Answer:
The correct answer is - 1.02 V
Explanation:
From the reduction-oxidation reaction:
Ni²⁺(aq) + 2 Fe²⁺(aq) → Ni(s) + 2 Fe³⁺(aq)
Ni²⁺ is reduced to Ni(s) while Fe²⁺ is oxidized to Fe³⁺. Thus, the half reactions are:
Reduction (cathode) : Ni²⁺(aq) + 2 e‑ → Ni(s) Eº= ‑0.25 V
Oxidation (anode) : 2 x (Fe²⁺ → Fe³⁺ + e-)(aq) Eº= -0.77 V
-------------------------------------
Ni²⁺(aq) + 2 Fe²⁺(aq) → Ni(s) + 2 Fe³⁺(aq)
In order to calculate the Eºcell, we have to add the reduction potential of the reaction in cathode (reduction) to the oxidation potential of the anode (oxidation):
Eºcell= Eºr + Eºo= (-0.25 V) + (-0.77 V) = - 1.02 V
There are 454 grams in one pound. How many pounds are in 700 grams
Answer:
1.543 pounds = 700 grams
How are Math, Physics, Chemistry, and Biology all related?
Answer:
- you have to do maths in all 3
- atoms make up everything even parts of a cel and theyre studied in chem and physics
- chemistry is used in biology by finding out what different substances are eg cytoplasm in a cell
a boy capable of swimming 2.1m/a in still water is swimming in a river with a 1.8 m/a current. At what angle must he swim in order to end up directly opposite his starting point?
Answer:
The boy must swim at an angle of 59°northwest to get to a position directly opposite his starting point.
Explanation:
To get to a point directly opposite his starting point, the boy must travel at an angle x, in a direction northwest of his starting point. The speed of the boy and the speed of the river current forms a right-angled triangle with an an opposite side of 1.8 m/a and a hypotenuse of 2.1 m/a having an angle x.
Sin x = opp/ hyp
Sin x = 1.8/2.1
x = sin⁻¹ (1.8/2.10
x = 58.99
x = 59°
Therefore, the boy must swim at an angle of 59° in the northwesterly direction to get to a position directly opposite his starting point.
Which of the following correctly summarizes the
relative composition of the lithosphere with
respect to inorganic and organic material?
A) inorganic >> organic
B) inorganic = organic
C) inorganic << organic
D) There is no organic matter in the lithosphere
Answer:
A
Explanation:
The lithosphere represents the layer of hardened/solid rock that makes up the hard part of the earth, including the brittle upper portion of the mantle and the crust. The lithosphere is broken into pieces that are referred to as plates. The pieces move to and away from each other in a process known as plate tectonics. The movement of plates accounts for the global locations of volcanoes, earthquakes, and mountain ranges.
The lithosphere is made of largely of inorganic materials known as silicates. The weathering of the solid rocks together with the interaction of living organisms gives rise to soil with an appreciable amount of organic materials.
The correct option is, therefore A.
0.25 L of aqueous solution contains 0.025g of HCLO4 (strong acid) what will be the Ph of the solution g
Answer:
The pH of the solution will be 3
Explanation:
The strength of acids is determined by their ability to dissociate into ions in aqueous solution. A strong acid is any compound capable of completely and irreversibly releasing protons or hydrogen ions, H⁺. That is, an acid is said to be strong if it is fully dissociated into hydrogen ions and anions in solution.
Being pH=- log [H⁺] or pH= - log [H₃O⁺] and being a strong acid, all the HClO₃ dissociates:
HClO₄ + H₂O → H₃O⁺ + ClO₄-
So: [HCLO₄]= [H₃O⁺]
The molar concentration is:
[tex]molar concentration=\frac{number of moles of solute}{volume solution}[/tex]
The molar mass of HClO₄ being 100 g / mole, then if 100 grams of the compound are present in 1 mole, 0.025 grams in how many moles are present?
[tex]moles of HClO_{4} =\frac{0.025 grams*1 mole}{100 grams}[/tex]
moles of HClO₄= 0.00025
Then:
[tex][HClO_{4}]=\frac{0.00025 moles}{0.25 L}[/tex]
[tex][HClO_{4}]=0.001 \frac{ moles}{ L}[/tex]
Being [HCLO₄]= [H₃O⁺]:
pH= - log 0.001
pH= 3
The pH of the solution will be 3
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.
