Answer:
Cyclopropylmethylene
How do the test variables (independent variables) and outcome variables (dependent variables) in an experiment compare? A. The test variables (independent variables) and outcome variables (dependent variables) are the same things. B. The test variable (independent variable) controls the outcome variable (dependent variable). C. The test variable (independent variable) and outcome variable (dependent variable) have no affect on each other. D. The outcome variable (dependent variable) controls the test variable (independent variable).
Answer:
I'm on the exact same queston
Answer:
The test variable (independent variable) controls the outcome variable (dependent variable)
Explanation:
its right on study island
How many atoms of Chlorine are in 1.00 mol of Chlorine gas?
6.022 x 10∧23
3.01 x 10∧23
6.022 x 10∧24
Answer:
6.02 × 10²³ atoms Cl₂
Explanation:
Avagadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
Step 1: Define
1.00 mol Cl₂ (g)
Step 2: Use Dimensional Analysis
[tex]1.00 \hspace{3} mol \hspace{3} Cl_2(\frac{6.02(10)^23 \hspace{3} atoms \hspace{3} Cl_2}{1 \hspace{3} mol \hspace{3} Cl_2} )[/tex] = 6.02 × 10²³ atoms Cl₂
3. A student took a calibrated 200.0 gram mass, weighed it on a laboratory balance, and
found it read 196.5 9. What was the student's percent error?
Answer:
The answer is 1.71 %Explanation:
The percentage error of a certain measurement can be found by using the formula
[tex]P(\%) = \frac{error}{actual \: \: number} \times 100\% \\ [/tex]
From the question
actual mass = 200 g
error = 200 - 196.59 = 3.41
We have
[tex]p(\%) = \frac{3.41}{200} \times 100 \\ = 1.705[/tex]
We have the final answer as
1.71 %Hope this helps you
5 advantages of storing oil underground in salt dome?
Answer:
Salt domes storage has advantages in cost, security, environmental risk, and maintenance. Salt formations offer the lowest cost, most environmentally secure way to store crude oil for long periods of time. Stockpiling oil in artificially-created caverns deep within the rock-hard salt costs historically about $3.50 per barrel in capital costs. Storing oil in above ground tanks, by comparison, can cost $15 to $18 per barrel - or at least five times the expense. Also, because the salt caverns are 2,000-4,000 feet below the surface, geologic pressures will sea; any crack that develops in the salt formation, assuring that no crude oil leaks from the cavern. An added benefit is the natural temperature differential between the top of the caverns and the bottom - a distance of around 2,000 feet; the temperature differential keeps the crude oil continuously circulating in the caverns, giving the oil a consistent quality.
Which pair of elements would most likely have a similar arrangement of outer
electrons and have similar chemical behaviors?
boron and aluminum
helium and fluorine
carbon and nitrogen
chlorine and oxygen
Answer:
Boron and Aluminum
Explanation:
If you write the electron configuration for boron and aluminum, you get:
[tex]1s^22s^22p^1[/tex] for boron and [tex]1s^22s^22p^63s^23p^1[/tex] for aluminum. Both have 3 valance electrons and has 2 electrons in a s-orbital and 1 in a p-orbital. These valance electron similarities are based on the column/group the elements are. Therefore, Boron and Aluminum have similar chemical behaviours and similar arrangement of outer/valance electrons.
I need help with this please
Thank you
Answer:
From fastest to slowest its: (4)A to B, (1)E to F, (3)C to D, (2)D to E
Explanation:
The steeper the line is the faster she went. D to E she didn't make any progress because the line is straight. Sry I'm terrible at explaining things.
What is the volume of a substance that has a mass of 59 g and a density of 1.98 g/mL?
(show all work)
Answer:
29.8
Explanation:
The formula for volume is mass/ density, so 59/1.98. 29.8 is the answer.
What volume of 1.27 M HCl is required to prepare 197.4 mL of 0.456 M HCl
Answer:
70.88 mL volume of 1.27 M of HCl is required.
Explanation:
Given data:
Initial volume = ?
