Answer:
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Explanation:
Answer:
a. differences in boiling points, fractional distillations,water,ethanol and methanol, 100 ,78,56, no
Explanation:
alcohol chemistry
Answer:
In the reaction 3H2 + N2 → ____ 2 NH3
D) no coefficient is needed
as it is balanced already.
Answer:
D
Explanation:
Answer:15 gl
Explanation:
Specific volume:
Explanation:
We can solve this problem by applying the equation of state for ideal gases, which states that:
where
p is the gas pressure
V is the gas volume
n is the number of moles
R is the gas constant
T is the absolute temperature
The equation can be re-arranged as
Here we want to find the specific volume, which is the volume of gas per number of moles, so we have to divide by n:
For the gas in this problem, we have:
is the pressure
, converting to Kelvin:
is the gas constant
Solving for Vs,
/mol
And since
1 feet = 0.3048 m
Then the specific volume in SI units is
/mol
Learn more about ideal gases:
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The conjugate bases of the following acids are as follows:
CONJUGATE BASE:
Learn more at: brainly.com/question/12883745?referrer=searchResults
Answer:
Answers are in the examples of Conjugate Base.
Explanation:
Conjugate Acid is a specie which is formed when a base abstract or picks a proton (H⁺).
Example:
NH₃ + HCl → NH₄⁺ + Cl⁻
In this example Ammonia (NH₃) is a base which after reacting with acid forms the conjugate acid (NH₄⁺).
Conjugate Base is a specie which is formed when an acid looses / donates its proton (H⁺) to a base.
Example:
(a)
B + HNO₃ → BH⁺ + NO₃⁻
In this example HNO₃ (Nitric Acid) is the acid which on reaction with base forms the conjugate base ( NO₃⁻) called nitrate.
(b)
B + H₂SO₄ → BH⁺ + HSO₄⁻
B + HSO₄⁻ → BH⁺ + SO₄⁻
In this example H₂SO₄ (Sulfuric Acid) is the acid which on reaction with base forms the conjugate bases ( HSO₄⁻ and SO₄⁻) respectively.
(c)
B + H₂S → BH⁺ + HS⁻
In this example H₂S (Hydrogen Sulfide) is the acid which on reaction with base forms the conjugate base HS⁻.
(c)
B + HCOOH → BH⁺ + HCOO⁻
In this example HCOOH (Formic Acid) is the acid which on reaction with base forms the conjugate base HCOO⁻ called formate ion.
The melting range results from the fact that different components of the impure material melt at different temperatures. Thermodynamically, the material minimizes its Gibbs free energy (ie maximizes its entropy) during this process, which determines the phase mixtures with the lowest energy states.
During the melting point test, when the first crystals of the compound begin to melt, mark this as the lower limit (low end point). When all the crystals have melted, mark this as the upper limit (high point).
Even a small amount of impurities lowers the melting point of a compound by a few degrees and extends the temperature range of the melting point. Because the impurity causes defects in the crystal lattice, it is easier to overcome intermolecular interactions.
Melting point is the temperature at which a certain solid changes from a solid state to a liquid or melts. The melting point of a substance is the temperature at which it changes its state from solid to liquid. At the melting point, the solid and liquid phases are in equilibrium. The melting point of a substance depends on pressure and is usually defined at a constant pressure, such as 1 atmosphere or 100 kPa. Molecular composition, attraction and impurities can all affect the melting point of substances.
To learn more about melting point, refer;
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Answer:
a pure substance has a specific melting point but a non-pure substance has a its melting point over a range
Explanation:
test of purity
Answer:
You should dilute to 414.1 mL
Explanation:
Let's apply the dilution formula to solve this:
M conc . Vol conc = M dil . Vol dil
Molarity for concentrated solution → 4.50 M
Volume for concentrated solution → 60 mL
Molarity for diluted solution → Unknown
Volume for diluted solution → Answer
Let's determine the molarity for diluted solution with data given.
We convert the mass of KI to moles
3.25 g . 1 mol/166 g = 0.0195 moles
Let's convert the volume from mL to L
30 mL . 1 L/1000mL = 0.030 L
Molarity (mol/L) = 0.0195 mol /0.030L = 0.652 M
We replace the data in the formula
4.50 M . 60 mL = 0.652 M . Volume for diluted solution
(4.50M . 60mL) / 0.652M = Volume for diluted solution
414.1 mL = Volume for diluted solution
Answer: 413mL
Explanation:
Molar Mass of KI = 39 + 127 = 166g/mol
Mass conc of KI = 3.25g
Number of mole = Mass conc/Molar Mass
Number of mole = 3.25/166 = 0.0196 mol
But recall
Molarity = number of mole / Volume
Number of mole = 0.0196 mol
Volume = 30mL = 30/1000 = 0.03L
Molarity = 0.0196/0.03
Molarity = 0.653M
C1 = 4.5M
V1 = 60mL = 60/1000 = 0.06L
C2 = 0.653M
V2 =?