A sample of ammonia gas was allowed to come to equilibrium at 400 K. 2NH3(g) <----> N2(g) 3H2(g) At equilibrium, it was found that the concentration of H2 was 0.0484 M, the concentration of N2 was 0.0161 M, and the concentration of NH3 was 0.295 M. What was the initial concentration of ammonia
Answer:
0.327 M
Explanation:
Step 1: Write the balanced equation
2 NH₃(g) ⇄ N₂(g) + 3H₂(g)
Step 2: Make an ICE chart
2 NH₃(g) ⇄ N₂(g) + 3 H₂(g)
I x 0 0
C -2y +y +3y
E x-2y y 3y
Step 3: Find the value of y
The concentration of N₂ at equilibrium is 0.0161 M. Then,
y = 0.0161
Step 4: Find the value of x
The concentration of NH₃ at equilibrium is 0.295 M. Then,
x-2y = 0.295
x-2(0.0161) = 0.295
x = 0.327
9
What might happen if acidic chemicals were emitted into
the air by factories? Choose the best answer.
A
The acid would destroy metallic elements in the air
B
The acid would be neutralized by bases within clouds
C
Acid rain might destroy ecosystems and farmland
D
Violent chemical reactions would take place within the
atmosphere
co search
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BI
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)
Calculate the pH of a buffer that is 0.13 M in lactic acid and 0.10 M in sodium lactate. Express your answer using two decimal places.
Answer:
pH of the buffer is 3.75
Explanation:
It is possible to find pH of a buffer using Hendersson-Hasselbalch equation:
pH = pKa + log [A⁻] / [HA]
Where [A⁻] is molar concentration of the conjugate base and [HA] concentration of the weak acid
In the lactic buffer, pKa = 3.86. Lactic acid is the weak acid and its conjugate base is tha lactate salt. H-H equation for this buffer is:
pH = 3.86 + log [Lactate] / [Lactic acid]
Replacing with the concentrations of the problem:
pH = 3.86 + log [0.10M] / [0.13M]
pH = 3.75
pH of the buffer is 3.75
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?
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%
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3. Identify the reagents you would use to convert 1-bromopentane into each of the following compounds: (a) Pentanoic acid (b) Hexanoic acid (c) Pentanoyl chloride (d) Hexanamide (e) Pentanamide (f) Ethyl hexanoate
Answer:
Explanation:
a )
CH₃CH₂CH₂CH₂CH₂Br + KOH ⇒ CH₃CH₂CH₂CH₂CH₂OH
CH₃CH₂CH₂CH₂CH₂OH + acidic potassium dichromate ⇒ CH₃CH₂CH₂CH₂COOH
b )
CH₃CH₂CH₂CH₂CH₂Br + KCN ⇒ CH₃CH₂CH₂CH₂CH₂CN Hydrolysis ⇒ CH₃CH₂CH₂CH₂CH₂COOH .
c )
CH₃CH₂CH₂CH₂CH₂Br + KOH ⇒ CH₃CH₂CH₂CH₂CH₂OH
CH₃CH₂CH₂CH₂CH₂OH + acidic potassium dichromate ⇒ CH₃CH₂CH₂CH₂COOH + SOCl₂ ( thionyl chloride ) ⇒ CH₃CH₂CH₂CH₂COCl
d )
CH₃CH₂CH₂CH₂CH₂Br + KCN ⇒ CH₃CH₂CH₂CH₂CH₂CN Hydrolysis ⇒ CH₃CH₂CH₂CH₂CH₂COOH + PCC ( NH₃ ) ⇒ CH₃CH₂CH₂CH₂CH₂CONH₂
e )
CH₃CH₂CH₂CH₂CH₂Br + KCN ⇒ CH₃CH₂CH₂CH₂CH₂CN Hydrolysis ⇒ CH₃CH₂CH₂CH₂CH₂COOH + C₂H₅OH ( Ethyl alcohol + H⁺ )⇒
CH₃CH₂CH₂CH₂CH₂COOC₂H₅ ( ethyl hexanoate )
Draw the major organic product that is expected when cyclopentanecarboxylic acid is treated with each of the following reagents:
a. NaOH
b. [H+]
Answer:
a. Sodium cyclopentanecarboxylate
b. No reaction
Explanation:
In this case, in the cyclopentanecarboxylic acid we have a carboxylic acid functional group. Therefore we have an "acid". The acids by definition have the ability to produce hydronium ions ([tex]H^+[/tex]).
