Answer:
Adding a catalyst increases the speed of the reaction.
Explanation:
The catalyst increases the reaction rate since what it does is to consume the reagents more easily in order to obtain the products faster.
The catalysts DO NOT ALTER THE QUANTITY OF PRODUCTS OR REAGENTS, that is to say that they do not add grams nor do they intervene in the masses of the chemical compounds of the reaction, they only speed up the procedure.
Name of th molecule
1. CH3CH2CHClCHBrCH3
2.C=C-CH3
CH3CH=CHCH2
Answer:
1: 2-bromo-3-chloropentane
Explanation:
find longest carbon chain =5
place the Br and Cl on the carbon chain
follow naming rules I guess
can someone assist me with this chemistry question please?
Answer:
490 J.
Explanation:
The following data were obtained from the question:
Mass (m) = 5 Kg
Height (h) = 10 m
Acceleration due to gravity (g) = 9.8 m/s²
Potential energy (P.E) =..?
Potential energy is the energy stored in a body by virtue of its location. Mathematically, it can be expressed as:
P.E = mgh
Where:
m is the mass of the object measured in kilograms (Kg).
g is the acceleration due to gravity and the value is 9.8 m/s².
h is the height to which the object is located measured in metre (m)
P.E is the potential energy measured in joule (J).
With the above formula, we can obtain the potential energy possed by the object as follow:
P.E = mgh
P.E = 5 x 9.8 x 10
P.E = 490 J
The,the potential energy possed by the object is 490 J
Give the concentration of each type of ion in the following solutions:
a. 0.50 M CO(NO3)2
b. 1 M Fe(C1O4)3
Answer:
a.
[tex]M_{Co^{2+}}=0.5M\\ \\M_{NO_3^{-}}=1.0M[/tex]
b.
[tex]M_{Fe^{3+}}=1.0M\\ \\M_{ClO_4^{-}}=3.0M[/tex]
Explanation:
Hello,
a. In this case, the ions are cobalt (II) and nitrate, for which, one mole of cobalt (II) nitrate contains one mole of cobalt (II) and two moles of nitrate (see subscripts), therefore, concentrations turn out:
[tex]M_{Co^{2+}}=0.5\frac{molCo(NO_3)_2}{L}* \frac{1molCo}{1molCo(NO_3)_2}=0.5M\\ \\M_{NO_3^{-}}=0.5\frac{molCo(NO_3)_2}{L}* \frac{2molNO_3^{-}}{1molCo(NO_3)_2}=1.0M[/tex]
b. In this case, the ions are iron (III) and chlorate, for which one mole of iron (III) is contained in one mole of iron (III) chlorate and three moles of chlorate are in one mole of iron (III) chlorate (see subscripts), therefore, the concentrations turn out:
[tex]M_{Fe^{3+}}=1.0\frac{molFe(ClO_4)_3}{L}* \frac{1molFe^{3+}}{1molFe(ClO_4)_3}=1.0M\\ \\M_{ClO_4^{-}}=0.5\frac{molFe(ClO_4)_3}{L}* \frac{3molClO_4^{-}}{1molFe(ClO_4)_3}=3.0M[/tex]
Regards.
At 25.0°C the Henry's Law constant for methane CH4 gas in water is ×1.410−3/Matm.
Calculate the mass in grams of CH4 gas that can be dissolved in 75.mL of water at 25.0°C and a CH4 partial pressure of 0.68atm. Round your answer to 2 significant digits.
Answer:
1.1 × 10⁻³ g
Explanation:
Step 1: Given data
Henry's Law constant for methane (k): 1.4 × 10⁻³ M/atm
Volume of water (=volume of solution): 75 mL
Partial pressure of methane (P): 0.68 atm
Step 2: Calculate the concentration of methane in water (C)
We will use Henry's law.
[tex]C = k \times P = 1.4 \times 10^{-3}M/atm \times 0.68atm = 9.5 \times 10^{-4}M[/tex]
Step 3: Calculate the moles of methane in 75 mL of water
[tex]\frac{9.5 \times 10^{-4}mol}{L} \times 0.075 L = 7.1 \times 10^{-5}mol[/tex]
Step 4: Calculate the mass corresponding to 7.1 × 10⁻⁵ mol of methane
The molar mass of methane is 16.04 g/mol.
[tex]7.1 \times 10^{-5}mol \times \frac{16.04g}{mol} = 1.1 \times 10^{-3} g[/tex]
Consider the following precipitation reaction occurring in aqueous solution:
3 SrCl2(aq)+2 Li3PO4(aq) →Sr3(PO4)2(s)+6 LiCl(aq)
Write the complete ionic equation and the net ionic equation for this reaction.
Answer:
[tex]3Sr^{+2}+6Cl^{-}+6Li^{+}+2PO_{4}^{3-}-->Sr_{3}(PO_{4})_{2}+6Li^{+}+6Cl^{-}\\\\3Sr^{+2}+2PO_{4}^{3-} --->Sr_{3}(PO_{4})_{2}[/tex]
First equation is the complete ionic equation.
