Answer:
ΔH = -59.6kJ/mol
Explanation:
The reaction that occurs between Ag⁺ and Cl⁻ ions is:
Ag⁺ + Cl⁻ → AgCl(s) + ΔH
To find ΔH we need to obtain moles of reaction and heat released in the reaction because ΔH is defined as heat released per mole of reaction.
Moles of reaction:
Moles of Ag⁺ and Cl⁻ added are:
Ag⁺: 0.100L * (0.100mol / L) = 0.01moles
Cl⁻: 0.100L * (0.200mol / L) 0 0.02 moles
That means limiting reactant is Ag⁺ and moles of reaction are 0.01 moles
Heat released:
To find heat released we must use coffe cup calorimeter equation:
Q = C*m*ΔT
Where C is specific heat of solution (4.18J/g°C), m is the mass of solution (200g because there are 100 + 100mL = 200mL and density of solution is 1g/mL) and ΔT is change in temperature (25.30°C - 24.60°C = 0.70°C).
Replacing:
Q = C*m*ΔT
Q = 4.18J/g°C * 200g * 0.70°C
Q = 585,2J
Is total heat released.
The calorimeter absorbs:
15.5J / °C * 0.7°C = 10.85
Thus, when 0.01 moles reacts, 585.2J + 10.85 = 596.05J are released (Heat released is heat abosrbed by calorimeter + Heat absorbed by water) and ΔH is:
ΔH = 596.05J / 0.01 moles =
ΔH = 59605J / mol =
ΔH = -59.6kJ/molAs heat is released, ΔH < 0.
6. Describe the energy inputs and outputs for the campfire. Use the law of conservation of energy to construct a valid qualitative equation that includes all the input and output energies involved.
Answer:
Energy in the campfire originates from the potential chemical energy of the wood, before it is burnt to warm and give light around the campfire.
Explanation:
For a camp fire, the energy input is in the form of the potential chemical energy, stored up in the firewood used to fuel the flame.
The energy output is in the form of heat energy that the campfire radiates all around, light energy given off from the flame, and a little bit of sound energy, heard in the cracking of the firewood as they burn in the flame.
chemical energy ⇒ heat energy + light energy + sound energy
The energy inputs of a campfire are logs and fuel, which is the stored potential energy in the form of chemical energy. The output of a campfire is light and heat energy.
What is energy?
Energy is an entity that can not be created, not destroyed. It's just transformed into other forms. There are different forms of energy present in the environment.
The energy input for a campfire comes in the form of the chemical potential energy that is stored in the firewood used to feed the blaze.
The campfire's energy production takes the form of heat energy that it radiates outward, light energy that it emits, and a little amount of sound energy from the crackling of the fuel as it burns.
Chemical potential energy ⇒ heat energy + light energy + sound energy.
Thus, logs and fuel, which is chemical energy that has been stored as potential energy, are the sources of energy for a campfire. A campfire produces light and heat energy.
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How do biotic factors impact ecological organization?
Answer:
plz mark as BRAINLIEST plz
Explanation:
The biotic factors in an ecosystem are the living organisms, such as animals. Biotic factors in an ecosystem are the participants in the food web, and they rely on each other for survival. ... These living organisms affect each other and influence the health of the ecosystem.
what do you mean by saying that the molality of a solution is 2?
Answer:
2 molal solution means that 2 moles of solute is dissolved in 1 kg of solvent.
