Answer:
Reaction of Chlorine with Hydrogen Chlorine and Hydrogen mixed together explodes when exposed to sunlight, which produces Hydrogen Chloride. In the dark away from sunlight, no reaction occurs, so light energy is required for a reaction. Cl2 + H2 = 2 HCl Reaction of Chlorine with Non-Metals Chlorine directly combines with most non-metals.
Explanation:
I hope this helps bro
If the rate of formation (also called rate of production) of compound C is 2M/s in the reaction A --->2C, what is the rate of consumption of A
Answer:
[tex]r_A=-1\frac{M}{s}[/tex]
Explanation:
Hello,
In this case, given the rate of production of C, we can compute the rate of consumption of A by using the rate relationships which include the stoichiometric coefficients at the denominators (-1 for A and 2 for C) as follows:
[tex]\frac{1}{-1} r_A=\frac{1}{2}r_C[/tex]
In such a way, solving the rate of consumption of A, we obtain:
[tex]r_A=-\frac{1}{2} r_C=-\frac{1}{2}*2\frac{M}{s}\\ \\r_A=-1\frac{M}{s}[/tex]
Clearly, such rate is negative which account for consumption process.
Regards.
PV = nRT. If P = 1 atm, V = 5.0 liter, R = 0.0821 L.atm/mol.K, and T = 293 K; what is the value of n?
Answer:
n = 0.207 mole
Explanation:
We have,
P = 1 atm
V = 5 liter
R = 0.0821 L.atm/mol.K
T = 293 K
We need to find the value of n. The relation is as follows :
PV = nRT
Solving for n,
[tex]n=\dfrac{PV}{RT}\\\\n=\dfrac{1\times 5}{0.0821 \times 293}\\\\n=0.207\ \text{mol}[/tex]
So, the value of n is 0.207 mol.
Complete the sentences describing the cell.
a. In the nickel-aluminum galvanic cell, the cathode is ____ .
b. Therefore electrons flow from___ to ____.
c. The ____ electrode loses mass, while the ____ electrode gains mass.
Answer:
a. In the nickel-aluminum galvanic cell, the cathode is nickel electrode.
b. Therefore electrons flow from the aluminium electrode to the nickel electrode.
c. The aluminium electrode loses mass, while the nickel electrode electrode gains mass.
Explanation:
Voltaic or galvanic cells are electrochemical cells in which spontaneous oxidation-reduction reactions. The two halves of the redox reaction are separate and electron transfer is required to occur through an external circuit for the redox reaction to take place. That is, one of the metals in one of the half cells is oxidized while the metal of the other half cell is reduced, producing an exchange of electrons through an external circuit. This makes it possible to take advantage of the electric current.
Given:
E ⁰N i ⁺² = − 0.23 V is the standard reduction potential for the nickel ion
E ⁰ A l ⁺³ = − 1.66 V is the standard reduction potential for the aluminum ion
The most negative potentials correspond to more reducing substances. In this case, the aluminum ion is the reducing agent, where oxidation takes place. In the anodic half cell oxidations occur, while in the cathode half cell reductions occur. So the aluminum cell acts as the anode while the nickel cell acts as the cathode.
So a. In the nickel-aluminum galvanic cell, the cathode is nickel electrode.
The metal that is oxidized gives electrons to the metal that is reduced through the outer conductor. Then the electrons flow spontaneously from the anode to the cathode.
Then b. Therefore electrons flow from the aluminium electrode to the nickel electrode.
Ni⁺², being the cathode, accepts electrons, becoming Ni (s) and depositing on the Ni electrodes.
So, c. The aluminium electrode loses mass, while the nickel electrode electrode gains mass.
Which balanced redox reaction is occurring in the voltaic cell represented by the notation of A l ( s ) | A l 3 ( a q ) | | P b 2 ( a q ) | P b ( s ) Al(s)|AlX3 (aq)||PbX2 (aq)|Pb(s)
The question is missing. Here is the complete question.
Which balanced redox reaction is ocurring in the voltaic cell represented by the notation of [tex]Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)}[/tex]?
