Answer:
1.3x10⁻⁸ mol/L
Explanation:
0.0013μmol, Calculate concentration in mol/L
To obtain concentration in mol/L we need to convert the μmoles to moles and mL to liters:
Moles silver(II) oxide:
0.0013μmol × (1mol / 1x10⁶μmol) = 1.3x10⁻⁹ moles
Liters solution:
100mL * (1L / 1000mL) = 0.1L
That means concentration in mol/L is:
1.3x10⁻⁹ moles / 0.1L =
1.3x10⁻⁸ mol/Ltwelve grams of sodium chloride wwere dissolved in 52 ml (52g) of distilled water, calculate the % sodium chloride in the solution
Answer:
The mass of sodium chloride in the mixture is 18.75%
Explanation:
Here, we want to calculate the percentage of sodium chloride in the mixture.
The total mass of the mixture is 52 g + 12 g = 64 g
So the percentage mass of sodium chloride will be;
mass of sodium chloride/ Total mass * 100%
That will be: 12/64 * 100 = 18.75%
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.
3 points
18) A student determines the density of gold to be 20.9g/L. The true
density of gold is 19.30g/L. What is the student's percent error?round
answer to 2 significant figures *
Answer:
The answer is 8.29 %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 density = 19.30g/L
error = 20.9 - 19.3 = 1.6
We have
[tex]p(\%) = \frac{1.6}{19.3} \times 100 \\ = 8.290155440...[/tex]
We have the final answer as
8.29 %Hope this helps you
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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1
An atom of element Q contains 19 electrons, 19 protons and 20 neutrons.
What is Q?
A calcium
B potassium
С
strontium
D
yttrium
Answer:
b)Potassium is the right answerAnswer:
B. Potassium
Explanation:
The element with 19 electrons, 19 protons, and 20 neutrons is potassium
Selenium has six valence electrons. What is the valence of selenium?
HELP ILL MARK BRAINLEST
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:
When converting an “ordinary” number that is greater than 1 to scientific notation, how many non-zero digits are to the LEFT of the decimal point when you are finished?
Answer: 0 I think
Explanation:
pretty sure its zero because i learned it last year but im in middle school so you might want to look it up.
If it takes 26.0 mL of 0.0250 M potassium dichromate to titrate 25.0 mL of a solution containing Fe2 , what is the molar concentration of Fe2
Answer:
Explanation:
moles of potassium dichromate = .0250 x .026 = 65 x 10⁻⁵ moles
1 mole of potassium dichromate reacts with 6 moles of Fe⁺²
65 x 10⁻⁵ moles of potassium dichromate will react with
6 x 65 x 10⁻⁵ moles of Fe⁺²
= 390 x 10⁻⁵ moles
390 x 10⁻⁵ moles are contained in 25 mL of solution
molarity of solution = 390 x 10⁻⁵ / 25 x 10⁻³
= 15.6 x 10⁻² M .
I don’t really understand, if anybody can help I’ll really appreciate it ! Thank you.
Answer:
175
Explanation:
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.
Atomic radius is....
O The tendency for an atom to attract electrons
The energy required to remove an electron
O The energy required to add an electron
O The distance from the nucleus to the last orbital
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
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.
Stephen learned that there are two forces that keep the moon in orbit around Earth. How do these forces keep the moon from flying off into space?
A. Gravity keeps the moon in motion, and inertia attracts the moon toward Earth.
B. Gravity attracts the moon toward Earth, and inertia keeps the moon in motion.
C. Gravity attracts the moon toward Earth, and the distance keeps it from going further away.
D. Mass weighs the moon down so it stays close to Earth, and inertia keeps the moon in motion.
Answer:
b
Explanation:
Answer:
Gravity attracts the moon Earth, and Inertia keeps the moon in motion.
Explanation:
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 are in 10 g of He
Answer:
6.7
⋅
10
23
atoms of H
Explanation:
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.
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
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.
