Answer:
Option A. The carbocation accepts a pair of electrons.
Explanation:
A carbocation is defined as a positively charged carbon, which is bound to 3 substituents. Since it has no electrons nonbonding, it only has six electrons in its valence shell. With only six electrons in its valence shell, a carbocation is a powerful electrophile (and Lewis acid) and can react with any nucleophile that is found.
Carbocations are proposed as intermediates in many organic reactions. They also work like free radicals, which are electron-deficient species.
Same as free radicals, the carbocations are stabilized by alkyl substituents.
If the net force acting on an object is 0 N, the forces are
Since the forces equal to 0, it's balanced because the object isn't moving.
How many moles of NaCl , if mixed with excess Pb2+ ions in solution, would be needed to form 45.5 g of PbCl2 ?
The number of mole of NaCl needed to react with excess Pb²⁺ to produce 45.5 g of PbCl₂ is 0.328 mole
We'll begin by calculating the number of mole in 45.5 g of PbCl₂. This can be obtained as follow:
Mass of PbCl₂ = 45.5 g
Molar mass of PbCl₂ = 207 + (35.5×2) = 278 g/mol
Mole of PbCl₂ =?Mole = mass / molar mass
Mole of PbCl₂ = 45.5 / 278
Mole of PbCl₂ = 0.164 mole Finally, we shall determine the number of mole of NaCl needed to produce 0.164 mole (i.e 45.5 g) of PbCl₂. This can be obtained as follow:2NaCl + Pb²⁺ —> PbCl₂ + 2Na⁺
From the balanced equation above,
2 moles of NaCl reacted to produce 1 mole of PbCl₂
Therefore,
Xmol of NaCl will react to produce 0.164 mole of PbCl₂ i.e
Xmol of NaCl = 2 × 0.164
Xmol of NaCl = 0.328 mole
Thus, the number of mole of NaCl needed for the reaction is 0.328 mole
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Answer:
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Explanation:
Given the following balanced equation, determine the rate of reaction with respect to [Cl2]. If the rate of Cl2 loss is 4.64 × 10-2 M/s, what is the rate of formation of NOCl? 2 NO(g) + Cl2(g) → 2 NOCl(g)
Answer:
[tex]r_{NOCl}=9.28x10^{-2}M[/tex]
Explanation:
Hello!
In this case, given the balanced chemical reaction:
[tex]2 NO(g) + Cl_2(g) \rightarrow 2 NOCl(g)[/tex]
Since there is 1:2 mole ratio between chlorine and NOCl, based on the rate proportions, we can write:
[tex]\frac{1}{-1}r_{Cl_2} =\frac{1}{2}r_{NOCl}[/tex]
It means that for the formation of NOCl, we obtain:
[tex]r_{NOCl}=\frac{2}{-1}r_{Cl_2} \\\\r_{NOCl}=\frac{2}{-1}(-4.64x10^{-2}M)\\\\r_{NOCl}=9.28x10^{-2}M[/tex]
Notice that chlorine is disappearing, which means its rate is negate.
Best regards!
how many atoms does sodium hypochlorite have?
Answer:
NaClO; the contained elements ae Na, Cl, O.
A 50.0 mL solution of Ba(OH)2 is combined with a 150 mL solution of 0.20 M HCl. If the resulting solution has a hydroxide ion concentration of 0.12 M, what was the concentration of Ba(OH)2 in the original solution?
