Answer:
Explanation:
we know that
ΔH=m C ΔT
where ΔH is the change in enthalpy (j)
m is the mass of the given substance which is water in this case
ΔT IS the change in temperature and c is the specific heat constant
we know that given mass=2.9 g
ΔT=T2-T1 =98.9 °C-23.9°C=75°C
specific heat constant for water is 4.18 j/g°C
therefore ΔH=2.9 g*4.18 j/g°C*75°C
ΔH=909.15 j
Which of the following is required for the flow of current in all systems?
a) the presence of ions
b) an electrical potential ofo
c) a closed circuit
d) a short circuit
Answer:
I would say c) a closed circuit.
Hope I was right.
When equal moles of an acid and a base are mixed, after reaction the two are compounds are said to be at the _______________. Select one: Indicator point Stoichiometric point Equilibrium point End point
Answer:
when equal moles of an acid and base are mixed,after reaction the two are compounds are said to be at the Equivalent point.
P.1 is a variant of SARS-CoV2. This is the so-called "Brazil variant". It has the amino acid the substitution N501Y in the spike protein on the virus surface. In cellular fluids at ~pH 7 the wild type spike protein has a net negative charge. This substitution will make the spike protein
more negative
less hydrophobic
able to absorb move UV light
less negative
There are various variants of Cov id virus. The Brazilian variant P also known as Gamma variant is the third variant of the original SARS-CoV2.
The correct answer is more negative
This variant has raised concerns since it has ability to spread more quickly then previous variants and this is more negative variant.
It is assumed that Cov id variant Gamma and Delta has ability to absorb move UV light but this is not proved yet and research is underway.
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The decomposition of ethylene oxide(CH₂)₂O(g) → CH₄(g) + CO(g)is a first order reaction with a half-life of 58.0 min at 652 K. The activation energy of the reaction is 218 kJ/mol. Calculate the half-life at 629 K.
Answer:
Half-life at 629K = 252.4min
Explanation:
Using Arrhenius equation:
[tex]ln\frac{K_1}{K_2} = \frac{Ea}{R} (\frac{1}{T_2} -\frac{1}{T_1})[/tex]
And as Half-life in a first order reaction is:
[tex]t_{1/2}=\frac{ln2}{K}[/tex]
We can convert the half-life of 58.0min to know K₁ adn replacing in Arrhenius equation find half-life at 629K:
[tex]58.0min=\frac{ln2}{K}[/tex]
K = 0.01195min⁻¹ = K₁
[tex]ln\frac{0.01195min^{-1}}{K_2} = \frac{218kJ/mol}{8.314x10^{-3}kJ/molK} (\frac{1}{629K} -\frac{1}{652K})[/tex]
[tex]ln\frac{0.01195min^{-1}}{K_2} =1.47[/tex]
[tex]\frac{0.01195min^{-1}}{K_2} =4.35[/tex]
K₂ = 2.75x10⁻³ min⁻¹
And, replacing again in Half-life expression:
[tex]t_{1/2}=\frac{ln2}{2.75x10^{-3}min^{-1}}[/tex]
Half-life at 629K = 252.4minThe half-life of the first-order reaction of ethylene oxide decomposition at 629 K is 251.1 min when the half-life at 652 K is 58.0 min and the activation energy is 218 kJ/mol.
