Answer:
Molar volume, or volume of one mole of gas , depends on pressure and temperature, and is 22.4 liters - at 0 °C (273.15 K) and 1 atm (101325 Pa), or STP (Standard Temperature and Pressure), for every gas which behaves similarly to an ideal gas. The ideal gas molar volume increases to 24.0 liters as the temperature increases to 20 °C (at 1 atm).
What mass of precipitate (in g) is formed when 250.0 mL of 0.150 M CuCl₂ is mixed with excess KOH in the following chemical reaction?
CuCl₂(aq) + 2 KOH(aq) → Cu(OH)₂(s) + 2 KCl(aq)
Answer:
3.6487g
CuCl2 moles reacted = (0.15×250)/1000
according to balanced chemical equation
precipitated Cu(OH)2 moles = Reacted CuCl2 moles
molar mass of Cu(OH)2 = 63.5+ (17+1)×2 = 97.5
mass of precipitate = (97.5 × 0.15×250)/1000
= 3.648g
……….is strong due to the ……………..between positive ions and negative delocalized electrons
Answer:
atom &bond
Explanation:
atom is strong due to the bond
How is the compound NH3 classified?
A. As a salt
B. As a base
C. As an acid
D. As ionic
Answer:
B
Explanation:
Ammonia is considered a base as it's pH is 11
Answer from Gauthmath
The compound NH3 (Ammonia) can be classified as a weak Base. Below you can learn more about Ammonia.
What is Ammonia (NH3)?Ammonia is a chemical compound which is derived from the combination of Nitrogen and Hydrogen. It is denoted by the chemical formula NH3.
Ammonia is a base and when it reacts with acids to gives out salts. Physically, It is a colorless gas with a distinct characteristic of a pungent smell.
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An antacid tablet weighing 1.30g was fully neutralized at 42.00 mL(an excess amount) of 0.250MHCl. 10.00 mL of 0.100 M NaOH was then used to back titrate the excess HCl. How many moles of acid did the antacid neutralize
Answer:
0.0095 moles of acid were neutralized by the antiacid
Explanation:
The antiacid is a base that neutralize the acid in stomach. To find the moles of acid neutralized we need to find the moles of acid added initially. This acid is added in excess, then, the moles of NaOH added reacts to neutralize the moles of acid in excess. The difference between initial moles of HCl and moles of NaOH needed to titrate the excess = Moles of HCl that were neturalized by the antiacid as follows:
Moles HCl added:
42.00mL = 0.04200L * (0.250mol/L) = 0.0105 moles HCl
Moles NaOH to titrate the excess:
10.00mL = 0.01000L * (0.10mol/L) = 0.0010 moles NaOH = Moles HCl in excess.
Moles of acid that were neutralized:
0.0105 moles - 0.0010 moles =
0.0095 moles of acid were neutralized by the antiacidplease see attachment
Answer:
I'll see it
Explanation:
9. Consider a magnesium atom with charge +2. How many overall electrons are on this particle?
Hint: Magnesium's atomic number is 12.
10
12
14
The sample concentration was measured at 50mg/ml. The loading concentration needs to be 10mg/ml. The final volume needs to be 25ul. What is the volume of sample needed and the amount of buffer needed to reach 25ul
Answer:
a) [tex]V_1=5ul[/tex]
b) [tex]v=20ul[/tex]
Explanation:
From the question we are told that:
initial Concentration [tex]C_1=50mg/ml[/tex]
Final Concentration [tex]C_2=10mg/ml[/tex]
Final volume needs [tex]V_2 =25ul[/tex]
Generally the equation for Volume is mathematically given by
[tex]C_1V_1=C_2V_2[/tex]
[tex]V_1=\frac{C_1V_1}{C_2}[/tex]
[tex]V_1=\frac{10*25}{50}[/tex]
[tex]V_1=5ul[/tex]
Therefore
The volume of buffer needed is
[tex]v=V_2-V_1\\\\v=25-5[/tex]
[tex]v=20ul[/tex]
Determine whether each of the examples represents a colligative property or a non-colligative property. boiling point elevation Choose... color Choose... freezing point depression Choose... vapor pressure lowering Choose... density Choose...
Answer:
boiling point elevation - colligative property
color - non-colligative property
freezing point depression - colligative property
vapor pressure lowering - colligative property
density - non-colligative property
Explanation:
A colligative property is a property that depends on the number of particles present in the system.
