K1 and K2 are related algebraically because once the values are inserted into the equilibrium equation, both equations will yield a denominator of 100.
17. According to Newton's first law, an object in motion will stay in what kind of
motion as long as the net forces acting on it are zero?
A. slowing motion in a straight line
B. accelerating motion in a straight line
C. constant motion in a straight line
D. constant motion in an unpredictable direction
Answer:
C
Explanation:
According to Newton's first law, an object in motion will stay in motion unless acted upon by an unbalanced force. The option which gives the same meaning is :
constant motion in a straight lineAccording to Newton's first law, an object in motion as long as the net forces acting on it are zero will stay in:
C. constant motion in a straight line
A constant motion in mechanics is a number that remains constant throughout the motion, effectively placing a restriction on the motion.
What is Newton's first law of motion?The foundation of classical mechanics is made up of the three laws of motion that Newton suggested. These three physical laws. In accordance with the three laws of motion, a body's connection to the forces operating on it and the motion that resulted from those forces are both described. As Galileo's law of inertia, Newton's First Law of Motion is also referred to. If a body is not forced by an outside force to change its condition, it will remain in its state of rest or uniform motion in a straight line. As a factor that has the potential to alter an object's condition, the law defines force and mentions it.In this way, Newton uses the inertia of rest, motion, and direction to describe his first law of motion.A body has an attribute called inertia that makes it resist changes to its state. A body's inertia of translational motion is measured by its mass.To learn more about Newton's first law of motion, refer to:
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Find the volume occupied by 128g of SO2.
The volume occupied by the given amount of sulfur dioxide will be 84 L.
please show working my dear citizen
which of the following measurements is equivalent to 5.461x10^-7m?
Answer:
B. 0.0000005461m
I used the method of moving the decimal.
A sample of gas occupies 10.0 L at 240°C under a pressure of
80.0 kPa. At what temperature would the gas occupy 20.0 L if
we increased the pressure to 107 kPa?
Answer: 1090°C
Explanation: According to combined gas laws
(P1 × V1) ÷ T1 = (P2 × V2) ÷ T2
where P1 = initial pressure of gas = 80.0 kPa
V1 = initial volume of gas = 10.0 L
T1 = initial temperature of gas = 240 °C = (240 + 273) K = 513 K
P2 = final pressure of gas = 107 kPa
V2 = final volume of gas = 20.0 L
T2 = final temperature of gas
Substituting the values,
(80.0 kPa × 10.0 L) ÷ (513 K) = (107 kPa × 20.0 L) ÷ T2
T2 = 513 K × (107 kPa ÷80.0 kPa) × (20.0 L ÷ 10.0 L)
T2 = 513 K × (1.3375) × (2)
T2 = 1372.275 K
T2 = (1372.275 - 273) °C
T2 = 1099 °C
1090 degree Celsius
hope it helps
What volume of 1.50 mol/L stock solution is needed to make 125 mL of 0.60 mol/L solution?
Chemistry 11 Solutions
978Ͳ0Ͳ07Ͳ105107Ͳ1Chapter 8 Solutions and Their Properties • MHR | 85
Amount in moles, n, of the NaCl(s):
NaCl
2.5 g
m
n
M
58.44 g
2
4.2778 10 m l
ol
o
/m
u
Molar concentration, c, of the NaCl(aq):
–2 4.2778 × 10 mol
0.100
0.42778 mol/L
0.43 mol
L
/L
n
c
V
The molar concentration of the saline solution is 0.43 mol/L.
Check Your Solution
The units are correct and the answer correctly shows two significant digits. The
dilution of the original concentrated solution is correct and the change to mol/L
seems reasonable.
Section 8.4 Preparing Solutions in the Laboratory
Solutions for Practice Problems
Student Edition page 386
51. Practice Problem (page 386)
Suppose that you are given a stock solution of 1.50 mol/L ammonium sulfate,
(NH4)2SO4(aq).
What volume of the stock solution do you need to use to prepare each of the
following solutions?
a. 50.0 mL of 1.00 mol/L (NH4)2SO4(aq)
b. 2 × 102 mL of 0.800 mol/L (NH4)2SO4(aq)
c. 250 mL of 0.300 mol/L NH4
+
(aq)
What Is Required?