With this in mind, for molecule a. we will have an acid-base reaction, because NaOH is a base. When we put together an acid and a base we will have as products a salt and water. In this case, the products are Sodium cyclopentanecarboxylate (the salt) and water.
For the second molecule, we have the hydronium ion ([tex]H^+[/tex]). This ion can not react with an acid. Because, the acid will produce the hydronium ion also, so a reaction between these compounds is not possible.
See figure 1
I hope it helps!
Chlorine monoxide and dichlorine dioxide are involved in the catalytic destruction of stratospheric ozone. They are related by the equation:
2ClO(g) ⇌ Cl2O2(g) for which Kc is 4.96×10^11 at 273 K.
For an equilibrium mixture in which [Cl2O2] is 6.00 x 10^-6M, what is [ClO]?
Answer:
[ClO] = 3.48×10¯⁹ M.
Explanation:
The following data were obtained from the question:
Equilibrium constant (Kc) = 4.96×10¹¹
Concentration of Cl2O2, [Cl2O2] = 6x10¯⁶ M.
Concentration of ClO, [ClO] =.?
The equation for the reaction is given below:
2ClO(g) ⇌ Cl2O2(g)
The equilibrium constant for a reaction is simply defined as the ratio of the concentration of product raised to their coefficient to the concentration of the reactant raised to their coefficient.
The equilibrium constant, Kc for the reaction is given by:
Kc = [Cl2O2] / [ClO]²
Thus, we can calculate the concentration of ClO, [ClO] as follow:
Kc = [Cl2O2] / [ClO]²
4.96×10¹¹ = 6x10¯⁶ / [ClO]²
Cross multiply
4.96×10¹¹ × [ClO]² = 6x10¯⁶
Divide both side by 4.96×10¹¹
[ClO]² = 6x10¯⁶ / 4.96×10¹¹
[ClO]² = 1.21×10¯¹⁷
Take the square root of both side
[ClO] = √ (1.21×10¯¹⁷)
[ClO] = 3.48×10¯⁹ M
Therefore, the concentration of ClO, [ClO] is 3.48×10¯⁹ M.
The tosylate of (2R,3S)-3-phenylbutan-2-ol undergoes an E2 elimination on treatment with sodium ethoxide. Draw the structure of the alkene that is produced.
Answer:
(R)-but-3-en-2-ylbenzene
Explanation:
In this reaction, we have a very strong base (sodium ethoxide). This base, will remove a hydrogen producing a double bond. We know that the reaction occurs through an E2 mechanism, therefore, the hydrogen that is removed must have an angle of 180º with respect to the leaving group (the "OH"). This is known as the anti-periplanar configuration.
The hydrogen that has this configuration is the one that placed with the dashed bond (red hydrogen). In such a way, that the base will remove this hydrogen, the "OH" will leave the molecule and a double bond will be formed between the methyl and the carbon that was previously attached to the "OH", producing the molecule (R) -but-3- en-2-ylbenzene.
See figure 1
I hope it helps!
The enthalpy change for a chemical reaction is: a. the temperature change b. the amount of heat given off or absorbed c. related to molar volume d. none of the above
Answer:
b. the amount of heat given off or absorbed
Explanation:
Hello,
In this case, we should take into account a formal definition of enthalpy change such as an energetic change that occurs in a system when matter is transformed by a given chemical reaction from reactants to products. Thus, such energetic change is macroscopically exhibited and it is related with either a temperature increase or decrease; it means that if a reaction exhibits a temperature increase, we say that heat was given off and if the temperature exhibits a decrease, we say that heat is absorbed. For that reason, answer is b. the amount of heat given off or absorbed.
Regards.
The complex ion Fe(CN)63- is paramagnetic with one unpaired electron. The complex ion Fe(SCN)63- has five unpaired electrons. Where does SCN- lie in the spectrochemical series with respect to CN-?