Freezing point depression, boiling point elevation and vapour pressure lowering are all colligative properties of solutions.
Colour and density do not depend on the number of particles present hence they are not colligative properties.
The boiling point elevation, freezing point depression, and vapor pressure lowering are colligative properties. And color and density are non-colligative properties.
Explanation:
The colligative properties are the properties depending upon the number of particles of solute not on the nature of the solute.Example of colligative properties:Vapor pressure loweringElevation boiling pointDepression in freezing pointOsmotic pressureThe non-colligative properties are the properties depending upon the nature of solute and solvent.Example of non-colligative properties :ViscositySurface tensionDensitySolubilitySo, from this, we can conclude that boiling point elevation, freezing point depression, and vapor pressure lowering are colligative properties. And color and density are non-colligative properties.
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What is the hydrogen atoms in 39.6g of ammonium sulphate,NH4 2SO4
What Volume of silver metal will weigh exactly 2500.0g. The density of silver
Answer:
cm3 = 2500.0 g / 10.5 g/cm3 = 238 cm3
How many moles of Al are needed to react exactly with 10.00 moles of Fe2O3 according to the following
equation?
Fe2O3 + 2 Al → Al2O3 + 2Fe
A) 15.0 moles
1
B) 20.0 moles
C) 30.0 moles
D) 60.0 moles
E) 35.0 moles
Answer:
Answer is B) 20.0 moles
Explanation:
From the equation,
1 mole of Fe2O3 = 2 moles of Al
therefore 10.0 moles of Fe2O3 = 10×2
= 20.0 moles.
A 18.0 L gas cylinder is filled with 6.20 moles of gas. The tank is stored at 33 ∘C . What is the pressure in the tank?
Express your answer to three significant figures and include the appropriate units.
Answer:
8.65 atm
Explanation:
Using ideal law equation;
PV = nRT
Where;
P = pressure (atm)
V = volume (L)
n = number of moles (mol)
R = gas law constant (Latm/molK)
T = temperature (K)
According to the information given in this question;
V = 18.0 L
n = 6.20 moles
R = 0.0821 Latm/molK
T = 33°C = 33 + 273 = 306K
P = ?
Using PV = nRT
P × 18 = 6.20 × 0.0821 × 306
18P = 155.76
P = 155.76/18
P = 8.65 atm
what is the IUPAC name of 2NaOH(s)
Answer:
NaoH= sodium hydroxide
What is Bose Einstein state of matter
Given its formula and Avogadro's Number (6.02 x 10^23 molecules/mol), deduce how many molecules are present in 3 x 10^-16 grams of TCDD. Type in only a number without using scientific notation.
Answer:
5 × 10⁵ molecules (500,000 molecules)
Explanation:
Step 1: Convert 3 × 10⁻¹⁶ g to moles
We will use the molar mass of TCDD (321.97 g/mol).
3 × 10⁻¹⁶ g × 1 mol/321.97 g = 9 × 10⁻¹⁹ mol
Step 2: Convert 9 × 10⁻¹⁹ mol to molecules
The required conversion factor is Avogadro's number (6.02 × 10²³ molecules/mol).
9 × 10⁻¹⁹ mol × 6.02 × 10²³ molecules/1 mol = 5 × 10⁵ molecules
By process of incineration, a mystery substance is empirically determined to contain 40.00% carbon by weight, 6.67% hydrogen, and 53.33% oxygen. Its molecular weight ranges between 55 and 62 g/mole. a. (6 points) Determine the chemical formula of this substance
Answer:
C₂H₄O₂
Explanation:
Step 1: Divide each percentage by the atomic mass of the element
C: 40.00/12.01 = 3.331
H: 6.67/1.01 = 6.60
O: 53.33/16.00 = 3.333
Step 2: Divide all the numbers by the smallest one
C: 3.331/3.331 = 1
H: 6.60/3.331 ≈ 2
O: 3.333/3.331 ≈ 1
The empirical formula is CH₂O, with a molecular weight of 12 g/mol + 2 × 1 g/mol + 16 g/mol = 30 g/mol. The molecular weight of the compound must be a product of 30, such as 60 (between 55 and 62 g/mol). Since we have to multiply by 2 (30 to 60) to get to the molecular weight of the compound, we also have to multiply the empirical formula by 2 to get the chemical formula of the compound.