Second equation is the net ionic equation.
Calculate the height of a column of water at 25 °C that corresponds to normal atmospheric pressure. The density of water at this temperature is 1.0 g/
Answer:
10.328 m
Explanation:
normal atmospheric pressure = 101325 Pa
density of water at 25 °C = 1.0 g/cm^3 = 1000 kg/m^3
pressure = pgh
where p = density
g = acceleration due to gravity = 9.81 m/s^2
h = height of column
imputing values, we have
101325 = 1000 x 9.81 x h
height of column h = 101325/9810 = 10.328 m
g The most common position for an double bond in an unsaturated fatty acid is delta _________(fill in the number).
Answer:
The most common position for an double bond in an unsaturated fatty acid is delta 9 (Δ⁹)
Explanation:
Unsaturated fatty acids are carboxylic acids which contains one or more double bonds. The chain length as well as the number of double bonds is written separated by a colon. The positions of the double bonds are specified starting from the carboxyl carbon, numbered as 1, by superscript numbers following a delta (Δ). For example, an 18-carbon fatty acid containing a single double bond between carbon number 9 and 10 is written as 18:1(Δ⁹).
In most monounsaturated fatty acids, the double bond is between C-9 and C-10 (Δ⁹), and the other double bonds of polyunsaturated fatty acids are generally Δ¹² and Δ¹⁵. This positioning is due to the nature of the biosynthesis of fatty acids. In the mammalian hepatocytes, double bonds are introduced easily into fatty acids at the Δ⁹ position, but cannot introduce additional double bonds between C-10 and the methyl-terminal end. However, plants are able to introduce these additional double bonds at the Δ¹² and Δ¹⁵ positions.
To condition the buret, add a small volume of ____________ to the buret and rotate the barrel ____________ so that the liquid makes contact with the full inner surface of the barrel. Complete this action _________ and discard each volume in a designated waste container.
Answer:
Acid, Horizontally, Drain off the acid by opening the stopcock
Explanation:
A buret is a calibrated glass appratus used to measure and deliver accurate volume of liquid, usually acids, in acid-base titrations.
Before usually the buret to carry out your titration reaction, it is advised to condition or prepare your buret for use.
To condition the buret, add a small volume of the acid to the buret and rotate the barrel horizontally to ensure that the liquid ( acid) makes contact with the inner surface of the barrel o the buret.
This procedure is done to wash off any previous acid or liquid that the buret had been used for in previous titrations.
Complete this action by opening the stopcock of the buret to drain off the acid and discarding each volume in a designated waste container.
Note that acids could be corrosive and dangerous to the skin and so should be handled with great care
Many free radicals combine to form molecules that do not contain any unpaired electrons. The driving force for the radical–radical combination reaction is the formation of a new electron‑pair bond. Consider the formation of hydrogen peroxide. 2OH(g)⟶H2O2(g) Write Lewis formulas for the reactant and product species in the chemical equation. Include nonbonding electrons.
Answer:
In the attached image the Lewis equation is shown where it is shown how two oxygens react with two hydrogens to meet the octet of the electrons.
Explanation:
Hydrogen peroxide is one of the most named chemicals since it is not only sold as "hydrogen peroxide" in pharmacies but it is also one of the great weapons of immune defense cells to defend ourselves against anaerobic bacteria.
The disadvantage of this compound is that when dividing it forms free oxygen radicals that are considered toxic or aging for our body.
In the attached image below, you will see the Lewis equation is shown there. There, you will see how two oxygens react with two hydrogens to come about the octet of the electrons.
When two or more atoms bond with each other, they often form a molecule. When two hydrogens and an oxygen share electrons through covalent bonds, a water molecule is formed.
The octet rule is known as when most atoms want to gain stability in their outer most energy level by filling themselves that is the S and P orbitals of the highest energy level with eight electron.
HOOH is the compound that is form. It is called Hydrogen peroxide. This because it is has reactive oxygen species and the simplest peroxide.
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A mineral that helps in clotting of blood________.
please tell me what is the answer of this question.
Answer:
Vitamin K
Explanation:
this is the answer
please help me I am begging you.. )))): PLEASE HELP ME ~~~~~~~~~~~~~~~~~~~~~~ A football player experiences acute pain in his knee. Which of the following methods can a doctor use to diagnose the reason for the pain? --_-_-____- A.) Use infrared radiation from warm objects to look inside the knees. B.) Use radio waves emitted by radioactive substances to look at bones. C.) Use radiations emitted by very hot objects to penetrate the skin and bones. D.) Use x‒ray radiation to see if there are any fractured bones.
Answer:
D. Use x-ray radiation to see if there are any fractured bones.
Explanation:
The football player may have fractured a bone while he was practicing or playing, so it is best for the doctor to check if the player broke his bone or fractured it.