Initial molarity = 1.27 M
Final volume = 197.4 mL
Final molarity = 0.456 M
Solution:
Formula:
M₁V₁ = M₂V₂
Now we will put the values in formula.
1.27 M × V₁ = 0.456 M × 197.4 mL
V₁ = 0.456 M × 197.4 mL/1.27 M
V₁ = 90.014M.mL/1.27 M
V₁ = 70.88 mL
70.88 mL volume of 1.27 M of HCl is required.
Check all of the boxes that are true about the proton:
it is outside the nucleus
it has a positive charge
it has no mass
it has a negative charge
it is inside the nucleus
it is the same as the atomic number
it is the same as the number of neutrons
75% of the isotopes have a mass
Answer:
it is outside the nucleus F
it has a positive charge T
it has no mass F
it has a negative charge F
it is inside the nucleus ...it is part OF the nucleus.
it is the same as the atomic number T
it is the same as the number of neutrons F
75% of the isotopes have a mass ima just guess cuz i dunno about this one...i think it matters on the atom element.
Explanation:
An insulated container is used to hold 47.0 g of water at 23.5°C. A sample of copper weighing 10.3 g is placed in a dry test tube and
heated for 30 minutes in a boiling water bath at 100.0°C. The heated test tube is carefully removed from the water bath with laboratory
tongs and inclined so that the copper slides into the water in the insulated container. Given that the specific heat of solid copper is
0.385 J/(g.°C), calculate the maximum temperature of the water in the insulated container after the copper metal is added.
Answer:
[tex]T_f=25.0\°C[/tex]
Explanation:
Hello.
In this case, considering that the sample of hot copper is submerged into the water and the container is isolated, the heat lost by the copper is gained by the water so we can write:
[tex]Q_{Cu}=-Q_w[/tex]
In terms of mass, specific heat and temperature we write:
[tex]m_{Cu}C_{Cu}(T_f-T_{Cu})=-m_wC_w(T_f-T_w)[/tex]
Whereas the final temperature is the same for both copper and water because they are in contact until thermal equilibrium is reached. In such a way, the required maximum temperature no more than the equilibrium temperature and is computed as shown below:
[tex]T_f=\frac{m_{Cu}C_{Cu}T_{Cu}+m_wC_wT_w}{m_{Cu}C_{Cu}+m_wC_w}[/tex]
Thus, plugging the given data in the formula, we obtain:
[tex]T_f=\frac{10.3g*0.385\frac{J}{g\°C}*100\°C +47.0g*4.184\frac{J}{g\°C}*23.5\°C }{10.3g*0.385\frac{J}{g\°C}+47.0g*4.184\frac{J}{g\°C}}\\\\T_f=25.0\°C[/tex]
Which is a small change considering the initial one, because the mass of water is greater than the mass of copper as well as for the specific heats.
Best regards!
The maximum temperature of the water in the insulated container after the copper metal is added is 25 °C
From the question given above above, the following data were obtained:
Mass of water (Mᵥᵥ) = 47 g
Temperature of water (Tᵥᵥ) = 23.5°C
Specific heat capacity of water (Cᵥᵥ) = 4.184 J/gºC
Mass of copper (M꜀) = 10.3 g
Temperature of copper (M꜀) = 100 °C
Specific heat capacity of copper (C꜀) = 0.385 J/gºC
Equilibrium temperature (Tₑ) =?The equilibrium temperature of the mixture can be obtained as follow:
Heat loss by copper = Heat gained by water
Q꜀ = Qᵥᵥ
M꜀C꜀(M꜀ – Tₑ) = MᵥᵥCᵥᵥ(Tₑ– Mᵥᵥ)
10.3 × 0.385 (100 – Tₑ) = 47 × 4.184 (Tₑ – 23.5)
3.9655 (100 – Tₑ) = 196.648 (Tₑ – 23.5)
Clear bracket
396.55 – 3.9655Tₑ = 196.648Tₑ – 4621.228
Collect like terms
396.55 + 4621.228 = 196.648Tₑ + 3.9655Tₑ
5017.778 = 200.6135Tₑ
Divide both side by 200.6135
Tₑ = 5017.778 / 200.613
Tₑ = 25 °CThus, the equilibrium temperature of the mixture is 25 °C. Therefore, the maximum temperature of the water in the insulated container is 25 °C
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Help :( the blue boxes are clickable
Answer:
I am sure they are why won't they
Which 2 letters have the most kinetic energy?
