Answer:
See explanation
Explanation:
Number of moles of naphthalene = 0.32g/128.1705 g/mol = 0.0025 moles
Molality = number of moles/ mass of Solvent in kilograms
Molality = 0.0025/0.025 Kg
Morality = 0.1 m
But
∆T= K × i × m
Where ∆T = boiling point elevation
i= number of particles (this is equal to 1 because naphthalene is molecular and not ionic)
m= molality of naphthalene = 0.1 m
K= boiling point elevation constant = 5.12 °C/m
∆T= 5.12 °C/m ×0.1 = 0.512°C
For freezing point depression
∆T= K× i × m
Where ∆T= freezing point depression
i= number of particles (this is equal to 1 because naphthalene is molecular and not ionic)
m= molality of naphthalene = 0.1 m
K= freezing point depression constant = 2.67 °C/m
∆T= 2.67 °C/m ×0.1 = 0.267°C
From Raoult's law;
∆P = XBPA°
Where;
∆P = vapour pressure lowering
XB = mole fraction of solute
PA° = vapour pressure of pure solvent
Number of moles of solvent = mass/molar mass = 25g/ 78 g/mol= 0.3205 moles
Total number of moles = number of moles of solute + number of moles of solvent = 0.0025 moles + 0.3205 moles = 0.323 moles
Mole fraction of solute = 0.0025 moles/0.323 moles = 0.0077
Vapour pressure of benzene = 100 torr
Therefore;
∆P = 0.0077 × 100torr = 0.77 torr
Hence;
∆P = 0.77 torr
Which of the following are decomposers?
O A. mice
O B. snakes
O C. fungi
OD. spiders
ASAP
A 50.0 g sample of an unknown substance, initially at 20.2 °C, was heated with 1.55 kJ of energy. The final temperature of the substance was 125.0 °C. Determine the specific heat of this substance.
Answer:
0.296j/g⁰c
Explanation:
we have the following information from this question before us.
mass iv substance = 50grams
we have initial temperature ti = 20.2⁰c
final temperature = 125⁰c
the energy that was provided = 155kj
we proceed with this formula
energy = mcΔT
1.55x10³ = 50 x c x (125-20.2)
1.55x10³ = c x 50gm x 104.8k
we divide through to get c
c = 1.55x10³/50g x 104.8
c = 0.296J/g⁰c
that is the specific heat of this substance.
thank you!
What is the heat gjoules transferred by a chemical reaction to the reservoir of a calorimeter containing 135g of dilute aqueous solution (c=4.184 J/g^ C) the reaction causes the temperature of the reservoir to rise from 23.0 27.0 degrees * C
Answer:
Required heat = 2,259.36 N (approx)
Explanation:
Given:
Mass = 135 g
Specific heat (c) =4.184 J/gC
Change in temperature ΔT = 27-23 = 4 c
Find:
Required heat
Computation:
Q = mcΔT
Q = (135)(4.184)(4)
Required heat = 2,259.36 N (approx)
The heat transferred by the chemical reaction to the reservoir of a calorimeter containing 135 g of dilute aqueous solution is 2259.36 J
To solve this question, we'll begin by calculating the change in the temperature of the reservoir. This can be obtained as follow:
Initial temperature (T₁) = 23 °C
Final temperature (T₂) = 27 °C
Change in temperature (ΔT) =?ΔT = T₂ – T₁
ΔT = 27 – 23
ΔT = 4 °CFinally, we shall determine the heat transferred to the reservoir.
Change in temperature (ΔT) = 4 °C
Mass (M) = 135 g
Specific heat capacity (C) = 4.184 J/gºC
Heat (Q) =?Q = MCΔT
Q = 135 × 4.184 × 4
Q = 2259.36 JTherefore, the heat transferred by the chemical reaction to the reservoir of a calorimeter containing 135 g of dilute aqueous solution is 2259.36 J
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A chemical engineer must calculate the maximum safe operating temperature of a high-pressure gas reaction vessel. The vessel is a stainless-steel cylinder that measures 41.0cm wide and 49.2cm high. The maximum safe pressure inside the vessel has been measured to be 3.70MPa. For a certain reaction the vessel may contain up to 2.50kg of dinitrogen difluoride gas.