Answer:
SCN- is a weak field ligand while CN- is a strong field ligand
Explanation:
The spectrochemical series is an arrangement of ligands according to their magnitude of crystal field splitting. Ligands that cause only a small degree of crystal field splitting are called weak field ligands while ligands that cause large crystal field splitting are called strong field ligands.
Strong field ligands often lead to the formation of low spin complexes with the least number of unpaired electrons while high spin complexes are formed by weak field ligands.
CN- is a strong field ligand as it lies towards the right hand side of the spectrochemical series.
SCN- is a weak field ligand hence it forms a high spin complex having the maximum number of unpaired electrons for Fe^3+, hence the answer.
SCN⁻ lies in the lower (weak field) region of the spectrochemical series while CN⁻ lies in the higher (stronger field) region.
CN⁻ is a strong field ligand with a large splitting constant, and it is high up in the spectrochemical series.
Conversely, SCN⁻ is a weak field ligand with a low splitting constant, and it is lower in the spectrochemical series.
Hence, SCN⁻ lies in the lower (weak field) region of the spectrochemical series while CN⁻ lies in the higher (stronger field) region.
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Gas is contained in a 9.00-L vessel at a temperature of 24.0°C and a pressure of 5.00 atm. (a) Determine the number of moles of gas in the vessel. mol (b) How many molecules are in the vessel? molecules
Answer:
a. Moles in the vessel = 1.85 moles of the gas
b. 1.11x10²⁴ molecules are in the vessel
Explanation:
a.It is possible to determine moles of a gas using the general law of gases:
PV = nRT
Where P is pressure: 5.00atm; V is volume = 9.00L; R is gas constant: 0.082atmL/molK; T is absolute temperature: 273.15K +24.0 = 297.15K
Computing the values:
PV / RT = n
5.00atm* 9.00L / 0.082atmL/molK*297.15K = n
Moles in the vessel = 1.85 moles of the gasb. With Avogadro's number we can convert moles of any compound to molecules thus:
Avogadro's number = 6.022x10²³ molecules / mole
1.85moles ₓ (6.022x10²³ molecules / mole) =
1.11x10²⁴ molecules are in the vesselA baseball has a mass of 0.145 kilograms. If acceration due to gravity is 9.8m/s,what is the weight of the baseball in newtons?
Answer:
I hope it works
Explanation:
As we know that
w=m*g
given m=0.145 , g=9.8
hence we get
w= (9.8)*(0.145)
w=1.421 m/sec 2
if its help-full thank hit the stars and brain-list it thank you
Nitric oxide reacts with oxygen to give nitrogen dioxide, an important reaction in the Ostwald process for the industrial synthesis of nitric acid: 2NO(g)+O2(g)⇌2NO2(g)
Part A If Kc=6.9×105 at 227 ∘C,
what is the value of Kp at this temperature? Express your answer using two significant figures. Kp =
Part BIf Kp=1.3×10−2 at 1000 K, what is the value of Kc at 1000 K? Express your answer using two significant figures. Kc =
Answer:
Kp=1.68×10⁴∆1.7×10⁴
Kc=1.06∆1.1
Explanation:
Value of Kp at 227°C is 2.86×10² and value of Kc at 1000 K is 1.56.
How are Kp and Kc related?Kp and Kc are related by the formula Kp=Kc(RT).For part A , Kp is calculated as,
Kp=6.9×10⁵×8.314×500=28.683×10² and for part B Kc is calculated as,
Kc=1.3×10[tex]^-2[/tex]/(8.314×1000)=1.56
Kc and Kp are equilibrium constants of a mixture of ideal gases. Kp is equilibrium constant when concentrations at equilibrium are in atmospheric pressure and Kc is equilibrium constant when concentrations are in molarity. The relation is only valid for gaseous mixtures. The relation between these two parameters is obtained through ideal gas equation.
Kc and Kp of reaction change with temperature of reaction but remain unaffected by change in concentration , pressure and presence of catalyst.
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#SPJ2
3,3-dibromo-4-methylhex-1-yne
Explanation:
see the attachment. hope it will help you...An 80L capacity steel cylinder contains H2 at a pressure of 110 atm and 30 ° C, after extracting a certain amount of gas, the pressure is 80 atm at the same temperature. How many liters of hydrogen (measured under normal conditions) have been extracted?
