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.
Hey. what is the difference between a boulder and a rock? ]
And bonus: what did the duck say when she bought lipstick?
Answer:
The difference between a rock and a boulder can be explained in terms of size and detachment. A rock is defined by geologists as an aggregate of minerals. A boulder is a type of rock, specifically a large detached one. All boulders are rocks, but not all rocks are boulders.
If 75.4 J of energy is absorbed by 0.25 mol of CCl4 at constant pressure, what is the change in temperature? The specific heat of CCl4 is 0.861 J/g·°C.
Answer:
ΔT = 2.28°C
Explanation:
Heat, H = 75.4J
Number of moles = 0.25 mol
Specific heat capacity, c = 0.861 J/g·°C
Change in temperature, ΔT = ?
These quantities are related by the following equation;
H = mc ΔT
Mass, m = Number of moles * Molar mass
m = 0.25mol * 153.82 g/mol
m = 38.455g
S back to the equation;
H = mc ΔT
Substituting the values;
75.4 = 38.455 * 0.861 * ΔT
ΔT = 75.4 / 33.11
ΔT = 2.28°C
The change in temperature is 2.28 °C
First, we will determine the mass of CCl₄ absorbed
From the given information,
Number of moles of CCl₄ absorbed = 0.25 mol
Using the formula
Mass = Number of moles × Molar mass
Molar mass of CCl₄ = 153.82 g/mol
∴ Mass of CCl₄ absorbed = 0.25 × 153.82
Mass of CCl₄ absorbed = 38.455 g
Now, using the formula
Q = mcΔT
Where Q is the quantity of heat
m is the mass
c is the specific heat of substance
and ΔT is the change in temperature
From the given information
Q = 75.4 J
c = 0.861 J/g.°C
Putting the parameters into the formula, we get
75.4 = 38.455 × 0.861 ×ΔT
75.4 = 33.109755 × ΔT
∴ ΔT = 75.4 ÷ 33.109755
ΔT = 2.28 °C
Hence, the change in temperature is 2.28 °C
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When 1604 J of heat energy is added to 48.9 g of hexane, C6H14, the temperature increases by 14.5 ∘C. Calculate the molar heat capacity of C6H14.
Answer:
THE MOLAR HEAT CAPACITY OF HEXANE IS 290.027 J/ C
Explanation:
1604 J of heat is added to 48.9 g of hexane
To calculate the molar heat capacity of hexane, it is important to note that the molar heat capacity of a substance is the measure of the amount of heat needed to raise 1 mole of a substance by 1 K.
Since 1604 J of heat = 48.9 g of hexane
Molar mass of hexane = 86 g/mol = 1 mole
then;
1604 J = 48.9 g
x = 86 g
x = 1604 * 86 / 48.9
x = 4205.4 J
Hence, 4205.4 J of heat will be added to 1 mole or 86 g of hexane to raise the temperature by 14.5 C.
In other words,
heat = molar heat capacity * temperature change
molar heat capacity = heat/ temperature change
Molar heat capacity = 4205.4 J / 14.5 C
Molar heat capacity = 290.027 J/C
The molar heat capacity of hexane is 290.027 J/ C
When titrating a strong acid with a strong base, after the equivalence point is reached, the pH will be determined exclusively by: Select the correct answer below:
A) hydronium concentration
B) hydroxide concentration
C) conjugate base concentration
D) conjugate acid concentration
Answer:
B) hydroxide concentration
Explanation:
Hello,
In this case, since we are talking about strong both base and acid, since the base is the titrant and the acid the analyte, once the equivalence point has been reached, some additional base could be added before the experimenter realizes about it, therefore, since the titrant is a strong base, it completely dissociates in hydroxide ions and metallic ions which allows us to compute the pOH of the solution by known the hydroxide ions concentration.
After that, due to the fact that the pH is related with the pOH as shown below:
pH=14-pOH
We can directly compute the pH.