Required:
Calculate the maximum safe operating temperature the engineer should recommend for this reaction. Write your answer in degrees Celsius. Be sure your answer has the correct number of significant digits.
Answer:
[tex]T=2.78x10^3 \°C[/tex]
Explanation:
Hello,
In this case, considering that the safe temperature may be computed via the ideal gas law as we now the pressure, mass and volume via the dimensions:
[tex]V=\pi r^2 h=\pi *(41.0cm)^2*49.2cm=2.60x10^5cm^3*\frac{1L}{1000cm^3} =260L[/tex]
The pressure in atm is:
[tex]P=3.70MPa*\frac{1x10^6Pa}{1MPa} \frac{1atm}{101325Pa} =36.5atm[/tex]
And the moles considering the mass and molar mass (66 g/mol) of dinitrogen difluoride (N₂F₂):
[tex]n_{N_2F_2}=2.50kg*\frac{1000g}{1kg}*\frac{1mol}{66g} =37.9mol[/tex]
In sich a way, by applying the ideal gas equation, which is not the best assumption but could work as an approximation due to the high temperature, the temperature, with three significant figures, will be:
[tex]T=\frac{PV}{nR}=\frac{36.5Pa*260L}{37.9mol*0.082\frac{atm*L}{mol*K} }\\ \\T=3053.6K-273.15\\\\T=2.78x10^3 \°C[/tex]
Best regards.
pls help! I'll give u 10 points and the brainlest answer!!!
Answer:
371km
Explanation:
the lower the more fuel there is
The indicator dilution method is a technique used to determine flow rates of fluids in channels for which devices like rotameters and orifice meters cannot be used (e.g., rivers, blood vessels, and large- diameter pipelines). A stream of an easily measured substance (the tracer) is injected into the channel at a known rate and the tracer concentration is measured at a point far enough downstream of the injection point for the tracer to be completely mixed with the flowing fluid. The larger the flow rate of the fluid, the lower the tracer concentration at the measurement point. A gas stream that contains 1.50 mole% CO2 flows through a pipeline, Twenty (20.0) kilograms of CO2 per minute is injected into the line. A sample of the gas is drawn from a point in the line 150 meters downstream of the injection point and found to contain 2.3 mole% CO2.(a) Estimate the gas flow rate (k mol/min) upstream of the injection point.(b) Eighteen seconds elapses from the instant the additional CO2 is first injected to the time the CO2 concentration at the measurement point begins to rise. Assuming that the tracer travels at the average velocity of the gas in the pipeline (i.e., neglecting diffusion of CO2), estimate the average velocity (m/s), if the molar gas density is 0.123k mol/m3 what is the pipe diameter?
Answer:
[a]. 55.5kmol/min.
[b]. The average velocity = 8.3m/s.
pipe diameter = 1.0954m.
Explanation:
So, from the question above we are given the following data or information or parameters which are going to help us in solving the above question;
=> ''A gas stream that contains 1.50 mole% CO2 flows through a pipeline''
=> ''Twenty (20.0) kilograms of CO2 per minute is injected into the line.''
=> '' . A sample of the gas is drawn from a point in the line 150 meters downstream of the injection point and found to contain 2.3 mole% CO2''
=> ''molar gas density is 0.123k mol/m3''
[a].The gas flow rate (k mol/min) upstream of the injection point can be calculated as follows.
- Determine the mass balance= j₁ +[ 20 /.123 × 44] = j₃. Thus, the flow gas rate is given as;
w₁ × 1.50 mole% CO2 + 3.6954= 2.3 mole% CO2 × w₃.
-3.610 = - 0.008w₁.
w₁ = 451.3 m³/min.
Thus, the flow rate = 0.123 × 451.3 = 55.51 kmol/min.