(a) [tex]Al_{(s)}+Pb^{2+}_{(aq)} ->Al^{3+}_{(aq)}+Pb_{(s)}[/tex]
(b) [tex]2Al^{3+}_{(aq)}+3Pb_{(s)} -> 2Al_{(s)}+3Pb^{2+}_{(aq)}[/tex]
(c)[tex]Al^{3+}_{(aq)}+Pb_{(s)} ->Al_{(s)}+Pb^{2+}_{(aq)}[/tex]
(d) [tex]2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}[/tex]
Answer: (d) [tex]2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}[/tex]
Explanation: Redox Reaction is an oxidation-reduction reaction that happens in the reagents. In this type of reaction, reagent changes its oxidation state: when it loses an electron, oxidation state increases, so it is oxidized; when receives an electron, oxidation state decreases, then it is reduced.
Redox reactions can be represented in shorthand form called cell notation, formed by: left side of the salt bridge (||), which is always the anode, i.e., its half-equation is as an oxidation and right side, which is always the cathode, i.e., its half-equation is always a reduction.
For the cell notation: [tex]Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)}[/tex]
Aluminum's half-equation is oxidation:
[tex]Al_{(s)} -> Al^{3+}_{(aq)}+3e^{-}[/tex]
For Lead, half-equation is reduction:
[tex]Pb^{2+}_{(aq)}+2e^{-} -> Pb_{(s)}[/tex]
Multiply first half-equation for 2 and second half-equation by 3:
[tex]2Al_{(s)} -> 2Al^{3+}_{(aq)}+6e^{-}[/tex]
[tex]3Pb^{2+}_{(aq)}+6e^{-} -> 3Pb_{(s)}[/tex]
Adding them:
[tex]2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}[/tex]
The balanced redox reaction with cell notation [tex]Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)}[/tex] is
[tex]2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}[/tex]
What is the formula of a compound if a sample of the compound contains 0.492 mol X, 0.197 mol Y, and 0.295 mol Z?
Answer:
X₅Y₂Z₃
Explanation:
The formula of a compound is determined as the whole number ratio between moles of each element present in the molecule.
The molecule is made from X, Y and Z. To fin the ratio we will divide the given moles in the moles of Y (0.197 moles), because is the element with the low number of moles.
X = 0.492 moles / 0.197 moles = 2.5
Y = 0.197 moles / 0.197 moles = 1
Z = 0.295 moles / 0.197 moles = 1.5
But, as the formula is given just with whole numbers, if we multiply each number twice:
X = 2.5*2 = 5
Y = 1*2 = 2
Z = 1.5*2 = 3
The formula is:
X₅Y₂Z₃At a given temperature the vapor pressures of benzene and toluene are 183 mm Hg and 59.2 mm Hg, respectively. Calculate the total vapor pressure over a solution of benzene and toluene with X Benzene
CHECK COMPLETE QUESTION BELOW
At a given temperature the vapor pressures of benzene and toluene are 183 mm Hg and 59.2 mm Hg, respectively. Calculate the total vapor pressure over a solution of benzene and toluene with Xbenzene = 0.580.
Answer:
The total vapor pressure is [tex]81.3 mmHg[/tex]
Explanation:
We will be making use of Dalton and Raoults equation in order to calculate the total pressure,
Which is [tex]PT= (PA × XA) +(PB ×XB)[/tex]
PT= total vapor pressure
From the question
Benene's Mole fraction = 0.580
then to get Mole fraction of toluene we will substract the one of benzene from 1. because total mole fraction is always 1.
= (1 - 0.580) = 0.420
Vapor pressure of benzene given = 183 mmHg
Vapor pressure of toluene given= 59.2 mmHg
If we substitute those value into above equation, we have
PT=(183×0.580)+(59.2×0.420)
=81.3mmHg
Therefore,, the total vapor pressure of the solution is 81.3 mmHg
After balancing the following reaction under acidic conditions, how many mole equivalents of water are required and on which side of the reaction do they appear?