Best regards.
oxide. b) Silicon dio
43. What is the nature of an enzyme?
a) Vitamin. b) Lipid. c) Carbohydrate. d) Protein
44. Name the enzyme which catalyzes the oxidation-reduction reaction?
a) Transaminase. b) Glutamine synthetase. c) Phosphofructokinase. d) Oxidoreductase
nontido
Answer:
43) protein
44) oxidoreductase
What would happen to the measured cell potentials if 30 mL solution was used in each half-cell instead of 25 mL
Answer:
The answer is "[tex]\bold{\log \frac{[0] mole}{[R]mole}}[/tex]"
Explanation:
[tex]E_{cell} =E_{cell}^{\circ} - \frac{0.0591}{n}= \log\frac{[0]}{[R]}\\[/tex]
In the above-given equation, we can see from [tex]E_{ceu}[/tex], of both oxidant [tex]conc^n[/tex]as well as the reactant were connected. however, weight decreases oxidant and reduction component concentration only with volume and the both of the half cells by the very same factor and each other suspend
[tex]\to \log \frac{\frac{\text{oxidating moles}}{25 \ ml}}{\frac{\text{moles of reduction}}{25 ml}} \ \ = \ \ \log \frac{\frac{\text{oxidating moles}}{30 \ ml}}{\frac{\text{moles of reduction}}{30 ml}} \\\\\\[/tex]
[tex]\to {\log \frac{[0] mole}{[R]mole}}[/tex]
The cell potential of the electrochemical reaction has been the same when the volume has been reduced from 30 mL to 25 mL in each half cells.
The cell potential has been given as the difference in the potential of the two half cells in the electrochemical reaction.
The two cells has been set with the concentration of solutions in the oxidation and reduction half cells.
Cell potential changeThe cell potential has been changed when there has been a change in the potential of the half cells.
The volume of 30 mL to the solution has been, resulting in the cell potential difference of x.
With the volume of 25 mL, there has been the difference in the potential being similar to the 30 mL solution, i.e. x.
Thus, the cell potential of the electrochemical reaction has been the same when the volume has been reduced from 30 mL to 25 mL in each half cells.
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The ceramic glaze on a red-orange Fiestaware plate is U2O3 and contains 50.1 grams of 238U, but very little 235U. (a) What is the activity of the plate (in Ci)
Answer:
The correct answer is 1.68 × 10⁻⁵ Ci
Explanation:
The activity of the uranium is determined by using the formula,
R = 0.693 N/t1/2 -------------- (i)
The number of atoms is, N = nNA
Here, NA is the Avogadro number and n is the number of moles. The value of n is m/M, that is, mass/molecular mass. Now the value of N becomes,
N = (m/M) NA
The m or mass of uranium given is 50.1 grams, and the molecular mass is 238 g/mol, now putting the values we get,
N = (50 g/238 g) (6.023 × 10²³) = 1.26 × 10²³
The half-life of 238U from year to second is,
t1/2 = (4.468 × 10⁸ year) (3.16 × 10⁷ s/ 1 year) = 1.412 × 10¹⁶ s
Substituting the values of t1/2 as 1.412 × 10¹⁶, and 1.26 × 10²³ for N in equation (i) we get,
R = 0.639 (1.26 × 10²³) / 1.412 × 10¹⁶ s
= 6.18 × 10⁶ Bq (2.7027 × 10⁻¹¹ Ci/1 Bq)
= 1.68 × 10⁻⁵ Ci
Hence, the activity of the plate is 1.68 × 10⁻⁵ Ci
Which of the following elements is in the same group as Sulfur (S)?
Answer:
PLEASE SHOW ME THE ELEMENTS OR I WOULD ENLIST ALL THE ELEMENTS.