[b]. The average velocity = 150/ 18 seconds = 8.3 m/s.
[c]. Pipe Diameter can be calculated as follows;
0.123 × 8.3 × [3.12 × (pipe diameter)² / 4] = 55.51/60.
[1.0209 × 3.12]/4 × [pipe diameter]² = 0.9252.
0.8014065 × [pipe diameter]² = 0.9252.
[pipe diameter]² = 0.9252/ .8014065.
[pipe diameter]² = 1.15443.
[pipe diameter] = √1.15443.
[pipe diameter] = 1.2 m² = 1.0954m.
A) The gas flow rate (k.mol/min) upstream of the injection point is;
55.51 k.mol/min
B) The average velocity and the diameter of the pipe are respectively;
8.33 m/s and 1.072 m
Gas Flow rate in thermodynamics
Due to the fact that there is no reaction or accumulation in the pipeline, we can say that the input of CO₂ is equal to it's output. Thus;
Upstream + Injection = Downstream
A) If we assume that x is the flow rate of gas (kmol/min), then there is the we can write the balance as:
0.015x + (20/44) = 0.023(x + 20/44)
Expanding gives us;
0.015x + 0.4545 = 0.023x + 0.01045
Solving for x gives;
x = 55.51 kmol/min
B) Formula for velocity is;
v = distance/time
Thus, since the length of the pipe is 150 m/s and time is 18 seconds, then velocity is;
v = 150/18
v = 8.33 m/s
Now, volumetric flow rate is calculated from;
Q = q' * ρ
where;
q' is gas flow rate
ρ is gas density
Thus;
Q = 55.51 * 0.123
Q = 451.3 m³/min = 7.522 m³/s
We can also express average velocity as;
v = Q/A
where A is area = πD²/4
Thus;
8.33 = 7.522/(πD²/4)
Making D the subject of the formula gives;
D = √((4 * 7.522)/(8.33π))
D = 1.072 m
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What is the process of cell eating called
Answer:
Phagocytosis
Explanation:
1) A hanglider flies with a horizontal velocity of 13 m/s when his wallet falls out. If the hanglider is 1450 m above the earth when the wallet falls out, find the horizontal distance the wallet travels before hitting the ground.
Answer:
The wallet is falling from a height of 1450 m
We are given that:
downward acceleration (a) = 10 m/s/s
downward displacement (s) = 1450 m
downward initial velocity (u) = 0 m/s (Since the wallet was dropped, it had no initial velocity)
horizontal initial velocity (v) = 13 m/s
Since there is no external force being applied on the wallet except the force of gravity, the wallet will keep moving at a constant velocity of 13 m/s
To calculate the horizontal distance travelled by the wallet, we need to know how long the wallet was airborne
To calculate how long the wallet was airborne, we have to find out that how long it took the wallet to hit the ground
Time taken for the wallet to hit the ground:
From the second equation of motion:
s = ut + 1/2 at² ------------(for vertical motion of the wallet)
1450 = (0)(t) + 1/2 (10)(t)²
1450 = 5t²
t² = 290
t = 17 (approx)
Time taken for the wallet to hit the ground = 17 seconds
Horizontal distance travelled by the wallet:
Since the wallet hits the ground after 17 seconds, it will move horizontally at a constant velocity of 13 m/s for 17 seconds
horizontal acceleration = 0 m/s/s
From the second equation of motion:
s = ut + 1/2at²
s = ut + 1/2 (0)t²
s = ut
here, u is the horizontal initial velocity of the wallet and the time taken by the wallet to hit the ground is 't'
s = (13)(17)
s = 221 m
Hence, the horizontal distance travelled by the wallet while falling from a height of 1450 m is 221 m
What is the density of a liquid with the same mass as in problem d, if it has a volume of 43.2 mL?
(part d: A liquid has a volume of 62.7 mL and a density of 2.59 g/mL. What is its mass?)