What volume (mL) of a 0.3428 M HCl(aq) solution is required to completely neutralize 23.55 mL of a 0.2350 M Ba(OH)2(aq) solution
Answer:
The volume required for complete neutralize is 32.29 mL
Explanation:
The computation of the volume required for complete neutralize is shown below:
As we know that, the balanced equation is
[tex]Ba(OH)_2 + 2Hcl\rightarrow Bacl_2 + 2H_2O[/tex]
Now
The number of moles of [tex]Ba(OH)_2[/tex] = n_1 = 1
And, the number of moles of Hcl = n_2 = 2
Therefore
The equation i.e. to be used to find out the volume is given below:
[tex]\frac{M_1V_1}{n_1} = \frac{M_2V_2}{n_2}[/tex]
[tex]V_2 = \frac{M_1V_1}{n_1} \times \frac{n_2}{M_2} \\\\ = \frac{0.2350 \times 23.55}{1} \times \frac{2}{0.3428} \\\\ = \frac{11.0685}{0.3428}[/tex]
= 32.29 mL
Hence, the volume is 32.29mL
Which of the following statements is true concerning the reaction given below?2H2S(g) + O2(g) → 2S(s) + 2H2O(g)a. The reaction is second-order in H2S(g) and first-order in O2(g).b. The reaction is first-order in H2S(g) and second-order in O2(g).c. The rate law is Rate = k[H2S]2[O2].d. The rate law is Rate = k[H2S][O2].e. The rate law may be determined only by experiment.
Answer:
The rate law may be determined only by experiment.
Explanation:
For a reaction, A + B ---> C, the rate law can only be determined from experimental data. Chemists determine the rate of reaction by carefully observing the changes in the concentration of species as the reaction progresses.
Hence, the rate law is not determined by inspection of the chemical reaction equation, it must be obtained from the experimental data, hence the answer given.
When cyclohexene is treated with KMnO4, H2O, the syn-1,2-diol is produced. What reaction occurs when benzene is similarly treated
Answer:
No reaction is observed
Explanation:
The benzene ring is aromatic. Being an aromatic ring, the benzene ring is remarkably stable to all reactions that destroy the aromatic ring.
Alkenes are oxidized to alkanols in the presence of KMnO4 but this reaction does not occur with benzene. However, substituted benzenes having hydrogen atoms attached to the substituent carbon atom can be oxidized to the corresponding carboxylic acid.
If the OH‑ ion concentration in an aqueous solution at 25.0 °C is 6.6 x 10‑4 M, what is the molarity of the H+ ion?
Answer:
1.5 × 10⁻¹¹ M
Explanation:
Step 1: Given data
Concentration of OH⁻ ([OH⁻]): 6.6 × 10⁻⁴ MTemperature: 25°CConcentration of H⁺ ([H⁺]): ?Step 2: Consider the self-ionization of water
H₂O(l) ⇄ H⁺(aq) + OH⁻(aq)
Step 3: Calculate the molar concentration of H⁺
We will use the equilibrium constant for the self-ionization of water (Kw).
Kw = 1.0 × 10⁻¹⁴ = [H⁺] × [OH⁻]
[H⁺] = 1.0 × 10⁻¹⁴ / [OH⁻]
[H⁺] = 1.0 × 10⁻¹⁴ / 6.6 × 10⁻⁴
[H⁺] = 1.5 × 10⁻¹¹ M
What is the percent by mass of the solution formed when 5.0 g of solute is dissolved in 40. g of water
Answer:
2
Explanation:
so let say 5.0/100×40
that is 2
thank you please follow me
The percent by mass is the estimation of the solute present in the solution or the substance. The percent by mass of 5 gm solute is calculated to be 11.11%.
What is the percent by mass?The percent by mass is defined as the division of the mass of the solute of the element by the mass of the compound or the solution as the whole. The formula for percent by mass is given as:
mass percent = (mass of solute ÷ total mass of compound or solution) x 100%
The mass of the solute has been the numerator and the mass of the solution has been the denominator that must be multiplied by 100 to get the percent by mass or mass percentage. It is represented using %.
Given,
Mass of solute = 5.00 grams
Mass of compound (solution) = solute + water
= 40.0 grams + 5.00 grams
= 45.00 grams
Substituting the values above as:
mass percent = (mass of solute ÷ total mass of compound or solution) x 100%
The percent by mass = (5.00 ÷ 45.00) x 100%
= 11.11 %
Therefore, 11.11 % is the percent by a mass of 5.00-gram solute dissolved in 40.0 grams of water.