The activation energy of a reaction is related to its rate constant as follows:
[tex] k = Ae^{-\frac{E_{a}}{RT}} [/tex] (1)
Where:
k: is the rate constant A: is the pre-exponential factor[tex]E_{a}[/tex]: is the activation energy of the reaction = 218 kJ/mol R: is the gas constant = 8.314 J/(K*mol)T: is the temperature
We can find the rate constant of the first-order reaction at 652 K with the half-life as follows:
[tex]k_{652} = \frac{ln(2)}{t_{1/2}_{(652)}}[/tex] (2)
Where [tex]t_{1/2}_{(652)}[/tex] is the half-life at 652 K= 58.0 min
Hence, the rate constant at 652 K is:
[tex] k_{652} = \frac{ln(2)}{58.0 min} = 0.012 min^{-1} [/tex]
Now, from equation (1) we can find the pre-exponential factor (A):
[tex]A = \frac{k_{652}}{e^{(-\frac{E_{a}}{RT_{1}})}} = \frac{0.012 \:min{-1}}{e^{(-\frac{218\cdot 10^{3} \:J/mol}{8.314 \:J/(K*mol)*652 \:K})}} = 3.51 \cdot 10^{15} min^{-1}[/tex]
With the pre-exponential factor we can calculate the rate constant at 629 K (eq 1):
[tex]k_{629} = 3.51 \cdot 10^{15} min^{-1}*e^{(-\frac{218 \cdot 10^{3} J/mol}{8.314 J/(K*mol)*629 K})} = 2.76 \cdot 10^{-3} min^{-1}[/tex]
Finally, the half-life at 629 K is (eq 2):
[tex] t_{1/2}_{629} = \frac{ln(2)}{2.76\cdot 10^{-3} min^{-1}} = 251.1 min [/tex]
Therefore, the half-life at 629 K is 251.1 min.
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A qualitative researcher may select a ______ case that is unusually rich in information pertaining to the research question.
A) critical
B) typical
C) deviant
D) rare
Answer:
D. Rare
Explanation:
Qualitative research has to do with non-numerical data and is used to understand the beliefs of a group of people which can be gotten from surveys, questionnaires, interviews, etc.
A qualitative researcher may select a RARE case that is unusually rich in information pertaining to the research question.
This is because he wants to get an insight into the why, how, where and when of that particular ase as it's not a usual occurrence.
Nitrogen gas can be prepared by passing gaseous ammonia over solid copper (II) oxide at high temperatures. If 18.1 g of Nh3 is reacted with 90.4 g of CuO, which is the limiting reactant? How many grams of N2 will be formed? Explain how you solved for your answers.
Balanced Equation: 2NH3 + 3CuO → 3Cu + N2 + 3H2O
I first converted the given grams of the reactants into moles, and then divided the moles by the coefficients in front of each of the reactant. The result with the smallest value will be the limiting reactant, and the value of CuO was the smallest, so it's the limiting reactant.
After figuring out which reactant is the limiting one, I took their given grams and converted it into moles, the divided it by the ratio of N2 to CuO (it's in the equation) to obtain the moles of N2, and then multiply it with the molar mass of N2 to get its mass in grams.
1.46 mol sample of nitrogen gas at a temperature of 14.0 °C is found to occupy a volume of 23.7 liters. The pressure of this gas sample is mm Hg
Explanation:
Since there are no gas constant with the unit of mmHg, you just have to use the gas constant for atm, and then convert that into mmHg
If you drip an ink drop into a cup of water and wait for a few seconds, all the water will be colored with the ink. This experiment is an example of facilitated diffusion ?
true
false
Answer:
false, it is not an example of facilitated diffusion
Answer:
TrueExplanation:
When a drop of ink added into the water gradually moves in the whole quantity of water due to this entire water turns into blue color. This is nothing but the diffusion of ink particles into the water molecules. This is because water, as well as ink molecules, are in random motion due to the motion of ink substance.
Match the tools with the advantages they offer to astronomers.
photography
space telescope
radio telescope
optical telescope
w
detects electromagnetic frequencies outside
the visible spectrum that reaches Earth
captures images that can be shared and
compared by scientists
obtains a magnified and clear view of a part
of the sky to observe celestial objects
allows access to images taken from outside
Earth's atmosphere
Answer:
- photography (captures images that can be shared and
compared by scientists)
- space telescope (allows access to images taken from outside
Earth's atmosphere)
- radio telescope (detects electromagnetic frequencies outside
the visible spectrum that reaches Earth)
- optical telescope (obtains a magnified and clear view of a part
of the sky to observe celestial objects)
Explanation:
Photography is used to capture still images based on the principle that some compounds react in the presence of optical energy.
Space telescope is a type of observatory telescope positioned in outer space to observe distant planets, galaxies and other astronomical objects. Space telescopes reduces the interference from ultraviolet frequencies, X-rays and gamma rays; as well as light pollution which ground-based observatories encounter.