Answer:
3.76
Explanation:
You start by finding the mass of part d. Mass= Density times Volume.In this case you would multiply 62.7 and 2.59 which is 162.393 rounded is 162.4. Then density= mass/ volume which is 162.4/ 43.2. density equals 3.76
calculate the mol fraction of ethanol and water in a sample of rectified spirit which contains 95% of ethanol by mass
Answer:
We are given that there is 95% ethanol by mass in rectified spirit
so, we can say that in a 100g sample, we have 95 grams of ethanol and 5 grams of water
we will find the number of moles of ethanol and water in 100g solution of rectified spirit and use that to calculate the mole fraction
Moles of Ethanol:
Molar mass of ethanol = 46 grams / mol
Number of moles = Given mass / molar mass
Number of moles = 95 / 46
Moles of Ethanol = 2 moles (approx)
Moles of Water:
Molar mass of water = 18 grams per mol
Number of moles = Given mass / molar mass
Moles of water = 5 / 18
Moles of water = 0.28 moles (approx)
Mole Fractions:
Mole fraction of a specific compound is the number of moles of that compound divided by the total number of moles in the solution
Mole fraction of Ethanol:
Moles of ethanol / (moles of ethanol + moles of water)
2 / (2 + 0.28)
2 / (2.28) = 0.9 (approx)
Mole fraction of Water:
Moles of water / (Moles of ethanol + moles of water)
0.28 / (2 + 0.28)
0.28 / (2.28) = 0.1 (approx)
A graduated cylinder is filled to an initial volume of 25mL. A solid object is dropped into the graduated cylinder. The final volume of the graduated cylinder is 60mL. The mass of the object is 140g. What is the density of the solid object?
Answer:
The answer is 4 g/mLExplanation:
The density of a substance can be found by using the formula
[tex]density = \frac{mass}{volume} \\ [/tex]
volume = final volume of water - initial volume of water
volume = 60 - 25 = 35 mL
From the question we have
[tex]density = \frac{140}{35} \\ [/tex]
We have the final answer as
4 g/mLHope this helps you
the de broglie equation predicts that the wavelength (in m) of a proton moving at 1000 m/s is what?
(h=6.63 x 10^-34J•s; mass od a photon = 1.673 x 10^-24
Explanation:
from
de Broglie wavelength= h/ mv
substitute the values
joules= mass(kg) × acceleration (m/s^2) × distance(m)= kgm^2/s^2.
sorry I don't have my calc with me
According to De Broglie's Hypothesis, matter has both particle and wave properties. The wavelength of De Broglie is given as λ = h p,
wavelength (in m) of a proton = 11.09 x[tex]10^ {-55[/tex] m.
What is de Broglie statement?According to De Broglie's Hypothesis, matter has both particle and wave properties. The wavelength of De Broglie is given as λ = h p, where represents the particle momentum, and can be written as: λ= h m v.The de Broglie hypothesis demonstrated that wave-particle duality was a fundamental principle shared by both radiation and matter, rather than an aberrant behavior of light.De Broglie's equation does not apply to larger particles; rather, it only applies to microscopic particles such as electrons, protons, and neutrons. De Broglie's equation is only valid for moving microscopic particles.λ= h m v.
= 6.63 x [tex]10^{-34}[/tex] x 1.673 x [tex]10^{-24[/tex]x 1000
= 11.09 x[tex]10^ {-55[/tex] m
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Given the following balanced equation:
9 Fe2O3 + 2 NH3 → 6 Fe3O4 + N2 + 3 H2O
A) how many grams of NH3 are needed to react with 125 grams of Fe2O3?
B) how many grams of Fe3O4 will be produced?
Answer:
What mass of sodium hydroxide is needed to react completely with 10.0 g of iron( III) chloride? The conversion sequence is: A. B.
Explanation:
make me as brain liest
Is lead a representative metal or transitional metal?
Answer:
It's a representative metal
Explanation: Transitional metals are metals of various chemical elements and have valence electrons—i.e., electrons that can participate in the formation of chemical bonds.