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A 14.584 g sample of CaCl2 was added to 12.125 g of K2CO3 and mixed in water. What is the limiting reactant and what is the theoretical yield of CaCO3
Answer:
The correct answer is 8.786 g CaCO₃
Explanation:
The balanced reaction is the following:
CaCl₂(ac) + K₂CO₃(ac) → CaCO₃(s) + 2 KCl(ac)
From the stoichiometry, 1 mol of CaCl₂ (111 g) reacts with 1 mol of K₂CO₃ (138 g) to form 1 mol CaCO₃(100 g) and 2 moles of KCl (149 g).
The stoichiometric ratio CaCl₂/K₂CO₃ is: 111 g/138 g= 0.80 g CaCl₂/K₂CO₃.
We have 14.584 g CaCl₂ and 12.125 g K₂CO₃, which gives a ratio of: 14.584g/12.125 g= 1.2 g CaCl₂/K₂CO₃.
0.8 ∠ 1.2 ⇒ K₂CO₃ is the limiting reactant
We use the limiting reactant to calculate the grams of CaCO₃ produced. For this, we know that from 138 g K₂CO₃ 100 g of CaCO₃ are produced. So, we multiply the amount of K₂CO₃ by this stoichiometric ratio to obtain the grams of CaCO₃ produced:
12.125 g K₂CO₃ x 100 g CaCO₃/138 g K₂CO₃= 8.786 g CaCO₃
Therefore, the theoretical yield of CaCO₃ is 8.786 g.
Is it possible for a piece of cork to have more mass than a piece of rock?
Answer:
Yes
Explanation:
It is entirely possible for a piece of cork to have more mass than a piece of rock. This would largely depend on the volume of the two objects. The density of a substance remains the same irrespective of the size of the substance.
Density = mass/volume
At constant density, it means that the mass of an object depends only on the volume of the object.
Hence, a piece of cork can weigh more than a piece of rock if the volume or size of the cork is bigger enough than that of the rock.
Draw line structures for the cis and trans configurations of CH3CH2CH=CHCH3.
Answer:
See attached picture.
Explanation:
Hello,
In this case, on the attached picture you will find the required line structures for the cis and trans configurations of the given compound (2-pentene). Take into account for the cis that the adjacent carbons to those having the double bond remain in the same plane, whereas for the trans one, the adjacent carbons remain in a different plane.
Regards-
(5.391 ´ 10-7) + (2.5531 ´ 10-6) =
Answer:
[tex](5.391 * 10^{-7}) + (2.5531 * 10^{-6}) = 3.0922 * 10^{-6}[/tex]
Explanation:
Given
[tex](5.391 * 10^{-7}) + (2.5531 * 10^{-6})[/tex]
Required
Solve
[tex](5.391 * 10^{-7}) + (2.5531 * 10^{-6})[/tex]
Remove the brackets
[tex]5.391 * 10^{-7} + 2.5531 * 10^{-6}[/tex]
Express [tex]10^{-7}[/tex] as [tex]10^{-1-6}[/tex]
[tex]5.391 * 10^{-1-6} + 2.5531 * 10^{-6}[/tex]
Apply law of indices
[tex]5.391 * 10^{-1} * 10^{-6} + 2.5531 * 10^{-6}[/tex]
[tex]0.5391 * 10^{-6} + 2.5531 * 10^{-6}[/tex]
Factorize:
[tex](0.5391 + 2.5531)10^{-6}[/tex]
[tex](3.0922)10^{-6}[/tex]
Remove bracket
[tex]3.0922 * 10^{-6}[/tex]
Hence:
[tex](5.391 * 10^{-7}) + (2.5531 * 10^{-6}) = 3.0922 * 10^{-6}[/tex]
Three different samples were weighed using a different type of balance for each sample. The three were found to have masses of 0.6160959 kg, 3.225 mg, and 5480.7 g. The total mass of the samples should be reported as?