MnO41- (aq) + Cl1- (aq) → Mn2+ (aq) + Cl2 (g)
a. 2 moles of H2O on the reactant side
b. 2 moles of H2O on the product side
c. 4 moles of H2O on the product side
d. 8 moles of H2O on the product side
e. 10 moles of H2O on the reactant side
Answer:
d. 8 moles of H2O on the product side
Explanation:
Hello,
In this case, we need to balance the given redox reaction in acidic media as shown below:
[tex]MnO_4^{1-} (aq) + Cl^{1-} (aq) \rightarrow Mn^{2+} (aq) + Cl_2 (g)\\\\(Mn^{7+}O^{2-}_4)^{1-} (aq) + Cl^{1-} (aq) \rightarrow Mn^{2+} (aq) + Cl_2 (g)\\\\\\\\(Mn^{7+}O^{2-}_4)^{1-} (aq)+8H^++5e^- \rightarrow Mn^{2+}+4H_2O\\\\2Cl^{1-}\rightarrow Cl_2^0+2e^-\\\\2*[(Mn^{7+}O^{2-}_4)^{1-} (aq)+8H^++5e^- \rightarrow Mn^{2+}+4H_2O]\\\\5*[2Cl^{1-}\rightarrow Cl_2^0+2e^-]\\\\\\\\2(Mn^{7+}O^{2-}_4)^{1-} (aq)+16H^++10e^- \rightarrow 2Mn^{2+}+8H_2O\\\\10Cl^{1-}\rightarrow 5Cl_2^0+10e^-\\[/tex]
Then, we add the half reactions:
[tex]2(Mn^{7+}O^{2-}_4)^{1-} (aq)+16H^++10Cl^{1-} \rightarrow 2Mn^{2+}+8H_2O+5Cl_2^0[/tex]
Thereby, we can see d. 8 moles of H2O on the product side.
Best regards.
Pentanone was treated with excess sodium cyanide in HCl (aq) followed by hydrogen gas has over Pd. This produced:________
A. 2-amino-1-hexanol
B. 1-amino-2-methylpentan-2-ol
C. 1-cyano-1-pentanol
D. 2-aminomethylpentan-1-ol
Answer:
B. 1-amino-2-methylpentan-2-ol
Explanation:
In this case, the first step, we have the attack of the nucleophile cyanide ([tex] CN^-[/tex] produced by sodium cyanide to the carbon on the carbonyl group (C=O) producing a negative charge in the oxygen.
Then HCl protonates the molecule to produce a cyanohydrin. This cyanohydrin can be reduced by the action of hydrogen gas ([tex]H_2[/tex]) in the presence of a catalyst ([tex]Pd[/tex]), producing an amino group. With this in mind, the final molecule is: 1-amino-2-methylpentan-2-ol.
See figure 1 to further explanations
I hope it helps!
When 5.58g H2 react by the following balanced equation, 32.8g H2O are formed. What is the percent yield of the reaction? 2H2(g)+O2(g)⟶2H2O(l)
A) 11.7%
B) 17.0%
C) 38.9%
D) 65.7%
Answer:
D) 65.7%
Explanation:
Based on the reaction:
2H2(g)+O2(g)⟶2H2O(l)
2 moles of hydrogen produce 2 moles of water assuming an excess of oxygen.
To find percent yield of the reaction we need to find theoretical yield (The yield assuming all hydrogen reacts producing water). With theoretical yield and actual yield (32.8g H₂O) we can determine percent yield as 100 times the ratio between actual yield and theoretical yield.
Theoretical yield:
Moles of 5.58g H₂:
5.58g H₂ ₓ (1 mol / 2.016g) = 2.768 moles H₂
As 2 moles of H₂ produce 2 moles of H₂O, if all hydrogen reacts will produce 2.768 moles H₂O. In grams:
2.768 moles H₂O ₓ (18.015g / mol) =
49.86g H₂O is theoretical yield
Percent yield:
Percent yield = Actual yield / Theoretical yield ₓ 100
32.8g H₂O / 49.86g ₓ 100 =
65.7% is percent yield of the reaction
D) 65.7%Solid cesium bromide has the same kind of crystal structure as CsCl which is pictured below: If the edge length of the unit cell is 428.7 pm, what is the density of CsBr in g/cm3.
Answer:
[tex]\mathbf {density \ d =4.4845 \ g/cm^3}[/tex]
Explanation:
Let recall the crystal structure of CsBr obtains a BCC structure. In a BCC structure, there exist only two atom per cell.