Answer:
2200 L
Explanation:
Ideal gas law:
PV = nRT,
where P is absolute pressure,
V is volume,
n is number of moles,
R is universal gas constant,
and T is absolute temperature.
The initial number of moles is:
(110 atm) (80 L) = n (0.0821 L atm / K / mol) (30 + 273.15) K
n = 353.58 mol
After some gas is removed, the number of moles remaining is:
(80 atm) (80 L) = n (0.0821 L atm / K / mol) (30 + 273.15) K
n = 257.15 mol
The amount of gas removed is therefore:
n = 353.58 mol − 257.15 mol
n = 92.43 mol
At normal conditions, the volume of this gas is:
PV = nRT
(1 atm) V = (92.43 mol) (0.0821 L atm / K / mol) (273.15 K)
V = 2162.5 L
Rounded, the volume is approximately 2200 liters.
For a particular reaction at 235.8 °C, ΔG=−936.92 kJ/mol , and ΔS=513.79 J/(mol⋅K) . Calculate ΔG for this reaction at −9.9 °C.
Answer:
-138.9 kJ/mol
Explanation:
Step 1: Convert 235.8°C to the Kelvin scale
We will use the following expression.
K = °C + 273.15 = 235.8°C + 273.15 = 509.0 K
Step 2: Calculate the standard enthalpy of reaction (ΔH°)
We will use the following expression.
ΔG° = ΔH° - T.ΔS°
ΔH° = ΔG° / T.ΔS°
ΔH° = (-936.92kJ/mol) / 509.0K × 0.51379 kJ/mol.K
ΔH° = -3.583 kJ (for 1 mole of balanced reaction)
Step 3: Convert -9.9°C to the Kelvin scale
K = °C + 273.15 = -9.9°C + 273.15 = 263.3 K
Step 4: Calculate ΔG° at 263.3 K
ΔG° = ΔH° - T.ΔS°
ΔG° = -3.583 kJ/mol - 263.3 K × 0.51379 kJ/mol.K
ΔG° = -138.9 kJ/mol
A compound is found to contain 26.94 % nitrogen and 73.06 % 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 104.02 g/mol. What is the molecular formula for this compound?
Answer:
THE EMPIRICAL FORMULA OF THE COMPOUND IS NF2
THE MOLECULAR FORMULA OF THE COMPOUND IS N2F4
Explanation:
To calculate the empirical formula for the compound, we:
1. Write out the percentage weight of each elements
N = 26.94%
F = 73.06 %
2. Divide each by its atomic mass
( N= 14, F = 19)
N = 26.94 / 14 = 1.924
F = 73.06 / 19 = 3.845
3. Divide each by the smaller of the values
N = 1.924 / 1.924 = 1
F = 3.845 / 1.924 = 1,998
4. Round up to a whole number and write the empirical formula
N= 1
F = 2
So the empirical formula of the compound is N F2
To calculate the molecular formula, we:
(N F2 )n = molecular weight
( 14 + 19*2) n = 104.02
52 n = 104.02
n = 2.000
The molecular formula of the compound will be:
(N F2)2 = N2F4
In conclusion, the empirical formula of the compound is NF2 and the molecular formula of the compound is N2F4
CI
Which of the following statements is INCORRECT?
(1)
(2)
the compound contains a o molecular orbital formed by the overlap of one carbon
sp2 hybrid orbital and one hydrogen sp3 hybrid orbital
the compound contains a T molecular orbital formed by the overlap of two
unhybridized carbon p atomic orbitals
the compound contains a polar C-Cl bond
each carbon atom of the C=C bond is sp2 hybridized
(3)
(4)
Answer:
The compound contains a o molecular orbital formed by the overlap of one carbon sp2 hybrid orbital and one hydrogen sp3 hybrid orbital.
Explanation:
Molecular orbital is function which describes wave like behavior of an electron in a molecule. The molecular orbital theory describes the electronic structure of molecule using quantum mechanics. Electrons are not assigned to individual bonds between atoms. The compound contains sp2 hybrid orbial which is polar C - CI bond.