Answer:
0.54M of Ba(OH)2
Explanation:
When Ba(OH)2 reacts with HCl, BaCl2 and H2O are produced as follows:
Ba(OH)2 + 2HCl → BaCl2 + 2H2O
The remanent hydroxide ion is because not all Ba(OH)2 reacts. Thus, we need to find moles of Ba(OH)2 that doesn't react and moles of Ba(OH)2 that reacts. The ratio between total moles and volume of the Ba(OH)2 solution = 0.050L is the molarity of the original solution
Moles of Ba(OH)2 that doesn't react:
50mL + 150mL = 0.200L * (0.12 mol OH- / L) = 0.024 moles OH-
2 moles of OH- are in 1 mole of Ba(OH)2:
0.024 moles OH- * (1mol Ba(OH)2 / 2 mol OH-) = 0.012 moles Ba(OH)2
Moles of Ba(OH)2 that react:
0.150L * (0.20mol / L) = 0.030 moles HCl
2 moles of HCl react per mole of Ba(OH)2:
0.030 moles HCl * (1mol Ba(OH)2 / 2 mol HCl) = 0.015 moles Ba(OH)2
Total moles:
0.012mol + 0.015mol = 0.027mol Ba(OH)2 in 50mL
0.027mol Ba(OH)2 / 0.0500L =
0.54M of Ba(OH)2
How many Stars are In the Milky way Galaxy
Answer:
100 thousand million stars
Explanation:
Answer:
100 to 400 billion stars
3. Theoretically how many grams of magnesium is required to produce to 5.0 g of
Magnesium oxide?
Answer:
3grams
Explanation:
The reaction for the production of Magnesium dioxide will be
Mg + O2 → MgO
we have 5g of MgO (molar mass 40g)
no of moles of MgO = 5/40 = 0.125
Using unitary method we have
1 mole of Mg require 1 mole of MgO
0.125 Mole of MgO = 0.125mole of Mg
n = given mass /molar mass
0.125 = mass / molar mass
mass = 0.125* 24 = 3grams
Which of these lead (II) salts will dissolve to the greatest extent in water?
a. PbSO4, Ksp = 1.7x10^-8
b. PbI2, Ksp = 6.5x10^-9
c. PbCrO4, Ksp = 1.8x10^-14
d. PbS, Ksp = 2.5x10^-27
e. Pb3(AsO4)2, Ksp = 4.0x10^-36
Answer:
a. PbSO4, Ksp = 1.7x10^-8.
Explanation:
Hello!
In this case, since the solubility product indicates how likely a solid is able to ionize and consequently dissolve in water, we can infer that the larger the solubility product Ksp, the more ions are able dissolve in water; therefore the proper answer goes with the largest Ksp, which is a. PbSO4, Ksp = 1.7x10^-8 since the power goes closer to 1 than the other options.
Best regards!
Which of the following covalent bonds is the most polar (i.e., highest percent ionic character)?
A) Al I
B) Si I
C) Al Cl
D) Si Cl
E) Si P
Answer:
answer d is correct that works
what is the mass of 16.6 x 10^23 molecules of barium nitrate
The mass of 16.6 × 10²³ molecules of barium nitrate is 718.7g
The number of moles of a substance can be calculated by dividing the mass by its molar mass. That is;no. of moles = mass(g) ÷ molar mass (g/mol)no. of moles of Ba(NO3)2 = 16.6 × 10²³ ÷ 6.02 × 10²³no. of moles = 2.75 × 10⁰no. of moles = 2.75moles. Molar mass of Ba(NO3)2 = 261.34g/molmass of Ba(NO3)2 = 261.34 × 2.75Mass of Ba(NO3)2 = 718.7gTherefore, the mass of 16.6 × 10²³ molecules of barium nitrate is 718.7g.Learn more at: https://brainly.com/question/15743584?referrer=searchResults
A 13.8 g of zinc is heated to 98.8 c in boiling water and then dropped onto a beaker containing 45.0 g of water at 25.o °C .when the water and metal come to thermal equilibrium the temperature is 27.1°C .what is the specific heat capacity
Answer:
[tex]C_{zinc}=0.400\frac{J}{g\°C}[/tex]
Explanation:
Hello!
In this case, by considering thermodynamics and the temperature of zinc, we can infer it is hot while water is cold, it means that the heat lost by the zinc is gained by the water and we can write:
[tex]Q_{Zn}=-Q_{water}[/tex]
In terms of mass, specific heat and temperatures we can write:
[tex]m_{Zn}C_{Zn}(T_{eq}-T_{Zn})=-m_{water}C_{water}(T_{eq}-T_{water})[/tex]
In such a way, by solving for the specific heat capacity of the zinc we write:
[tex]C_{zinc}=\frac{-m_{water}C_{water}(T_{eq}-T_{water})}{m_{Zn}(T_{eq}-T_{Zn})}[/tex]
Thus, by plugging in the given values we obtain:
[tex]C_{zinc}=\frac{-45.0g*4.184\frac{J}{g\°C}* (27.1\°C-98.8\°C)}{13.8g*(27.1\°C-98.8\°C)}\\\\C_{zinc}=0.400\frac{J}{g\°C}[/tex]
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how fast is a formula 1 car? (For my chemistry project)
Answer:
it's 200 mph fast........