The density d of CsBr in g/cm³ can be calculated by using the formula:
[tex]\mathtt{ density \ d = \dfrac{z \times molar\ mass \ (M)}{ edge \ length \ (a) \ \times avogadro's \ number \ (N)}}[/tex]
where;
z = 1 mole of CsBr
edge length = 428.7 pm = (4.287 × 10⁻⁸)³ cm
molar mass of CsBr = 212.81 g/mol
avogadro's number = 6.023 × 10²³
[tex]\mathtt{ density \ d = \dfrac{1 \times 212.81}{(4.287 \times 10^{-8})^3 \times 6.023 \times 10^{23}}}[/tex]
[tex]\mathtt{ density \ d = \dfrac{ 212.81}{47.4540533}}[/tex]
[tex]\mathbf {density \ d =4.4845 \ g/cm^3}[/tex]
Refer to the figure.
30. How many planes are shown in the figure?
31. How many planes contain points B, C, and E?
32. Name three collinear points.
3. Where could you add point G on plane N
so that A, B, and G would be collinear?
4. Name a point that is not coplanar with
A, B, and C.
5. Name four points that are coplanar.
BN
Answer:
30. 5 planes are shown
31. 1 plane
32. CEF
33. on line AB
34. E or F
35. ABCD or BCEF or CDEF or ACEF
Explanation:
30. Each of the surfaces of the rectangular pyramid is a plane. There are 5 planes.
__
31. 3 points define one plane only.
__
32. The only points shown on the same line segment are points E, F, and C.
__
33. If G is to be collinear with A and B, it must lie on line AB.
__
34. The only points shown that are not on plane N are points E and F. Either of those will do.
__
35. There are three planes that have 4 points shown on them. The four points that are on the same plane are any of ...
ABCDBCEFCDEFPlane ACEF is not shown on the diagram, but we know that those 4 points are also coplanar. (Any point not on line CE, together with the three points on that line, will define a plane with 4 coplanar points.)
Each energy sublevel contains __________ number of electrons. For example, sublevel D can hold up to _______ electrons. A. the same, 10 B. the same, 14 C. a different, 6 D. a different, 10
Answer:
Each energy sublevel contains a different number of electrons. For example, sublevel D can contain up to 10 electrons
Explanation:
The atoms are surrounded by propellers that within each propeller there is a certain number of electrons, these electrons jump from orbit to orbit according to the amount of energy they have. The four levels that make up the electronic cloud that surrounds an atom are: s p d f.
When these electrons change orbit or level they release energy in the form of light, which is known as a photon.
A compound is found to contain 31.42 % sulfur, 31.35 % oxygen, and 37.23 % fluorine by weight. To answer the questions, enter the elements in the order presented above. 1. What is the empirical formula for this compound? 2. The molecular weight for this compound is 102.1 g/mol. What is the molecular formula for this compound?
Explanation:
To obtain the empirical and molecular formula of this compound from the percent composition of the elements, we follow the steps below;
Step 1: Divide the percentage composition by the atomic mass
Sulphur = 31.42 / 32 = 0.9819
Oxygen = 31.35 / 16 = 1.9594
Flourine = 37.23 / 19 = 1.9595
Step 2: Divide by the lowest number
Sulphur = 0.9819 / 0.9819 = 1
Oxygen = 1.9594 / 0.9819 ≈ 2
Flourine = 1.9595 / 0.9819 ≈ 2
This means the ratio of the elements is 1 : 2: 2
The empirical formular (simplest formular of a compound) of the compound is;
SO₂F₂
To obtain the molecular formular (Actual formular of a compound);
(SO₂F₂)n = 102.1
Inserting the atomic masses and solving for n;
(102)n = 102.1
n ≈ 1
The molecular formular is; (SO₂F₂)₁ = SO₂F₂
15. How many moles of carbon tetrachloride (CCI) is represented by 543.2 g of carbon tetrachloride? The atomic weight of carbon is 12.01
and the atomic weight of chlorine is 35.45.