Explanation:
Group 6A (or VIA) of the periodic table are the chalcogens: the nonmetals oxygen (O), sulfur (S), and selenium (Se), the metalloid tellurium (Te), and the metal polonium (Po)
how many moles of oxygen atoms are present in 0.4 moles of oxygen gas
Answer:
Each molecule of O2 is made up of 2 oxygen atoms. So 1 mole of O2 molecules is made up of 2 moles of oxygen atoms. Therefore 1 mole of oxygen gas contains 2 moles of oxygen atoms. And 0.4 moles of oxygen gas contains 0.8 moles of oxygen atoms.
There are 0.8 moles of oxygen atoms in 0.4 moles of oxygen gas.
Oxygen gas (O₂) consists of two oxygen atoms bonded together. Therefore, to determine the number of moles of oxygen atoms present in a given amount of oxygen gas, we can simply multiply the number of moles of oxygen gas by the number of oxygen atoms per molecule, which is 2.
Given that we have 0.4 moles of oxygen gas, we can calculate the number of moles of oxygen atoms as follows:
Number of moles of oxygen atoms = Number of moles of oxygen gas × Number of oxygen atoms per molecule
= 0.4 moles × 2
= 0.8 moles
Therefore, there are 0.8 moles of oxygen atoms present in 0.4 moles of oxygen gas.
This calculation is based on the stoichiometry of oxygen gas, which indicates that each molecule of O₂ contains two oxygen atoms. By considering the mole ratio between oxygen gas and oxygen atoms, we can determine the number of moles of oxygen atoms in a given quantity of oxygen gas.
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Calculate the moles of Iron (Fe) in 3.8 x 10^{21} atoms of Iron. Please show your work
Answer: 6.31×10⁻³ moles Fe
Explanation:
To calculate moles when given atoms, we need to use Avogadro's number.
Avogadro's number: 6.022×10²³ atoms/mol
[tex](3.8*10^2^1 atoms)*\frac{mol}{6.022*10^2^3 atoms} =6.31*10^-^3 mols[/tex]
The atoms cancel out, and we are left with moles. There are 6.31×10⁻³ moles Fe.
Which term describes the repeated arrangement of the same molecule?
Answer:
1. C
Carbon atoms
2. C
CH3CH2OH
3. C
extended structure
AND 4. B
One grey sphere, four white spheres, one red sphere
TOOK THE QUIZ. YOUR WELCOME :))
The term which describe the repeated arrangement of same molecules is extended structure. Thus option D is correct.
What is extended structure?Extended structure can be defined as a structure in which the subunits are arranged in a repeating pattern and occur in a consistent ratio.
Sodium chloride and diamond are some of the example of extended structure.
The expanded or extended structure departs from the bilayer configuration, with well-separated hydrophilic and hydrophobic layers made up of propagating chains along the calixarene cavity axis are some characteristics.
Thus, the term which describe the repeated arrangement of same molecules is extended structure. Thus option D is correct.
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Your question is incomplete but probably your complete question was
Which term describes the repeated arrangement of the molecule?
A. Bonds
B. Atoms
C. Molecular Model
D. Extended Structure
Given the balanced chemical equation for the decomposition for INO, and the rate of disappearance of INO, write the expressions for the rates of appearance of I2 and NO.
2INO(g) → I2(g) + 2NO(g)
Reactant: Product(I2): Product(NO):
-∆[INO]/2∆t = ??/??
Answer:
rate of disappearance of -0.5d[INO]/dt
rate of appearance of I2 = d[I2]/dt
rate of appearance of No = 0.5*d[NO]/dt
Explanation:
According to chemical equilibrium, d[I₂]/dt and d[NO]/dt is the expressions for the rates of appearance of I₂ and NO respectively.
What is chemical equilibrium?Chemical equilibrium is defined as the condition which arises during the course of a reversible chemical reaction with no net change in amount of reactants and products.A reversible chemical reaction is the one wherein the products as soon as they are formed react together to produce back the reactants.
At equilibrium, the two opposing reactions which take place take place at equal rates and there is no net change in amount of the substances which are involved in the chemical reaction.At equilibrium, the reaction is considered to be complete . Conditions which are required for equilibrium are given by quantitative formulation.
