Lesson 17 Conservation of Momentum grade 9 I want the answer.
Answer:
1) 1.5 kg
2) 0.6 kg
Explanation:
[tex](3)(30) = 60m, m = 90/60 = 1.5[/tex] kg
[tex](0.8)(15) = 20m, m = 12/20 = 0.6[/tex] kg
At the start of a reaction, there are 0.0249 mol N2,
3.21 x 10-2 mol H2, and 6.42 x 10-4 mol NH3 in a
3.50 L reaction vessel at 375°C. If the equilibrium constant, K, for the reaction:
N2(g) + 3H2(g)= 2NH3(g)
is 1.2 at this temperature, decide whether the system is at equilibrium or not. If it is not, predict in which direction, the net reaction will proceed.
Answer:
Explanation:
The reaction is given as:
[tex]N_{2(g)} + 3H_{2(g)} \to 2NH_{3(g)}[/tex]
The reaction quotient is:
[tex]Q_C = \dfrac{[NH_3]^2}{[N_2][H_2]^3}[/tex]
From the given information:
TO find each entity in the reaction quotient, we have:
[tex][NH_3] = \dfrac{6.42 \times 10^{-4}}{3.5}\\ \\ NH_3 = 1.834 \times 10^{-4}[/tex]
[tex][N_2] = \dfrac{0.024 }{3.5}[/tex]
[tex][N_2] = 0.006857[/tex]
[tex][H_2] =\dfrac{3.21 \times 10^{-2}}{3.5}[/tex]
[tex][H_2] = 9.17 \times 10^{-3}[/tex]
∴
[tex]Q_c= \dfrac{(1.834 \times 10^{-4})^2}{(0.0711)\times (9.17\times 10^{-3})^3} \\ \\ Q_c = 0.6135[/tex]
However; given that:
[tex]K_c = 1.2[/tex]
By relating [tex]Q_c \ \ and \ \ K_c[/tex], we will realize that [tex]Q_c \ \ < \ \ K_c[/tex]
The reaction is said that it is not at equilibrium and for it to be at equilibrium, then the reaction needs to proceed in the forward direction.
Please help
Apply your knowledge and understanding of equilibrium constant in solving the following problems:
The equilibrium constant Kc for the reaction below is 170 at 500 K.
Determine whether the reaction mixture is at equilibrium when the concentrations of the components at this temperature are as follows:
[N2]=1.50
[H2]=1.00
[NH3]=8.00
If it is not at equilibrium, state and explain in which direction the reaction will proceed.Multi Line Text.
2()+32() ⇄ 23()
Answer:
The reaction will proceed to the right to attain the equilibrium.
Explanation:
Step 1: Write the balanced equation
2 N₂(g) + 3 H₂(g) = 2 NH₃(g)
Step 2: Calculate the reaction quotient
The reaction quotient (Qc) is calculated in the same way as the equilibrium constant (Kc) but it uses the concentrations at any time.
Qc = [NH₃]² / [N₂]² × [H₂]³
Qc = 8.00² / 1.50² × 1.00³ = 28.4
Since Qc ≠ Kc, the reaction is not at equilibrium.
Since Qc < Kc, the reaction will proceed to the right to attain the equilibrium.
Combustion of a 1.031-g sample of a compound containing only carbon, hydrogen,
and oxygen produced 2.265 g of CO2(g) and 1.236 g of H2O(g). What is the
empirical formula of the compound?
3 국
Molar masses in g/mol: CO2 = 44.01; H20 = 18.02; C = 12.01; H = 1.01
C3H50
Answer:
C₃H₈O
Explanation:
From the question given above, the following data were obtained:
Mass of compound = 1.031 g
Mass of CO₂ = 2.265 g
Mass of H₂O = 1.236 g
Empirical formula =?
Next, we shall determine the mass of carbon, hydrogen and oxygen in the compound. This can be obtained as follow:
For carbon, C:
Mass of CO₂ = 2.265 g
Molar mass of CO₂ = 44.01 g/mol
Molar mass of C = 12.01 g/mol
Mass of C =?