O A. 11.4 moles
O B.3.53 moles
C. 5.42 moles
D. 8.35x10 moles
Answer:
well, first off. the formula for carbon tetrachloride is CCl4
We need to find the molar mass of carbon tetrachloride
1(Mass of C) + 4(mass of chlorine)
1(12) + 4(35.5)
12 + 142
154 g/mol
Number of moles of CCl3 in 543.2g CCl3
n = given mass / molar mass
n = 543.2/153
n = 3.53 moles
always remember to brainly the questions you find helpful
Answer:
3.53 moles
Explanation:
How much work (in Joules) is required to expand the volume of a pump from 0.00 L to 2.50 L against an external pressure of 1.10 atm
Answer:
W = 278.64375 Joules
Explanation:
The information given in this problem are;
Initial volume = 0L
Final volume = 2.50L
ΔV = 2.50 - 0 = 2.50 L
External pressure, P = 1.10 atm
Work = ?
These parameters are related by the equation;
w = - P ΔV
W = - (1.10 )(2.50)
W = 2.75 L atm
Upon conversion to joules;
1 liter atmosphere is equal to 101.325 joule
W = 278.64375 Joules
A battery is an example of a(n) _________. A. anode B. voltaic cell C. cathode D. electrolytic cell
Answer:
The answer is D) Electrolytic cell
Explanation:
An electrolytic cell is a device used for the decomposition by the electrical current of ionized substances called electrolytes.
When the two electrodes are connected by a wire, electrical energy is produced, and a flow of electrons takes place from the electrode.
These cells are the closest thing to a galvanic battery.
Answer:
b. voltaic cell
Explanation:
Founders Education answer. had to take this quiz 4 times
Experiment:
Part I: Voltaic Cell
Assume that you are provided with the following materials:
Strips of metallic zinc, metallic copper, metallic iron
1M aqueous solutions of ZnSO4, CuSO4, FeSO4, and aqueous iodine (I2)
Other required materials to create Voltaic cells such as beakers, porous containers, graphite rods, a voltmeter, and a few wires with alligator clips.
In this modified version of the lab, after thoroughly studying the lab hand out and watching the videos, identify 4 different combinations of Voltaic cells that are possible to be created with the above materials.
For each cell created, include the following details.
Which electrode was the anode and which was the Cathode?
The anode and cathode half reactions.
Balanced equation for each cell you propose to construct.
Calculated Eocell
Short hand notation (line notation) for each cell (be sure to include the inactive electrode if needed)
Answer:
Here are four possible voltaic cells.
Explanation:
1. Standard reduction potentials
E°/V
I₂(s) + 2e⁻ ⟶ 2I⁻(aq); 0.54
Cu²⁺(aq) + 2e⁻ ⟶ Cu(s); 0.34
Fe²⁺(aq) + 2e⁻ ⟶ Fe(s); -0.41
Zn²⁺(aq) + 2e⁻ ⟶ Zn(s); -0.76
2. Possible Voltaic cells
(a) Zn/I₂
E°/V
Anode: Zn(s) ⟶ Zn²⁺(aq) + 2e⁻; 0.76
Cathode: I₂(s) + 2e⁻ ⟶ 2I⁻(aq); 0.54
Cell: Zn(s) + I₂(s) ⟶ Zn²⁺(aq) + 2I⁻(aq); 1.30
Zn(s)|Zn²⁺(aq)∥I⁻(aq)|I₂(s)|C(s, graphite)
Zn is the anode; graphite is the cathode.
(b) Zn/Cu²⁺
E°/V
Anode: Zn(s) ⟶ Zn²⁺(aq) + 2e⁻; 0.76
Cathode: Cu²⁺(aq) + 2e⁻ ⟶ Cu(s); 0.34
Cell: Zn(s) + Cu²⁺(s) ⟶ Zn²⁺(aq) + Cu(s); 1.10
Zn(s)|Zn²⁺(aq)∥Cu²⁺(aq)|Cu(s)
Zn is the anode; Cu is the cathode.
(c) Zn/Fe²⁺
E°/V
Anode: Zn(s) ⟶ Zn²⁺(aq) + 2e⁻; 0.76
Cathode: Fe²⁺(aq) + 2e⁻ ⟶ Fe(s); -0.41
Cell: Zn(s) + Fe²⁺(s) ⟶ Zn²⁺(aq) + Fe(s); 0.35
Zn(s)|Zn²⁺(aq)∥Fe²⁺(aq)|Fe(s)
Zn is the anode; Fe is the cathode.