A radio telescope is a specialized antenna and radio receiver used to receive radio waves from astronomical radio sources in the sky. Radio telescope is used to study radio frequencies emitted by astronomical objects, that fall outside the visible light spectrum.
An optical telescope is used to gather and focuses light, from a far distant object. Optical telescope is used within the visible light spectrum of the electromagnetic spectrum, to create a magnified image, for direct view, or to make a photograph, or to collect data through electronic image sensors.
Answer:
- photography (captures images that can be shared and
compared by scientists)
- space telescope (allows access to images taken from outside
Earth's atmosphere)
- radio telescope (detects electromagnetic frequencies outside
the visible spectrum that reaches Earth)
- optical telescope (obtains a magnified and clear view of a part
of the sky to observe celestial objects)
Explanation:
Write chemical equations for the acid-base reactions that occur when:
a. Solutions of HC₂H₃O₂ (acetic acid) and KOH are mixed.
b. Solutions of HCN (hydrocyanic acid) and Na₂CO₃ are mixed.
Acetic acid is basically known as CH_3COOH.
Chemical Equation:-
[tex]\boxed{\sf {CH_3COOH\atop Acetic\:acid}+{KOH\atop Potassium\:Hydroxide}\longrightarrow {CH_3COOK\atop Potassium\: Acetate}+{H_2O\atop Water}}[/tex]
This kinda reaction is called Neutralization Reaction.Solution:-2
[tex]\boxed{\sf {HCN\atop Hydrogen\:Cyanate}+{Na_2CO_3\atop Sodium\:Carbonate}\longrightarrow {NaCN\atop Sodium\:Cyanate}+{H_2O\atop water}+{CO_2\atop Carbon\:dioxide}}[/tex]
Hydro cyanic acid is also called HydrogenCyanate.Calculate the enthalpy change (∆H) for the reaction- N2(g) + 3 F2(g) –––> 2 NF3(g) given the following bond enthalpies: N≡N 945 kJ/mol F–F 155 kJ/mol N–F 283 kJ/mol
Answer:
– 844 kJ/mol.
Explanation:
The following data were obtained from the question:
N2(g) + 3 F2(g) –––> 2 NF3(g)
Enthalpy of N≡N (N2) = 945 kJ/mol
Enthalpy of F–F (F2) = 155 kJ/mol
Enthalpy of N–F3 (NF3) = 283 kJ/mol
Enthalpy change (∆H) =?
Next, we shall determine the enthalpy of reactant.
This is illustrated below:
Enthalpy of reactant (Hr) = 945 + 3(155)
Enthalpy of reactant (Hr) = 945 + 465
Enthalpy of reactant (Hr) = 1410 kJ/mol
Next, we shall determine the enthalpy of the product.
This is illustrated below:
Enthalpy of product (Hp) = 2 x 283
Enthalpy of product (Hp) = 566 kJ/mol
Finally, we shall determine the enthalpy change (∆H) for the reaction as follow:
Enthalpy of reactant (Hr) = 1410 kJ/mol
Enthalpy of product (Hp) = 566 kJ/mol
Enthalpy change (∆H) =?
Enthalpy change (∆H) = Enthalpy of product (Hp) – Enthalpy of reactant (Hr)
Enthalpy change (∆H) = 566 – 1410
Enthalpy change (∆H) = – 844 kJ/mol
Answer:
– 844 kJ/mol.
Explanation:
The following data were obtained from the question:
N2(g) + 3 F2(g) –––> 2 NF3(g)
Enthalpy of N≡N (N2) = 945 kJ/mol
Enthalpy of F–F (F2) = 155 kJ/mol
Enthalpy of N–F3 (NF3) = 283 kJ/mol
Enthalpy change (∆H) =?
Next, we shall determine the enthalpy of reactant.
This is illustrated below:
Enthalpy of reactant (Hr) = 945 + 3(155)
Enthalpy of reactant (Hr) = 945 + 465
Enthalpy of reactant (Hr) = 1410 kJ/mol
Next, we shall determine the enthalpy of the product.