Mass of C = molar mass of C / molar mass of CO₂ × mass of CO₂
Mass of C = 12.01/44.01 × 2.265
Mass of C = 0.618 g
For hydrogen, H:
Mass of H₂O = 1.236 g
Molar mass of H₂O = 18.02 g/mol
Molar mass of H₂ = 2 × 1.01 = 2.02 g/mol
Mass of H =?
Mass of H = Molar mass of H₂ / Molar mass of H₂O × Mass of H₂O
Mass of H = 2.02/18.02 × 1.236
Mass of H = 0.139 g
For oxygen, O:
Mass of compound = 1.031 g
Mass of C = 0.618 g
Mass of H = 0.139 g
Mass of O =?
Mass of O = Mass of compound – (mass of C + mass of H)
Mass of O = 1.031 – ( 0.618 + 0.139)
Mass of O = 1.031 – 0.757
Mass of O = 0.274 g
Finally, we shall determine the empirical formula. This can be obtained as follow:
C = 0.618 g
H = 0.139 g
O = 0.274 g
Divide by their molar mass
C = 0.618 / 12.01 = 0.051
H = 0.139 / 1.01 = 0.138
O = 0.274 / 16 = 0.017
Divide by the smallest
C = 0.051 / 0.017 = 3
H = 0.138 / 0.017 = 8
O = 0.017 / 0.017 = 1
Therefore, the empirical formula of the compound is C₃H₈O
Based on the data provided, the empirical formula of the compound is C₃H₈O
What is empirical formula?The empirical formula of compound is its simplest formula showing the mole ratio of the elements in the compound.
First, the mass of carbon, hydrogen and oxygen in the compound is determined first as follows:
For carbon, C:
Mass of CO₂ = 2.265 g
Molar mass of CO₂ = 44.01 g/mol
Molar mass of C = 12.01 g/mol
Mass of C = 12.01/44.01 × 2.265
Mass of C = 0.618 g
For hydrogen, H:
Mass of H₂O = 1.236 g
Molar mass of H₂O = 18.02 g/mol
Molar mass of H₂ = 2 × 1.01 = 2.02 g/mol
Mass of H = 2.02/18.02 × 1.236
Mass of H = 0.139 g
For oxygen, O:
Mass of compound = 1.031 g
Mass of C = 0.618 g
Mass of H = 0.139 g
Mass of O = Mass of compound – (mass of C + mass of H)
Mass of O = 1.031 – ( 0.618 + 0.139)
Mass of O = 0.274 g
To determine the empirical formula of a compound, the mole ratio of the elements are determined as follows:
Mole ratio = reacting mass/molar massMole ratio of the elements:
Carbon Hydrogen Oxygen
0.618 g/12.01 0.139 / 1.01 0.274 / 16
0.051 0.138 0.017
Divide by the smallest ratio to convert to whole numbers
0.051 / 0.017 0.138 / 0.017 0.017 / 0.017
3 : 8 : 1
Therefore, the empirical formula of the compound is C₃H₈O
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what is an example of a change in genetic traits of an organism do to human affect
Answer:
A person's skin color, hair color, dimples, freckles, and blood type are all examples of genetic variations that can occur in a human population.
Explanation:
The blending together of some genes is called:
Answer:
its called molding
Explanation:
chemical properties of citric acid
Answer:
Citric Acid is a weak acid with a chemical formula C6H8O7.
...
Properties of Citric Acid – C6H8O7.