Using the provided table, determine the enthalpy for the reaction
2 NH3 (g) + 3 N20 (g) 4 N2 (g) + 3 H20 (1)
Answer:
ΔH°r = -1009.8 kJ
Explanation:
Let's consider the following balanced reaction.
2 NH₃(g) + 3 N₂O(g) ⇒ 4 N₂(g) + 3 H₂O(l)
We can calculate the standard enthalpy of the reaction (ΔH°r) using the following expression.
ΔH°r = [4 mol × ΔH°f(N₂(g)) + 3 mol × ΔH°f(H₂O(l))] - [2 mol × ΔH°f(NH₃(g)) + 3 mol × ΔH°f(N₂O(g))]
ΔH°r = [4 mol × 0 kJ/mol + 3 mol × (-285.8 kJ/mol)] - [2 mol × (-46.2 kJ/mol) + 3 mol × 81.6 kJ/mol]
ΔH°r = -1009.8 kJ
Therapeutic drugs generally need to have some hydrophobic and hydrophilic components to be able to effectively reach their target organs and tissues given there are aqueous and nonaqueous parts of the body. The degree to which a compound is hydrophobic and hydrophilic can be determined by measuring its relative solubility in water and octanol, C8H17OH, and water. To do this, a sample of the compound is added to a mixture of water and octanol and mixed well. Water and octanol are immiscible so after the mixture settles, the concentration of the compound in water and the concentration of the compound in octanol is measured. The ratio of the concentrations is called the partition ratio:
The question is incomplete as some part is missing:
concentration in octanol Partition Ratio = concentration in water
a) What are the intermolecular forces of attraction between octanol molecules? Explain.
b) Which of the intermolecular forces of attraction identified in (a) account for most of the interactions between octanol molecules? Explain. Use the immiscibility in water and the data included in figures 1 and 2 as evidence to support your answer.
c) Would a compound with a partition coefficient less than one be more hydrophobic or more hydrophilic than one with a partition coefficient greater than 10? Explain.
d) Would nonane (figure 2) be more soluble in water or octanol? Explain.
e) Draw another structure for a compound with the same chemical formula as nonane (CH20) that has a lower boiling point. Explain.
f) Are any of the C atoms in the structure you drew for CH20 sp?hybridized? Explain.
Octanol Boiling point = 195°C Figure 2 Nonane (CH20) Boiling point = 151°C
Answer:
1. The forces between octanol molecules would be attractive. These forces include Vanderwaal forces, H-bonds due to the presence of highly polar O-H group.
2. H-bonding ahould account for most of the attractive forces. The O-H bond should behave like and dipole, oxygen of one molecule attracts the hydrogen of the neighbouring molecule forming D-H...A links throughout (D stands for donor of H-Bond and A for acceptor for H-Bond).
3. Partition coefficient less than 1 will be more hydrophilic, generally drugs with low partition coefficients are regarded as hydrophilic. As parition coefficient of 10 mean more of the solute is dispersed in octanol as compared to water.
4. Nonane is non polar, so it would not dissolve in water. It follows the rule like dissolves like. Polar substances dissolve in polar solvents. 1-octanol is able to bind with water through hydrogen bonds thus its soluble in water but nonane doesn't. Nonane will forms a different layer from water.
5) no all carbons in 2-methyloctane are single bonded. Thus sp3 hybrid. A sp2 hybridised carbon would have a double bond C=C.
Question 23
1 pts
When solutions of AgNO3 and NaOH react, the balanced molecular equation is:
2 AgNO3(aq) + 2NaOH(aq) O--> Ag2O(s) + 2 NaNO3(aq) + H20(1)
How much Ag2O is produced when 0.200 g of AgNO3 and 0.200 g of NaOH react?
a. 0.127 g
c. 0.273 g
b. 0.136 g
d. 0.400 g
OB
OC
OA
OD
Answer:
Option B. 0.136 g
Explanation:
The balanced equation for the reaction is given below:
2AgNO3(aq) + 2NaOH(aq) —> Ag2O(s) + 2NaNO3(aq) + H2O(l)
Next, we shall determine the masses of AgNO3 and NaOH that reacted and the mass of Ag2O produced from the balanced equation. This is illustrated below:
Molar mass of AgNO3 = 108 + 14 + (16x3) = 170g/mol
Mass of AgNO3 from the balanced equation = 2 x 170 = 340g
Molar mass of NaOH = 23 + 16 + 1 = 40g/mol
Mass of NaOH from the balanced equation = 2 x 40 = 80g
Molar mass of Ag2O = (108x2) + 16 = 232g/mol
Mass of Ag2O from the balanced equation = 1 x 232 = 232g
Summary:
From the balanced equation above,
340g of AgNO3 reacted with 80g of NaOH to produce 232g of Ag2O.