Factors which affect chemical equilibrium are change in concentration , change in pressure and temperature and presence of catalyst.
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For the reaction CO2(g) + H2(g)CO(g) + H20(g)
∆H°=41.2 kJ and ∆S°=42.1 J/K
The standard free energy change for the reaction of 1.96 moles of Co2(g) at 289 K, 1 atm would be_________KJ.
This reaction is (reactant, product)___________ favored under standard conditions at 289 K.
Assume that ∆H° and ∆S° are independent of temperature.
Answer:
The ΔG° is 29 kJ and the reaction is favored towards reactant.
Explanation:
Based on the given information, the ΔH°rxn or enthalpy change is 41.2 kJ, the ΔS°rxn or change in entropy is 42.1 J/K or 42.1 * 10⁻³ kJ/K. The temperature given is 289 K. Now the Gibbs Free energy change can be calculated by using the formula,
ΔG° = ΔH°rxn - TΔS°rxn
= 41.2 kJ - 289 K × 42.1 × 10⁻³ kJ/K
= 41.2 kJ - 12.2 kJ
= 29 kJ
As ΔG° of the reaction is positive, therefore, the reaction is favored towards reactant.
What do you predict the chemical formula for the compound formed between calcium and sulfur?
Answer:
calcium donates two vanence electrons to sulfur atom to form Ca2+ ion and an S2+ - ion
Calculate the [H+] and pH of a 0.0010 M acetic acid solution. The Ka of acetic acid is 1.76×10−5. Use the method of successive approximations in your calculations.
Answer:
[tex][H^+]=0.000123M[/tex]
[tex]pH=3.91[/tex]
Explanation:
Hello,
In this case, dissociation reaction for acetic acid is:
[tex]CH_3COOH\rightleftharpoons CH_3COO^-+H^+[/tex]
For which the equilibrium expression is:
[tex]Ka=\frac{[CH_3COO^-][H^+]}{[CH_3COOH]}[/tex]
Which in terms of the reaction extent [tex]x[/tex] could be written as:
[tex]1.74x10^{-5}=\frac{x*x}{[CH_3COOH]_0-x}=\frac{x*x}{0.0010M-x}[/tex]
Thus, solving by using a solver or quadratic equation we obtain:
[tex]x_1=0.000123M\\\\x_2=-0.000141M[/tex]
And clearly the result is 0.000123M, which also equals the concentration of hydronium ion in solution:
[tex][H^+]=0.000123M[/tex]
Now, the pH is computed as follows:
[tex]pH=-log([H^+])=-log(0.000123)\\\\pH=3.91[/tex]
Best regards.
Given that π = n M R T, rearrange the equation to solve for V
Answer:
V= n/M
Explanation:
From;
π = nRT/V = MRT
Where;
n= number of moles
R= gas constant
T= absolute temperature
M= molar mass
V= volume of the solution
π= osmotic pressure
Thus;
nRT/V = MRT
nRT = VMRT
V= nRT/MRT
V= n/M
A critical reaction in the production of energy to do work or drive chemical reactions in biological systems is the hydrolysis of adenosine triphosphate, ATP, to adenosine diphosphate, ADP, as described by the reaction
ATP(aq)+H2O(l)→ADP(aq)+HPO2−4(aq) for which ΔGrxn=30.5 kJ/mol at 37.0∘C and pH 7.0.