This is illustrated below:
Enthalpy of product (Hp) = 2 x 283
Enthalpy of product (Hp) = 566 kJ/mol
Finally, we shall determine the enthalpy change (∆H) for the reaction as follow:
Enthalpy of reactant (Hr) = 1410 kJ/mol
Enthalpy of product (Hp) = 566 kJ/mol
Enthalpy change (∆H) =?
Enthalpy change (∆H) = Enthalpy of product (Hp) – Enthalpy of reactant (Hr)
Enthalpy change (∆H) = 566 – 1410
Enthalpy change (∆H) = – 844 kJ/mol
Explanation:
What are some geographic features that could be found in the hydrosphere?
Lakes, oceans, glaciers, clouds, etc. It categorizes all forms of water on earth.
hydro = water
Answer:
Lakes, streams, ground water, polar ice caps, glaciers, water vapor, and rivers!
Explanation:
The hydrosphere is made up of all the water on Earth. So anything that is water, like oceans, can be found in the hydrosphere:)
Check 0/1 ptRetries 5 Element R has three isotopes. The isotopes are present in 0.0825, 0.2671, and 0.6504 relative abundance. If their masses are 97.62, 109.3, and 138.3 respectively, calculate the atomic mass of element R.
Answer:
Atomic mass = 127.198 amu
Explanation:
The average atomic mass is obtained by summing the masses of the isotopes each multiplied by its abundance.
Atomic mass = (97.62 * 0.0825) + (109.3 * 0.2671) + (138.3 * 0.6504)
Atomic mass = 8.05365 + 29.19403 + 89.95032
Atomic mass = 127.198 amu
Compare and contrast the quantum mechanical model and Niels Bohr of the atom
Answer:
the quantum mechanical model VS. Neils Bohr of the atom
Explanation:
The Bohr model and quantum model are models that explain the structure of an atom. ... The key difference between Bohr and quantum model is that Bohr model states that electrons behave as particles whereas quantum model explains that the electron has both particle and wave behavior.
A solution containing a unknown ionic compound, vigorously bubbles when hydrochloric acid (HCl) is added to the solution. This might indicate that the solution contains which anion?
Answer:
CO3^2-
Explanation:
In qualitative analysis, we try to use chemical reactions to determine the composition of an unknown substance. The addition of certain reagents to the unknown solution gives certain results that show the presence or absence of certain species from the unknown sample.
When dilute HCl is added to an unknown sample and effervescence is observed, then the unknown sample must contain CO3^2- or HCO3^-. The presence of these species is confirmed if the gas evolved is passed through limewater and the gas turns limewater milky.
Please Help! Use Hess’s Law to determine the ΔHrxn for: Ca (s) + ½ O2 (g) → CaO (s) Given: Ca (s) + 2 H+ (aq) → Ca2+ (aq) + H2 (g) ΔH = 1925.9 kJ/mol 2 H2 (g) + O2 (g) → 2 H2O (l) ΔH = −571.68 kJ/mole CaO (s) + 2 H+ (aq) → Ca2+ (aq) + H2O (l) ΔH = 2275.2 kJ/mole ΔHrxn =
Answer:
ΔHrxn = -635.14kJ/mol
Explanation:
We can make algebraic operations of reactions until obtain the desire reaction and, ΔH of the reaction must be operated in the same way to obtain the ΔH of the desire reaction (Hess's law). Using the reactions:
(1)Ca(s) + 2 H+(aq) → Ca2+(aq) + H2(g) ΔH = 1925.9 kJ/mol
(2) 2H2(g) + O2 g) → 2 H2O(l) ΔH = −571.68 kJ/mole
(3) CaO(s) + 2 H+(aq) → Ca2+(aq) + H2O(l) ΔH = 2275.2 kJ/mole
Reaction (1) - (3) produce:
Ca(s) + H2O(l) → H2(g) + CaO(s)
ΔH = 1925.9kJ/mol - 2275.2kJ/mol = -349.3kJ/mol
Now this reaction + 1/2(2):
Ca(s) + ½ O2(g) → CaO(s)
ΔH = -349.3kJ/mol + 1/2 (-571.68kJ/mol)
ΔHrxn = -635.14kJ/molFor the cell notation shown, which species is undergoing reduction?