C1V1 = C2V2
4.5 x 0.06 = 0.653 x V2
V2 = (4.5 x 0.06) / 0.653
V2 = 0.413L
Converting to mL
0.413 x 1000 = 413mL
Answer: (a) 0.1M
(b) pH = 1
Explanation:
Moles = Molarity × Volume me(L)
The initial number of moles in the flask = 0.5M × 0.010L = 0.005moles
After the addition of 40 mL of in the flask again, the total volume in the flask = 10 mL + 40 mL = 50 mL
Concentration = Moles/Volume(L)[H3O+]
=
0.005moles/0.050L = 0.1M
pH = −log[H3O+] = −log(0.1) = 1
Answer:
The approximate H3O+ concentration before the titration begins is 0.1 M
pH of the solution in the flask is 1
Explanation:
Initial concentration of HCl = 0.5 mol/L
Initial number of mols in Erlenmeyer flask = (0.5 mol/L)(0.010 L) = 0.005 mol
After addition of 40 mL of volume to the flask again, the total volume in the flask = 50 mL
Therefore - the resultant [H3O+] concetration in the solution = 0.005 mol / 0.050 L = 0.1 M (before titration begins)
Hence - pH = -log[H3O+] = -log (0.1) = 1
Answer:
The density of the crystal is 10.22 g/cm³
Explanation:
Step 1: Data given
Molybdenum (Mo) has a BCC crystal structure
atomic radius = 0.1363 nm
atomic weight of 95.94 g/mol
Step 2: Calculate density of a bcc crystal
Density ρ= (nA *Mo)/(Vc*Na)
⇒For BCC, n = 2 atoms/unit cell, and realizing that V c = a ³
a = 4R/√3
Vc =(4R/√3)³
AMo = atomic weight = 95.94 g/mol
ρ = (n*A Mo)/((4R/√3)³ *Na)
⇒with n = 2 atoms/unit cell
⇒ with AMo = 95.94 g/mol
⇒ with Vc =(4R/√3)³ = (4*0.1363*10-7 cm)³ /(√3)³
⇒with Na = 6.022*10^23 atoms/mol
ρ =10.22 g/cm³
The density of the crystal is 10.22 g/cm³
Answer:
The density of the molybdenum BCC crystal structure is 10.22 g/cm^3
Explanation:
Density (D) of BCC crystal structure is given as:
D = (n × AW) ÷ (12.32r^3 × Na)
n is the number of atoms of molybdenum per unit cell = 2
AW is the atomic weight of molybdenum = 95.94 g/mol
r is the atomic radius of molybdenum = 0.1363 nm = 0.1363×10^-9 m = 0.1363×10^-9 × 100 = 1.363×10^-8 cm
Na is Avogadro's number = 6.022×10^23 atoms/mol
D = (2×95.94) ÷ [12.32 × (1.363×10^-8)^3 × 6.022×10^23] = 191.88 ÷ 18.78 = 10.22 g/cm^3
Answer:
Composition of Titanium in weight percent is 89.97 wt%
Composition of Aluminum in weight percent is 6.03 wt%
Composition of Vanadium in weight percent is 4.00 wt%
Explanation:
To solve this problem, we use the following formula
where;
is composition of Titanium in percentage
is mass of titanium
is mass of aluminum
is mass of vanadium
Step (1) : Composition of Titanium in weight percent
= (100*218)/(218+14.6+9.7)
= 89.97 wt%
Step (2) : Composition of Aluminum in weight percent
= (100*14.6)/(218+14.6+9.7)
= 6.03 wt%
Step (2) : Composition of Vanadium in weight percent
= (100*9.7)/(218+14.6+9.7)
= 4.00 wt%
Answer:
% titanium = 89.97 %
% aluminium = 6.03 %
% vanadium = 4.00 %
Explanation:
Step 1: Data given
An alloy contains 218.0 kg titanium, 14.6 kg of aluminium and 9.7 kg of vanadium
Step 2: Calculate total mass
Total mass = mass titanium + mass aluminium + lass vanadium
Total mass = 218.0 + 14.6 + 9.7
Total mass = 242.3 kg
Step 3: Calculate weight percent
Weight percent = (mass / total mass) * 100%
% titanium = (218.0 kg / 242.3 kg)*100%
% titanium = 89.97 %
% aluminium = (14.6 kg / 242.3 kg) * 100 %
% aluminium = 6.03 %
% vanadium = (9.7 kg/ 242.3 kg) *100 %
% vanadium = 4.00 %