C6H8O7 Citric Acid
Molecular Weight/ Molar Mass 192.124 g/mol
Density 1.66 g/cm³
Boiling Point 310 °C
Melting Point 153 °C
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What is the Ksp expression for the dissociation of calcium oxalate?Immersive Reader
(4 Points)
Ksp=[Ca⁺²] x [C₂O₄⁻²]
Ksp=[Ca⁺²]² x [C₂O₄⁻²]
Ksp=[Ca⁺²]⁴ x [C₂O₄⁻²]
Ksp=[Ca⁺²] x [C₂O₄⁻²]²
Answer:
Ksp = [Ca⁺²] × [C₂O₄⁻²]
Explanation:
Step 1: Write the balanced reaction for the dissociation of calcium oxalate
CaC₂O₄(s) ⇄ Ca⁺²(aq) + C₂O₄⁻²(aq)
Step 2: Write the expression for the solubility product constant (Ksp) of calcium oxalate
The solubility product constant is the equilibrium constant for the dissociation reaction, that is, it is equal to the product of the concentrations of the products raised to their stoichiometric coefficients divided by the product of the concentrations of the reactants raised to their stoichiometric coefficients. It doesn't include solids nor pure liquids because their activities are 1.
Ksp = [Ca⁺²] × [C₂O₄⁻²]
What is the electronegativity difference
between sodium and chlorine?
Answer: 2.23 is the difference.
Explanation: Sodium has an electronegativity of 0.93 and Chlorine has an electronegativity of 3.16, so when Sodium and Chlorine form an ionic bond, in which the chlorine takes an electron away from sodium, forming the sodium cation, Na+, and the chloride anion, Cl-.
1. Calculate and interpret the equilibrium constant. Using the reaction below.
The equilibrium concentrations 0.60 M for E, 0.80 M for F, and 1.30 M for G. (Note: E, F, and G are all gases.) Do not include your solution.
Answer:
kc = [G]² / [E] [F], kc = [1.30M]² / [0.60M] [0.80M]
Explanation:
The reaction is:
E + F ⇄ 2G
The equilibrium constant, kc, must be written as the ratio of the molar concentrations of products over reactants. Each concentration powered to its coefficient.
For the reaction of the problem, kc is:
kc = [G]² / [E] [F]Replacing the given concentrations:
kc = [1.30M]² / [0.60M] [0.80M]Estimate the volume of a solution of 5M NaOH that must be added to adjust the pH from 4 to 9 in 100 mL of a 100 mM solution of a phosphoric acid?
Answer:
3mL of 5M NaOH must be added to adjust the pH to 7.20
Explanation:
When NaOH is added to phosphoric acid, H₃PO₄, the reaction that occurs are:
NaOH + H₃PO₄ ⇄ NaH₂PO₄ + H₂O pKa1 = 2.15
NaOH + NaH₂PO₄ ⇄ Na₂HPO₄ + H₂O pKa2 = 7.20
NaOH + Na₂HPO₄ ⇄ Na₃PO₄ + H₂O pKa3 = 12.38
We can adjust the pH at 7.20 = pKa2 if NaH₂PO₄ = Na₂HPO₄. To make that, we must convert, as first, all H₃PO₄ to NaH₂PO₄ and the half of NaH₂PO₄ to Na₂HPO₄. To solve this question we need to find the moles of phophoric acid in the initial solution. 1.5 times these moles are the moles of NaOH that must be added to fix the pH to 7.20:
Moles H₃PO₄:
100mL = 0.100L * (0.100mol / L) = 0.0100 moles H₃PO₄
Moles NaOH:
0.0100 moles H₃PO₄ * 1.5 = 0.0150 moles NaOH
Volume NaOH:
0.0150 moles NaOH * (1L / 5moles) = 3x10⁻³L 5M NaOH are required =
3mL of 5M NaOH must be added to adjust the pH to 7.203 mL of 5 Molar NaOH is required to adjust the pH of phosphoric acid.
What is pH?It is the negative log of the concentration of Hydrogen ions in the solution.
To calculate the volume of NaOH first, calculate the moles of NaOH and H₃PO₄.
Moles of H₃PO₄.