(d) Fe/I₂
E°/V
Anode: Fe(s) ⟶ Fe²⁺(aq) + 2e⁻; 0.41
Cathode: I₂(s) + 2e⁻ ⟶ 2I⁻(aq); 0.54
Cell: Zn(s) + I₂(s) ⟶ Zn²⁺(aq) + 2I⁻(aq); 0.95
Fe(s)|Fe²⁺(aq)∥I⁻(aq)|I₂(s)|C(s, graphite)
Fe is the anode; graphite is the cathode.
How did Jesseca Kusher create her new material?
Answer:
Jesseca Kusher, an 18-year-old researcher from Spartansburg, S.C., invented a paint-on coating for roofing shingles. Her formula could reduce a home's cooling costs and possibly cut ozone pollution in urban areas...
SUPPORT ME ...........
Answer:
Jesseca created mixtures containing graphite, gypsum, and mica that could be painted on roof shingles.
Explanation:
Hope this helped!!
What is the pH of 10.0 mL solution of 0.75 M acetate after adding 5.0 mL of 0.10 M HCl (assume a Ka of acetic acid of 1.78x10-5)
Answer:
5.90
Explanation:
Initial moles of CH3COO- = 10.0/1000 x 0.75 = 0.0075 mol
Moles of HCl added = 5.0/1000 x 0.10 = 0.0005 mol
CH3COO- + HCl => CH3COOH + Cl-
Moles of CH3COO- left = 0.0075 - 0.0005 = 0.007 mol
Moles of CH3COOH formed = moles of HCl added = 0.0005 mol
pH = pKa + log([CH3COO-]/[CH3COOH])
= -log Ka + log(moles of CH3COO-/moles of CH3COOH)
= -log(1.78 x 10^(-5)) + log(0.007/0.0005)
= 5.90
Answer:
The correct answer is 5.895.
Explanation:
The reaction will be,
CHCOO⁻ + H+ ⇔ CH₃COOH
Both the HCl and the acetate are having one n factor.
The millimoles of CH₃COO⁻ is,
= Volume in ml × molarity = 10 × 0.75 = 7.5
The millimoles of HCl = Volume in ml × molarity = 5 × 0.1 = 0.5
Therefore, 0.5 will be the millimoles of CH₃COOH formed, now the millimoles of the CH₃COO⁻ left will be, 7.5-0.5 = 7.0
The volume of the solution is, 10+5 = 15 ml
The molarity of CH₃COO⁻ is, millimoles / volume in ml = 7/15
The molarity of CH₃COOH is 0.5/15
pH = pKa + log[CH₃COO⁻]/[CH₃COOH]
= 4.74957 + 1.146
= 5.895
How many molecules are there in 3.5 moles of carbon dioxide? A. 63.21 x 10^23 B. 21.07 x 10^23 C. 42.14 x 10^23 D. 6.02 x 10^23
Answer:
B. 21.07 x 10²³ molecules
Explanation:
Avogadro's Number: 6.022 x 10²³
Step 1: Set up equation
[tex]3.5 mols CO_2(\frac{6.022(10^{23}) moleculesCO_2}{1 mol CO_2})[/tex]
Step 2: Multiply and cancel out units
3.5(6.022 x 10²³) = 21.07 x 10²³ molecules CO₂
Step 3: Convert to proper scientific notation
≈ 2.11 x 10²³ molecules CO₂
When the nuclide bismuth-210 undergoes alpha decay:
The name of the product nuclide is_____.
The symbol for the product nuclide is_____
Fill in the nuclide symbol for the missing particle in the following nuclear equation.
_____ rightarrow 4He+ 234Th
2 90
Write a balanced nuclear equation for the following:
The nuclide radium-226 undergoes alpha emission.
Explanation:
An atom undergoes alpha decay by losing a helium atom.
So when bismuth undergoes alpha decay, we have;
²¹⁰₈₃Bi --> ⁴₂He + X
Mass number;
210 = 4 + x
x = 206
Atomic number;
83 = 2 + x
x = 81
The element is Thallium. The symbol is Ti.