W
X
Y
Z
Answer:
See Explanation (X and Z)
Explanation:
The question has missing details as the attachment that illustrates the graph is missing.
I'll answer this question using the attached.
Kinetic energy increases as height decreases
Base on the attachment, from order of highest height to the least, we have:
W
Y
Z
X
So, we can conclude that X has the highest kinetic energy and it is immediately followed by Z
Hence:
X and Z answers the question
What is the gravitational potential energy, in joules, of a 75 kg person that is 1000.0
meter above the ground? Gravitational acceleration = 9.81 m/s2
Answer:
In this example, a 3 kilogram mass, at a height of 5 meters, while acted on by Earth's gravity would have 147.15 Joules of potential energy, PE = 3kg * 9.81 m/s 2 * 5m = 147.15 J. 9.81 meters per second squared (or more accurately 9.80665 m/s 2 ) is widely accepted among scientists as a working average value for Earth's gravitational pull.
Explanation:
Write the net ionic equation for the reaction of aqueous solutions of ammonium chloride and iron(III) hydroxide.
Answer:
[tex]Fe(OH)_3(s)\rightarrow Fe^{3+}(aq)+3OH^-(aq)[/tex]
Explanation:
Hello.
In this case, for the reaction between aqueous solutions of ammonium chloride and iron (III) hydroxide, we have the following complete molecular reaction:
[tex]3NH_4Cl(aq)+Fe(OH)_3(s)\rightarrow 3NH_4OH+FeCl_3[/tex]
And the full ionic equation, taking into account that the iron (III) hydroxide cannot be dissolved as it is insoluble in water:
[tex]3NH_4^+(aq)+3Cl^-(aq)+Fe(OH)_3(s)\rightarrow 3NH_4^+(aq)+3OH^-(aq)+Fe^{3+}(aq)+3Cl^-(aq)[/tex]
Finally, the net ionic equation, considering that spectator ions are NH₄⁺, Cl⁻ as they are both the left and right side, therefore, the net ionic equation is:
[tex]Fe(OH)_3(s)\rightarrow Fe^{3+}(aq)+3OH^-(aq)[/tex]
Best regards.
The net ionic equation for the reaction of aqueous solutions should be Fe(Oh)3 ➡Fe3+(aq) + 3OH^-(aq).
Net ionic equation:
When the reaction lies between the between aqueous solutions of ammonium chloride and iron (III) hydroxide
So, here the total reaction should be
3NH4Cl(aq) + Fe(OH)3(s) ➡ 3NH4OH + FeCl3
So, here net ionic equation, considering that spectator ions are NH₄⁺, Cl⁻ since they are both the left and right sides.
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How many atoms are in 10 g of He
Answer:
6.7
⋅
10
23
atoms of H
Explanation:
The image below shows a model of the atom. Which subatomic particle does the arrow in
the image below identify?
?
A. electron
B. neutron
C. orbital
D. proton
The correct answer is A. Electron
Explanation:
The model of this atom depicts the nucleus of this in the center of the model, this section of the atom contains sub-particles known as protons and neutrons. Moreover, in the atom, the nucleus is surrounded by three sub-particles that orbit or move around the nucleus. These sub-particles are the electrons; these differ from other sub-particles because they have a negative charge and they are not part of the nucleus. Also, these move around the nucleus is orbits, although they move similarly to waves. According to this, the correct answer is A.
Please help me ! Thank you
Answer:
10
Explanation:
What is the acceleration of a 7 kg mass if a force of 68.6 N is used to move it toward earth
Answer:
acceleration = force/mass
= (68.6+mg)/7
= 19.6 m/s²
Explanation:
9.8 m/s² is the acceleration acting on 7 kg mass if force of 68.6 N is used to move it towards earth.