Required:
Calculate the value of ΔGrxn in a biological cell in which [ATP]=5.0 mM, [ADP]=0.60 mM, and [HPO2−4]=5.0 mM
Answer:
ΔG = -49.64 KJ/mol
Explanation:
The actual free energy change of the reaction under the given conditions, ΔG is given by the formula below;
ΔG = ΔG'° + RT ln([ADP][HPO₂⁴⁻] / [ATP])
where ΔG'° = -30.5 KJ/mol; R = 8.315 J/mol.K; T = 37°C = 310 K; [ADP] = 5.0 mM = 0.005 M; [HPO₂⁴⁻] = 0.60 mM = 0.0006 M; [ATP] = 5.0 mM = 0.005 M
ΔG = -30.5 KJ/mol + (8.315 J/mol.K)(310 K) ln {(0.005)(0.0006)/(0.005)}
ΔG = -30.5 KJ/mol + (2.58 KJ/mol * -7.4186)
ΔG = -30.5 KJ/mol - 19.14 KJ/mol
ΔG = -49.64 KJ/mol
The two common properties of all solids are fixed _____ and _____.
Answer:
shape
volume
Hope this helps! （づ￣3￣）づ╭❤～
Explanation:
A compound is found to contain 11.21 % hydrogen and 88.79 % oxygen by mass. What is the empirical formula for this compound?
Answer:
H₂O
Explanation:
The empirical formular of the compound is obtained using the following steps;
Step 1: Divide the percentage composition by the atomic mass
Hydrogen = 11.21 / 1 = 11.21
Oxygen = 88.79 / 16 = 5.55
Step 2: Divide by the lowest number
Hydrogen = 11.21 / 5.55 = 2.02 ≈ 2
Oxygen = 5.55 / 5.55 = 1
This means the ratio of the elements is 2 : 1
The empirical formular (simplest formular of a compound) of the compound is;
H₂O
Answer:Empirical formula ======== H₂O
Explanation:The empirical formula of a compound shows the whole number ratio for each atom in a compound.
To find empirical formula. we follow the below steps
The total mass of the compound here is 100 grams, that is (11.21% of hydrogen + 88.79% of oxygen) we can then assume 11.21 grams of hydrogen and 88.79grams of oxygen
Hydrogen Oxygen
1.composition by mass 11.21 88.79
molecular weight 1.007g/mol 15.990g/mol
2.Divide composition by mass 11.21/1.007 88.79/15.99
by each molecular weight to get 11.13 5.553
no of moles
3 Divide by the least number of moles
to get atomic ratio 11.13/5.553 5.553/5.553
2.004 1.00
4.Convert to whole numbers 2 1
Empirical formula ======== H₂O
A 2.87g sample of carbon reacts with hydrogen to form 3.41g of car fuel. What is the empirical formula of the car fuel?
Answer:
The empirical formulae for the car fuel is C4H9
If for a particular process, ΔH=308 kJmol and ΔS=439 Jmol K, in what temperature range will the process be spontaneous?
Answer:
The process will be spontaneous above 702 K.
Explanation:
Step 1: Given data
Standard enthalpy of the reaction (ΔH°): 308 kJ/molStandard entropy of the reaction (ΔS°): 439 J/mol.KStep 2: Calculate the temperature range in which the process will be spontaneous
The reaction will be spontaneous when the standard Gibbs free energy (ΔG°) is negative. We can calculate ΔG° using the following expression.
ΔG° = ΔH° - T × ΔS°
When ΔG° < 0,
ΔH° - T × ΔS° < 0
ΔH° < T × ΔS°
T > ΔH°/ΔS°
T > (308,000 J/mol)/(439 J/mol.K)
T > 702 K
The process will be spontaneous above 702 K.
Glucose is soluble in water. Why is cellulose, which is made up of glucose, insoluble in water?
.....................
Cellulose is insoluble in water which is made up of glucose because it possesses high inter and intramolecular hydrogen bonding between the hydroxyl groups of the neighboring chains.
What is the function of cellulose?Due to being insoluble in nature, it serves as the fundamental component of the cell membrane in the plant. This is why the cell wall of plant cells are made of cellulose.
The chains of the cellulose are strongly bonded to each other. So, it is very difficult for the molecules of water to rupture or destruct these bonds between the chains. It is a hydrogen bond cross-linked polymer and more complex than glucose.