CH₂O × 2 = C₂H₄O₂
20. An oxide of osmium (symbol Os) is a pale yellow solid. If 2.89 g of the compound contains 2.16 g of osmium, what is its empirical formula?
The empirical formula is OsO₄ :
Explanation:
Osmium oxide contains osmium and oxygen only.
Thus, we shall determine the mass of oxygen in osmium oxide. This can be obtained as follow:
Mass of compound = 2.89 g
Mass of Os = 2.16 g
Mass of O =?Mass of O = (Mass of compound) – (Mass of Os)
Mass of O = 2.89 – 2.16
Mass of O = 0.73 g
Finally, we shall determine the empirical formula of the compound. This can be obtained as follow:
Mass of Os = 2.16 g
Mass of O = 0.73 g
Empirical formula =..?Os = 2.16 g
O = 0.73 g
Divide by their molar mass of
Os = 2.16 / 190 = 0.011
O = 0.73 / 16 = 0.046
Divide by the smallest
Os = 0.011 / 0.011 = 1
O = 0.046 / 0.011 = 4
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State whether the error introduced by each of the following problems would result in a high or a low value for the Cu recovery or would not affect the results. Explain. a. Some of the copper nitrate solution splashes out of the beaker in step 1. _____________
Answer:
Low value for copper recovery
Explanation:
The percentage recovery is obtained from;
Percent recovery = amount of substance you actually collected / amount of substance you were supposed to collect × 100
Note that the fact that some of the copper nitrate solution splashed out of the beaker means that some amount copper has been lost from the system. This loss of copper leads to a lower value of copper recovered from solution.
Determine whether the statement about identifying a halide is true: Regardless of any concentration of ammonium solution, the precipitates in the reaction solution of my unknown halide after 0.1M AgNO3 remain because my unknown halide solution contains Br. Select one: True False
Answer:
False
Explanation:
The statement ; Regardless of any concentration of ammonium solution the precipitate of unknown halide after 0.1M AgNO3 will remain is FALSE
This is Because the remaining concentration of AgNO3 is dependent on the solubility of Ag⁺
Based on the standard EMF series and your knowledge of half-reactions, determine the cell potential and spontanei ty of a cell that consists of a pure cobalt electrode in a solution of Co^2+ ions; the other half is a lead electrode immersed in a Pb^2+ solution.
Pb +2e- Pb Sn +2e Sn Ni 2e Ni Co 2e -0.126 -0.136 -0.250 -0.277 Co
a. +0.403, spontaneous
b. -0.403, nonspontaneous
c. +0.151, spontaneous
d. -0.151, nonspontaneous
Answer:
+0.151, spontaneous
Explanation:
Given that;
Co^2+(aq) + 2e ---->Co(s) -0.28 V
Pb^2+(aq) + 2e ---->Pb(s). -0.13 V
Hence Co is the anode and Pb is the cathode
E°cell = E°cathode - E°anode
So;
E°cell = -0.13 V - (-0.28 V)
E°cell = 0.15 V
The cell reaction is spontaneous since E°cell is positive.
Suppose you are using distillation to separate cyclohexane and toluene. The boiling point of cyclohexane is Choose... oC and the boiling point of toluene is Choose... oC. Therefore, the liquid collected first should be Choose... .
Answer:
81°C
111°C
cyclohexane
Explanation:
Distillation is a process of separating two liquids based on differences in Bolling point. For two substances having different boiling points, they are collected as they are converted into vapour, condensed and move down the condenser one after the other.
Since the boiling point of cyclohexane is less than that of toluene, cyclohexane is collected first before toluene.
A 70.0‑g piece of metal at 80.0 °C is placed in 100 g of water at 22.0 °C contained in a calorimeter. The metal and water come to the same temperature at 24.6 °C. How much heat did the metal give up to the water?
Answer:
1087.84 J
Explanation:
From the question given above, the following data were obtained:
Mass of metal (Mₘ) = 70 g
Temperature of metal (Tₘ) = 80 °C
Mass of water (Mᵥᵥ) = 100 g
Temperature of water (Tᵥᵥ) = 22 °C
Equilibrium temperature (Tₑ) = 24.6 °C
Heat lost by metal (Qₘ) =?