You need to calculate the initial volume, V1, of (NH4)2SO4(aq) stock solution
needed to prepare each given dilute solution.
The dilution gives the relationship between the molarity and the volume of the solution. The volume of stock solution with a molarity of 1.50 mol/L is 50 mL.
What is dilution?Dilution is said to be the addition of more volume to the concentrated solution to make it less in molar concentration. This tells about the inverse and indirect relationship between the volume and the molar concentration of the solution.
Given,
Initial volume = V₁
Initial molar concentration (M₁) = 1.50 mol/L
Final volume (V₂) = 125 mL = 0.125 L
Final molar concentration (M₂)= 0.60 mol/L
The dilution is calculated as:
M₁V₁ = M₂V₂
V₁ = M₂V₂ ÷ M₁
Substituting the values in the above formula as
V₁ = M₂V₂ ÷ M₁
V₁ = (0.60 mol/L × 0.125 L) ÷ 1.50 mol/ L
V₁ = 0.05 L
= 50 mL
Therefore, 50 mL of stock solution is needed to make a 0.60 mol/L solution.
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During electrophilic aromatic substitution, a resonance-stabilized cation intermediate is formed. Groups, already present on the benzene ring, that direct ortho/para further stabilize this intermediate by participating in the resonance delocalization of the positive charge. Assume that the following group is present on a benzene ring at position 1 and that you are brominating the ring at positon 4. Draw the structure of the resonance contributor that shows this group actively participating in the charge delocalization.
-----OCH3
Answer:
See explanation and image attached
Explanation:
Aromatic compounds undergo electrophilic aromatic substitution reactions in which the aromatic ring is maintained.
Substituted benzenes may be more or less reactive towards electrophilic aromatic substitution than benzene depending on the nature of the substituent present in the ring.
Substituents that activate the ring towards electrophilic substitution such as -OCH3 are ortho-para directing.
The major products of the bromination of anisole are p-bromoanisole and o-bromoanisole. The resonance structures leading to these products are shown in the image attached.
Given the following balanced reaction: 2Na(s) + F2(g) --> 2NaF(s)
a) How many moles of NaF will be made from 2.6 moles of F2?
b) How many moles of NaF will be made from 4.8 moles of Na?
Answer:
yes it is corrwect iyt is absolitle correct
Explanation:
Part A
When the following liquids are poured into the same container, they separate as shown in the image. Based on the data
in the table below, what caused the order of the layers?
rubbing alcohol
vegetable oil
water
corn syrup
Mass
Liquid
corn syrup
water
Volume Used
95 cm
90 cm
85 cm
105 cm?
130.158
90.00 8
77.358
81.908
Density
1.37 g/cm
1 g/cm
0.91 g/cm
0.78 g/cm
vegetable oil
rubbing alcohol
I
B
X
Font Sizes
A- A -
E 3
Answer: The layers are ordered by density, with the least dense layer on top, and the densest layer on the bottom.
Explanation:
Plato
The density of toluene (C7H8) is 0.867 and that of thiophene (C4H4S) is 1.065 g/ml. A solution is made by dissolving 10.00g thiophene in 250.00ml of toluene. a)Calculate the molarity of the solution
b)Assuming the volume are addictive ,calculate the molarity of the solution
Answer:
Calcular la molaridad de una solución que se preparó disolviendo 14 g de KOH en suficiente
agua para obtener 250 mL de solución. (masa molar del KOH = 56 g/mol).
Resolución: de acuerdo a la definición de “molaridad” debemos calcular primero, el número de mol de soluto (KOH) que
se han disuelto en el volumen dado, es decir, “se transforma g de soluto a mol de soluto” por medio de la masa molar,
así:
56 g de KOH 14 g de KOH
----------------- = ------------------- X = 0,25 mol de KOH
1 mol X
Ahora, de acuerdo con la definición de molaridad, el número de mol debe estar contenido en 1000 mL (o 1 L) de
solución, que es el volumen estándar para esta unidad de concentración, lo que se determina con el siguiente planteamiento:
0,25 mol X
----------------------- = ------------------------- X = 1 mol de KOH
250 mL de solución 1000 mL de solución
Explanation:
What volume of each solution contains 0.14 mol of KCl? Express your answer using two significant figures.