Answer:
6.096799125kg
Explanation:
According to the question, three different samples weighed using different types of balance had masses: 0.6160959 kg, 3.225 mg, and 5480.7 g.
Based on observation, the mass units in the three measurements are different but must be uniform in order to find the total mass. Hence, we need to convert to the standard unit (S.I unit of mass), which is kilograms (kg)
Since 1kg equals 1,000,000mg
Hence, 3.225mg will be 3.225/1000000
= 0.000003225kg
Also, 1kg equals 1000g
Hence, 5480.7g will be 5480.7/1000
= 5.4087kg
Hence, the total mass of the three samples (now in the same unit) are:
5.4807kg + 0.000003225kg + 0.6160959 kg
= 6.096799125kg
The distance between the nuclei of two iron atoms is about 4 angstroms.a. Trueb. False
Answer:
The correct option is b. false
Explanation:
The distance between the nucleus of an atom and it's outermost shell is called is atomic radius. The atomic radius of an Iron atom (Fe) is 0.126 nm or 1.26 angstrom. The distance between the nuclei of two Iron atoms will be 1.26 × 2 = 2.52 angstroms.
Since 2.52 angstroms is lower than 4 angstroms, the correct option is false
What happens to the molecules of a liquid when it cools
Answer:
As the molecules of a liquid are cooled they slow down. As the molecules slow down they take up less volume. Taking up less room because of the molecules lower energy causes the liquid to contract.
Explanation:
Atoms of elements at the top of a group on the periodic table are smaller than the atoms of elements at the bottom of the group. How does this help explain the difference in the reactivity of metals within a group?
Answer:
a
Explanation:
a
A 147-g piece of metal has a density of 7.00 g/ml. 50- ml graduated cylinder contains 20.0 ml of water what is the final volume after the metal is added to the graduated cylinder
Answer:
The final volume was 41
Explanation:
m = 147 grams
d = 7.00 g/mL
V = x - 20
=========
d = m/V
=========
7 = 147 / (x - 20)
Multiply both sides by x - 20
7*(x - 20) = 147
Divide both sides by 7
x - 20 = 147 / 7
x - 20 = 21
Add 20 to both sides
x = 21 + 20
x = 41
The final volume was 41
1.How did the electrolyte properties of the NaCl solution compare as you added more salt to the water
Explanation:
The dissociation of NaCl increases. This happens because of the Le Chatalier's principle which says that, the rate of reaction shift to product side when more product is added to the reaction. Therefore, its electrolytic properties are enhanced.
Which of the following species is not capable of acting as an Arrhenius acid in aqueous solution?
A) CHCl_3
B) HNO_3
C) H_2SO_4
D) H_3O^+
E) HClO_4
Answer:
CHCl_3
Explanation:
An Arrhenius acid produces H^+ in solution. If we look at the options provided, all the other species are acids that contain at least one replaceable hydrogen ion hence they possess H^+ and can act as Arrhenius acids in solution.
However, CHCl_3 does not contain a replaceable hydrogen ion hence it does not function as an Arrhenius acid.
The species that should not be capable of acting as an Arrhenius acid in an aqueous solution is option A. CHCl_3.
What is Arrhenius acid?it is a compound that rised the concentration of hydrogen ion (H +) in an aqueous solution. Here the CHCl_3 should not generate the proton at the time when it should be mixed with water since it creates two immiscible layer. Moreover, the HNO3, H2SO4, H3O+, and HCIO4 should produce proton in aqueous.
Therefore, the option A should be considered.
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The displacement by CH3CO2- on (a) bromoethane or (b) bromocyclohexane. ___ Submit AnswerTry Another Version
Answer:
The displacement by [tex]CH_{3}CO_{2}^{-}[/tex] on bromoethane.
Explanation:
Given that,
The displacement by [tex]CH_{3}CO_{2}^{-}[/tex] on bromoethane or bromocyclohexane.
We know that,
Bromoethane is not stable. It can easily break. But bromocyclohexane is more stable obstruct and very strong to displace.