Next, we shall determine the limiting reactant. This can be obtained as follow:
From the balanced equation above,
340g of AgNO3 reacted with 80g of NaOH.
Therefore, 0.2g of AgNO3 will react with = (0.2 x 80)/340 = 0.047g of NaOH.
From the calculations made above, only 0.047g out of 0.2g of NaOH given, reacted completely with 0.2g of AgNO3. Therefore, AgNO3 is the limiting reactant and NaOH is the excess reactant.
Now, we can calculate the mass of Ag2O produced from the reaction of 0.2g of AgNO3 and 0.2g of NaOH.
In this case, we shall use the limiting reactant because it will produce the maximum yield of Ag2O as all of it is used up in the reaction.
The limi reactant is AgNO3 and the mass of Ag2O produced can be obtained as follow:
From the balanced equation above,
340g of AgNO3 reacted to produce 232g of Ag2O.
Therefore, 0.2g of AgNO3 will react to produce = (0.2 x 232)/340 = 0.136g of Ag2O.
Therefore, 0.136g of Ag2O was produced from the reaction.
Determine the rate of a reaction that follows the rate law:
rate = k[A]”[B]", where:
k= 1.5
[A] = 1 M
[B] = 3 M
m = 2
n = 1
Answer:
k= 1.5
[A] = 1 M
[B] = 3 M
m = 2
n = 1
Explanation:
rate = k[A]”[B]"
The rate of the reaction is 4.5 mol L⁻¹s⁻¹.
What is meant by rate of a reaction ?Rate of a reaction is defined as the change in concentration of any one of the reactants or products of the reaction, in unit time.
Here,
The concentration of A, [A] = 1 M
The concentration of B, [B] = 3 M
The partial order with respect to A, m = 2
The partial order with respect to B, n = 1
The rate constant of the reaction, k = 1.5
The rate of the reaction,
r = k[A]^m [B}^n
r = 1.5 x 1² x 3
r = 4.5 mol L⁻¹s⁻¹
Hence,
The rate of the reaction is 4.5 mol L⁻¹s⁻¹.
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If an electron has a principal quantum number (n) of 7 and an angular momentum quantum number (l) of 1, the subshell designation is ________
Answer:
7p
Explanation:
principal quantum number is 7
n=7( principle shell)
angular momentum quantum number gives sub shell
l = 1 means it is p orbital
so answer is 7p orbital
Question 4 (2 points)
CuO(s) + H2(g)
Cu(s) +
H2O(1)
Balance the equation
Answer:
CuO(s) + H₂(g) --> Cu(s) + H₂O(l)
Explanation:
It is already balanced. You can see that the values of the elements of the reactants are equal to the values of the elements of the products.
A substance used as a cleaner and a fuel is 37.48% C, 49.93% O and 12.58% H by mass. A 0.2804-g sample of the substance occupies a volume of 250.0 mL when it is vaporized at 75o C and 1.00 atm of pressure.
R = 0.0821 L atm/ mol K
a) This compound can be made by combining gaseous carbon monoxide with hydrogen gas (with this compound as the only product). What is the maximum mass of this compound that can be prepared if 8.0 kg of hydrogen gas react with 59.0 kg of carbon monoxide gas?
b) If 59.6 kg of the product is actually produced, given the reaction described in (a), what is the percent yield?
c) This compound (the substance you identified in part a) is a potential replacement for gasoline. The products of the complete combustion of this fuel are the same as those for the complete combustion of a hydrocarbon (CO2 and H2O). Calculate the volume of CO2 produced at 27o C and 766 mmHg when 1.00 gallon of this fuel is completely combusted. The density of the fuel is 0.7914 g/mL. 1 gallon = 3.785 liters
d) A claim was made that this fuel is better for the environment because it produces less CO2 per gallon than gasoline, which can be represented by the formula C8H18 (octane). Is this claim true? Octane has a density of 0.6986 g/mL
Answer:
Explanation:
We shall find out the molecular formula of the substance .
Ration of number of atoms of C , O and H
= [tex]\frac{37.48}{12} :\frac{49.93}{16} :\frac{12.58}{1}[/tex]
= 3.12 : 3.12 : 12.58
= 1 : 1 : 4
volume of gas at NTP
= 250 x 273 / 350 mL .
= 195 mL .
Molecular weight of the substance = .2804 x 22400 / 195 g
= 32. approx
Let the molecular formula be
(COH₄)n
n x 32 = 32
n = 1
Molecular formula = COH₄
The compound appears to be CH₃OH
a )
CO + 2H₂ = CH₃OH
28g 4g 32g
59 8
For 8 kg hydrogen , CO required = 56 kg
CO is in excess . hydrogen is the limiting reagent .
mass of product formed
= 32 x 8 / 4
= 64 kg
b )
percentage yield = product actually formed / product to be formed theoretically x 100
= 59.6 x 100 / 64
= 93.12 %
c )
2CH₃OH + 3O₂ = 2CO₂ + 4H₂O .
