Answer:
3.89 ×10^-5
Explanation:
Since they are gaseous reactants, we obtain the equilibrium constant from the given partial pressures;
p(H2O) = 0.0900 bar
p(H2) = 0.00100 bar
p(O2) = 0.00350 bar
The equation of the reaction is;2H2O(g)⇄2H2(g)+O2(g)
Kp= p(H2) . p(O2)/p(H2O)
Kp= 0.00100 × 0.00350/0.0900
Kp= 3.89 ×10^-5
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:
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
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⁺
……….is strong due to the ……………..between positive ions and negative delocalized electrons
Answer:
atom &bond
Explanation:
atom is strong due to the bond
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.
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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When the equation,
O2 + __C 10H 22 →
CO2 +
H2O is balanced, the coefficient of O2 is?
Please help!
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]
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
How many nitrate ions are present in the following aqueous solution?
5.08 L of a solution containing 2.68 x 1021 formula units of lithium nitrate per liter.
When you have the number, determine its log (base 10) and enter that value with 3 decimal places
Answer:
22.1
Explanation:
Step 1: Calculate the number of formula units of lithium nitrate
A 5.08 L solution contains 2.68 × 10²¹ formula units per liter.
5.08 L × 2.68 × 10²¹ formula units/1 L = 1.36 × 10²² formula units
Step 2: Calculate the number of nitrate ions
Lithium nitrate dissociates completely in water according to the following equation.
LiNO₃(aq) ⇒ Li⁺(aq) + NO₃⁻(aq)
The molar ratio of LiNO₃ to NO₃⁻ is 1:1. The number of nitrate ions is 1/1 × 1.36 × 10²² = 1.36 × 10²².
Then,
log (1.36 × 10²²) = 22.1
The number of nitrate ions are present in the following aqueous solution is 13.6x10²¹, and log(base 10) value of this is 22.2.
What are strong electrolytes?
Those salts which are completely dissociate into their ions in the solution form will known as strong electrolyte.
Lithium nitrate is a strong electrolyte and it shows complete dissociation as:
LiNO₃ → Li⁺ + NO₃⁻
Given that,
1 liter of solution containing = 2.68 x 10²¹ formula unit
5.08 L of solution containing = 5.08 x 2.68 x 10²¹ = 13.6x10²¹ formula unit
So, number of nitrate ion in given solution = 13.6x10²¹ formula unit
Log(13.6x10²¹) = 22.1
Hence, required values are 13.6x10²¹ and 22.1.
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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
Suppose a 250.mL flask is filled with 1.7mol of H2 and 0.90mol of I2. The following reaction becomes possible:
+H2gI2g 2HIg
The equilibrium constant K for this reaction is 5.51 at the temperature of the flask.
Calculate the equilibrium molarity of I2. Round your answer to two decimal places.
Explanation:
here's the answer. I just plug the expression into my calculator and find the intercept to avoid the quadratic formula
b. Sodium has an emission spectrum with two visible wavelengths, both very close to 590 nm. If you had a light source that contained a mixture of sodium and hydrogen, what color filter would you use to measure only the hydrogen spectrum
Answer:
Blue or Purple color filter
Explanation:
Given that Sodium has an emission spectrum with wavelength ≈ 590nm and for a wavelength of 590nm the color is yellow.
Hence To filter out the color ( yellow ) to enable the measurement of the Balmer series of hydrogen spectrum, we have to use a filter that possess the complementary color of yellow ( i.e. purple(RYB color model) or blue (RGB additive color model )
therefore color filter to be used = Blue or Purple
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 antiacidHow 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.
100 mL of 0.2 mol/L sodium carbonate solution and 200 mL of 0.1 mol/L calcium nitrate solution are mixed together. Calculate the mass of calcium carbonate that would precipitate and the concentration of the sodium nitrate solution that will be produced.
Answer:
Explanation:
Na2CO3+Ca(NO3)2=CaCO3+2NaNO3
nNa2CO3=0.02
nCa(NO3)2=0.02
mCaCO3=0.02*100=2 gram
nNaNo3=0.04
Cm=2/15
From the calculation, the mass of the product is 2 g.
What is a reaction?A chemical reaction occurs when two more substances are mixed together. In this case, the reaction is shown by; Ca(NO3)2 + Na2CO3 ----> CaCO3(s) + 2NaNO3.
Number of moles of Na2CO3 = 100/1000 L * 0.2 mol/L = 0.02 moles
Number of moles of Ca(NO3)2 = 200/1000 L * 0.1 mol/L = 0.02 moles
Since the reaction is equimolar, amount of the product = 0.02 moles * 100 g/mol = 2 g
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What is Bose Einstein state of matter
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.
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.
Give 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
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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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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Valproic acid, used to treat seizures and bipolar disorder, is composed of C, H, and O. A 0.165-g sample is combusted to produce 0.166 g of water and 0.403 g of carbon dioxide. What is the empirical formula for valproic acid
Answer:
The empirical formula is C4H8O
Explanation:
Step 1: Data given
Valproic acid is composed of C, H, and O
Mass of the sample = 0.165 grams
Mass of water = 0.166 grams
Mass of CO2 = 0.403 grams
Molar mass of water ( H2O) = 18.02 g/mol
Molar mass of CO2 = 44.01 g/mol
Atomic mass of C = 12.01 g/mol
Atomic mass of O = 16.0 g/mol
Atomic mass H = 1.01 g/mol (H2 = 2.02 g/mol)
Step 2: The equation
CxHyOz + O2 → CO2 + H2O
Step 3: Calculate the number of carbon in the sample
The carbon comes from CO2
Mass C = (12.01 g/mol/44.01 g/mol) * 0.403 grams
Mass C = 0.110 grams
Step 4: Calculate mass of hydrogen in the sample
The hydrogen comes from H2O
Mass H = (2.02/18.02) * 0.166 grams
MAss H = 0.0186 grams
Step 5: Calculate mass of O
The mass of O in the sample = Mass of sample - mass of H - mass of C
The mass of O = 0.165 grams - 0.110 grams - 0.0186 grams
The mass of O = 0.0364 grams
Step 6: Calculate moles
Moles C = 0.110 grams / 12.01 g/mol = 0.00916 moles
Moles H = 0.0186 / 1.01 = 0.0184 moles
Moles O = 0.0364/16.0 = 0.00228 moles
Step 7: Calculate empirical formula
We divide by the smallest amount of moles
C: 0.00916/ 0.00228 = 4
H: 0.0184/0.00228 = 8
O: 0.00228/0.00228 = 1
The empirical formula is C4H8O
What is the hydrogen atoms in 39.6g of ammonium sulphate,NH4 2SO4
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
please see attachment
Answer:
I'll see it
Explanation:
what is the IUPAC name of 2NaOH(s)
Answer:
NaoH= sodium hydroxide
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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Complete (predict the products, write correct formulas) and balance the following reaction:
Gaseous hydrochloric acid reacts with solid calcium hydroxide. Include states of matter and use subscripts as needed to receive full credit for a correct answer.
Explanation:
since hydrochloric acid is an acid and calcium hydroxide is a base, we know that this is an acid base reaction. The ions will then dissociatiate and bond with one another.
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