Facilitated diffusion is a form of passive transport and involves the movement of molecules (like glucose) that are too large to just go through the cell membrane without help. These larger molecules move through the cell membrane from High concentration to Low concentration with the help of _____________________________.
Answer: Membrane Protiens
Explanation:
In facilitated diffusion, molecules diffuse across the plasma membrane with assistance from membrane proteins, such as channels and carriers. A concentration gradient exists for these molecules, so they have the potential to diffuse into (or out of) the cell by moving down it.
List the steps in the water cycle, in the correct order?
What is the volume of 11.2 g of O2 at 7.78 atm and 415 K?
Answer:
1.53 L
Explanation:
Step 1: Given data
Mass of oxygen (m): 11.2 gPressure (P): 7.78 atmTemperature (T): 415 KIdeal gas constant (R): 0.0821 atm.L/mol.KStep 2: Calculate the moles (n) corresponding to 11.2 g of oxygen
The molar mass of oxygen is 32.00 g/mol.
11.2 g × (1 mol/32.00 g) = 0.350 mol
Step 3: Calculate the volume of oxygen
We will use the ideal gas equation.
P × V = n × R × T
V = n × R × T / P
V = 0.350 mol × (0.0821 atm.L/mol.K) × 415 K / 7.78 atm
V = 1.53 L
1.547 grams of hydrated MgSO4 is heated in a crucible. After heating, 0.7554 grams of anhydrous MgSO4 remains in the crucible. How many waters of hydration were attached to the MgSO4
Answer:
7
Explanation:
Let x represent the number of moles of water in the hydrated salt i.e MgSO₄.xH₂O
The following data were obtained from the question:
Mass of MgSO₄.xH₂O = 1.547 g
Mass of anhydrous MgSO₄ = 0.7554 g
Mole of H₂O = x =?
Next, we shall determine the mass of water, H₂O in the hydrated salt, MgSO₄.xH₂O. This can be obtained as follow:
Mass of MgSO₄.xH₂O = 1.547 g
Mass of anhydrous MgSO₄ = 0.7554 g
Mass of H₂O =?
Mass of H₂O = (Mass of MgSO₄.xH₂O) – (Mass of anhydrous MgSO₄)
Mass of H₂O = 1.547 – 0.7554
Mass of H₂O = 0.7916 g
Finally, we shall determine the value of the x as illustrated below:
Mass of MgSO₄.xH₂O = 1.547 g
Molar mass of MgSO₄.xH₂O = 24 + 32 + (16×4) + x[(2×1) + 16]
= 24 + 32 + 64 + x(2 + 16)
= 120 + 18x
Mass of H₂O = 0.7916 g
Molar mass of xH₂O = 18x
Molar Mass of xH₂O/ Molar mass of MgSO₄.xH₂O = mass of xH₂O /Mass of MgSO₄.xH₂O
18x/ 120 + 18x = 0.7916/1.547
Cross multiply
0.7916 (120 + 18x) = 18x × 1.547
94.992 + 14.2488x = 27.846x
Collect like terms
94.992 = 27.846x – 14.2488x
94.992 = 13.5972x
Divide both side by 13.5972
x = 94.992 / 13.5972
x = 7
Thus, the formula for the hydrated salt, MgSO₄.xH₂O is MgSO₄.7H₂O
Number of moles of water, H₂O in the hydrated salt MgSO₄.7H₂O is 7.
The number of moles of attached water molecules is 7.
Mass of hydrated MgSO4 = 1.547 grams
Mass of anhydrous MgSO4 = 0.7554 grams
Number of moles of hydrated MgSO4 = 1.547 grams/120 + 18x
Number of moles of anhydrous MgSO4 = 0.7554 grams /120
Number of moles of anhydrous salt = Number of moles of hydrated salt
0.7554 grams /120 = 1.547 grams/120 + 18x
0.7554(120 + 18x) = 1.547 × 120
90.6 + 13.6x = 185.6
185.6 - 90.6 /13.6 = x
x = 7
The number of moles of attached water molecules is 7.