1.8 M KCl
Answer:
Solution given:
1 mole of KCl[tex]\rightarrow [/tex]22.4l
1 mole of KCl[tex]\rightarrow [/tex]74.55g
we have
0.14 mole of KCl[tex]\rightarrow [/tex]74.55*0.14=10.347g
74.55g of KCl[tex]\rightarrow [/tex]22.4l
10.347 g of KCl[tex]\rightarrow [/tex]22.4/74.55*10.347=3.11litre
volume of each solution contains 0.14 mol of KCl contain 3.11litre.
[tex]\:[/tex]
1 mole of KCl → 22.4l
1 mole of KCl → 74.55g
we have
0.14 mole of KCl → 74.55*0.14=10.347g
74.55g of KCl → 22.4l
10.347 g of KCl → 22.4/74.55*10.347=3.11litre
volume of each solution contains 0.14 mol of KCl contain 3.11litre.
What must happen to uranium before it can be used as a fuel source?
Answer: Uranium enrichment. Uranium is used to fuel nuclear reactors; however, uranium must be enriched before it can be used as fuel. Enriching uranium increases the amount of uranium-235 (U235) that can sustain the nuclear reaction needed to release energy and produce electricity at a nuclear power plant.
It is advised that the bromobenzene solution be added slowly to the magnesium-ether solution so that it isn't present in a high concentration, thus reducing the amount of by-product formed. What is the by-product
Answer:
Biphenyl
Explanation:
The reaction of bromo benzene with magnesium-ether solution yields a Grignard reagent.
The byproduct of this reaction is biphenyl. It is formed when two unreacted bromobenzene molecules are coupled together.
Hence, It is advised that the bromobenzene solution be added slowly to the magnesium-ether solution so that it isn't present in a high concentration, thus reducing the amount of biphenyl by-product formed.
What is the balanced form of the following equation?
Br2 + S2O32- + H2O → Br1- + SO42- + H+
Answer:
5 Br₂ + S₂O₃²⁻ + 5 H₂O ⇒ 10 Br⁻ + 2 SO₄²⁻ + 10 H⁺
Explanation:
We will balance the redox reaction through the ion-electron method.
Step 1: Identify both half-reactions
Reduction: Br₂ ⇒ Br⁻
Oxidation: S₂O₃²⁻ ⇒ SO₄²⁻
Step 2: Perform the mass balance, adding H⁺ and H₂O where appropriate
Br₂ ⇒ 2 Br⁻
5 H₂O + S₂O₃²⁻ ⇒ 2 SO₄²⁻ + 10 H⁺
Step 3: Perform the charge balance, adding electrons where appropriate
2 e⁻ + Br₂ ⇒ 2 Br⁻
5 H₂O + S₂O₃²⁻ ⇒ 2 SO₄²⁻ + 10 H⁺ + 10 e⁻
Step 4: Make the number of electrons gained and lost equal
5 × (2 e⁻ + Br₂ ⇒ 2 Br⁻)
1 × (5 H₂O + S₂O₃²⁻ ⇒ 2 SO₄²⁻ + 10 H⁺ + 10 e⁻)
Step 5: Add both half-reactions
5 Br₂ + S₂O₃²⁻ + 5 H₂O ⇒ 10 Br⁻ + 2 SO₄²⁻ + 10 H⁺
15.27
The following equilibria were attained at 823 K:
COO(s) + H2() Co(s) + H2O(g) K = 67
COO(s) + CO(8) = Co(s) + CO2(8) K = 490
Based on these equilibria, calculate the equilibrium con-
stant for
H2(g) + CO2(g) = CO(g) + H2O(g) at 823 K.