64 g 2 x 22.4 L
Gram of gas in 1 gallon of fuel
= .7914 x 3785
= 2995.5 g
CO₂ produced at NTP by 2995.5 g CH₃OH
= 2 x 22.4 x 2995.5 / 64 L
= 2096.85 L
At 27° C and 766 mm Hg , this volume is equal to
2096.85 x 300 x 760 / 273 x 766
= 2286.18 L .
d )
C₈H₁₈ = 8CO₂
114g 8 x 22.4 L
gram of fuel per unit gallon
= .6986 x 3785
= 2644.2g
gram of CO₂ produced by 1 gallon of fuel at NTP
= 8 x 22.4 x 2644.2 / 114
= 4156.5 L
So it produces more CO₂ .
Which of the following is NOT a type of crystal structure? A. None of these B. Metallic C. Ionic D. Macromolecular (giant covalent)
Answer:
A. None of these.
Explanation:
A crystal structure is an arrangement of atoms or ions in a repeating three-dimensional array.
B. is wrong. Metal atoms, such as gold, arrange themselves into a crystal structure.
C. is wrong. Ionic solids, such as sodium chloride, arrange themselves into a crystal structure.
D. is wrong. Macromolecules (network solids), such as diamond, arrange themselves into a crystal structure.
The correct answer is None of these.
What is a crystal structure?A crystal structure is a three-dimensional collection of atoms or ions that repeats itself.Metal atoms(gold), Ionic solids (sodium chloride), and Macromolecules(network solids) arrange themselves into a crystal structure.Learn more about crystal structure here:-
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A 1.555-g sample of baking soda decomposes with heat to produce 0.991 g Na2CO3. Refer to Example Exercise 14.l and show the calculation for the theoretical yield of Na2CO3.
What is the percent yield of sodium carbonate, Na2CO3?
6. A 1473-g unknown mixture with baking soda is heated and has a mass loss of 0.325 g. Refer to Example Exercise 14.2 and show the calculation for the percentage NaHCOs in the mixture.
Answer:
a) 101%
b)59.7%
Explanation:
The equation for the thermal decomposition of baking soda is shown;
2NaHCO3 → Na2CO3 + H2O + CO2
Number of moles of baking soda= mass/molar mass= 1.555g/84.007 g/mol = 0.0185 moles
From the reaction equation;
2 moles of baking soda yields 1 mole of sodium carbonate
0.0185 moles of baking soda will yield = 0.0185 moles ×1 /2 = 9.25 ×10^-3 moles of sodium carbonate.
Therefore, mass of sodium carbonate= 9.25 ×10^-3 moles × 106gmol-1= 0.9805 g of sodium carbonate. This is the theoretical yield of sodium carbonate.
%yield = actual yield/theoretical yield ×100
% yield = 0.991/0.9805 ×100
%yield = 101%
Since ;
2NaHCO3 → Na2CO3 + H2O + CO2
And H2O + CO2 ---> H2CO3
Hence I can write, 2NaHCO3 → Na2CO3 + H2CO3
Molar mass of H2CO3= 62.03 gmol-1
Molar mass of baking soda= 84 gmol-1
Therefore, mass of baking soda=
0.325/62.03 × 2 × 84 = 0.88 g of NaHCO3
% of NaHCO3= 0.88/1.473 × 100 = 59.7%
The decomposition reaction of baking soda is a reaction in which water and carbon dioxide ae given off as gaseous products.
5. The theoretical yield of Na₂CO₃ is approximately 0.9809 gramsThe percentage yield of sodium carbonate is approximately 101.02%.6. Percentage of NaHCO₃ in the mixture is approximately 59.76%.Reasons:
Mass of baking soda = 1.555 g
Mass of Na₂CO₃ produced = 0.991 g
Required:
Calculation for the theoretical yield
Solution:
Theoretical yield (mass) of Na₂CO₃ produced is found as follows;
Molar mass of Na₂CO₃ = 105.9888 g/mol
Molar mass of NaHCO₃ = 84.007 g/mol
[tex]\displaystyle 1.555 \, g \, NaHCO_3 \times \frac{1 \, mol \, NaHCO_3}{84.007 \, g \, NaHCO_3} \times \frac{1 \, mol \, Na_2CO_3}{2 \, mol \, NaHCO_3} \times 105.9888 \ g \approx 0.9809 \, g \, Na_2CO_3[/tex]
The theoretical yield of Na₂CO₃ ≈ 0.9809 grams.
The percentage yield is given as follows;
[tex]\displaystyle Percentage \ yield = \mathbf{\frac{Actual \, Yield}{Theorectical \, Yield} \times 100 \%}[/tex]
The percentage yield of Na₂CO₃ is therefore;
[tex]\displaystyle Percentage \ yield \ of \ Na_2CO_3= \frac{0.991}{0.9809} \times 100 \% \approx \underline{ 101.02 \%}[/tex]
(Some baking soda may remain if the reaction is not completed)
6. Mass of the unknown mixture of baking soda = 1473 g
Mass loss from the mixture = 0.325 g
Required:
The percentage of NaHCO₃ in the mixture.