For the second part;
X --> ⁴₂He + ²³⁴₉₀Th
Mass number;
x = 4 + 234 = 238
Atomic Number;
x = 2 + 90 = 92
The balanced nuclear equation is;
²³⁸₉₂U --> ⁴₂He + ²³⁴₉₀Th
Describe the similarities between H3O and NH3. Compare/contrast their shapes and polarities within the context of your answer. These molecules are called isoelectronic. Why
Answer:
Explanation:
[tex]H_3O^+[/tex] also known as hydronium ion is formed as a result of the reaction between an hydrogen proton and a water molecules.
i.e [tex]\mathtt{H^+ + H_2O \to H_3O^+}[/tex]
(molecular geometry for the hydronium ion shows that the lewis structure of hydronium ion possess a three hydrogen ion bonded to a central atom known as oxygen. The oxygen possess a lone pair with a positive ion. So we have three hydrogen atoms and a lone pair attached to the oxygen. We can now say that there are four groups as the steric number in which one of them is a lone pair. This give rise to the trigonal pyramidal shape of the [tex]H_3O^+[/tex] (hydronium ion) with a bond angle of about 109,5°
Similarly, [tex]NH_3[/tex] on the other hand also known as ammonia has a shape that can be also determined by the Lewis structure.
IN ammonia, there are three hydrogen and a lone pairs of electron spreading out as far away from each other from the centre nitrogen. In essence, the valence shell electron pair around hydrogens tend to repel each other. Hence, giving it a trigonal pyramidal shape.
From above the similarities between H3O and NH3 is in their molecular geometry in which both H3O and NH3 have the same shape.
These molecules are called isoelectronic. Why?
Isoelectronic molecules are molecules having the same number of electrons and same electronic configuration structure. As a result H3O and NH3 possess the same number of electrons in the same orbitals and they also posses the same structure.
Which of the following provides a characteristic of
MgO(s) with a correct explanation?
Choose 1 answer:
А
It is hard because its ions are held together by strong
electrostatic attractions.
B
It is malleable because its atoms can easily move past
one another without disrupting the bonding.
It is a poor conductor of electricity because its
electrons are tightly held within covalent bonds and
lone pairs.
It has a high melting point because its molecules
interact through strong intermolecular forces.
Answer:
А It is hard because its ions are held together by strong electrostatic attractions.
B It is malleable because its atoms can easily move past one another without disrupting the bonding.
Explanation:
These are correct explanations of the properties of magnesium.
C is wrong. Mg is a good conductor of electricity and it has metallic bonds.
D is wrong. Mg has no molecules. It has no intermolecular forces.
the ka of hypochlorous acid (hclo) is 3.0 x10^-8 at 25.0°C. What is the % of ionization of hypochlorous
Answer:
0.14%
Explanation:
The computation of % is shown below:
As we know that
HClO <=> H+ + ClO-
I 0.015 0 0
C -a +a +a
E 0.015-a a a
Now
[tex]Ka = \frac{[H+][ClO-]}{[HClO]}[/tex]
[tex]= \frac{a^{2}}{(0.015 - a)} \\\\= 3.0 \times 10^{-8}[/tex]
[tex]a^{2} + 3.0 \times 10^{-8}a - 4.5 \times 10^{-10} = 0[/tex]
Now Solves the quadratic equation i.e.
[tex]a = 2.120 \times 10^{-5}[/tex]
[tex][H+] = a = 2.120 \times 10^{-5} M[/tex]
So,
% ionization is
[tex]= \frac{[H+]}{[HClO]}_{initial} \times 100\%\\\\= 2.120 \times 10^{-5}\div0.015 \times 100\%[/tex]
= 0.14%
Hence, the percentage of hypochlorous ionization is 0.14%
explain how the liquid in a thermometer changes so that it can be used to measure a temprature
Answer:
The liquid that is often used in thermometers is chrome.
It is khwon for raising its volule when the temperature raises and vice-versa. ● the temperature and the volume are proprtional to each other so using Mathematics, scientists have figured out a way to benefit from it to make a thermometer.