Therefore, due to high inter and intramolecular hydrogen bonding between the hydroxyl groups of the neighboring chains, cellulose is insoluble in water.
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What is the concentration of A after 50.7 minutes for the second order reaction A → Products when the initial concentration of A is 0.250 M? (k = 0.117 M⁻¹min⁻¹)
Answer:
0.101 M
Explanation:
Step 1: Given data
Initial concentration of A ([A]₀): 0.250 MFinal concentration of A ([A]): ?Time (t): 50.7 minRate constant (k): 0.117 M⁻¹.min⁻¹Step 2: Calculate [A]
For a second-order reaction, we can calculate [A] using the following expression.
1/[A] = 1/[A]₀ + k × t
1/[A] = 1/0.250 M + 0.117 M⁻¹.min⁻¹ × 50.7 min
[A] = 0.101 M
.) At 500 oC, cyclopropane, C3H6, rearranges to form propene. The reaction is first order with a rate constant of 6.7 x 10-4 s-1. If the initial concentration of C3H6 is 0.0500 M, (a) what is the molarity of C3H6 after 30 min
Answer:
0.015 M
Explanation:
For a first order reaction;
ln[A] =ln[A]o - kt
[A] = final concentration
[A]o =initial concentration
k= rate constant
t= time taken
ln[A] =ln[A]o - kt
ln[A] = ln(0.0500) - 6.7 x 10-4 (30 × 60)
ln[A] = -2.9957 - 1.206
ln[A] = -4.202
e^ln[A] = e^(-4.202)
A= 0.015 M
When the following molecular equation is balanced using the smallest possible integer coefficients, the values of these coefficients are: potassium hydrogen sulfate (aq) potassium hydroxide (aq) potassium sulfate (aq) water (l)\
Answer:
Explanation:
Answer:
1, 1, 1, 1
Explanation:
potassium hydrogen sulfate + potassium hydroxide ⟶ potassium sulfate + water(l)
KHSO₄ + KOH ⟶ K₂SO₄ + H₂O
1. Put a 1 in front of the most complicated-looking formula (K₂SO₄?):
KHSO₄ + KOH ⟶ 1K₂SO₄ + H₂O
2. Balance S:
We have fixed 1 S on the right. We need 1 S on the left. Put a 1 in front of KHSO₄ to fix it.
1KHSO₄ + KOH ⟶ 1K₂SO₄ + H₂O
3. Balance K:
We have fixed 2 K on the right and 1 K on the left. We need 1 more K on the left. Put a 1 in front of KOH.
1KHSO₄ + 1KOH ⟶ 1K₂SO₄ + H₂O
4. Balance O
We have fixed 4 O on the right and 5 O on the left. We need 1 more O on the right. Put a 1 in front of H₂O.
1KHSO₄ + 1KOH ⟶ 1K₂SO₄ + 1H₂O
Every formula has a coefficient. The equation should be balanced.
5. Check that atoms balance:
[tex]\begin{array}{ccc}\textbf{Atom} & \textbf{On the left} & \textbf{On the right}\\\text{K} & 2 &2\\\text{H} & 2 & 2\\\text{S} & 1 & 1\\\text{O}&5&5\\\end{array}[/tex]
It checks.
The coefficients are 1, 1, 1, 1.
A 1 liter solution contains 0.436 M hypochlorous acid and 0.581 M potassium hypochlorite. Addition of 0.479 moles of barium hydroxide will: (Assume that the volume does not change upon the addition of barium hydroxide.)