What is force?Force is defined as a cause which is capable of changing the motion of an object. It can cause an object which has mass to change it's velocity. It is also simply a push or a pull . It has both magnitude as well as direction.Hence, it is a vector quantity.
It has SI units of Newton and is represented by'F'.Newton's second law states that force which acts on an object is equal to momentum which changes with time. If mass of object is constant, acceleration is directly proportional to net force acting on an object.
The concepts which related to force are thrust and torque .Thrust increases the velocity of an object and torque produces change in rotational speed of an object.
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Please help
What is an element
Answer:
An element is atoms with the same number of protons.
Explanation:
Protons, electrons, and neutrons.
An electrochemical cell has an Eocell of 1.50 V and transfers 1 mole of electrons. The Faraday constant is 96,485 C/mol e-.
What is the value of delta G?
Answer:
ΔG° = -1.45 × 10⁵ J
Explanation:
Step 1: Given data
Standard cell potential (E°cell): 1.50 VMoles of electrons transferred (n): 1 mol e⁻Faraday constant (F): 96,485 C/mol e⁻Step 2: Calculate the standard Gibbs free energy change (ΔG°)
We will use the following expression.
ΔG° = -n × F × E°cell
ΔG° = -1 mol e⁻ × 96,485 C/mol e⁻ × 1.50 V
ΔG° = -1.45 × 10⁵ J
By apply Gibbs's free energy, the value of delta G is equal to -144727.5 Joules.
Given the following data:
Faraday constant = 96,485 C/mol e-[tex]E^{ \circ}_{cell}[/tex] = 1.50 VoltsNumber of moles of electrons = 1 molTo determine the value of delta G, we would apply Gibbs's free energy:
Mathematically, Gibbs's free energy is given by the formula:
[tex]\Delta G^\circ = -nFE^{ \circ}_{cell}[/tex]
Where:
F is Faraday constant.n is the number of moles.[tex]E^{ \circ}_{cell}[/tex] is the electromotive force.Substituting the given parameters into the formula, we have;
[tex]\Delta G^\circ = -1 \times 96485 \times 1.50[/tex]
[tex]\Delta G^\circ = -144727.5[/tex]
Delta G = -144727.5 Joules
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Selenium has six valence electrons. What is the valence of selenium?
HELP ILL MARK BRAINLEST
How many moles would be in 24.23 grams of SrSO4?
Use two digits past the decimal for all values.
Answer:
about 0.13 mol
Explanation:
To find number of mols when given grams you first have to find the molar mass of the compound. This is done by adding up the atomic masses of the element in the compound. So Sr= 88 g/mol S=32 g/mol and O=16 g/mol. Then 88+32+(16x4)=184. Then using this you can convert from grams to mols by dividing the grams by the molar mass. So, 24.23/184 equals about 0.13 mol.
1) The speed constant for the second order reaction in the gas phase
It varies with the temperature according to the table below. Calculate the activation energy for the process, according to Arhhenius' equation
Answer:
41.7 kJ/mol
Explanation:
ln(k) = ln(A) − Eₐ/(RT)
Pick any two points. I'll choose 100°C and 400°C.
When T = 100°C = 373 K, k = 1.10×10⁻⁹ L/mol s:
ln(1.10×10⁻⁹) = ln(A) − Eₐ/(R × 373)
When T = 400°C = 673 K, k = 4.40×10⁻⁷ L/mol s:
ln(4.40×10⁻⁷) = ln(A) − Eₐ/(R × 673)
Subtract the two equations and solve:
ln(4.40×10⁻⁷) − ln(1.10×10⁻⁹) = -Eₐ/(R × 673) + Eₐ/(R × 373)
5.991 = 0.00120 Eₐ/R
Eₐ/R = 5013.4
Eₐ = 41700 J/mol
Eₐ = 41.7 kJ/mol
The pKb values for the dibasic base B are pKb1=2.10 and pKb2=7.54. Calculate the pH at each of the points in the titration of 50.0 mL of a 0.60 M B(aq) solution with 0.60 M HCl(aq).