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What element has 6 protons,7 neutrons and 6 electrons
Answer:
corbon 12
Explanation:
carbon atoms with the usual 6 nutrons have a mass number of 12
6 protons plus 7 neutrons equals 13 for carbon nuclei with 7 neutrons, these atoms make up the carbon-13 element.
What is carbon-13?
A naturally occurring stable isotope of carbon with a nucleus comprising six protons and seven neutrons is carbon-13 (13C). It makes up roughly 1.1% of all the native carbon on Earth and is one of the environmental isotopes.
Since they are carbon atoms with extra neutrons, carbon 12, 13, and 14: You could also say that the difference in mass between these isotopes is molecular mass.
Therefore, molecular mass is determined by adding the number of protons and neutrons (if you have 1 mole of each isotope, carbon 14 would have the greatest mass).
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A scientist plans to find out the cause of decrease in bird population in a location. He interviews 50 people who work in the local oil mine and 45 of them believe that the mine does not affect the birds. The scientist concludes that 90 percent of people in the location believe that the local oil mine does not cause decrease in bird population. Why is the scientist's conclusion most likely unreliable? (2 points)
Group of answer choices
a. the results of the investigation cannot be repeated
b. the source of information could be biased
c. bird population cannot be measured accurately
d. the investigation is done by only one person
Answer:
Explanation:
If 1.4434 moles of H2O are produced, how many moles of N2 will also be produced?
Answer:
1 gram of N2 is equal to 0.035697202053303 mole. i dont know the rest but intried to help
. hopefully this will give you someone else a starting point. goodluck
Examples 36
A cylinder is 350cm³ long and two gases A and B have relative molecular weights of 64 and 16 respectively. If the two gases, at the same temperature and pressure, are released simultaneously at both ends determine the distance from one end at which the gases meet.
2x = 350- x
3x = 350
x = 350/x
therefore,
x = 116.67cm.
so, (350 - x)cm³ = (350- 166.67)cm³ = 233.33cm³
Hence distance covered by gas A is 116.67cm that by B is 233.33cm
Identify the change of state occurring in each situation. your friend's perfume producing an aroma that fills the room making ice cream from milk in an ice cream maker formation of frost on a cold windshield overnight wax dripping from a candle naphthalene solid (found in moth balls) producing fumes in the closet to deter moths your breath fogging up on your car windshield on a cold winter day
Answer:
The solution to this question can be defined as follows:
Explanation:
[tex]W_{ax}[/tex] Naphthalene falls off from candle something, strong fumes-sublimation produces. On a rainy day in winter, the breath fogging up on your cor's windshield. Brost formation on an overnight deposition on a frozen windshield. Making milk ice cream freezing the perfuming scent that fills the creation of the room.For each of the following molecules draw the Lewis structure on a separate sheet of paper. MAKE SURE TO FOLLOW THE RULES FROM CLASS (ie do not break the octet rule unless necessary to connect all the atoms). Then based on your structure indicate:
the total number of valence electrons.
the electronic and molecular shapes (choose from: linear, trigonal planar, bent, tetrahedral, trigonal pyramidal, trigonal bipyramidal, seesaw, T-shaped, octahedral, square pyramidal, or square planar).
whether or not the molecule is polar (Y/N).
Note: The central atom is the first atom listed, except for HCN, H2CO, and OCN-, where carbon is the central atom (underlined).
Formula Valence electrons Electronic Shape Molecular Shape Polar (Y/N)
HCN
PH3
CHCl3
NH4+
H2CO
SO42-
SeF2
CO2
O2
ClO4-
HBr
PF5
BeH2
PO43-
BH3
Br3-
Answer:
Kindly check the explanation section.
Explanation:
Without mincing words let us dive right into the solution to the question above, taking each compound at a time.
NB: Kindly Check attachment for the Lewis Structure of each of the chemical compounds.
Therefore, the number of valence electrons, electronic shape, molecular shape and whether the molecules are polar(Polarity) is given below for each chemical compound.
(1). Compound: HCN
(a). number of valence electrons = 10.
