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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twelve 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%
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:
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:
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
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:
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.
Hydrogen has 3 isotopes. Hydrogen 1, Hydrogen 2 and Hydrogen 3. What is the difference between these 3 is isotopes
Answer:
Number of neutrons
Explanation:
All have one single proton. Hydrogen has no neutrons. Hydrogen 2 or deuterium has 1 neutron. Hydrogen 3 or tritium has 2 neutrons.
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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H₂C=CH-CH₂-CH=CH₂
how many single and double bonds
Answer:
2 double bonds 2 single bonds
what is so unique about water that hydrogen bonding becomes possible
Answer:
Water molecules are polar, so they form hydrogen bonds. This gives water unique properties, such as a relatively high boiling point, high specific heat, cohesion, adhesion and density.
EXPLANATION :
So it is unusual for water to be a liquid at room temperature! Water is liquid at room temperature so it's able to move around quicker than it is as solid, enabling the molecules to form fewer hydrogen bonds resulting in the molecules being packed more closely together.
A volume of 80.0 mL of a 0.690 M HNO3 solution is titrated with 0.790 M KOH. Calculate the volume of KOH required to reach the equivalence point. Express your answer to three significant figures and include the appropriate units.
Given :
A volume of 80.0 mL of a 0.690 M [tex]HNO_3[/tex] solution is titrated with 0.790 M KOH.
To Find :
The volume of KOH required to reach the equivalence point.
Solution :
We know, at equivalent point :
moles of [tex]HNO_3[/tex] = moles of KOH
[tex]M_{HNO_3}V_{HNO_3}=M_{KOH}V_{KOH}\\\\0.690\times 80 = 0.790\times V_{KOH}\\\\V_{KOH}=\dfrac{0.690\times 80 }{ 0.790}\ ml\\\\V_{KOH}=69.87\ ml[/tex]
Therefore, volume of KOH required is 69.87 ml.
Hence, this is the required solution.
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
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.
Which of the following is a good definition of matter?
O A. Anything that is made up of light and gravity
O B. Anything that has mass and takes up space
O C. Anything that produces heat and mass
O D. Anything that has energy and creates heat
Answer:
B
Explanation:
I did the question before and got it right.
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
Use the data in the Successive Ionization Energies and Electron Affinity tables to determine the following. (Assume the values in the Successive Ionization Energies table are given to the ones place.) (a) the electron affinity of Ar2+ kJ/mol (b) the electron affinity of Si + kJ/mol (c) the ionization energy of Cl − kJ/mol (d) the ionization energy of Cl kJ/mol (e) the electron affinity of Cl + kJ/mol
Answer:
Use the data in the Successive Ionization Energies and Electron Affinity tables to determine the following. (Assume the values in the Successive Ionization Energies table are given to the ones place.)
(a)
the electron affinity of Ar2+
kJ/mol
(b)
the electron affinity of S+
kJ/mol
(c)
the ionization energy of Cl−
kJ/mol
(d)
the ionization energy of Cl
kJ/mol
(e)
the electron affinity of Cl+
kJ/mol
Successive Ionization Energies in Kilojoules per Mole for the Elements in Period 3 General increase 13 14 Element 11 12 I. 16
Explanation:
A sample of an ideal gas has a volume of 2.23 L at 289 K and 1.05 atm. Calculate the pressure when the volume is 1.08 L and the temperature is 304 K. P= atm
Answer:
2.28 atm
Explanation:
V₁ = 2.33L, V₂ = 1.08L
T₁ = 289K, T₂ = 304K
P₁ = 1.05 atm, P₂ = ?
Where V₁ and V₂ are initial and final volume respectively
T₁ and T₂ are initial and final temperature respectively
P₁ and P₂ are initial and final pressure respectively
The formula to be used here is the general gas equation:
P₁V₁/T₁=P₂V₂/T₂
1.05 × 2.23/289 = P₂ × 1.08/304
P₂ × 1.08 × 289 = 1.05 × 2.23 × 304
P₂ = (1.05 × 2.23 ×304) ÷ (1.08 × 289)
P₂ = 711.82 ÷ 312.12
P₂ = 2.28 atm
If equal volumes of a strong base and a weaker acid are mixed together, what would you expect the pH of the resulting salt to be
Answer:
Above 7
Explanation:
The equivalence point of any titration can be read off from the appropriate titration curve.