What are three ways that we can identify when a chemical reaction is occurring?
If 75J of heat are applied to 8.4 L of a gas at 732 mmHg and 298K, what is the final temperature, in K, of the gas? Cp for an ideal gas is 20.79 J/mol*K.
Answer:
309 K
Explanation:
Step 1: Convert the pressure to atm
We will use the conversion factor 1 atm = 760 mmHg.
732 mmHg × 1 atm/760 mmHg = 0.963 atm
Step 2: Calculate the moles (n) of the ideal gas
We will use the ideal gas equation.
P × V = n × R × T
n = P × V/R × T
n = 0.963 atm × 8.4 L/0.0821 atm.L/mol.K × 298 K
n = 0.33 mol
Step 3: Calculate the temperature change
We will use the following expression.
Q = n × Cp × ΔT
ΔT = Q/n × Cp
ΔT = 75 J/0.33 mol × 20.79 J/mol.K
ΔT = 11 K
Step 4: Calculate the final temperature
T = 298 K + 11 K = 309 K
Robert was changing the oil in his truck. He dumped the used oil on the ground in his yard. He didn't know it, but Robert was hurting the environment because the oil-
HELP FAST
Answer:
goes deep into the ground and pollutes the groundwater.
Which of these mixtures would NOT be separated by using simple distillation?
Ink and water
Salt and water
Sand and water
There is a golden role of solubility, polar solute dissolve in polar solvent and non polar solute dissolve in non polar solvent. Therefore, the correct option is option B that is Salt and water.
What is solution?Solutions are a homogeneous mixture of two or more substances. A solution is a homogeneous mixture of solvent and solute molecules. Solvent is a substance that is in large amount in solution. solute is the substance which is in small amount in a solution. There are two types of mixture that is homogeneous and heterogeneous. Solution is a homogeneous solution.
Salt and water because salt is transparent and will dissolve into the water among given solute, salt NaCl is a polar solute which will dissolve in polar solvent that is water.
Therefore, the correct option is option B that is Salt and water.
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A 2kg metal cylinder is supplied with 1600J of energy to heat it from 5*C to 13*C. What is the SHC of the metal?
Answer:
100Jkg/°C
Explanation:
Given parameters:
Mass of metal = 2kg
Amount of heat energy = 1600J
Initial temperature = 5°C
Final temperature = 13°C
Unknown:
Specific heat capacity of the metal = ?
Solution:
Specific heat capacity of a body is the amount of heat needed to raise the temperature of unit mass of a body by 1°C.
H = m x C x (T₂ - T₁ )
H is the amount of heat
m is the mass
C is the unknown specific heat capacity
T is the temperature
Insert the parameters and solve;
1600 = 2 x C x (13 - 5)
1600 = 16C
C = 100Jkg/°C
A container holds 3.41×10−3mol of carbon dioxide (CO2). After the addition of 8.41×10−4mol of carbon dioxide, the volume of the container increases to 95.2mL, with the temperature and pressure remaining constant. What was the initial volume of the container?
Answer:
76.366
Explanation:
Knewton answer is 76.4
Answer:
76.4mL
Explanation:
First, calculate the final number of moles of carbon dioxide (n2) in the container.
n2 = n1+n added = 3.41×(10^-3) mol+8.41×(10^-4) mol = 4.25×(10^-3) mol
Rearrange Avogadro's law to solve for V1.
V1 =(V2×n1)/n2
Substitute in the known values for n1, V2, and n2.
V1 = (95.2mL×(3.41×10^-3 mol))/(4.25×(10^-3) mol)=76.4mL
So the initial volume is 76.4mL.
A graduated cylinder initially has 32.5 mL of water in it. After a 75.0 g piece of lead (Pb) is added to the graduated cylinder, the water level rises to the 39.1 mL mark. What is the the volume of the piece of lead.
