1: 1 ratio to form a neutral ionic compound?
a) [tex]O^{2-}[/tex]
b) [tex]F^{-}[/tex]
c) [tex]N^{3-}[/tex]
d) [tex]S^{2-}[/tex]
Answer: b) [tex]F^{-}[/tex]
Explanation:
For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.
Here potassium is having an oxidation state of +1 called as cation and thus is an anion must have an oxidation state of -1 if they have to combine in 1: 1 ratio to give neutral ionic compound.
Thus the anion has to be [tex]F^-[/tex] which combines with [tex]K^+[/tex] in 1: 1 ratio to give [tex]KF[/tex]
Two football players are running toward each other. One football player has a mass of 105 kg and is running at 8.6 m/s. The other player has a mass of 90 kg and is running at -9.0 m/s. What is the momentum of the system after the football players collide? 93 kg · m/s 1,713 kg · m/s. 810 kg · m/s. 903 kg · m/s.
Answer:
Total momentum of both player after collision =93 Kg m/s
Explanation:
According to law of conservation of momentum
For an isolated system of bodies , momentum of bodies before and after collision remains same.
momentum is given by mass* velocity
_________________________________________
Here the isolated system of bodies are
two football players.
Momentum of player before collision
Momentum of player 1 = 105*8.6 = 903 Kg m/s
Momentum of player 2 = 90*-9 = -810 Kg m/s
Total momentum of both player before collision = 903 + (-810) = 93 Kg m/s
as by conservation of
Total momentum of both player before collision = Total momentum of both player after collision
Total momentum of both player after collision =93 Kg m/s
Answer:A is the Answer
Explanation:
Balance the following
Na+02-→ Na20
Al+O2 ->Al2O3
H2+12+ ->HI
Mg+H2O → Mg(OH)2+H2
Ca+O2 -> Cao
Answer:
1. Na + O2 → Na2O (Balanced)
2. 4Al + 3O2 → 2(Al2O3) (Balanced)
3. H2 + i2 → 2HI (Balanced)
4. Mg + 2H2O → Mg(OH)2+ H2 (Balanced)
5. 2Ca +O2 → 2CaO (Balanced)
A 25.0-mL sample of 0.100M Ba(OH)2(aq) is titrated with 0.125 M HCl(aq).
How many milliliters of the titrant will be needed to reach the equivalence point?
Answer:
20.0
Explanation:
NaOH = (25.0) (0.100m) \ 0.125M = 20.0mL
Given 3.4 grams of x compound with a molar mass of 85 g and 4.2 grams of y compound with a molar mass of 48 g How much of compound xy can be generated 2x + y2 = 2xy
Answer:
[tex]4.36~g~XY[/tex]
Explanation:
In this case, we can start with the reaction:
[tex]2X + Y_2~->~2XY[/tex]
If we check the reaction, we will have 2 X and Y atoms on both sides. So, the reaction is balanced. Now, the problem give to us two amounts of reagents. Therefore, we have to find the limiting reagent. The first step then is to find the moles of each compound using the molar mass:
[tex]3.4~g~X\frac{1~mol~X}{85~g~X}=0.04~mol~X[/tex]
[tex]4.2~g~Y_2\frac{1~mol~Y_2}{48~g~Y_2}=0.0875~mol~Y_2[/tex]
Now, we can divide by the coefficient of each compound (given by the balanced reaction):
[tex]\frac{0.04~mol~X}{1}=~0.04[/tex]
[tex]\frac{0.0875~mol~Y_2}{2}=0.04375[/tex]
The smallest value is for "X", therefore this is our limiting reagent. Now, if we use the molar ratio between "X" and "XY" we can calculate the moles of XY, so:
[tex]0.04~mol~X\frac{2~mol~XY}{2~mol~X}=0.04~mol~XY[/tex]
Finally, with the molar mass of "XY" we can calculate the grams. Now, we know that 1 mol X = 85 g X and 1 mol [tex]Y_2[/tex] = 48 g [tex]Y_2[/tex] (therefore 1 mol Y = 24 g Y). With this in mind the molar mass of XY would be 85+24 = 109 g/mol. With this in mind:
[tex]0.04~mol~XY\frac{109~g~XY}{1~mol~XY}=4.36~g~XY[/tex]
I hope it helps!
1. In this experiment, the procedure instructs you to dissolve solid potassium hydrogen tartrate (KHT) in two different solvents. What are these two solvents? (2 pts)
Answer:
Water
Explanation:
Solid potassium hydrogen tartrates (KHT) is soluble in water. This is especially at room temperature.
The solvent for KHT is water.
The amount of space an object takes up is called _____. gravity weight mass volume
A meteorologist filled a weather balloon with 3.00L of the inert noble gas helium. The balloon's pressure was 765 torr. The balloon was released to an altitude with a pressure of 530 torr. What was the volume (L) of the weather balloon
Answer:
4.33 L
Explanation:
Step 1: Given data
Initial volume of the balloon (V₁): 3.00 L
Initial pressure of the balloon (P₁): 765 torr
Final volume of the balloon (V₂): ?
Final pressure of the balloon (P₂): 530 torr
Step 2: Calculate the final volume of the balloon
If we consider Helium to behave as an ideal gas, we can calculate the final volume of the balloon using Boyle's law.
[tex]P_1 \times V_1 = P_2 \times V_2\\V_2 = \frac{P_1 \times V_1}{P_2} = \frac{765torr \times 3.00L}{530torr} = 4.33 L[/tex]
In laboratory experiment, a NOVDEC Student was
required to prepare 500 cm3 of Im Solution of
glucose (c6, H12,06) Determine the
i Molar
Mass
ii) Amount of ghicoseB. In in moles in the Solrition
[ C= 12, H = 10, 0=16]
Answer:
i. Molar mass of glucose = 180 g/mol
ii. Amount of glucose = 0.5 mole
Explanation:
The volume of the glucose solution to be prepared = 500 [tex]cm^3[/tex]
Molarity of the glucose solution to be prepared = 1 M
i. Molar mass of glucose ([tex]C_1_2H_6O_6[/tex]) = (6 × 12) + (12 × 1) + (6 × 16) = 180 g/mol
ii. mole = molarity x volume. Hence;
amount (in moles) of the glucose solution to be prepared
= 1 x 500/1000 = 0.5 mole
The volume of a sample of oxygen is 300mL when the pressure is 1 atm and the temperature is 27 C . At what temperature is the volume 1.00 L and the pressure.500 atm?
Answer:
T2 = 500K
Explanation:
Given data:
P1 = 1atm
V1 = 300ml
T1= 27 + 273 = 300K
T2 = ?
V2 = 1.00ml
P2 = 500atm
Apply combined law:
P1xV1//T1 = P2xV2/T2 ...eq1
Substituting values into eq1:
1 x 300/300 = 500 x 1/T2
Solve for T2:
300T2 = 500 x 300
300T2 = 150000
Divide both sides by the coefficient of T2:
300T2/300 = 150000/300
T2 = 500K