[tex]\rm moles \ of \ H_3PO_4 = 100\rm \ mL = 0.100\rm \ L \times (0.100 \ mol / L)\\\\\rm moles \ of \ H_3PO_4 = 0.01[/tex]
The moles of NaOH:
[tex]\rm Moles \ of \ NaOH =0.01 \ moles \ H_3PO_4\times 1.5 \\\\\rm Moles \ of \ NaOH= 0.0150[/tex]
The volume of NaOH:
[tex]\rm Volume\ of \ NaOH = \rm 0.0150\ moles\ NaOH \times (1 \ L / 5 \ moles) \\\\\rm Volume\ of \ NaOH = 3\times 10^{-3} L[/tex]
Therefore, 3 mL of 5 Molar NaOH is required to adjust the pH of phosphoric acid.
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a. An aqueous solution of Mn(NO3)2 is very pale pink, but an aqueous solution of K4[Mn(CN)6] is deep blue. Explain why the two differ so much in the intensities of their colors.
b. Predict which of the following compounds would be colorless in aqueous solution:
a. K2[Co(NCS)4]
b. Zn(NO3)2
c. [Cu(NH3)4]Cl2
d. CdSO4
e. AgClO3
f. Cr(NO3)2
Answer:
See Explanation
Explanation:
The colour of many transition metal complexes stem from transitions of electrons between energy levels. These transitions are governed by the spin selection rules and the colour is determined by the magnitude of crystal field splitting.
According to the spin selection rules, transitions in which ΔS = 0 are forbidden. Hence, a Mn^2+high spin compound is expected to be colourless. However, contrary to the spin selection rules Mn^2+high spin compounds do exhibit transitions in which the intensity is only about one-hundredth of the intensity of the spin allowed transitions. Thus many Mn^2+ high spin compounds such as Mn(NO3)2 are very pale pink or off white.
Note also that the crystal field stabilization energy of Mn^2+ which is a d^5 low spin ion is zero hence the very pale colour observed.
K4[Mn(CN)6] is deep blue as a result of charge transfer. Also, the compound exhibits an observed crystal field stabilization energy because it is a d^5 low spin compound hence the observed colour. Its low spin nature is because the cyanide ion is a strong field ligand hence it causes a greater magnitude of crystal filed splitting.
The following compounds are colourless;
Zn(NO3)2
CdSO4
AgClO3
One thing that is common to all the compounds listed above is that they are all d^10 compounds. This means that they all possess completely filled d-orbitals hence they are colourless.
In 1773 Benjamin Franklin, investigating the calming effect of oil on the water of a pond near London, observed that a teaspoonful of oil would cover about half an acre of the surface. From this information, estimate the height or thickness of an oil molecule in the floating film of oil. Then estimate the cross sectional area of an oil molecule in the floating film You may assume that Franklin used a lamp oil, like camphene, C10H16.
Required:
Calculate the thickness (in nm) of the surface film and the surface area (in nm2) occupied by an oleic acid molecule on water.
Answer:
a. 2.44 nm
b. 15.33 nm²
Explanation:
a. Calculate the thickness (in nm) of the surface film
Since the volume of a teaspoon equals V = 4.93 cm³ and the area of the oil film is half an acre = 1/2 × 4046.86 m² = 2023.43 m²
The volume of the oil film, V' equals the volume of the oil in the teaspoon, V
V' = Ah where A = cross-sectional area of oil film = 2023.43 m² = 2023.43 × 10⁴ cm² and h = height of oil film
So, V' = V
Ah = V
h = V/A
= 4.93 cm³/2023.43 × 10⁴ cm²
= 0.00244 × 10⁻⁴ cm
= 2.44 × 10⁻⁷ cm
= 2.44 × 10⁻⁷ cm × 1m/100 cm
= 2.44 × 10⁻⁹ m
= 2.44 nm
b. The surface area (in nm2) occupied by an oleic acid molecule on water.