The equilibrium constant for the reaction is K = 0.137
We obtain the equilibrium constant considering the following equilibria and their constants:
COO(s) + H₂(g) → Co(s) + H₂O(g) K₁ = 67
COO(s) + CO(g) → Co(s) + CO₂(g) K₂ = 490
We write the first reaction in the forward direction because we need H₂(g) in the reactants side:
(1) COO(s) + H₂(g) → Co(s) + H₂O(g) K₁ = 67
Then, we write the second reaction in the reverse direction because we need CO₂(g) in the reactants side. Thus, the equilibrium constant for the reaction in the reverse direction is the reciprocal of the constant for the reaction in the forward direction (K₂):
(2) Co(s) + CO₂(g) → COO(s) + CO(g) K₂ = 1/490
From the addition of (1) and (2), we obtain:
COO(s) + H₂(g) → Co(s) + H₂O(g) K₁ = 67
+
Co(s) + CO₂(g) → COO(s) + CO(g) K₂ = 1/490
-------------------------------------------------
H₂(g) + CO₂(g) → CO(g) + H₂O(g)
Notice that Co(s) and COO(s) are removed that appear in the same amount at both sides of the chemical equation.
Now, the equilibrium constant K for the reaction that is the sum of other two reactions is calculated as the product of the equilibrium constants, as follows:
K = K₁ x K₂ = 67 x 1/490 = 67/490 = 0.137
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A solution is made by dissolving 5.84 grams of NaCl in enough distilled water to give a final volume of 1.00 L. What is the molarity of the solution
Group of answer choices
0.0250 M
0.400 M
0.100 M
1.00 M
Answer:
Explanation:
1. A solution is made by dissolving 5.84g of NaCl is enough distilled water to a give a final volume of 1.00L. What is the molarity of the solution? a. 0.100 M b. 1.00 M c. 0.0250 M d. 0.400 M 2. A 0.9% NaCl (w/w) solution in water is a. is made by mixing 0.9 moles of NaCl in a 100 moles of water b. made and has the same final volume as 0.9% solution in ethyl alcohol c. a solution that boils at or above 100°C d. All the above (don't choose this one) 3. In an exergonic process, the system a. gains energy b. loses energy c. either gains or loses energy d. no energy change at all
Answer:
[tex]\boxed {\boxed {\sf 0.100 \ M }}[/tex]
Explanation:
Molarity is a measure of concentration in moles per liter.
[tex]molarity = \frac{moles \ of \ solute}{liters \ of \ solution}}[/tex]
The solution has 5.84 grams of sodium chloride or NaCl and a volume of 1.00 liters.
1. Moles of SoluteWe are given the mass of solute in grams, so we must convert to moles. This requires the molar mass, or the mass of 1 mole of a substance. These values are found on the Periodic Table as the atomic masses, but the units are grams per mole, not atomic mass units.
We have the compound sodium chloride, so look up the molar masses of the individual elements: sodium and chlorine.
Na: 22.9897693 g/mol Cl: 35.45 g/molThe chemical formula (NaCl) contains no subscripts, so there is 1 mole of each element in 1 mole of the compound. Add the 2 molar masses to find the compound's molar mass.
NaCl: 22.9897693 + 35.45 = 58.4397693 g/molThere are 58.4397693 grams of sodium chloride in 1 mole. We will use dimensional analysis and create a ratio using this information.
[tex]\frac {58.4397693 \ g\ \ NaCl} {1 \ mol \ NaCl}[/tex]
We are converting 5.84 grams to moles, so we multiply by that value.
[tex]5.84 \ g \ NaCl *\frac {58.4397693 \ g\ NaCl} {1 \ mol \ NaCl}[/tex]
Flip the ratio. It remains equivalent and the units of grams of sodium chloride cancel.
[tex]5.84 \ g \ NaCl *\frac {1 \ mol \ NaCl}{58.4397693 \ g\ NaCl}[/tex]
[tex]5.84 *\frac {1 \ mol \ NaCl}{58.4397693 }[/tex]
[tex]0.09993194823 \ mol \ NaCl[/tex]
2. MolarityWe can use the number of moles we just calculated to find the molarity. Remember there is 1 liter of solution.
[tex]molarity= \frac{moles \ of \ solute}{liters \ of \ solution}[/tex]
[tex]molarity= \frac{ 0.09993194823 \ mol \ NaCl}{1 \ L}[/tex]
[tex]molarity= 0.09993194823 \ mol \ NaCl/L[/tex]
3. Units and Significant FiguresThe original measurements of mass and volume have 3 significant figures, so our answer must have the same. For the number we calculated, that is the thousandths place. The 9 in the ten-thousandths place tells us to round the 9 to a 0, but then we must also the next 9 to a 0, and the 0 to a 1.