Complete Question
The pKb values for the dibasic base B are pKb1=2.10 and pKb2=7.54. Calculate the pH at each of the points in the titration of 50.0 mL of a 0.60 M B(aq) solution with 0.60 M HCl(aq).
(a) before addition of any HCl (b) after addition of 25.0 mL of HCl
Answer:
a The value is [tex]pH =12.81[/tex]
b [tex]pH = 11.9[/tex]
Explanation:
From the question we are told that
The first pKb value for B is [tex]pK_b_1 = 2.10[/tex]
The second pKb value for B is [tex]pK_b_2 = 7.54[/tex]
The volume is [tex]V = 50.0 mL =[/tex]
The concentration of B is [tex][B] = 0.60 M[/tex]
The concentration of [tex]C_A = 0.60 M[/tex]
Generally the reaction equation showing the first dissociation of B is
[tex]\ce{B_{(aq) } + H_2O _{(l)} <=> BH^+ _{(aq)} + OH^- _{(aq)} }[/tex]
Here the ionic constant for B is mathematically represented as
[tex]K_i = \frac{[BH^+] [OH^-]}{[B]}[/tex]
Let denot the concentration of [BH^+] as z and since [tex][BH^+] = [OH^-][/tex] then [tex][OH^-][/tex] is also z
So [B] = 0.60 - z
Here [tex]K_i[/tex] is ionic constant for the first reaction of a dibasic base B and the value is
[tex]K_i = 7.94 *10^{-3}[/tex]
So
[tex] 7.94 *10^{-3}= \frac{z^2}{ 0.60 - z}[/tex]
=> [tex]z^ 2 + 0.00794 z - 0.00476[/tex]
using quadratic formula to solve this equation
[tex]z = 0.0651[/tex]
Hence the concentration of [tex]OH^{-}[/tex] is [tex][OH^-] =0.0651[/tex]
Generally [tex]pOH = -log [OH^-][/tex]
=> [tex]pOH = -log (0.065)[/tex]
=> [tex]pOH = 1.187 [/tex]
Generally the pH is mathematically represented as
[tex]pH = 14 - 1.187[/tex]
[tex]pH =12.81[/tex]
Generally the volume of [tex]HCl[/tex] at the second dissociation of the base B is [tex] 50 mL [/tex]
The volume of the [tex]HCl[/tex] half way to the first dissociation of the base is 25mL
Now the pOH at half way to the first dissociation of the base is
[tex]pOH = -log(K_i)[/tex]
=> [tex]pOH = -log(0.00794)[/tex]
=> [tex]pOH = 2.100[/tex]
Generally the pH after addition of 25.0 mL of HCl is
[tex]pH = 14 - 2.100[/tex]\
=> [tex]pH = 11.9[/tex]
The first dissociation's equation is as follows:
[tex]B(aq) + H_2O(l) \leftrightharpoons BH^{+} (aq) + OH^{-}(aq) \\\\[/tex]
Constant of base ionization
[tex]\to K_{bl}=\frac{[BH^{+}][OH^{-}]}{[B]}\\\\ \to 7.94\times 10^{-3} = \frac{x\times x}{(0.95- x)} \\\\\to 7.94\times 10^{-3} = \frac{x^2}{(0.95- x)} \\\\\to x^2=7.94\times 10^{-3} (0.95-x) \\\\\to x^2=7.543\times 10^{-3} - 7.94\times 10^{-3} x) \\\\\to x^2=7.543\times 10^{-3} - 7.94\times 10^{-3} x) \\\\\to x = 0.0830\ M\\\\[/tex]
So,
[tex]\to [OH^{-}] = 0.0830\ M\\\\[/tex]
The second dissociation of the base equation is
[tex]BH^{+}\ (aq) + H_20\ (l) \leftrightharpoons BH_2^{2+}\ (aq) + OH^{-}\ (aq) \\\\[/tex]
Constant of base ionization
[tex]\to K_{bl}=\frac{[BH^{+}][OH^{-}]}{[B]}\\\\ \to 3.2 \times 10^{-8} =\frac{y \times (0.0830+y)}{(0.0830- y)}\\\\[/tex]
[tex]\to y= \frac{ 3.2 \times 10^{-8} \times (0.0830- y)}{ (0.0830+y)} \\\\ \to y= \frac{ 3.2 \times 10^{-8} \times (0.0830- y)}{ (0.0830+y)} \\\\ \to y = 3.2\times 10^{-8}[/tex]
So,
[tex]\to [OH^{-}] = 0.0830\ M \\\\\to pOH = 1.08 \\\\\to pH = 14.00 - pOH = 12.92\\\\[/tex]
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a compound has a molecular formular of C12H24O6.What is the compound's empirical formula
Answer:
The empirical formula for C12 H24 O6 is C2 H4 O.