Answer:
39.1-32.5 and you will find your answer it always like that, you subtract your starting point from your ending point
Explanation:
The volume of the piece of lead is 6.6 mL
From the question given above, the following data were obtained:
Volume of water = 32.5 mLMass of lead (Pb) = 75 gVolume of water + Lead = 39.1 mLVolume of lead =?The volume of the lead can be obtained as follow:
Volume of lead = (Volume of water + Lead) – (Volume of water)
Volume of lead = 39.1 – 32.5
Volume of lead = 6.6 mL
Therefore the volume of the lead is 6.6 mL
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Describe the potential energy of a chemical reaction and how the energy is accessed?
Explanation:
The energy released due to breakage of bond or we can say that protons and electrons go from a structure of higher potential energy to lower potential energy. During this change, potential energy is converted to kinetic energy in form of heat.
Generally chemical potential energy is accessed in form heat, which the difference between the potential energies of reactant and product.
What are five minerals found on space rocks
Answer:
oxygen, silicon, iron, calcium, sodium
Explanation:
these are all found in space rocks known as meteorites. hope this helps :)
which have
different numbers of electrons?
Answer:
These are called isotopes. They have the same number of protons (and electrons), but different numbers of neutrons. Different isotopes of the same element have different masses. Mass is the word for how much substance (or matter) something has.
What is the difference between Mega and Milli symbols?
Answer:
Mega = 10^6
Milli = 10^-3
Explanation:
Mega is whatever times 10 to 6th power.
Milli is whatever times 10 to the -3rd power.
Suppose the current flowing from a battery is used to electroplate an object with silver. Calculate the mass of silver that would be deposited by a battery that delivers 1.65 A·hr of charge.
Answer:
m = 0.00659 kg = 6.59 g
Explanation:
From Faraday's Law of Electrolysis, we know that:
m = ZQ
where,
m = mass of silver deposited = ?
Q = charge supplied = (1.65 A-hr)(3600 s/1 hr) = 5940 C
Z = electrochemical equivalent of silver = 1.18 x 10⁻⁶ kg/C
Therefore,
m = (1.11 x 10⁻⁶ kg/C)(5940 C)
m = 0.00659 kg = 6.59 g
The mass of silver that would be deposited by a battery is 6.65 grams
The precipitation of Ag requires the removal of one electron. The reduction process for silver electrode at the cathode is as follows:
[tex]\mathbf{Ag^+ + e^- \to Ag(s)}[/tex]
The current flowing in the battery = 1.65 A = 1.65 C/sThe time at which the current is flowing = 1 hr = 3600sec∴
The charge Q = Current (I) × time (t)Charge Q = 1.65 C/s × 3600 sCharge (Q) = 5940 CIn one mole of an electron, the charge carried = 96500 C
Recall that:
The atomic mass of silver (Ag) = 108 g
∴
The mass of silver that would be deposited in a 5940 C can be computed as:
[tex]\mathbf{=5940\ C \times \dfrac{108 \ g }{96500 \ C}}[/tex]
= 6.65 grams
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Which statement describes how phase changes can be diagrammed as a substance is heated?
The phase is on the y-axis and the temperature is on the x-axis.
The temperature is on the y-axis and the phase is on the x-axis.
The time is on the y-axis and the temperature is on the x-axis.
The temperature is on the y-axis and the time is on the x-axis.
Answer:
The temperature is on the y-axis and the time is on the x-axis.
Explanation:
Which statement describes how phase changes can be diagrammed as a substance is heated? (D is the answer)
The phase is on the y-axis and the temperature is on the x-axis.
The temperature is on the y-axis and the phase is on the x-axis.
The time is on the y-axis and the temperature is on the x-axis.
The temperature is on the y-axis and the time is on the x-axis.
Which statement describes the appearance of a temperature-vs.-time graph? (C is the answer)
A horizontal line shows that the temperature increases at a constant rate over time.
A vertical line shows that the temperature decreases at a constant rate over time.
Horizontal lines where the temperature is constant during phase changes connect upward-sloping lines where the temperature increases.