Answer:
option (A) is right answer
Answer:
c
Explanation:
trust me I took the exam
Explain why only the lone pairs on the central atom are taken into consideration when predicting molecular shape
Answer:
Lone pairs cause more repulsion than bond pairs
Explanation:
A lone pair takes up more space around the central atom than bond pairs of electrons. This is because, a lone pair is attracted to only one nucleus while bond pairs are attracted to two nuclei.
Hence the repulsion between lone pairs is far greater than the repulsion between bond pairs or repulsion between a lone pair and a bond pair. The presence of a lone pair therefore distorts a molecule away from the ideal shape predicted on the basis of the valence shell electron pair repulsion theory.
Lone pairs are found to decrease the observed bond angles in a molecule.
Which one of the following compound does not undergo an aldol addition reaction in presence of aqueous sodium hydroxide?
a. butanal
b. 2-methylbutanal
c. 3-methylpentanal
d. 2, 2-dimethylbutanal
The compound does not undergo an aldol addition reaction in presence of aqueous sodium hydroxide is 2, 2-dimethylbutanal.
What is aldol reaction?The Aldol Reaction occurs when the enolate of an aldehyde or ketone combines with the carbonyl of another molecule at the aplha-carbon under basic or acidic circumstances to produce beta-hydroxy aldehyde or ketone.
For the formation of enolate ion, compound should contain alpha hydrogen in it and among the given compound only 2, 2-dimethylbutanal will not have alpha hydrogens.
Butanal, 2-methylbutanal and 3-methylpentanal will have aplha hydrogens in it so that it takes part in the aldol reaction.Hence 2, 2-dimethylbutanal does not undergo an aldol addition reaction.
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PLEASE HELP!! 40 POINTS
Answer:
1) 6.524779402×10^(-17)
2)521.1g
3)113
Explanation:
Answer: 1) 6.524779402×10^(-17)
2)521.1g
Explanation:
How many grams of magnesium chloride can be produced from 2.30 moles of chlorine gas reacting w excess magnesium Mg(s)+Cl2(g)->MgCl2(s)
The mass of magnesium chloride produced from 2.30 moles of chlorine gas is 218.99 grams.
How to calculate moles in stoichiometry?Stoichiometry refers to the study and calculation of quantitative (measurable) relationships of the reactants and products in chemical reactions.
According to this question, magnesium reacts with chlorine gas to form magnesium chloride as follows:
Mg + Cl₂ → MgCl₂
Based on the above chemical equation, 1 mole of chlorine gas forms 1 mole of magnesium chloride.
This means that 2.30 moles of chlorine gas will 2.30 moles of magnesium chloride.
Next, we convert moles of magnesium chloride to mass as follows:
molar mass of magnesium chloride = 95.211g/mol
mass of magnesium chloride = 95.211 × 2.30 = 218.99 grams.
Therefore, 218.99 grams of magnesium chloride will be formed.
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Chlorine monoxide and dichlorine dioxide are involved in the catalytic destruction of stratospheric ozone. They are related by the equation:
2ClO(g) ⇌ Cl2O2(g) for which Kc is 4.96×10^11 at 273 K.
For an equilibrium mixture in which [Cl2O2] is 6.00 x 10^-6M, what is [ClO]?
Answer:
[ClO] = 3.48×10¯⁹ M.
Explanation:
The following data were obtained from the question:
Equilibrium constant (Kc) = 4.96×10¹¹
Concentration of Cl2O2, [Cl2O2] = 6x10¯⁶ M.
Concentration of ClO, [ClO] =.?
The equation for the reaction is given below:
2ClO(g) ⇌ Cl2O2(g)
The equilibrium constant for a reaction is simply defined as the ratio of the concentration of product raised to their coefficient to the concentration of the reactant raised to their coefficient.