Solution:
The chemical in the mass loss from heating the NaHCO₃ = H₂CO₃
Molar mass of H₂CO₃ = 62.03 g/mol
[tex]\displaystyle \mathrm{Number \ of \ moles \ of \ H_2CO_3 \ produced} = \frac{0.325 \, g}{62.03 \, g/mol} \approx 5.2394 \times 10^{-3} \ moles[/tex]
The chemical reaction is presented as follows;
2NaHCO₃(s) [tex]\underrightarrow {\Delta \ Heated}[/tex] Na₂CO₃(s) + H₂CO₃(g)2 moles of NaHCO₃ produces 1 mole of H₂CO₃The number of moles of NaHCO₃ in the mixture is therefore;
2 × 5.2394 × 10⁻³ moles ≈ 1.04788 × 10⁻² molesMass of NaHCO₃ in the mixture is therefore
Mass of NaHCO₃ = 1.04788 × 10⁻² moles × 84.007 g/mol = 0.88029 g[tex]\displaystyle Percentage \ of \ NaHCO_3 \ in \ the \ mixture \ = \mathbf{ \frac{Mass \ of \ NaHCO_3}{Mass \ of \ mixture} \times 100}[/tex]
Which gives;
[tex]\displaystyle Percentage \ of \ NaHCO_3 \ in \ the \ mixture \ = \ \frac{0.88029 \, g}{1.473 \, g} \times 100 \approx \underline{ 59.76 \%}[/tex]Learn more here:
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The next few questions will walk you through solving the following problem: At a given temperature, a 5.0M solution of hydrazine (N2H4) as a pH of 11.34. Hydrazine is base.
A. What is the concentration of hydroxide ion at equilibrium?
B. What is the pK for hydrazine reacting with water at this temperature?
Answer:
A. [OH⁻] = 2.188x10⁻³M
B. pKb = 6.02
Explanation:
When hydrazine is in equilbrium with water, its reaction is:
N₂H₄(aq) + H₂O(l) ⇄ HN₂H₄⁺(aq) + OH⁻(aq)
Where Kb, is defined as the ratio between concentrations in equilibrium of the species, thus:
Kb = [HN₂H₄⁺] [OH⁻] / [N₂H₄]
A. From pH, you can find [OH⁻], thus:
pH = -log [H⁺]
11.34 = -log [H⁺]
4.57x10⁻¹² = [H⁺]
As 1x10⁻¹⁴ = [OH⁻] [H⁺]
1x10⁻¹⁴ / 4.57x10⁻¹² = [OH⁻]
[OH⁻] = 2.188x10⁻³MB. Concentrations in equilibrium of the species are:
[N₂H₄] = 5.0M - X
[HN₂H₄⁺] = X
[OH⁻] = X
Where X is reaction coordinate
As [OH⁻] = 2.188x10⁻³M
X = 2.188x10⁻³M
Replacing:
[N₂H₄] = 5.0M - 2.188x10⁻³M = 4.9978M
[HN₂H₄⁺] = 2.188x10⁻³M
[OH⁻] = 2.188x10⁻³M
Replacing in Kb expression:
Kb = [HN₂H₄⁺] [OH⁻] / [N₂H₄]
Kb = [2.188x10⁻³M] [2.188x10⁻³M] / [4.9978M]
Kb = 9.577x10⁻⁷
pKb is defined as -log Kb
pKb = -log 9.577x10⁻⁷
pKb = 6.02
What is the freezing point of an aqueous solution that boils at 105.9 ∘C? Express your answer using two significant figures.
Answer:
THE FREEZING POINT OF THE AQUEOUS SOLUTION IS - 7.3 °C
Explanation:
To solve this problem, we must know the following variables:
Normal boiling point of water (solvent) = 100 °C
The molar boiling point elevation constant of water = 1.51 °C /m
Normla freezing point of water ( solvent) = 0 °C
The molar freezing point depression constant = 1.86 °C /m
The boiling point of the aqueous solution = 105.9 °C
Molarity = xM
Change in boiling point = boiling point of solution - boiling point of water
Change in boiling point = 105.9 - 100 °C
= 5.9 °C
From the formula:
Change in boiling point = i * Kb * M
Re- arranging the formula by making M the subject of the equation, we have:
M = change in boiling point / Kb
i = 1
M = 5.9 °C / 1.51 °C/m
M = 3.907 M
Then, we calculate the freezing point:
Change in freezing point = i * Kb * M
= 1 * 1.86 °C/m * 3.907 M
= 7.267 °C
Hence, the freezing point = freezing point of water - change in freezing point
Freezing point = 0 °C - 7.267 °C
Freezing point = - 7.267 °C
Freezing point = -7.3 °C
Determine the radius of an Al atom (in pm) if the density of aluminum is 2.71 g/cm3 . Aluminum crystallizes in a face centered cubic structure with an edge leng
Answer:
143pm is the radius of an Al atom
Explanation:
In a face centered cubic structure, FCC, there are 4 atoms per unit cell.