A titration curve is a plot of the pH of analyte against the volume of titrant added.
For a strong base and weak acid, the equivalence point lies above 7.
The pH of the resulting salt to be pH> 7 .
What does Equivalence point tell?The equivalence point of any titration can be read off from the appropriate titration curve. A titration curve is a plot of the pH of analyte against the volume of titrant added. It is a point in titration at which the amount of titrant added is just enough to completely neutralize the analyte solution. At the equivalence point in an acid-base titration.
For a strong base and weak acid, the equivalence point lies above 7.
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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 .
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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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.
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
Which accurately represents these building blocks of matter from the smallest to the largest?
atom -- molecule or compound
O molecule -- atom - element
compound - molecule -- element
molecule atom or element
Answer:
A - Atom ---> molecule or compound.
Phosphoric acid, which is commonly used as rust inhibitor, food additive and etching agent for dental and orthopedic use, can be synthesized using a two-step thermal process. In the first step, phosphorus and oxygen react to form diphosphorus pentoxide: P4(l)+5O2(g-2 P20s(g) In the second step, diphosphorus pentoxide and water react to form phosphoric acld P20(9)+3 H200 2H,PO40) Write the net chemical equation for the production of phosphoric acid from phosphorus, oxygen and water.
Answer:
P₄(l) + 5 O₂(g) + 6 H₂O(l) ⇒ 4 H₃PO₄(aq)
Explanation:
Phosphoric acid is synthesized using a two-step thermal process.
In the first step, phosphorus and oxygen react to form diphosphorus pentoxide. The corresponding chemical equation is:
P₄(l) + 5 O₂(g) ⇒ 2 P₂O₅(g)
In the second step, diphosphorus pentoxide and water react to form phosphoric acid. The corresponding chemical equation is:
P₂O₅(g) + 3 H₂O(l) ⇒ 2 H₃PO₄(aq)
We can get the net chemical equation by adding the first step, the second step multiplied by 2, and canceling what is repeated on both sides.
P₄(l) + 5 O₂(g) + 2 P₂O₅(g) + 6 H₂O(l) ⇒ 2 P₂O₅(g) + 4 H₃PO₄(aq)
P₄(l) + 5 O₂(g) + 6 H₂O(l) ⇒ 4 H₃PO₄(aq)
One way to represent a substance is with a chemical formula. In the formula CO2, what do the symbols Cand o refer to?
Answer:
C is for carbon and O is for oxygen
Sodium carbonate, also known as soda ash, is used in glassmaking. It is obtained from a reaction between sodium chloride and calcium carbonate; calcium chloride is the other product. Calculate the percent yield of sodium carbonate if 92.6 g is collected when 112. g of sodium chloride reacts with excess calcium carbonate.
Answer:
The percentage yield of sodium carbonate is 91.47%
Explanation:
we start by writing the reaction equation:
2NaCl + CaCO3 ——-> Na2CO3 + CaCl2
From the reaction we can see that 2 moles of sodium chloride produced 1 mole of sodium carbonate
Let us calculate the actual number of moles of sodium chloride produced from 112 g of it
Mathematically,
number of moles = mass/molar mass
Molar mass of sodium chloride is 23 + 35.5 = 58.5 g/mole
So the number of moles of sodium chloride produced will be 112/58.5 = 1.91 moles
The number of moles of sodium carbonate produced is half of this = 1.91/2 = 0.955
The mass of sodium carbonate produced from 0.955 moles of it will be;
number of moles * molar mass
The molar mass of sodium carbonate is 106 g/mol
So the number of moles is = 0.955 * 106 = 101.23 g
Mathematically;
percentage yield = actual yield/theoretical yield * 100%
Percentage yield = 92.6/101.23 * 100% = 91.47%
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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What does this diagram represent?
Answer:
Linear molecule with two domains
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.
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