I add a 50. g piece of Al (c = 0.88 J/g-deg) that is at 225°C to 100. mL of water at 20°C. What is the final temperature of the water in °C? The density of water is approximately 1g/mL.
Answer:
THE FINAL TEMPERATURE OF WATER IS -4.117 °C
Explanation:
Mass of the aluminium = 50 g
c = 0.88 J/g C
Initial temperature of aluminium = 225 °C
Volume of water = 100 ml
Density of water = 1 g/ml
Mass of water = density * volume of water
Mass of water = 1 * 100 = 100 g of water
Initial temperature of water = 20 C
It is worthy to note that the heat of a system is constant and conserved as no heat is lost or gained by a closed system,
So therefore,
heat lost by aluminium = heat gained by water
H = mass * specific heat capacity * temeprature change
So:
m c ( T2- T1) = m c (T2-T1)
50 * 0.88 * ( T2 - 225) = 100 * 4.18 *( T2 - 20)
44 ( T2 - 225 ) = 418 ( T2 - 20)
44 T2 - 9900 = 418 T2 - 8360
-9900 + 8360 = 418 T2 - 44 T2
-1540 = 374 T2
T2 = - 4.117
So therefore the final temperature of water is -4.117 °C
A small amount of solid calcium hydroxide is shaken vigorously in a test tube almost full of water until no further change occurs and most of the solid settles out. The resulting solution is:______.
Answer:
Lime water, [tex]Ca(OH)_{2}_({aq} )[/tex] is formed.
Explanation:
Lime-water is a clear and colourless dilute solution of aqueous calcium hydroxide salt.
Small amounts of calcium hydroxide salt, [tex]Ca(OH)_{2}_(s)[/tex] is sparsely soluble at room temperature when dispersed vigorously. if in excess, a white suspension called 'milk of lime'is formed.
I hope this explanation is helpful.
Of the following substances, an aqueous solution of ________ will form basic solutions. NH4Br Pb(NO3)2 K2CO3 NaF
Answer:
K2CO3 and NaF
Explanation:
In order to ascertain which salt would form a basic solution we have to identify the classification of each of the salts.
- NH4Br: is the salt of a weak base (NH3) and a strong acid (HBr). This means that it would form an acidic solution.
- Pb(NO3): This is a normal salt, hence would not form a basic solution.
- K2CO3: This is salt that forms a strongly alkaline/basic solution.
- NaF: it is the salt of a strong base, NaOH, and a weak acid, HF. This means this would form a basic solution.
The compounds capable to form basic solutions are[tex]\rm \bold {K_2CO_3 }[/tex] and NaF. Thus, options C and D are correct.
The basic solution has been given with the presence of a high number of hydroxide ions, while the acidic solution has been the presence of hydrogen ions.
The solution has been considered as basic when the compound has been constituted of a strong base. The constituents of the following compounds have been:
Ammonium bromide: The basic part is ammonia, and is a weak base. Thus, forms an acidic solutionLead nitrate: The compound is salt and results in a neutral solution.Potassium carbonate: The base has been carbonate, and a strong base. Thus forms the basic solution.Sodium fluoride: The fluoride has been the basic part and has been a constituent of a strong base. It has been capable of forming a basic solution.The compounds capable to form basic solutions are[tex]\rm \bold {K_2CO_3 }[/tex] and NaF. Thus, options C and D are correct.
For more information about the basic solution, refer to the link:
https://brainly.com/question/3595168
Which of the following processes release energy? a. ball rolling down a hill b. formation of copper (II) oxide from copper and oxygen c. formation of ice from liquid water d. condensation of water on a wind shield of a car
Answer:
d. condensation of water on a wind shield of a car
Explanation:
Condensation involves the conversion of moist air into liquid.
Gas has a higher energy compared to liquid. This is why Gas particles move at random motion and faster in relation to solid and liquid particles due to the high energy content.
The conversion of the gas to liquid means that there was loss or release of energy which validates the answer.
Assume that 33.0 mL of a 0.10 M solution of a weak base B that accepts one proton is titrated with a 0.10 M solution of the monoprotic strong acid HX.How many moles of have been added at the equivalence point?n = ? mol