First, you need to obtain the mass of an unit cell using molar mass of Aluminium and thus, obtain edge length and knowing Edge = √8R you can find the radius, R, of an Al atom.
Mass of an unit cell
As 1 mole of Al weighs 26.98g. 4 atoms of Al weigh:
4 atoms × (1mole / 6.022x10²³atoms) × (26.98g / mole) = 1.792x10⁻²²g
Edge length
As density of aluminium is 2.71g/cm³, the volume of an unit cell is:
1.792x10⁻²²g × (1cm³ / 2.71g) = 6.613x10⁻²³cm³
And the length of an edge of the cell is:
∛6.613x10⁻²³cm³ = 4.044x10⁻⁸cm = 4.044x10⁻¹⁰m
Radius:
As in FCC structure, Edge = √8 R, radius of an atom of Al is:
4.044x10⁻¹⁰m = √8 R
1.430x10⁻¹⁰m = R.
In pm:
1.430x10⁻¹⁰m ₓ (1x10¹²pm / 1m) =
143pm is the radius of an Al atomThe radius of the atom of Al in the FCC structure has been 143 pm.
The FCC lattice has been contributed with atoms at the edge of the cubic structure.
The FCC has consisted of 4 atoms in a lattice.
The mass of the unit cell of Al can be calculated as:[tex]\rm 6.023\;\times\;10^2^3[/tex] atoms = 1 mole
4 atoms = [tex]\rm \dfrac{4}{6.023\;\times\;10^2^3}[/tex] moles
The mass of 1 mole Al has been 26.98 g/mol.
The mass of [tex]\rm \dfrac{4}{6.023\;\times\;10^2^3}[/tex] moles = [tex]\rm \dfrac{4}{6.023\;\times\;10^2^3}[/tex] moles × 26.98 g
The mass of 1 unit cell of Al has been = 1.792 [tex]\rm \bold{\times\;10^-^2^2}[/tex] g.
The volume of the Al cell can be calculated as:Density = [tex]\rm \dfrac{mass}{volume}[/tex]
Volume = Density × Mass
The volume of Al unit cell = 2.71 g/[tex]\rm cm^3[/tex] × 1.792 [tex]\rm \times\;10^-^2^2[/tex] g
The volume of Al cell = 6.613 [tex]\rm \times\;10^-^2^3[/tex] [tex]\rm cm^3[/tex]
The volume of the cube has been given as:Volume = [tex]\rm edge\;length^3[/tex]
6.613 [tex]\rm \times\;10^-^2^3[/tex] [tex]\rm cm^3[/tex] = [tex]\rm edge\;length^3[/tex]
Edge length = [tex]\rm \sqrt[3]{6.613\;\times\;10^-^2^3}[/tex] cm
Edge length = 4.044 [tex]\rm \times\;10^-^8[/tex] cm
Edge length = 4.044 [tex]\rm \bold{\times\;10^-^1^0}[/tex] m.
In an FCC lattice structure, the radius of the atom can be given by:Edge length = [tex]\rm \sqrt{8\;\times\;radius}[/tex]
4.044 [tex]\rm \bold{\times\;10^-^1^0}[/tex] m = [tex]\rm \sqrt{8\;\times\;radius}[/tex]
Radius = 1.430 [tex]\rm \bold{\times\;10^-^1^0}[/tex] m.
1 m = [tex]\rm 10^1^2[/tex] pm
1.430 [tex]\rm \bold{\times\;10^-^1^0}[/tex] m = 143 pm.
The radius of the atom of Al in the FCC structure has been 143 pm.
For more information about the FCC structure, refer to the link:
https://brainly.com/question/14934549
A solution is prepared by adding 6.24 g of benzene (C 6H 6, 78.11 g/mol) to 80.74 g of cyclohexane (C 6H 12, 84.16 g/mol). Calculate the mole fraction and molality of benzene in this solution.
Answer:
[tex]x_B=0.0769[/tex]
[tex]m=0.990m[/tex]
Explanation:
Hello,
In this case, we can compute the mole fraction of benzene by using the following formula:
[tex]x_B=\frac{n_B}{n_B+n_C}[/tex]
Whereas n accounts for the moles of each substance, thus, we compute them by using molar mass of benzene and cyclohexane:
[tex]n_B=6.24g*\frac{1mol}{78.11g}=0.0799mol\\ \\n_C=80.74g*\frac{1mol}{84.16g} =0.959mol[/tex]
Thus, we compute the mole fraction:
[tex]x_B=\frac{0.0799mol}{0.0799mol+0.959mol}\\ \\x_B=0.0769[/tex]
Next, for the molality, we define it as:
[tex]m=\frac{n_B}{m_C}[/tex]
Whereas we also use the moles of benzene but rather than the moles of cyclohexane, its mass in kilograms (0.08074 kg), thus, we obtain:
[tex]m=\frac{0.0799mol}{0.08074kg}=0.990mol/kg[/tex]
Or just 0.990 m in molal units (mol/kg).