Since the height of the oil film equals the diameter of the oil molecule, and the molecule is assumed to be a sphere of radius, r. Its surface area is thus A = 4πr²
r = h/2 = 2.44 nm/2 = 1.22 nm
A = 4πr²
A = 4π(1.22 nm)²
A = 4.88π nm²
A = 15.33 nm²
What volume of solution is needed to prepare a 0.50M solution of sodium hydroxide using 2.0 g of solute? Show your work
Answer:
0.10 L
Explanation:
First we convert the 2.0 grams of solute (meaning sodium hydroxide) into moles, using its molar mass:
2.0 g ÷ 40 g/mol = 0.05 mol NaOHThen we can calculate the required volume of solution, using the definition of molarity:
Molarity = moles / litersliters = moles / molarity0.05 mol / 0.50 M = 0.10 Lhow to solve x² in differential
Answer:
x² = mutiphy by them self
Explanation:
0=4
Balance this equation
H₂sicl2+ H₂O → H8Si4O4 + HCl
Balanced Equation is
4H2SiCl2+4H2O → H8Si4O4 + 8HCl
0.350 moles of H2O is equivalent to (blank amount) molecules of water.
Answer:
is has around 1/5 lyrics of water
Answer:
[tex]1 \: mole \: = \: 6.02 \times {10}^{23} \: molecules \\ 0.350 \: moles \: = \: (0.350 \times 6.02 \times {10}^{23} ) \: molecules \\ = 2.107 \times {10}^{23} \: molecules[/tex]
What is Climate Change and how can it effect plants?
Answer:
Climate change will make plants—and us—thirstier. The combined effects of increased CO2 levels and warmer temperatures will increase water consumption by vegetation. That will lead to water declines in streams and rivers like the Ashepoo River in South Carolina
Explanation:
If the plant population decreased, the amount of carbon in the atmosphere would _______.
Increase
Stay the same
Decrease
Answer:
increase answer
Explanation:
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A 24.0 gram sample of copper was
ncated from 25.0°C to 500.0°C 43783
of heat were absorbed, what is the
Specific_heat of copper?
Answer:
3.84 J/g°C
Explanation:
Using the formula as follows:
Q = m × c × ∆T
Where;
Q = amount of heat (J)
m = mass of substance
c = specific heat of copper
∆T = change in temperature (°C).
Based on the provided information;
Q = 43783J
m = 24g
∆T = 500°C - 25°C = 475°C
c = ?
Using Q = m × c × ∆T
43783 = 24 × c × 475
43783 = 11400c
c = 43783 ÷ 11400
c = 3.84 J/g°C
How long will it take a 500-W heater to raise the temperature of 400 g of water from 15.0 °C to 98.0
°C?
Explanation:
E=(98-15)×400×4.2
E=139440J
t=E/P
E=139440/500=278.88s
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Chemical bond stretching frequencies depend on two major factors, namely atomic weight and bond stiffness. The frequency absorbed is directly proportional to the bond strength (stiffness) but inversely proportional to the atomic weight. Which of the following bonds will give the highest absorption frequency value?
A. C single bond C.
B. C double bond C.
C. All would be the same frequency.
D. C triple bond C.
Answer:
D. C triple bond C.
Explanation:
The absorption frequency value is highest for carbon-carbon triple bond. The stretching frequencies are much higher than the bending frequency. It mainly depends upon the strength of the bonds as well as masses of the bonded atoms.
The triple bonds have higher stretching strength than the double bonds and double bonds have higher strength than single carbon to carbon bonds.
The triple bonds has a higher bond order, i.e. 3. So it has maximum stretching bonds.
A balloon can be inflated with 4.23 Liters of O2gas at STP. How much will the balloon weigh in grams?
Answer: 6.04 g
Explanation: In STP conditions, amoint of substance
n = V / Vm = 4.23 l / 22.41 l/mol = 0.188755 mol
Molar mass M(O2) = 32 g/mol and
mass m = nM
What mass of sucrose is needed to make 300.0 mL of a 0.5 M solution? (molar mass=
342.34 g/mol)
The enthalpy of solution of NaCl in water is about 3.88 kJ/mol. However, the solubility of NaCl in water is relatively high. Which statement about the entropy of the solution process explains why NaCl dissolves in water even though the process is endothermic?
a. the entropy increases when NaCl dissolves in water
b. the entropy remains the same when NaCl dissolves in water
c. the entropy decreases when NaCl dissolves in water
d. entropy has nothing to do with the solution process
We have that the statement that explains why NaCl dissolves in water even though the process is endothermic the given as,
The entropy increases when NaCl dissolves in water.Option A
HeatGenerally ,Sodium chloride is crystalline salt, when its dissolved is H20 , it dissociates its ion, the ions are scattered through out the solution which results in increased in randomness .