[tex]molarity \approx 0.100 \ mol \ NaCl/L[/tex]
1 mole per liter is 1 molar or M. We can convert the units.
[tex]molarity \approx 0.100 \ M \ NaCl[/tex]
The molarity of the solution is 0.100 M.
There are three isotopes of carbon. They have mass number of 12, 13 and 14. The average atomic mass of carbon is 12.0107 amu. What does this say about the relative abundances of the three isotopes?
Answer:
lots more of the carbon 12 than the others
havent calculated it percentage-wise but you can see its very close to 12 meaning it is of far greater abundance that carbon 13 and 14
Explanation:
Consider the reaction: NaNO3(s) + H2SO4(l) NaHSO4(s) + HNO3(g) ΔH° = 21.2 kJ
How much heat must absorbed by the reaction system to convert 100g of NaNO3 into NaHSO4(s)?
Answer:
endet nach selam nw
4gh7
Consider the Fischer ester synthesis of methyl benzoate from benzoic acid and methanol in the presence of sulfuric acid as a catalyst. A reaction was performed in which 3.3 g of benzoic acid was reacted with excess methanol to make 1.7 g of methyl benzoate. Calculate the theoretical yield and percent yield for this reaction.
Answer:
46.2%
Explanation:
Number of moles benzoic acid reacts = 3.3g/122.12 g/mol = 0.027 moles
Since the reaction is 1:1, 0.027 moles of methyl benzoate is formed.
Hence;
Theoretical yield of methyl benzoate = 0.027 moles × 136.15 g/mol = 3.68 g
% yield = actual yield/theoretical yield × 100
% yield = 1.7 g/3.68 g × 100
% yield = 46.2%
What separates the inner planets from the outer planets?
a. Main asteroid belt
b. Main comet belt
c. Kuiper belt
d. Outer planet belt
please help this is for SCIENCE test I need help
Answer:
main asteroid belt separates the inner planets from the outer planets
2. Write the chemical equation for the reaction NaOH Sodium Hydroxide AgNO3 Silver Nitrate
Answer:
AgNO3 + NaOH = AgOH + NaNO3.
Explanation:
Balancing Strategies: In this reaction, the products are initially NaNO3 + AgOH. However the AgOH would break down into Ag2O and H2O. This would give us NaNO3 + Ag2O + H2O as our products for the overall reaction.
Balancing Strategies: In this reaction, the products are initially NaNO3 + AgOH. However the AgOH would break down into Ag2O and H2O. This would give us NaNO3 + Ag2O + H2O as our products for the overall reaction.However, the equation balanced here is the initial reaction which produces AgOH and NaNO3.
A tree is an example
of a vascular plant that
is
because it
has deep roots.
A. tall
B. tiny
C. small
Dyshort
How long do spent fuel rods remain dangerously radioactive?
Answers
A.
The rods are no longer radioactive because the radioisotopes are used up.
B.
Spent fuel rods remain radioactive for several years after the fuel is exhausted.
C.
It takes tens of thousands of years for the radioisotopes in the rods to decay to safe levels.
D.
It is impossible to determine how long it will take for the radioisotopes to decay because they last too long.
Answer:
c
Explanation:
it takes 10,000 years to just reduce down the decay
Calculate the no. of moles in 15g of CaCl2
Answer:
[tex]\boxed {\boxed {\sf 0.14 \ mol \ CaCl_2}}[/tex]
Explanation:
We are asked to calculate the number of moles of 15 grams of calcium chloride (CaCl₂).
To convert from grams to moles, we use the molar mass, or the mass of 1 mole of a substance. Molar masses are found on the Periodic Table because they are equivalent to the atomic masses, but the units are grams per mole instead of atomic mass units.
Look up the individual elements in the compound: calcium and chloride.
Ca: 40.08 g/mol Cl: 35.45 g/molNotice the chemical formula has a subscript of 2 after Cl or chlorine. There are 2 moles of chlorine in every 1 mole of calcium chloride. We must multiply chlorine's molar mass by 2 before adding calcium's molar mass.
Cl₂: 35.45 * 2 = 70.9 g/mol CaCl₂= 40.08 + 70.9 = 110.98 g/molWe will convert using dimensional analysis, so we must create a ratio using the molar mass.