(b). electronic shape =linear.
(c). molecular shape = linear.
(d). Polarity = Y.
(2). Compound: PH3
(a). number of valence electrons = 8.
(b). electronic shape = Tetrahedral.
(c). molecular shape = Trigonal Pyramidal.
(d). Polarity = Y.
(3). Compound: CHCl3.
(a). number of valence electrons = 26.
(b). electronic shape = tetrahedral.
(c). molecular shape = tetrahedral.
(d). Polarity = Y.
(4). Compound: NH4^+
(a). number of valence electrons = 8
(b). electronic shape = tetrahedral
(c). molecular shape = tetrahedral
(d). Polarity = Y.
(5). Compound: H2CO
(a). number of valence electrons = 12.
(b). electronic shape = Trigonal planar.
(c). molecular shape = Trigonal planar
(d). Polarity = Y.
(6). Compound: SO4^2-
(a). number of valence electrons = 32.
(b). electronic shape = Tetrahedral.
(c). molecular shape = Tetrahedral.
(d). Polarity = N.
(7). Compound: SeF2.
(a). number of valence electrons = 20.
(b). electronic shape = Tetrahedral.
(c). molecular shape = bent.
(d). Polarity = Y.
(8). Compound: CO2.
(a). number of valence electrons = 16.
(b). electronic shape = linear.
(c). molecular shape = linear.
(d). Polarity = N.
(9). Compound: O2
(a). number of valence electrons = 32.
(b). electronic shape = Trigonal planar.
(c). molecular shape = Linear.
(d). Polarity = N.
(10). Compound: ClO4-.
(a). number of valence electrons = 32.
(b). electronic shape = Tetrahedral.
(c). molecular shape = Tetrahedral.
(d). Polarity = N.
(11). Compound: HBr.
(a). number of valence electrons = 8.
(b). electronic shape = Linear.
(c). molecular shape = Linear.
(d). Polarity = Y.
(12). Compound: PF5.
(a). number of valence electrons = 40.
(b). electronic shape = Trigonal Bipyramidal.
(c). molecular shape = Trigonal Bipyramidal.
(d). Polarity = N.
(13). Compound: BeH2.
(a). number of valence electrons = 4.
(b). electronic shape = Linear.
(c). molecular shape = Linear.
(d). Polarity = N.
(14). Compound: PO4^3-.
(a). number of valence electrons = 32.
(b). electronic shape = Tetrahedral.
(c). molecular shape = Tetrahedral.
(d). Polarity = N.
(15). Compound: BH3.
(a). number of valence electrons = 6.
(b). electronic shape = Trigonal planar.
(c). molecular shape = Trigonal planar.
(d). Polarity = N
(16). Compound: Br3-.
(a). number of valence electrons = 32.
(b). electronic shape = Trigonal Bipyramidal.
(c). molecular shape = Linear.
(d). Polarity = N.
A student dissolves of sodium hydroxide in of water in a well-insulated open cup. He that observes the temperature of the water rise from to over the course of minutes. Use this data, and any information you need from the ALEKS Data resource,
NaOH(s) rightarrow Na+(aq) + OH-(aq)
You can make any reasonable assumptions about the physical properties of the solution. Note for advanced students: it's possible the student did not do the experiment carefully, and the values you calculate may not be the same as the known and published values for this reaction.
1. Is this reaction exothermic, endothermic, or neither.
2. If you said the reaction was exothermic or endothermic, calculate the amount of neat that was released or absorbed by the reaction in this case.
3. Calculate the reaction enthalpy AW per mole of NaOH.
Answer:
This question is incomplete
Explanation:
This question is incomplete but the answer to (1) and steps involved in calculating (2) and (3) will be discussed in the attachment
1. The reaction is an exothermic reaction because heat was given off. Exothermic reaction is a reaction in which heat is released from a reaction into the surroundings. And according to the question, the student observed that the temperature of the water increased during the course of the minutes.