1.78 L of gas is at a pressure of 735 torr. What is the volume in liters (L) when the pressure decreases to 700.0 torr
Explanation:
using boyles law
p1v1=p2v2
735 x 1.76 = 700 x V2
1293.6 = 700 x V2
V2 = 1293.6/700
V2 = 1.85L
If 2.9g of water is heated from 23.9C to 98.9C, how much heat (in calories) was added to the water?
Answer:
Explanation:
we know that
ΔH=m C ΔT
where ΔH is the change in enthalpy (j)
m is the mass of the given substance which is water in this case
ΔT IS the change in temperature and c is the specific heat constant
we know that given mass=2.9 g
ΔT=T2-T1 =98.9 °C-23.9°C=75°C
specific heat constant for water is 4.18 j/g°C
therefore ΔH=2.9 g*4.18 j/g°C*75°C
ΔH=909.15 j
Which of the following do we need to know in order to calculate pH during an acid-base titration of a strong monoprotic acid with a strong monoprotic base? Select all that apply
a. the concentration of the acid
b. the concentration of the base titrant
c. the initial volume of the acid solution
d. the volume of the titrant used
Answer:
the volume of the titrant used
Explanation:
Acid-base titrations are usually depicted on special graphs referred to as titration curve. A titration curve is a graph that contains a plot of the volume of the titrant as the independent variable and the pH of the system as the dependent variable.
Hence, a titration curve is a graphical plot showing the pH of the analyte solution plotted against the volume of the titrant as the reaction is in progress. The titration curve is drawn by plotting data obtained during a titration, that is, volume of the titrant added (plotted on the x-axis) and pH of the system (plotted on the y-axis).
Rectangular cube 3.2 m length 1.2 m in height and 5 m in length is split into two parts. The container has a movable airtight divider that divides its length as necessary. Part A has 58 moles of gas and part B has 165 moles of a gas.
Required:
At what length will the divider to equilibrium?
Answer:
The length the divider is to equilibrium from Part A = 1.30 m and from Part B = 3.70 m
Explanation:
Given that:
A rectangular cube with 3.2 m breadth, 1.2 m height and 5 m in length is splitted into two parts.
The diagrammatic expression for the above statement can be found in the attached diagram below.
The container has a movable airtight divider that divides its length as necessary.
Part A has 58 moles of gas
Part B has 165 moles of a gas.
Thus, the movable airtight divider will stop at a length where the pressure on it is equal on both sides.
i.e
[tex]\mathtt{P = P_A = P_B}[/tex]
Using the ideal gas equation,
PV = nRT
where, P,R,and T are constant.
Then :
[tex]\mathsf{\dfrac{V_A}{n_A}= \dfrac{V_B}{n_B}}[/tex]
[tex]\mathsf{\dfrac{L_A \times B \times H}{n_A}= \dfrac{L_B \times B \times H}{n_B}}[/tex] --- (1)
since Volume of a cube = L × B × H
From the question; the L = 5m
i,e
[tex]\mathsf{L_A +L_B}[/tex] = 5
[tex]\mathsf{L_A = 5 - L_B}[/tex]
From equation (1) , we divide both sides by (B × H)
Then :
[tex]\mathsf{\dfrac{L_A }{n_A}= \dfrac{L_B }{n_B}}[/tex]
[tex]\mathsf{\dfrac{5-L_B}{58}= \dfrac{L_B }{165}}[/tex]
By cross multiplying; we have:
165 ( 5 - [tex]\mathsf{L_B}[/tex] ) = 58 (
825 - 165[tex]\mathsf{L_B}[/tex] = 58
825 = 165[tex]\mathsf{L_B}[/tex] +58
825 = 223[tex]\mathsf{L_B}[/tex]
[tex]\mathsf{L_B}[/tex] = 825/223
[tex]\mathsf{L_B}[/tex] = 3.70 m
[tex]\mathsf{L_A = 5 - L_B}[/tex]
[tex]\mathsf{L_A = 5 - 3.70}[/tex]
[tex]\mathsf{ L_A}[/tex] = 1.30 m
The length the divider is to equilibrium from Part A = 1.30 m and from Part B = 3.70 m
Which of the following represents six molecules of water? 6HO 2 6H 2O 1 6H 2O H 6O
Answer:
6H20 represents six molecules of water
Answer:
6H20 represents six molecules of water
Explanation:
