Answers
since we are given the moles for Co and O, we'll divide both of those moles by the lowest mole quantity, which is, in this case, 2.59. After dividing, we see that the ratio of O to Co is 2:1. So, for every 1 Co atom, there has to be 2 O atoms. we can then insert the 2 in for OH to satisfy this ratio.
Related Questions
A 1.0 L buffer solution is 0.250 M HC2H3O2 and 0.050 M LiC2H3O2. Which of the following actions will destroy the buffer?
A. adding 0.050 moles of NaOH
B. adding 0.050 moles of LiC2H3O2
C. adding 0.050 moles of HC2H3O2
D. adding 0.050 moles of HCl
E. None of the above will destroy the buffer.
Answers
Answer:
D
Explanation:
Addition of 0.05 M HCl, will react with all of the C2H3O2- from LiAc which will give 0.05 M more HAc. So there will be no Acetate ion left to make the solution buffer. Hence, the correct option for the this question is d, which is adding 0.050 moles of HCl.
The action that destroys the buffer is option c. adding 0.050 moles of HCl.
What is acid buffer?It is a solution of a weak acid and salt.
Here, The buffer will destroy at the time when either HC2H3O2 or NaC2H3O2 should not be present in the solution.
The addition of equal moles of HCl finishly reacts with equal moles of NaC2H3O2. Due to this, there will be only acid in the solution.
Since
moles of HC2H3O2 = 1*0.250 = 0.250
moles of NaC2H3O2 = 1*0.050 = 0.050.
moles of HCl is added = 0.050
Now
The reaction between HCl and NaC2H3O2
[tex]HCl + NaC_2H_3O_2 \rightarrow HC_2H_3O_2 + NaCl[/tex]
Now
BCA table is
NaC2H3O2 HCl HC2H3O2
Before 0.050 0.050 0.250
Change -0.050 -0.050 +0.050
After 0 0 0.300
Now, the solution contains the acid (HC2H3O2 ) only.
Therefore addition of 0.050 moles of HCl will destroy the buffer.
Learn more about moles here: https://brainly.com/question/24817060
For the following set of volume/temperature data, calculate the missing quantity after the change is made. Assume that the pressure and the amount of gas remain constant.
V=2.91 L at 23.0 °C
V= 4.20 L at ? °C
Answers
Answer:
155 °C
Explanation:
Step 1: Given data
Initial volume (V₁): 2.91 LInitial temperature (T₁): 23.0°CFinal volume (V₂): 4.20 LFinal temperature (T₂): ?Step 2: Convert the initial temperature to Kelvin
We will use the following expression.
K = °C + 273.15 = 23.0°C + 273.15 = 296.2 K
Step 3: Calculate the final temperature
Assuming an ideal gas behavior, we can calculate the final temperature using Charles' law.
V₁/T₁ = V₂/T₂
T₂ = V₂ × T₁/V₁
T₂ = 4.20 L × 296.2 K/2.91 L
T₂ = 428 K
Step 4: Convert the final temperature to Celsius
We will use the following expression.
°C = K - 273.15 = 428 - 273.15 = 155 °C
2) What is the concentration (M) of CH3OH in a solution prepared by dissolving 11.7 g of CH3OH in sufficient water to give exactly 230 mL of solution?
Answers
To determine the moles of CH3OH, divide the weight in grams of CH3OH by the molecular weight
11.7g / 32g/mol = 0.366 mol CH3OH
0.366 mol CH3OH / .230 L = your molarity
Which option is an isotope with 6 protons and 8 neutrons? (1 point)
O oxygen-6
o carbon-14
o oxygen-14
o carbon-8
Answers
Answer:
Carbon-14
Explanation:
It is because it have 6 protons and 8 neutrons which are added together to form an atomic mass of 14.
The mass number of an isotope is the sum of the number of protons and number of neutrons hence the correct answer is carbon-14.
Isotopes are atoms of the same element that has the same number of protons but different number of neutrons.
The atomic number is the same as the number of protons in the atom. Since there are six protons in the atom, the element must be carbon.
There are eight neutrons meaning that the mass number is 8 + 6 = 14
Therefore, the isotope is carbon-14.
Learn more: https://brainly.com/question/13214440
How many neutrons does Carbon- 14 and Carbon -15 have? *
Answers
Answer: 8 for both
Explanation:
What is the pH of a solution made by mixing 15.00 mL of 0.10 M acetic acid with 15.00 mL of 0.10 M KOH? Assume that the volumes of the solutions are additive. K a = 1.8 ×× 10-5 for CH3CO2H.
Answers
Answer:
pH = 8.72
Explanation:
This is like a titration of a weak acid and a strong base, in this case, we are at the equivalence point plus we have the same mmoles of acid and base. We have completely neutralized the acid.
CH₃COOH + OH⁻ ⇄ CH₃COO⁻ + H₂O
0.1M . 15 mL 0.1M . 15 mL
We only have (0.1M . 15 mL) mmoles of acetate ion. → 1.5 mmoles
As this compound acts like a base, we propose this equilibrium:
CH₃COO⁻ + H₂O ⇄ CH₃COOH + OH⁻ Kb
We need to work with Kb and we know, that Kw = Ka. Kb so, Kb = Kw/Ka
Kb = 1×10⁻¹⁴ /1×10 ⁻⁵ = 5.55×10⁻¹⁰
Concentration of CH₃COO⁻ → 1.5 mmol / 30mL (volumes of the solutions are additive) = 0.05M
So: [CH₃COOH] . [OH⁻] / [CH₃COO⁻] = Kb
x²/ 0.05-x = 5.55×10⁻¹⁰
We can avoid the quadractic equation because Kb is so small
[OH⁻] = √(5.55×10⁻¹⁰ . 0.05) = 5.27×10⁻⁶
pOH = - log [OH⁻] → 5.28
pH = 14 - pOH = 8.72
The pH of a solution made by mixing 15.00 mL of 0.10 M acetic acid should be 8.72.
Calculation of the pH of the solution:Since the following equation should be used.
CH₃COOH + OH⁻ ⇄ CH₃COO⁻ + H₂O
0.1M . 15 mL 0.1M . 15 mL
Now
(0.1M . 15 mL) mmoles of acetate ion. → 1.5 mmoles
So,
CH₃COO⁻ + H₂O ⇄ CH₃COOH + OH⁻ Kb
Now
Kw = Ka. Kb
Kb = Kw/Ka
And,
Kb = 1×10⁻¹⁴ /1×10 ⁻⁵
= 5.55×10⁻¹⁰
Now
[CH₃COOH] . [OH⁻] / [CH₃COO⁻] = Kb
x²/ 0.05-x = 5.55×10⁻¹⁰
Now
[OH⁻] = √(5.55×10⁻¹⁰ . 0.05) = 5.27×10⁻⁶
pOH = - log [OH⁻] → 5.28
pH = 14 - pOH
= 8.72
Hence, The pH of a solution made by mixing 15.00 mL of 0.10 M acetic acid should be 8.72.
Learn more about an acid here: https://brainly.com/question/4519963
For the following reaction, 22.8 grams of diphosphorus pentoxide are allowed to react with 13.5 grams of water . diphosphorus pentoxide(s) water(l) phosphoric acid(aq) What is the maximum mass of phosphoric acid that can be formed
Answers
Answer:
[tex]m_{H_3PO_4}=31.5gH_3PO_4[/tex]
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
[tex]P_2O_5(s)+3H_2O(l)\rightarrow 2H_3PO_4(aq)[/tex]
Thus, since the diphosphorus pentoxide to water molar ratio is 1:3 and we are given the mass of both of them, for the calculation of the maximum mass phosphoric acid that is yielded, one could first identify the limiting reactant, for which we compute the available moles of diphosphorus pentoxide (molar mass 142 g/mol):
[tex]n_{P_2O_5}=22.8gP_2O_5*\frac{1molP_2O_5}{142gP_2O_5}=0.161molP_2O_5[/tex]
And the moles of diphosphorus pentoxide that are consumed by 13.5 g of water (molar mass 18 g/mol):
[tex]n_{P_2O_5}^{consumed}=13.5gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{1molP_2O_5}{3molH_2O} =0.25molP_2O_5[/tex]
Hence, since less moles of diphosphorus pentoxide are available, we sum up it is the limiting reactant, therefore, the maximum mass of phosphoric acid (molar mass 98 g/mol) is computed by considering the 1:2 molar ratio between them as follows:
[tex]m_{H_3PO_4}=0.161molP_2O_5*\frac{2molH_3PO_4}{1molP_2O_5} *\frac{98gH_3PO_4}{1molH_3PO_4} \\\\m_{H_3PO_4}=31.5gH_3PO_4[/tex]
Regards.
Hello, I am a bit stuck on this. Could someone help?
Answers
in non- equilibrium cooling, carbon is allowed to diffuse its end position. true /false
Answers
Answer:
False because it is false
If the amount of radioactive iodine-123, used to treat thyroid cancer, in a sample decreases from 3.2 to 0.4 mg in 39.6 h, what is the half-life of iodine-123?
Answers
Answer:
Half life = 13.197 hour
Explanation:
Given:
Old amount (A₀) = 3.2
New amount (A) = 0.4
Radiation decay time (t) = 39.6 hour
Half life = T(1/2)
Find:
Half life = T(1/2) = T
Computation:
A = A₀[tex]e^{-(\frac{0.693t}{T} )}[/tex]
[tex]e^{-(\frac{0.693t}{T} )}[/tex] = 0.4 / 3.2
-[27.4428 / T] = In (0.125)
-[27.4428 / T] = -2.0794
[27.4428 / T] = 2.0794
T = 13.197
Half life = 13.197 hour
According to the following reaction, how many grams of ammonia will be formed upon the complete reaction of 31.2 grams of hydrogen gas with excess nitrogen gas ? nitrogen(g) + hydrogen(g) ammonia(g)
Answers
Answer:
176.8 g of ammonia, NH3.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
N2 + 3H2 —> 2NH3
Next, we shall determine the mass of H2 that reacted and the mass of NH3 produced from the balanced equation. This is illustrated below:
Molar mass of H2 = 2x1 = 2 g/mol
Mass of H2 from the balanced equation = 3 x 2 = 6 g
Molar mass of NH3 = 14 + (3x1) = 17 g/mol
Mass of NH3 from the balanced equation = 2 x 17 = 34 g.
From the balanced equation above,
6 g of H2 reacted to produce 34 g of NH3.
Finally, we shall determine the mass of ammonia, NH3 produced by reacting 31.2 g of H2.
This can be obtained as follow:
From the balanced equation above,
6 g of H2 reacted to produce 34 g of NH3.
Therefore, 31.2 g of H2 will react to produce = (31.2 x 34)/6 = 176.8 g of NH3.
Therefore, 176.8 g of ammonia, NH3 were obtained from the reaction.
When cyclopentane undergo free-radical substitution with bromine (Br2 /Heat) the product:
a. Bromo cyclopentene
b. Bromo cyclopentane + HBr
c. Bromo cyclopentene + HBr
Answers
Explanation:
b. Bromo cyclopentane + HBr
A 1.362 g sample of an iron ore that contained Fe3O4 was dissolved in acid with all of the iron being reduced to iron (II). The solution was acidified with sulfuric acid and titrated with 39.42 mL of 0.0281 M KMnO4, which oxidized the iron (II) to iron (III) while reducing the permanganate to manganese (II). Generate the balanced net ionic equation for the reaction. What is the mass percent of iron in this iron ore sample?
Answers
Answer:
a. MnO₄⁻ + 8H⁺ + 5Fe²⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O
b. 18.17% of Fe in the sample
Explanation:
a. In the reaction, Fe²⁺ is oxidized to Fe³⁺ and permanganate, MnO₄⁺ reduced to Mn²⁺, thus:
Fe²⁺ → Fe³⁺ + 1e⁻
MnO₄⁻ + 5e⁻ + 8H⁺ → Mn²⁺ + 4H₂O
5 times the iron and suming the manganese reaction:
MnO₄⁻ + 5e⁻ + 8H⁺ + 5Fe²⁺ → 5Fe³⁺ + 5e⁻ + Mn²⁺ + 4H₂O
MnO₄⁻ + 8H⁺ + 5Fe²⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂Ob. Moles of permanganate in the titration are:
0.03942L × (0.0281 moles / L) = 1.108x10⁻³ moles of MnO₄⁻
Based on the reaction, 1 mole of permanganate reacts with 5 moles of iron, if 1.108x10⁻³ moles of MnO₄⁻ reacts, moles of iron are:
1.108x10⁻³ moles of MnO₄⁻ × (5 moles Fe²⁺ / 1 mole MnO₄⁻) =
4.431x10⁻³ moles of Fe²⁺. Molar mass of Fe is 55.845g/mol. 4.431x10⁻³ moles of Fe²⁺ are:
4.431x10⁻³ moles of Fe²⁺ ₓ (55.845g / mol) =
0.2474g of Fe you have in your sample.Percent mass is:
0.2474g Fe / 1.362g sample ₓ 100 =
18.17% of Fe in the sampleThe mass percent of iron in the sample is 22.6%.
The net ionic equation of the reaction is;
5Fe^2+(aq) + 8H^+(aq) + MnO4^- -----> 5Fe^3+(aq) + Mn^2+(aq) + 4H2O(l)
Number of moles of MnO4^- = 39.42/1000 L × 0.0281 M = 0.0011 moles
If 5 moles of Fe^2+ reacts with 1 mole of MnO4^-
x moles of Fe^2+ reacts with 0.0011 moles
x = 5 moles × 0.0011 moles/1 mole
x = 0.0055 moles
Mass of Fe^2+ = 0.0055 moles × 56 g/mol = 0.308 g
Mass percent of iron = 0.308 g/ 1.362 g × 100/1
= 22.6%
Learn more: https://brainly.com/question/10643807
what is the bond energy required to break one mole of carbon-carbon bonds
Answers
Answer:
100 kcal of bond energy
4. Given that the enthalpy of reaction for a system at 298 K is -292 kJ/mol and the entropy for that system is 224 J/mol K, what's the free energy for the system?
A.-87,793 kJ
B.-358 kJ
C.-225 kJ
D. -66,751 kJ
Answers
Answer:
[tex]\Delta G=-359\frac{kJ}{mol}[/tex]
Explanation:
Hello,
In this case, we must remember that the Gibbs free energy is defined in terms of the enthalpy, temperature and entropy as shown below:
[tex]\Delta G=\Delta H -T\Delta S\\[/tex]
In such a way, for the given data, we obtain it, considering the conversion from J to kJ for the entropy in order to conserve the proper units:
[tex]\Delta G=-292\frac{kJ}{mol} -(298)(224\frac{J}{mol}*\frac{1kJ}{1000J} )\\\\\Delta G=-359\frac{kJ}{mol}[/tex]
Best regards.
Answer:
B- 358 kj
Explanation: I took the test
The substance formed on addition of water to an aldehyde or ketone is called a hydrate or a/an:_______
A) vicinal diol
B) geminal diol
C) acetal
D) ketal
Answers
Answer:
B) geminal diol
Explanation:
Hello,
In this case, considering the attached picture, you can see that the substance resulting from the hydrolysis of an aldehyde or a ketone is a geminal diol since the two hydroxyl groups are in the same carbon. Such hydrolysis could be carried out in either acidic or basic conditions depending upon the equilibrium constant.
Regards.
Which of the following is the balanced reaction, given the rate relationships below.
a. rate = − 13 Δ[CH4] Δt = − 12 Δ[H2O] Δt = − Δ[CO2] Δt = 14 Δ[CH3OH] Δt
b. rate = − 12 Δ[N2O5] Δt = 12 Δ[N2] Δt = 15 Δ[O2] Δt
c. rate = − 12 Δ[H2] Δt = − 12 Δ[CO2] Δt = − Δ[O2] Δt = 12 Δ[H2CO3] Δt
Answers
Answer:
a. [tex]3CH_4+2H_2O+CO_2\rightarrow 4CH_3OH[/tex]
b. [tex]2N_2O_5\rightarrow 2N_2 + 5O_2[/tex]
c. [tex]2H_2+2CO_2+O_2\rightarrow 2H_2CO_3[/tex]
Explanation:
Hello,
In this case, since those rate relationships have the stoichiometric coefficient at the denominators of the fractions preceding each ratio and the negative terms account for reactants and positive for products, we have:
a. [tex]3CH_4+2H_2O+CO_2\rightarrow 4CH_3OH[/tex]
b. [tex]2N_2O_5\rightarrow 2N_2 + 5O_2[/tex]
c. [tex]2H_2+2CO_2+O_2\rightarrow 2H_2CO_3[/tex]
Best regards.
How would you expect cesium, Cs, to react with water? explaining your reasoning.
Answers
Answer:
Caesium reacts rapidly with water to form a colourless basic solution of caesium hydroxide (CsOH) and hydrogen gas (H2).
Explanation:
The reaction continues even when the solution becomes basic. The resulting solution is basic because of the dissolved hydroxide. The reaction is exothermic.
Which of the following is a characteristic of a scientific practice?
Answers
If a salt is formed by combining NH3 (Kb=1.8×10−5) and CH3COOH (Ka=1.8×10−5), an aqueous solution of this salt would be:
Answers
Answer:
Neutral
Explanation:
pKa of acid = -log Ka
= -log (1.8 x 10^-5)
= 4.74
pKb of base = -log Kb
= 4.74
pKa of acid = pKb of base
salt pH formula : pH = 7 + 1/2 [pKa -pKb ]
here pKa = pKb
so pH = 7
the salt it is CH3COONH4 exactly neutral solution .
If a salt is formed by combining NH₃ (Kb=1.8×10⁻⁵) and CH₃COOH (Ka=1.8×10⁻⁵), an aqueous solution of this salt would be neutral.
What information does pH convey?pH of any solution tells about the acidity or basicity or neutral nature of the solution.
pH of any solution is directly proportional to the acid dissociation constant value (Ka) and base dissociation constant (Kb). In the question it is given that,
Value of Kb for NH₃ = 1.8×10⁻⁵
Value of Ka for CH₃COOH = 1.8×10⁻⁵
Ka & Kb values for the base and acid is same means it dissociates with same extent. So the aqueous solution of this acid and base is a neutral in nature as they have same number of acid and base ions in it.
Hence resultant solution will be a neutral solution .
To know more about neutral solution, visit the below link:
https://brainly.com/question/13805901
g Which ONE of the following pure substances will exhibit hydrogen bonding? A) methyl fluoride, FCH3 B) dimethyl ether, CH3C–O–CH3 C) formaldehyde, H2C=O D) trimethylamine, N(CH3)3 E) hydrazine, H2N-NH2
Answers
Answer:
C) formaldehyde, H2C=O.
Explanation:
Hello,
In this case, given that the hydrogen bondings are known as partial intermolecular interactions between a lone pair on an electron rich donor atom, particularly oxygen, and the antibonding molecular orbital of a bond between hydrogen and a more electronegative atom or group. Thus, among the options, C) formaldehyde, H2C=O, will exhibit hydrogen bonding since the lone pair of electrons of the oxygen at the carbonyl group, are able to interact with hydrogen (in the form of water).
Best regards.
what is the mass of cerrusite would contain 35g of lead?
Answers
Answer:
i believe its 42
Explanation:
What is the maximum number of electrons in the following energy level?
n = 4
Answers
Answer: 32 electrons
Explanation:
Name the alkyne. Spelling and punctuation count.
Answers
The name of a compound is determined by IUPAC nomenclature
The International Union of Pure and Applied Chemistry have designated a universal rule for the nomenclature of organic compounds generally known as IUPAC nomenclature.
The whole idea of the IUPAC nomenclature is to have a universally agreed pattern of naming organic compounds according to their structure.
The compound is always named in such a way that the substituents and the functional group receives the lowest number.
In this case, the functional group is alkyne hence the name of the compound ends in -yne. The methyl group is substituted on position four.
Putting all these together the compound should be named as; 4-methylhept-2-yne
For IUPAC nomenclature of an inorganic compound
see:
https://brainly.com/question/11587934
What must happen to uranium before it can be used as a fuel source?
Answers
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.
Fireworks are chemical reactions that release energy. Which of these phenomena are caused by chemical reactions that release energy? If you’re not sure, make a guess.
Answers
Answer:
All chemical reactions involve energy. Energy is used to break bonds in reactants, and energy is released when new bonds form in products. Endothermic reactions absorb energy, and exothermic reactions release energy. The law of conservation of energy states that matter cannot be created or destroyed.
Al diluir 25 g de sal de mesa en 250ml de agua, ¿En cuántos °C aumenta el punto de ebullición de la disolución formada? ( Ke = 0,52 °C/molal , PM NaCl = 58,44 g/mol)
Answers
Answer:
ΔT=[tex]0.87^{\circ}C[/tex]
Explanation:
Para esta pregunta debemos recordar la ecuación que nos permite calcular el aumento ebulliscopico (aumento del punto de ebullición):
ΔT=[tex]Kb*m[/tex]
Donde ΔT es el valor del aumento del punto de ebullición. Kb es la constante ebulloscopica para el agua ([tex]0.512\frac{Kg~^{\circ}C}{mol}[/tex]) y m es la molalidad ([tex]m=\frac{mol}{Kg~ of~ solvente}[/tex]).
Por lo tanto el primer paso es calcular la molalidad de la solución. Para lo cual tendremos que calcular las moles de sal en los 25 g. Si queremos hacer esto debemos recordar que la formula de sal de mesa es NaCl y que la masa molar de NaCl es 58.44 g/mol. Por lo tanto:
[tex]25~g~NaCl\frac{1~mol~NaCl}{58.44~g~NaCl}~=~0.42~mol~NaCl[/tex]
Ahora bien, también debemos saber los Kg de agua en la solución. Por lo que podemos usar la densidad del agua (1 g/mL) para convertir de mL a g y luego hacer la conversión a Kg:
[tex]250~mL\frac{1~g}{1~mL}\frac{1~Kg}{1000~g}~=~0.25~Kg[/tex]
Finalmente para el calculo de la molalidad podemos dividir los dos valores:
[tex]m=\frac{0.42~mol}{0.25~Kg}=1.68[/tex]
Con el valor de la molalidad se puede calcular ΔT al reemplazar los valores:
ΔT=[tex]1.68~\frac{mol}{Kg}*0.52\frac{Kg~^{\circ}C}{mol}=0.87^{\circ}C[/tex]
Si la temperatura de ebullición normal del agua es 100 ºC. Podemos calcular la temperatura final si adicionamos ΔT:
Temperatura final = 100 + 0.87 = 100.87 ºC
Espero sea de ayuda!