A hydrocarbon molecule contains carbon and hydrogen atoms in equal numbers. Its molar mass is 130.18 g/mol. What is the molecular formula for the hydrocarbon
Answer:
The molecular formula of the hydrocarbon is C10H10
Explanation:
Here, we are interested in finding the molecular formula for the hydrocarbon.
Since there are equal number of moles of carbon and hydrogen, then we have the molecular formula looking like;
CnHn
Kindly recall that the atomic mass of carbon is 12 a.m.u while that of hydrogen is 1 amu
so calculating the atomic mass of the compound, we have;
12(n) + 1(n) = 130.18
13n = 130.18
n = 130.18/13
n = 10.01
So the molecular formula will be C10H10
Answer:
C10H10
Explanation:
A hydrocarbon is a binary compound of carbon and hydrogen. Hence a hydrocarbon is a compound of the general formula (CH)n
Thus;
(12 + 1) n = 130.18
n= 130.18/13
n = 10
Hence the molecular formula of the compound is C10H10
Which of these four elements is the most reactive metal?
Answer:
Rubidium
Answer: Rubidium is the most reactive metal. Explanation: Metals are the elements that looses electrons and thus, their chemical reactivity will be the tendency to loose electrons.
Explanation:
What is the reason behind an atom being neutral?
The pH of a solution of 19.5 g of malonic acid in 0.250 L is 1.47.The pH of a 0.300 M Solution of sodium hydrogen malonate is 4.26. What are the values of Ka1and Ka2
Answer:
[tex]Ka_1=1.61x10^{-3}[/tex]
[tex]Ka_2=1.01x10^{-8}[/tex]
Explanation:
Hello.
In this case, since the stepwise dissociation of malonic acid which is a diprotic acid that we are going to symbolize by H₂A, is:
[tex]H_2A\rightleftharpoons H^++HA^-;Ka_1\\\\HA^-\rightleftharpoons H^++A^-;Ka_2[/tex]
The first ionization has the following equilibrium expression:
[tex]Ka_1=\frac{[H^+][HA^-]}{[H_2A]}[/tex]
Whereas the concentration of H⁺ equals the concentration of HA⁻ and is computed via the pH:
[tex][H^+]=[HA^-]=10^{-pH}=10^{-1.47}=0.0339M[/tex]
Next, we compute the molarity of the 19.5 g of malonic acid (molar mass = 104.06 g/mol) as shown below:
[tex][H_2A]=\frac{19.5g/(104.06 g/mol)}{0.250L}=0.750M[/tex]
Thus, Ka1 turns out:
[tex]Ka_1=\frac{(0.0339)(0.0339)}{0.750-0.0339}=1.61x10^{-3}[/tex]
Now, for the second ionization, since the 0.300-M sodium hydrogen malonate is the source of HA⁻, and the pH is 4.26, we can compute the concentration of both H⁺ and A⁻² again by considering the pH:
[tex][H^+]=[A^-^2]=10^{-4.26}=5.50x10^{-5}M[/tex]
Therefore Ka2 turns out:
[tex]Ka_2=\frac{[H^+][A^{-2}]}{[HA^-]}=\frac{(5.50x10^{-5})(5.50x10^{-5})}{0.300-(5.50x10^{-5})}\\ \\Ka_2=1.01x10^{-8}[/tex]
Best regards!
A block of aluminum weighing 100g is cooled from 108.4°C to 68.2°C
with the release of 1080 joules of heat. From this data, calculate the
specific heat of aluminum
Answer:
Specific heat of aluminum = 0.27 j/g.°C
Explanation:
Given data:
Mass of Al = 100 g
Initial temperature = 108.4°C
Final temperature = 68.2°C
Heat released = -1080 j
Specific heat of aluminum = ?
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 68.2°C - 108.4°C
ΔT = - 40.2°C
Q = m.c. ΔT
-1080 j = 100 g ×c ×- 40.2°C
-1080 j = -4020 g.°C ×c
c = -1080 j/-4020 g.°C
c = 0.27 j/g.°C