Consider the balanced equation below. Upper P Upper C l Subscript 3 Baseline + Upper C l Subscript 2 Baseline right arrow Upper P Upper C l Subscript 5. What is the mole ratio of PCl3 to PCl5? 1:1 2:1 3:5 5:3
Answer : The mole ratio of [tex]PCl_3[/tex] to [tex]PCl_5[/tex] is 1 : 1.
Explanation :
Balanced chemical reaction : It is a chemical reaction in which the number of atoms of individual elements present on reactant side must be equal to the product side.
The balanced chemical reaction is:
[tex]PCl_3+Cl_2\rightarrow PCl_5[/tex]
By the stoichiometry of the reaction we can say that 1 mole of [tex]PCl_3[/tex] reacts with 1 mole of [tex]Cl_2[/tex] to give 1 mole of [tex]PCl_5[/tex].
From this we conclude that the mole ratio of [tex]PCl_3[/tex] to [tex]PCl_5[/tex] is 1 : 1.
Hence, the mole ratio of [tex]PCl_3[/tex] to [tex]PCl_5[/tex] is 1 : 1.
Answer:
A
Explanation:
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.
Help :( the blue boxes are clickable
Answer:
I am sure they are why won't they
The solubility of silver(I)phosphate at a given temperature is 2.43 g/L. Calculate the Ksp at this temperature. After you calculate the Kspvalue, take the negative log and enter the (pKsp) value with 2 decimal places.
Answer:
Kps = 3.07 x 10⁻⁸
pKsp= 7.51
Explanation:
First, we calculate the molar solubility of silver(I)phosphate (Ag₃PO₄) from the solubility in g/L by using its molar mass (418.6 g/mol):
2.43 g/L x 1 mol/418.6 g = 5.8 x 10⁻³ mol/L= s
Now, we have to write the ICE chart for the aqueous equilibrium of Ag₃PO₄ as follows:
Ag₃PO₄(g) ⇄ 3 Ag⁺(aq) + PO₄³⁻
I 0 0
C +3s +s
E 3s s
Ksp = [Ag⁺]³[PO₄³⁻]= (3s)³s= 27s⁴
Since s=5.8 x 10⁻³ mol/L, we calculate Ksp:
Ksp= 27(5.8 x 10⁻³ mol/L)⁴= 3.07 x 10⁻⁸
The pKsp value is:
pKsp= - log Ksp = -log (3.07 x 10⁻⁸) = 7.51
What does this diagram represent?
Answer:
Linear molecule with two domains
Explanation:
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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How many liters of chlorine gas at 25°C and 0.950 atm can be produced by the reaction of 12.0 g of MnO2 with excess HCl(aq) according to the following chemical equation?
MnO2(s) + 4HCl(aq) → MnCl2(aq) + 2H2O(l) + Cl2(g)
Answer:
3.55 L.
Explanation:
We'll begin by calculating the number of mole in 12 g of MnO2. This can be obtained as follow:
Molar mass of MnO2 = 55 + (16×2)
= 55 + 32
= 87 g/mol
Mass of MnO2 = 12 g
Mole of MnO2 =...?
Mole = mass /Molar mass
Mole of MnO2 = 12 / 87
Mole of MnO2 = 0.138 mole
Next, we shall determine the number of mole Cl2 produced from the reaction. This is illustrated below:
The balanced equation for the reaction is given below:
MnO2(s) + 4HCl(aq) → MnCl2(aq) + 2H2O(l) + Cl2(g)
From the balanced equation above,
1 mole of MnO2 reacted to produce 1 mole of Cl2.
Therefore, 0.138 mole of MnO2 will also produce 0.138 mole of Cl2.
Finally, we shall determine the volume of Cl2 gas obtained from the reaction. This can be obtained as shown below:
Temperature (T) = 25 °C = 25 °C + 273 = 298 K
Pressure (P) = 0.950 atm
Number of mole (n) = 0.138 mole
Gas constant (R) = 0.0821 atm.L/Kmol
Volume (V) =.?
PV = nRT
0.950 × V = 0.138 × 0.0821 × 298
Divide both side by 0.950
V = (0.138 × 0.0821 × 298) / 0.950
V = 3.55 L
Therefore, 3.55 L of chlorine gas were obtained from reaction.
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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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
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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