Bromocyclohexane is a ring and we can not break easily of a ring.
So, bromocyclohexane does not displace by [tex]CH_{3}CO_{2}^{-}[/tex].
Hence, The displacement by [tex]CH_{3}CO_{2}^{-}[/tex] on bromoethane.
how do you make it so that 43200 ml have only 3 significant figures? I will give brainiest pls help
Answer:
3 sig figs
Explanation:
one of the rules for sig figs is a when a whole number has zeros if they are sandwiched between whole number for ex. 34007 they would all be sig figs. hope this helps!
btw all the zeros behind the decimal are significant for ex. 0.4800
Answer:
you could make it 43.2 x 10^2
Explanation:
A wooden block has the following measured dimensions: height 1.25 cm; width 2.5 cm and length of 15.956 cm. Calculate its volume in ml with the proper number of significant figures.
Answer:
[tex]V=50mL[/tex]
Explanation:
Hello,
In this case, by knowing that the volume of an object is computed by considering its dimensions, width, length and height, for the given measurements, we obtain:
[tex]V=W*H*L=1.25cm*2.5cm*15.956cm\\\\V=49.86cm^3[/tex]
Moreover, since one cubic centimetre equals one millilitre, the required volume is:
[tex]V=49.86cm^3*\frac{1mL}{1cm^3}\\ \\V=49.86mL[/tex]
Finally, since 2.5 cm has the fewest significant figures (2), the proper result is:
[tex]V=50mL[/tex]
Regards.
Which family contains elements with a full octet of valence electrons?
A. The actinides
B. The halogens
C. The alkali metals
D. The noble gases
The family of elements that contains elements with a full octet of valence electrons is D. The noble gases. These elements have achieved stability by completely filling their outer electron shells with 8 electrons (except for helium, which has 2).
Noble gases are located in Group 18 of the periodic table and include elements such as helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), and radon (Rn). These elements have a completely filled outer electron shell, also known as a full octet. A full octet means that the outermost energy level of the atom contains 8 electrons, except for helium which has only 2 electrons.
Having a full octet of valence electrons makes noble gases highly stable and unreactive. This stability is due to the fact that the atoms of noble gases have achieved the same electron configuration as the nearest noble gas element.
For example, helium has a full outer shell with 2 electrons, which is the same electron configuration as the nearest noble gas, neon. Neon and the other noble gases have 8 electrons in their outermost shell, fulfilling the octet rule.
In contrast, the other options mentioned:
A. The actinides: The actinides are a series of elements in the periodic table that have their valence electrons in the 5f orbital. They do not have a full octet of valence electrons.
B. The halogens: The halogens are located in Group 17 of the periodic table and include elements such as fluorine (F), chlorine (Cl), bromine (Br), iodine (I), and astatine (At). These elements have 7 valence electrons and are highly reactive, seeking to gain one electron to achieve a full octet.
C. The alkali metals: The alkali metals are located in Group 1 of the periodic table and include elements such as lithium (Li), sodium (Na), potassium (K), rubidium (Rb), cesium (Cs), and francium (Fr). These elements have 1 valence electron and are highly reactive, seeking to lose this electron to achieve a full octet.
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Heptane and water do not mix, and heptane has a lower density (0.684 g/mL.) than water (1.00 g/
mL). A 100-ml graduated cylinder with
an inside diameter of 3.08 cm contains 37.8 g of heptane and 34.7 g of water. What is the combined height of the two liquid layers in
the cylinder? The volume of a cylinder is r’h, wherer is the radius and h is the height.
cm
Answer:
Explanation:
volume of heptane= mass / density
volume of heptane = 37. 8 / .684
= 55.26 mL
volume of water = 34.7 / 1
= 34.7 mL or cc.