Leo carefully pipets 50.0 mL of 0.500 M NaOH into a test tube. She places the test tube
into a small beaker to keep it from spilling and then pipets 75.0 mL of 0.250 M HCl into
another test tube. When Leo reaches to put this test tube of acid into the beaker along
with test tube of base she accidentally knocks the test tubes together hard enough to
break them and their respective contents combine in the bottom of the beaker. Is the
solution formed from the contents of the two test tubes acidic or basic? What is the pH of
the resulting solution?
Please answer below questions one by one to assist you receive full credits
(Alternatively, you can discard my hints below, solve the problem using your own way
and send me the picture/copy of your complete work through email)
The mole of NaOH before mixing is
mol (save 3 significant figures)
The mole of HCl before mixing is
mol (save 4 significant figures)
After mixing, the solution is
(choose from acidic or basic)
The total volume of mixture is
L (save 3 significant figures)
The concentration of [OH-] is
M (save 3 significant figures)
The concentration of [H'l is
M (save 3 significant figures)
Answers
Let's consider the neutralization reaction between HCl and NaOH.
NaOH + HCl ⇒ NaCl + H₂O
To determine the pH of the resulting mixture, we need to determine the reactant in excess. First, we will calculate the reacting moles of each reactant.
NaOH: 0.0500 L × 0.500 mol/L = 0.0200 mol
HCl: 0.0750 L × 0.250 mol/L = 0.0188 mol
Now, let's determine the reactant in excess and the remaining moles of that reactant.
NaOH + HCl ⇒ NaCl + H₂O
Initial 0.0200 0.0188
Reaction -0.0188 -0.0188
Final 1.20 × 10⁻³ 0
The volume of the mixture is 50.0 mL + 75.0 mL = 125.0 mL. Then, 1.20 × 10⁻³ moles of NaOH are in 125.0 mL of solution. The concentration of NaOH is:
[NaOH] = 1.20 × 10⁻³ mol/0.1250 L = 9.60 × 10⁻³ M
NaOH is a strong base according to the following equation.
NaOH ⇒ Na⁺ + OH⁻
The concentration of OH⁻ is 1/1 × 9.60 × 10⁻³ M = 9.60 × 10⁻³ M.
The pOH is:
pOH = -log [OH⁻] = -log 9.60 × 10⁻³ = 2.02
We will calculate the pH using the following expression.
pH = 14.00 - pOH = 14.00 - 2.02 = 11.98
The pH is 11.98. Since pH > 7, the solution is basic.
You can learn more about neutralization here: https://brainly.com/question/16255996
What is the molecular mass of this substance? a. 31.02 u b. 63.02 u c. 47.02 u d. 126.04 u e. 110.01 u
Answers
Answer:
Molecular mass of HNO₃ = 63.015 g/mol
Explanation:
Note: The given question is incomplete , the given substance is nitric acid
Given:
Nitric acid (HNO₃)
Find:
Molecular mass
Computation:
Molecular mass of HNO₃ = 1.008 + 14.007 + 3(16)
Molecular mass of HNO₃ = 1.008 + 14.007 + 48
Molecular mass of HNO₃ = 63.015 g/mol
Provide two real world examples of habitats and niches within ecosystems? Explain the interdependence that exists within those ecosystems.
Answers
Answer:
An ecosystem encompasses living organisms and the nonliving elements of their environments. Hence, the components of an ecosystem include animals, plants, microorganisms, rocks, soil, minerals, atmosphere, and the surrounding water masses. An ecosystem can be huge, cutting across several nations, or it can be relatively small, such as the body of an animal, which is home to numerous microorganisms.
We can describe ecosystems under two headings – natural ecosystems and unnatural ecosystems. Unnatural ecosystems, which include agricultural and urban areas, are greatly modified and maintained by human activity. Conversely, natural ecosystems are self-sufficient, balanced ecological units, with a high proportion of native biodiversity and minimal human disruption.
The natural ecosystem is broad. It is divided into two major groups – terrestrial and aquatic ecosystems. These are further divided into many other smaller types of ecosystems as outlined in this article.
Explanation:
Hope it helps.. if yes, plz mark me as brainliest