Therefore, the entropy increases when NaCl dissolves in water,
Option A
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Which of the following
describes the zone of the
ocean where no light reaches?
A. up to 200 meter depth and includes
photosynthetic plants, sea anemones,
sponges, crabs, and clams
B. the "twilight zone" between 200-1000
meters deep and includes whales and octopi
and little life
C. permanent darkness below 1000 meters
with bioluminescent bacteria, bottom
feeders, and angler fish
Answer:
Bathypelagic
54% of the ocean lies in the Bathypelagic (aphotic) zone into which no light penetrates. This is also called the midnight zone and the deep ocean. Due to the complete lack of sunlight, photosynthesis cannot occur and the only light source is bioluminescence.
Explanation:
The small surface zone that has light is the photic zone. The entire rest of the ocean does not have light and is the aphotic zone.
Permanent darkness below 1000 meters with bioluminescent bacteria, bottom feeders, and angler fish is where no light reaches.
What is Darkness?This is referred to the state of being dark as a result of absence of light in the area.
The light ray penetration decreases with increase in depth thereby making areas below 1000 meters dark with bioluminescent bacteria, bottom feeders, and angler fish which is why option C was chosen.
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Three biologically important diatomic species, either because they promote or inhibit life, are
(a) CO,
(b) NO3 and
(c) CN-.
The first binds to hemoglobin, the second is a chemical messenger, and the third interrupts the respiratory electron transfer chain. Their biochemical action is a reflection of their orbital structure. Deduce their ground state electron configurations.
Answer:
See Explanation
Explanation:
We can write the molecular orbital configuration of molecules in the same way as we write the orbital electron configuration of atoms. The valence electrons in the molecule are filled into molecular orbitals of appropriate energy in accordance to the Aufbau principle.
For CO;
σ2s2, σ*2s2, Π2py2, Π2pz2, σ2px2
For NO;
σ2s2, σ*2s2, Π2px2, Π2py2, σ2pz2, Π*2px1
For CN-;
σ2s2, σ*2s2, Π2px2, Π2py2, σ2pz2
These are the ground state electron configurations of these molecules.
How did oxygen first enter Earth's atmosphere?
A. Meteorite impacts
B. Breakdown of Precambrian rocks
C. Biological processes
D. Volcanic outgassing
Answer:
I think c biological processes
when 18.0 g H20 is mixed with 33.5 g Fe, which is the limiting reactant?
Answer:
si.mple fe
Explanation:
ghhsshzhzbbzhh
A hot metal plate at 150°C has been placed in air at room temperature. Which event would most likely take place
over the next few minutes?
O Molecules in both the metal and the surrounding air will start moving at lower speeds.
O Molecules in both the metal and the surrounding air will start moving at higher speeds.
O The air molecules that are surrounding the metal will slow down, and the molecules in the metal will speed up.
O The air molecules that are surrounding the metal will speed up, and the molecules in the metal will slow down.
Answer: The event air molecules that are surrounding the metal will speed up, and the molecules in the metal will slow down would most likely take place over the next few minutes.
Explanation:
A process in which heat is evolved is called an exothermic process.
When hot metal plate at [tex]150^{o}C[/tex] is placed in air at room temperature then heat is given off by the metal plate due to which there will occur a decrease in kinetic energy of its molecules.
As a result, molecules in the metal will slow down.
Whereas heat is absorbed by the air molecules from the metal due to which kinetic energy of air molecules will increase.
Thus, we can conclude that the event air molecules that are surrounding the metal will speed up, and the molecules in the metal will slow down would most likely take place over the next few minutes.