Horizontal lines where the temperature increases are connected by upward-sloping lines where the temperature is constant for each phase.
A mixture of neon and xenon gases, at a total pressure of 739 mm Hg, contains 0.919 grams of neon and 19.1 grams of xenon. What is the partial pressure of each gas in the mixture?_______g Xe
Answer:
Partial pressure of neon = 175 mmHg
Partial pressure of xenon = 564 mmHg
Explanation:
The partial pressure of a gas in a mixture can be calculated as the product of the mole fraction of the gas (Xi) and the total pressure (Pt), as follows:
Pi = Xi Pt
The total pressure is 739 mmHg ⇒ Pt = 739 mmHg
In order to calculate the mole fraction of each gas, we have to first calculate the number of moles of each gas (n) by dividing the mass of the gas into the molar mass (MM):
For neon gas (Ne):
MM(Ne) = 20.18 g/mol
n(Ne)= mass/MM = 0.919 g x 1 mol/20.18 g = 0.045 mol Ne
For xenon gas (Xe):
MM(Xe) = 131.3 g/mol
n(Xe)= mass/MM = 19.1 g x 1 mol/131.3 g = 0.145 mol Xe
Now, we calculate the mole fraction (X) by dividing the number of moles of the gas into the total number of moles (nt):
nt= moles Ne + moles Xe = 0.045 mol + 0.145 mol = 0.190 mol
X(Ne) = moles Ne/nt = 0.045 mol/0.190 mol = 0.237
X(Xe) = moles Xe/nt = 0.145/0.190 mol = 0.763
Finally, we calculate the partial pressures of Ne and Xe as follows:
P(Ne) = X(Ne) x Pt = 0.237 x 739 mmHg = 175 mmHg
P(Xe) = X(Xe) x Pt = 0.763 x 739 mmHg = 564 mmHg
Scientists who are investigating what has happened over the years to the people who were exposed to radiation during the 1986 Chernobyl nuclear disaster are conducting what type of study
Answer:
Observational study
Explanation:
An investigation of what happened over the years to the people who were exposed to radiation during the 1986 Chernobyl nuclear disaster would be termed an observational study because it will require observing subjects and finding out if there is a kind of correlating factor in the subjects that could indicate the level of exposure to the nuclear radiation.
There are basically two types of research studies:
1. Observational studies
2. Experimental studies
Observational studies involve just the observation of subjects and trying to find out some correlation factors. This kind of study could be prospective when it deals with what could happen in the future based on certain factors or retrospective when it considers what has already happened. An observational study can be cross-sectional, case-only, case-control, or cohort study.
Experimental studies are also referred to as randomized or controlled trial studies. Here, subjects are grouped and each group receives a different treatment in order to isolate the effect of a particular factor or variable relating to the treatment.
1. A baseball pitcher won 75% of the games he pitched. If he pitched 43 ballgames, how many games
did he win?
Answer:
He won 32 games.
Explanation:
Another way of expressing 75% is 75/100.
To calculate how many games the baseball pitcher won, we multiply the total of games played (43, in this case) by the percent won:
43 * 75/100 = 32.25 ≅ 32
So the baseball pitcher won 32 games.
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A weather balloon with a volume of 3.40774 L
is released from Earth’s surface at sea level.
What volume will the balloon occupy at an
altitude of 20.0 km, where the air pressure is
10 kPa?
Answer in units of L.
Answer: The volume occupied at an altitude of 20.0 km is 34.5289 L
Explanation:
Boyle's Law: This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.
[tex]P\propto \frac{1}{V}[/tex] (At constant temperature and number of moles)
where,
[tex]P_1[/tex] = initial pressure of gas = 101.325 kPa ( sea level)
[tex]P_2[/tex] = final pressure of gas = 10 kPa
[tex]V_1[/tex] = initial volume of gas = 3.40774 L
[tex]V_2[/tex] = final volume of gas = ?
Now put all the given values in the above equation, we get the final pressure of gas.
[tex]101.325\times 3.40774=10\times V_2[/tex]
[tex]V_2=34.5289L[/tex]
Therefore, the volume occupied at an altitude of 20.0 km is 34.5289 L