Answer:
Exceed the buffer capacity and Raise the pH by several units
Explanation:
Options are:
Raise the pH slightly
Lower the pH slightly
Raise the pH by several units
Lower the pH by several units
Not change the pH
Exceed the buffer capacity
The hypochlorous acid, HClO, is in equilibrium with Hypochlorite ion (From potassium hypochlorite, ClO⁻) producing a buffer. Using H-H equation, pH of initial buffer is:
pH = pKa + log [ClO⁻] / [HClO]
pKa for hypochlorous acid is 7.53
pH = 7.53 + log [0.581M] / [0.436M]
pH = 7.65
Barium hydroxide reacts with HClO producing more ClO⁻, thus:
Ba(OH)₂ + 2HClO → 2ClO⁻ + 2H₂O
As 0.479 moles of Barium hdroxide are added. For a complete reaction you require 0.479mol * 2 = 0.958 moles of HClO
As you have just 0.436 moles (Volume = 1L),
The addition will:
Exceed the buffer capacityThe Ba(OH)₂ that reacts is:
0.436 moles HClO * (1mole (Ba(OH)₂ / 2 mol HClO) = 0.218 moles Ba(OH)₂ and will remain:
0.479 mol - 0.218 mol = 0.261 moles Ba(OH)₂
As 1 mole of Ba(OH)₂ contains 2 moles of OH⁻, moles of OH⁻ and molarity is:
0.261 moles* 2 = 0.522 moles OH⁻ = [OH⁻]
pOH = -log [OH⁻]
pOH = 0.28
And pH = 14 - pOH:
pH = 13.72
Thus, after the addition the pH change from 7.65 to 13.62:
Raise the pH by several units
16. The concentration of a solution of potassium hydroxide is determined by titration with nitric
acid. A 30.0 mL sample of KOH is neutralized by 42.7 mL of 0.498 M HNO3. What is the
concentration of the potassium hydroxide solution?
Answer:
[tex]M_{base}=0.709M[/tex]
Explanation:
Hello,
In this case, since the reaction between potassium hydroxide and nitric acid is:
[tex]KOH+HNO_3\rightarrow KNO_3+H_2O[/tex]
We can see a 1:1 mole ratio between the acid and base, therefore, for the titration analysis, we find the following equality at the equivalence point:
[tex]n_{acid}=n_{base}[/tex]
That in terms of molarities and volumes is:
[tex]M_{acid}V_{acid}=M_{base}V_{base}[/tex]
Thus, solving the molarity of the base (KOH), we obtain:
[tex]M_{base}=\frac{M_{acid}V_{acid}}{V_{base}} =\frac{0.498M*42.7mL}{30.0mL}\\ \\M_{base}=0.709M[/tex]
Regards.
Complete the following reactions
need answers?
i) P + [H]
ii) O2 + [H]
iii) LiH + H2O
iv) CaC2 +H2O
v) NaH + H2O
vi) CaH2 H2O
Answer:
this reaction will produce phosphineP+H---->PH
this reaction will produce waterO2+H---->H2O
this reaction will produce lithium hydroxide and hydrogen gasLiH+H2O---->LiOH+H2
this reaction will produce calcium hydroxide and acetylene gasCaC2+H2O---->Ca(OH)2+C2H2
this reaction will produce sodium hydroxide and hydrogen gasNaH+H2O---->NaOH+H2
this type of reaction will produce calcium hydroxide and hydrogen gasCaH2+H2O---->CaOH+H2
I hope this helps
Name the following compound from the concise formula:______.
CH3CH(CH3)CHCHCH(CH3)CH2CH3
A. 2,4-dimethyl-3-heptene
B. 2,5-dimethyl-3-heptene
C. 3,5-dimethyl-3-heptene
D. 2,5-dimethyl-4-heptene
Answer:
B. 2,5-dimethyl-3-heptene
Explanation:
Answer:
B. 2,5-dimethyl-3-heptene
Explanation:
An unknown gas diffuses 5 times slower than that of H2.The moleculer mass of unknown gas is??
Answer:
50.
Explanation:
We can write Graham's Law of Diffusion as:
(Rate 1)^2 = Molecular Mass 2
-------------- -------------------------
(Rate 2)^2 Molecular Mass 1
So using the Given Information:
1^2 / (1/5)^2 = Molecular Mass of unknown gas / 2, so:
25 = M/2
M = 50.