NOTE: Specific heat capacity of water (Cᵥᵥ) = 4.184 J/gºC
Heat lost by metal (Qₘ) = Heat gained by water (Qᵥᵥ)
Qₘ = Qᵥᵥ
Thus, we shall determine the heat gained by water. This can be obtained as follow:
Qᵥᵥ = MᵥᵥCᵥᵥ(Tₑ – Tᵥᵥ)
Qᵥᵥ = 100 × 4.184 (24.6 – 22)
Qᵥᵥ = 418.4 × 2.6
Qᵥᵥ = 1087.84 J
Thus, the heat gained by water is 1087.84 J.
Heat lost by metal (Qₘ) = Heat gained by water (Qᵥᵥ)
Qₘ = Qᵥᵥ
Qᵥᵥ = 1087.84 J
Qₘ = 1087.84 J
Therefore, the heat lost by the metal is 1087.84 J
A 70.0‑g piece of metal at 80.0 °C is placed in 100 g of water at 22.0 °C contained in a calorimeter. After reaching a temperature of 24.6 °C, the heat given up by the metal to the water is -1.08 kJ.
What is a calorimeter?A calorimeter is an object used for calorimetry, or the process of measuring the heat of chemical reactions or physical changes as well as heat capacity.
A 70.0‑g piece of metal at 80.0 °C is placed in 100 g of water at 22.0 °C contained in a calorimeter. The final temperature of the system is 24.6 °C.
Let's use the following expression to calculate the heat absorbed by the water.
Qw = c × m × ΔT
Qw = (4.184 J/g.°C) × 100 g × (24.6 °C - 22.0 °C) = 1.08 kJ
where,
Qw is the heat absorbed by the water.c is the specific heat capacity of water.m is the mass of water.ΔT is the change in the temperature for water.According to the law of conservation of energy, the sum of the heat absorbed by the water and the heat released by the metal (Qm) is zero.
Qw + Qm = 0
Qm = -Qw = -10.8 kJ
A 70.0‑g piece of metal at 80.0 °C is placed in 100 g of water at 22.0 °C contained in a calorimeter. After reaching a temperature of 24.6 °C, the heat given up by the metal to the water is -1.08 kJ.
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A reaction was performed, and the dichloromethane solvent was dried by adding magnesium sulfate drying agent. When the reaction flask was shaken, it was observed that the magnesium sulfate clumped together at the bottom of the flask. What does this observation indicate
The clumping of magnesium sulfate means that the wrong kind of drying agent have been used for the sample.
What is a drying agent?A drying agent is also referred to as a desiccant. It is a substance that is used to remove moisture from a sample. We must recall that the drying agent to be used must not react with the sample.
Since the magnesium sulfate was found to clump together at the bottom of the flask, it means that the wrong kind of drying agent have been used for the sample.
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When the equation,
O2 + __C 10H 22 →
CO2 +
H2O is balanced, the coefficient of O2 is?
Please help!
lution: What is the molarity of 245 g of H, SO4 dissolved in 1.00 L of solution?
Answer:
Cm = n/V
n(H2SO4) = 245/98 = 2.5 mol
Cm(H2SO4) = 2.5/1 = 2.5 M
Explanation:
name a factor tht affects the value of electron affinity
Answer:
Atomic sizeNuclear chargesymmetry of the electronic configurationGive four examples illustrating each of the following terms. a. homogeneous mixture b. heterogeneous mixture c. compound e. physical change d. element f. chemical change
1. homogenous: sugar solution
2. heterogeneous: sand solution
3. compound: water
4. physical change: ice melting
5. element: hydrogen
6. chemical change: burning fire
Determine the number of water molecules in 0.2830g Na.
Answer:
7.38*10^21
Explanation:
2Na+2H20=2NaOH+H2
nNa=0.0123
number of water moles: 0.012*6*10^23=7.38*10^21
Consider the constitutional isomers 2-methylbut-1-ene, 2-methylbut-2-ene, and 3-methylbut-1-ene. When each of these alkenes is subjected to catalytic hydrogenation (H2, Pt), a single product results. Which of the following best describes the structural relationship among these products?
a. the product are cis-trans isomers.
b. the product are identical.
c. the product are constitutional isomers.
d. the product are enantiomers.
e. the product are diastereomers.
Answer:
Explanation:
I am almost sure that the products are identical.
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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