[tex]\frac {110.98 \ g \ CaCl_2}{ 1 \ mol \ CaCl_2}[/tex]
We are converting 15 grams of calcium chloride to moles, so we must multiply the ratio by this value.
[tex]15 \ g \ CaCl_2 *\frac {110.98 \ g \ CaCl_2}{ 1 \ mol \ CaCl_2}[/tex]
Flip the ratio so the units of grams of calcium chloride cancel.
[tex]15 \ g \ CaCl_2 *\frac { 1 \ mol \ CaCl_2}{110.98 \ g \ CaCl_2}[/tex]
[tex]15 *\frac { 1 \ mol \ CaCl_2}{110.98}[/tex]
[tex]\frac { 15}{110.98} \ mol \ CaCl_2[/tex]
[tex]0.1351594882\ mol \ CaCl_2[/tex]
The original measurement of grams (15) has 2 significant figures, so our answer must have the same. For the number we calculated, that is the hundredth place. The 5 in the thousandth place tells us to round the 3 up to a 4.
[tex]0.14 \ mol \ CaCl_2[/tex]
15 grams of calcium chloride is approximately 0.14 moles of calcium chloride.
Carbonic acid (H₂CO₃) is a polyprotic acid. When carbonic acid dissolves in water, which is higher, the concentration of HCO₃- ions or the concentration of CO₃²- ions?
Please explain!
The concentration of CO₃²⁻ ions will be higher
To explain, I want you to imagine H₂CO₃ in water.
we know that it will lose 2 of it's protons, and form 2 ions
The ion which is more stable will have a higher concentration because that ion will refuse to react with anything else, so once something turns into that specific ion, it's not going back... unless there's a more stabler ion possible
In this case, the 2 ions formed are: HCO₃⁻ and CO₃⁽²⁻⁾, drawing the structures of both the ions tells us that both of them have resonance, but the CO₃⁽²⁻⁾ ion has more resonance structures and hence is more stable
Balance the following skeleton reaction and identify the oxidizing and reducing agents: Include the states of all reactants and products in your balanced equation. You do not need to include the states with the identities of the oxidizing and reducing agents.
NO_2(g) rightarrow NO_3^-(aq) +NO_2^- (aq) [basic]
The oxidizing agent is:______.
The reducing agent is:_______.
Answer:
a. 2NO₂ (g) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻ (aq) + H₂O (l)
b. i. NO₂⁻ is the oxidizing agent
ii. NO₃⁻ is the reducing agent.
Explanation:
a. Balance the following skeleton reaction
The reaction is
NO₂ (g) → NO₃⁻ (aq) + NO₂⁻ (aq)
The half reactions are
NO₂ (g) → NO₃⁻ (aq) (1) and
NO₂ (g) → NO₂⁻ (aq) (2)
We balance the number of oxygen atoms in equation(1) by adding one H₂O molecule to the left side.
So, NO₂ (g) + H₂O (l) → NO₃⁻ (aq)
We now add two hydrogen ions 2H⁺ on the right hand side to balance the number of hydrogen atoms
NO₂ (g) + H₂O (l) → NO₃⁻ (aq) + 2H⁺ (aq)
The charge on the left hand side is zero while the total charge on the right hand side is -1 + 2 = +1. To balance the charge on both sides, we add one electron to the right hand side.
So, NO₂ (g) + H₂O (l) → NO₃⁻ (aq) + 2H⁺ (aq) + e⁻ (4)
Since the number of atoms in equation two are balanced, we balance the charge since the charge on the left hand side is zero and that on the right hand side is -1. So, we add one electron to the left hand side.
So, NO₂ (g) + e⁻ → NO₂⁻ (aq) (5)
We now add equation (4) and (5)
So, NO₂ (g) + H₂O (l) → NO₃⁻ (aq) + 2H⁺ (aq) + e⁻ (4)
+ NO₂ (g) + e⁻ → NO₂⁻ (aq) (5)
2NO₂ (g) + H₂O (l) + e⁻ → NO₃⁻ (aq) + NO₂⁻ (aq) + 2H⁺ (aq) + e⁻ (4)
2NO₂ (g) + H₂O (l) → NO₃⁻ (aq) + NO₂⁻ (aq) + 2H⁺ (aq)
We now add two hydroxide ions to both sides of the equation.