Answer:
We are given the formula of the compound:
C12H24O6
The empirical formula of a molecular formula is the lowest whole number ratio between the number of atoms of each element
The ratio of C to H to O in the given formula is :
12 : 24 : 6
we notice that all 3 of the numbers have 6 in common. Dividing all three of the numbers by 6, we get:
2 : 4 : 1
Hence, the ratio of Carbon to Hydrogen to Oxygen in the empirical formula of the given compound is 2 : 4 : 1 ,
Empirical Formula = C2H4O
Consider the reaction of 30.0 mL of 0.235 M BaI₂ with 20.0 mL of 0.315 M Na₃PO₄.
Which of the following compounds would be the precipitate that forms?
a) Bal2
b) Na3PO4
c) Ba3(PO4)2
d)Nal
Answer:
C
Explanation:
Because sodium is basically always soluble with any compound, it is between a and c. a is part of the reactant so it cant be A. So C.
The statement, that describes the compounds would be the precipitate that forms in the reaction is "Ba3(PO4)2"
What is precipitate?Precipitate is a solid generated by a change in a solution, usually due to a chemical reaction or a change in temperature that reduces a solid's solubility.
What is compound?The combination of more than one element will be identified ad compound.
When cations and anions in aqueous solution combine to create an insoluble ionic solid called a precipitate, double displacement reactions occur, resulting in the formation of a solid form residue. Except for salts of Group 1 metals and ammonium, salts of phosphates and carbonates ions are insoluble, according to the solubility flow chart. The production of a solid white precipitate is used to demonstrate the insoluble nature of barium phosphate.
Hence the correct option is c.
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I don’t really understand, if anybody can help I’ll really appreciate it ! Thank you.
Answer:
175
Explanation:
PLEASE HELP!!!
what was the volume of air that has a volume of 6.00L at 120870 Pa, if the original pressure was 250020 Pa?
Answer:
The answer is 2.90 LExplanation:
In order to find the original pressure , we use the formula for Boyle's law which is
[tex]P_1V_1 = P_2V_2[/tex]
where
P1 is the initial pressure
P2 is the final pressure
V1 is the initial volume
V2 is the final volume
Since we are finding the original volume
[tex]V_1 = \frac{P_2V_2}{P_1} \\[/tex]
From the question
P1 = 250020 Pa
P2 = 120870 Pa
V2 = 6 L
We have
[tex]V_1 = \frac{120870 \times 6}{250020} = \frac{725220}{250020} \\ = 2.90064794...[/tex]
We have the final answer as
2.90 LHope this helps you
What happens when the elements in group 2 react with water?
Answer:
The Group 2 metals become more reactive towards the water as you go down the Group.
Explanation:
These all react with cold water with increasing vigour to give the metal hydroxide and hydrogen. ... You get less precipitate as you go down the Group because more of the hydroxide dissolves in the water. Summary of the trend in reactivity.
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what’s the most abundant isotope of lawrencium
Answer:
266Lr
Thirteen isotopes of lawrencium are currently known; the most stable is 266Lr with a half-life of 11 hours, but the shorter-lived 260Lr (half-life 2.7 minutes) is most commonly used in chemistry because it can be produced on a larger scale.
Explanation:
hopefully that helps you