The equilibrium constant, Kc for the reaction is given by:
Kc = [Cl2O2] / [ClO]²
Thus, we can calculate the concentration of ClO, [ClO] as follow:
Kc = [Cl2O2] / [ClO]²
4.96×10¹¹ = 6x10¯⁶ / [ClO]²
Cross multiply
4.96×10¹¹ × [ClO]² = 6x10¯⁶
Divide both side by 4.96×10¹¹
[ClO]² = 6x10¯⁶ / 4.96×10¹¹
[ClO]² = 1.21×10¯¹⁷
Take the square root of both side
[ClO] = √ (1.21×10¯¹⁷)
[ClO] = 3.48×10¯⁹ M
Therefore, the concentration of ClO, [ClO] is 3.48×10¯⁹ M.
Are acids harmful to work with.
Answer:
yes it is
Explanation:
because there are some acid which really harm skin.
Which of the following could be considered a scientific statement?
(A) I believe there is life on other planets.
(B) I observed that bees prefer red flowers.
(C) I think cake tastes much better than cookies.
(D) I consider yellow a cheerful color.
In the reaction 2 AgI + HgI 2 → Ag 2HgI 4, 2.00 g of AgI and 3.50 g of HgI 2 were used. What is the limiting reactant?
Answer:
AgI I the limiting reactant.
Explanation:
The balanced equation for the reaction is given below:
2AgI + HgI2 → Ag2HgI4
Next, we shall determine masses AgI and HgI2 that reacted from the balanced equation.
This is illustrated below:
Molar mass of Agl = 108 + 127 = 235 g/mol
Mass of AgI from the balanced equation = 2 x 235 = 470 g
Molar mass of HgI2 = 201 + (2x127) = 455 g/mol
Mass of HgI2 from the balanced equation = 1 x 455 = 455 g
From the balanced equation above,
470 g of AgI reacted with 455 g of HgI2.
Finally, we shall determine the limiting reactant as follow:
From the balanced equation above,
470 g of AgI reacted with 455 g of HgI2.
Therefore, 2 g of AgI will react with
= (2 x 455)/470 = 1.94 g of HgI2.
From the calculations made above, we can see that only 1.94 g out of 3.50 g of HgI2 given is needed to react completely with 2 g of AgI.
Therefore, AgI I the limiting reactant.
Three structural isomers have the formula C5H12.C5H12. Draw and name the isomers using IUPAC names. Draw the isomer with five carbon atoms in main chain.
Answer:
Explanation:
Answer in attached file .
g . Calculate the molar concentration for each of the following solutions. (a) 1.50 g NaCl in 100.0 mL of solution (b) 1.50 g K2Cr2O7 in 100.0 mL of solution (c) 5.55 g CaCl2 in 125 mL of solution (d) 5.55 g Na2SO4 in 125 mL of solution
Answer:
(a) [tex]M=0.257M[/tex]
(b) [tex]M=0.0510M[/tex]
(c) [tex]M =0.500M[/tex]
(d) [tex]M= 0.391M[/tex]
Explanation:
Hello,
In this case, since the molarity or molar concentration of a solution is computed by dividing the moles of solute by the volume of solution in liters, we proceed as follows:
(a) The molar mass of sodium chloride is 58.45 g/mol and the volume in liters is 0.100 L, therefore, the molarity is:
[tex]M=\frac{1.50gNaCl}{0.100L} *\frac{1molNaCl}{58.45gNaCl} =0.257M[/tex]
(b) The molar of potassium dichromate is 294.2 g/mol and the volume in liters is 0.100 L, therefore, the molarity is:
[tex]M=\frac{1.50gK_2Cr_2O_7}{0.100L} *\frac{1molK_2Cr_2O_7}{294.2gK_2Cr_2O_7} =0.0510M[/tex]
(c) The molar of calcium chloride is 111 g/mol and the volume in liters is 0.125 L, therefore, the molarity is:
[tex]M=\frac{5.55gCaCl_2}{0.100L} *\frac{1molCaCl_2}{111gCaCl_2} =0.500M[/tex]
(d) The molar of sodium sulfate is 142 g/mol and the volume in liters is 0.125 L, therefore, the molarity is:
[tex]M=\frac{5.55gNa_2SO_4}{0.100L} *\frac{1molNa_2SO_4}{142gNa_2SO_4} = 0.391M[/tex]
Best regards.