So, 2NO₂ (g) + H₂O (l) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻ (aq) + 2H⁺ (aq) + 2OH⁻ (aq)
The hydrogen ion and the hydroxide ion become a water molecule
2NO₂ (g) + H₂O (l) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻ (aq) + 2H₂O (l)
2NO₂ (g) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻ (aq) + H₂O (l)
So, the required reaction is
2NO₂ (g) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻ (aq) + H₂O (l)
b. Identify the oxidizing agent and reducing agent
Since the oxidation number of oxygen in NO₂ is -2. Since the oxidation number of NO₂ is zero, we let x be the oxidation number of N.
So, x + 2 × (oxidation number of oxygen) = 0
x + 2(-2) = 0
x - 4 = 0
x = 4
Since the oxidation number of oxygen in NO₂⁻ is -1. Since the oxidation number of NO₂⁻ is -1, we let x be the oxidation number of N.
So, x + 2 × (oxidation number of oxygen) = 0
x + 2(-2) = -1
x - 4 = -1
x = 4 - 1
x = 3
Also, the oxidation number of oxygen in NO₃⁻ is -1. Since the oxidation number of NO₃⁻ is -1, we let x be the oxidation number of N.
So, x + 2 × (oxidation number of oxygen) = -1
x + 3(-2) = -1
x - 6 = -1
x = 6 - 1
x = 5
i. The oxidizing agent
The oxidation number of N changes from +4 in NO₂ to +3 in NO₂⁻. So, Nitrogen is reduced and thus NO₂⁻ is the oxidizing agent
ii. The reducing agent
The oxidation number of N changes from +4 in NO₂ to +5 in NO₃⁻. So, Nitrogen is oxidized and thus and NO₃⁻ is the reducing agent.
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Uhm cell parts and functions
A cell is the structural and fundamental unit of life. The study of cells from its basic structure to the functions of every cell organelle is called Cell Biology. Robert Hooke was the first Biologist who discovered cells
two types of cell
1) Prokaryotes
2) Eukaryotes
Characteristics of Cells
1) Cells provide structure and support to the body of an organism.
2) The cell interior is organised into different individual organelles surrounded by a separate membrane.
3) The nucleus (major organelle) holds genetic information necessary for reproduction and cell growth
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A cell is defined as the fundamental, structural and functional unit of all life.
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what is meant by density
Answer:
The degree of compactness of a substance
Để xác định hàm lượng Cu trong hợp kim Cu-Zn người ta làm như sau: Hòa
tan hoàn toàn 2,068g mẫu hợp kim Cu-Zn trong lượng dư axit HNO3, thu được dung
dịch X. Đun đuổi axit dư, điều chỉnh tới pH 3 thu được 100mL dung dịch Y. Lấy
10mL dung dịch Y, thêm KI dư, rồi chuẩn độ dung dịch tạo thành bằng dung dịch
Na2S2O3 0,1M thì thấy hết 15,0 mL. Viết các phương trình phản ứng xảy ra. Tính
hàm lượng Cu trong mẫu hợp kim trên.
Balance the redox reaction Al(s) + MnO4^- (aq) --> MnO2 (s) + Al(OH)4^- (aq) in aqueous basic solution
Answer:
Al + MnO4- + 2H2O → Al(OH)4- + MnO2
Explanation:
First of all, we out down the skeleton equation;
Al + MnO4- → MnO2 + Al(OH)4-
Secondly, we write the oxidation and reduction equation in basic medium;
Oxidation half equation:Al + 4H2O + 4OH- → Al(OH)4- + 4H2O + 3e-
Reduction half equation:MnO4- + 4H2O + 3e- → MnO2 + 2H2O + 4OH-
Thirdly, we add the two half reactions together to obtain:
Al + MnO4- + 8H2O + 4OH- + 3e- → Al(OH)4- + MnO2 + 6H2O + 3e- + 4OH-
Lastly, cancel out species that occur on both sides of the reaction equation;
Al + MnO4- + 8H2O→ Al(OH)4- + MnO2 + 6H2O
The simplified equation now becomes;
Al + MnO4- + 2H2O → Al(OH)4- + MnO2