If l₁ be the length of heptane layer in the graduated cylinder
volume = cross sectional area x length or height of layer
π r² x l where r is radius of bore of the cylinder , l is height of liquid inside cylinder .
for heptane
π r² x l₁ = 55.26
3.14 x 1.54² x l₁ = 55.26
l₁ = 7.42 cm
for water
π r² x l₂ = 34.7
3.14 x 1.54² x l₂ = 34.7
l₂ = 4.65 cm
Combined height = l₁ + l₂
= 7.42 + 4.65
= 12.07 cm .
What is the Ksp of Al(OH)3 if the concentration of OH– in a saturated solution of Al(OH)3 is 1.6 × 10–8 M?
Answer:
[tex]Ksp=2.2x10^{-32}[/tex]
Explanation:
Hello,
In this case, since the dissociation of aluminum hydroxide is:
[tex]Al(OH)_3(s)\rightleftharpoons Al^{3+}+3OH^-[/tex]
The equilibrium expression is:
[tex]Ksp=[Al^{3+}][OH^-]^3[/tex]
Thus, given the concentration of hydroxyl ions in the solution, and the 3:1 mole ratio with the aluminum ions, the concentration of those turn out:
[tex][Al^{3+}]=1.6x10^{-8}\frac{molOH^-}{L}*\frac{1molAl^{3+}}{3molOH^-} =5.3x10^{-9}M[/tex]
Therefore, the solubility product, Ksp turns out:
[tex]Ksp=(5.3x10^{-9}M)[1.6x10^{-8}M]^3\\\\Ksp=2.2x10^{-32}[/tex]
Regards.
HF:______
a) dispersion force
b) dipole-dipole force
c) hydrogen bond
Answer:
HF: hydrogen bond
Explanation:
The hydrogen bond is a type of interaction that is established between a molecule that presents hydrogen with another that presents an atom with a high electronegativity, such as oxygen, fluorine or nitrogen (O, F, N), joining covalently due to their opposite charges.
The hydrogen atom has a positive charge and is known as a donor atom. The oxygen, fluorine or nitrogen atom is the bond acceptor atom.
So:
HF: hydrogen bond
If you have 24 grams of an unstable atom with a half life of 5,000 years, how many grams would be left after 15,000 years
Answer:
3 g.
Explanation:
The following data were obtained from the question:
Original amount (N₀) = 24 g
Half life (t½) = 5000 years
Time (t) = 15000 years
Amount remaining (N) =?
Next, we shall determine the rate constant K. This can be obtained as follow:
Half life (t½) = 5000 years
Rate constant (K) =.?
K = 0.693 / t½
K = 0.693 / 5000
K = 1.386×10¯⁴ / years
Finally, we shall determine the amount remaining (N) after 15000 years as follow:
Original amount (N₀) = 24 g
Time (t) = 15000 years
Rate constant (K) = 1.386×10¯⁴ / years
Amount remaining (N) =?
Log (N₀/N) = kt/2.3
Log (24/N) = 1.386×10¯⁴ × 15000/2.3
Log (24/N) = 2.079/2.3
Log (24/N) = 0.9039
24/N = antilog (0.9039)
24/N = 8.01
Cross multiply
24 = N × 8.01
Divide both side by 8.01
N = 24/8.01
N = 3 g
Therefore, the amount remaining after 15000 years is 3 g.
The radius of a nickel atom is 125 pm. How many nickel atoms would have to be laid side by side to span a distance of 1.74 mm
Answer:
6960000 Nickel atoms
Explanation:
radius = 125 pm = 125 x 10^-12 m
diameter of each atom = 125 pm x 2 = 250 pm = 250 x 10^-12 m
span of distance to be achieved = 1.74 mm = 1.74 x 10^-3 m
number of nickel atoms = (1.74 x 10^-3 m)/(250 x 10^-12 m) = 6960000 Nickel atoms
Please helpppp
Answer separately
1) 2) 3) 4) 5)
1) 4.5 mL
2) 12 mL
3) 82 mL
4) 110 mL
5) 330 mL
Which statement is true?
Answer:
where is statements if you give statements then ican answer okk