Calculate the mass of naphthalene required to react stoichiometrically with the moles of 2-bromo-2-methylpropane that you have placed in the flask. Dispense that quantity of naphthalene into the weighing dish. There is some imprecision in delivering the powder onto the weigh boat, but a mass within 0.040 g of the calculated stoichiometric quantity will be sufficient for this experiment. Note that you can use a combination of the 1 g and the 0.1 g buttons to add the required mass. If you overshoot the goal, you can discard the weighing boat and get a new one. RECord the sample mass dispensed below. mass of naphthalene (g)
The reaction of 2-bromo-2-methylpropane and naphthalene is stoichiometrically in a mole ratio of 1:1.
Notice that in the question, the moles of 2-bromo-2-methylpropane placed in the flask wasn't mentioned. if I assume that it was 1.24 moles then;
1 mole of naphthalene reacts with 1 mole of 2-bromo-2-methylpropane
x moles of naphthalene reacts with 1.24 mole of 2-bromo-2-methylpropane
x = 1 * 1.24/1
= 1.24 moles of naphthalene
Molar mass of naphthalene = 128.2 g/mol
Mass of naphthalene = 128.2 g/mol * 1.24 moles of naphthalene
Mass of naphthalene =159 g
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How should the magnetic field es be drawn for the magnets shown below!
Answer:
The magnetic field shows (D)
Explanation:
Because like poles repel eachother
The magnetic field is represented by two poles of a magnetic with magnetic filed lines from the poles. The image C represents the magnetic filed .
What is magnetic field?A magnetic filed is a generated by strong filed lines from a magnet with two poles. A magnet have two poles namely magnetic south and north poles.
The magnetic south poles of of a magnet is always attracted towards the magnetic north poles. Magnetic north poles of a magnet is generally shown in the left side and the field lines will be outward towards the geographic north pole.
The magnetic field lines from the south pole is inward and attracted towards the north. Therefore, right figure representing the magnetic filed is option C.
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Calculate the pH of a 0.20 M NH3/0.20 M NH4Cl buffer after the addition of 25.0 mL of 0.10 M HCl to 65.0 mL of the buffer.
Answer:
Explanation:
For pH of a buffer solution , the formula is
pH = pKa + log [ Base ] / [ conjugate acid ]
= pKa + log [ NH₃ ] / [ NH₄⁺ ]
Ka = Kw / Kb
Kb for NH₄OH = 1.8 x 10⁻⁵
Ka = 10⁻¹⁴ / 1.8 x 10⁻⁵
= 5.6 x 10⁻¹⁰
pH = - log ( 5.6 x 10⁻¹⁰ ) + log 0.2 / 0.2
= 10 - log 5.6
= 9.25
Effect of addition of HCl
H⁺ of HCl will react with NH₃ to produce NH₄⁺
25 mL of .1 HCl = 2.5 mM of HCl
25 mL of .1 NH₄⁺ = 2.5 mM of NH₄⁺
65 mL of .2 M NH₃ = 13 mM of NH₃
65 mL of .2 M NH₄⁺ = 13 mM of NH₄⁺
NH₃ + H⁺ = NH₄⁺
NH₄⁺ formed = 2.5 + 13 mM
15.5 mM of NH₄⁺
NH₃ = 13 mM
Concentration of NH₃ = 13 / 90
Concentration of NH₄⁺ = 15.5 / 90
pH of final buffer mixture
= 9.6 + log 13 / 15.5
= 9.25 - .076
= 9.174
The pH value is mathematically given as
pH= -6.332.
What is the pH of a 0.20 M NH3/0.20 M NH4Cl buffer?Question Parameters:
the pH of a 0.20 M NH3/0.20 M NH4Cl buffer
the addition of 25.0 mL of 0.10 M HCl to 65.0 mL of the buffer.
Generally, the equation for the Chemical Reaction is mathematically given as
HCl + NH3 --> NH4^+ + Cl^-
Therefore
pH= pka + log(13/14).
pH= -6.3 + log 0.93.
pH= -6.3+ (-0.032).
pH= -6.332.
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