Best regards.
Considering the definition of mole fraction and molality:
the mole fraction of benzene is 0.077.the molality of benzene is 0.9908 [tex]\frac{moles}{kg}[/tex].You know that:
Mass of benzene = 6.24 gramsMass of cyclohexane= 80.74 gramsMolar mass of benzene= 78.11 g/moleMolar mass of cyclohexane= 84.16 g/moleMole fractionThe molar fraction is a way of measuring the concentration that expresses the proportion in which a substance is found with respect to the total moles of the solution.
Being the molar mass and the mass in the solution of each compound, the number of moles of each compound can be calculated as:
Benzene: [tex]\frac{mass of benzene}{molar mass of benzene} =\frac{6.24 grams}{78.11 \frac{grams}{mole} } = 0.08 moles[/tex]Cyclohexane:[tex]\frac{mass of cyclohexane}{molar mass of ciclohexane} =\frac{80.74 grams}{84.16\frac{grams}{mole} } = 0.96 moles[/tex]So, the total moles of the solution can be calculated as:
Total moles = 0.08 moles + 0.96 moles = 1.04 moles
Finally, the mole fraction of benzene can be calculated as follow:
[tex]\frac{number moles of benzene}{total moles} =\frac{0.08 moles}{1.04 moles} = 0.077[/tex]
Finally, the mole fraction of benzene is 0.077.
MolalityMolality is the ratio of the number of moles of any dissolved solute to kilograms of solvent.
The Molality of a solution is determined by the expression:
[tex]Molality=\frac{number of moles of solute}{kilograms of solvent}[/tex]
In this case, you know:
number of moles of solute= 0.08 moles Mass of solvent = 80.74 g = 0.08074 kg (being 1000 g=1 kg)Replacing:
[tex]Molality benzene=\frac{0.08 moles}{0.08074 kg}[/tex]
Molality benzene= 0.9908 [tex]\frac{moles}{kg}[/tex]
Finally, the molality of benzene is 0.9908 [tex]\frac{moles}{kg}[/tex].
Learn more about:
mole fraction brainly.com/question/14434096?referrer=searchResults brainly.com/question/10095502?referrer=searchResults molality brainly.com/question/20366625?referrer=searchResults brainly.com/question/4580605?referrer=searchResults
Classify each molecule by whether its real bond angles are the same as or different than its model (ideal) bond angles. In other words, do the bond angles change when you switch between Real and Model mode at the top of the page?
The question is incomplete; the complete question is: Classify each molecule by whether its real bond angles are the same as or different than its model (ideal) bond angles. In other words, do the bond angles change when you switch between Real and Model mode at the top of the page? Same (angles do not change) Different (angles change) Answer Bank | H2O | CO2, SO2, XeF2, BF3 CIF3, NH3, CH4, SF4, XeF4, BrF5, PCI5,SF6
Answer:
Compounds whose real bond angle are the same as ideal bond angle;
SF6, BF3, CH4, PCI5
Compounds whose real bond angles differ from ideal bond angles;
H2O, CO2, SO2, XeF2, CIF3, NH3, SF4, XeF4, BrF5
Explanation:
According to the valence shell electron pair repulsion theory (VSEPR), molecules adopt various shapes based on the number of electron pairs on the valence shell of the central atom of the molecule. The electron pairs usually orient themselves as far apart in space as possible leading to various observed bond angles.
The extent of repulsion of lone pairs is greater than that of bond pairs. Hence, the presence of lone pairs on the valence shell of the central atom in the molecule distorts the bond angles of molecules away from the ideal bond angles predicted on the basis of valence shell electron pair repulsion theory.
For instance, methane is a perfect tetrahedron having an ideal bond angle of 109°28'. Both methane and ammonia are based on a tetrahedron, however, the presence of a lone pair of electrons on nitrogen distorts the bond angle of ammonia to about 107°. The distortion of lone pairs in water is even more as the bond angles of water is about 104°.
Write the empirical formula
Answer:
[tex]Fe(C_{2}H_{3}O_{2})_{3}\\Fe (OH)_{3}\\\\NH_{4}(C_{2}H_{3}O_{2})\\\\NH_{4}OH[/tex]
Explanation:
[tex]Fe^{3+}(C_{2}H_{3}O_{2})^{-}_{3}--->Fe(C_{2}H_{3}O_{2})_{3}\\Fe^{3+} (OH^{-})_{3}--->Fe (OH)_{3}\\\\NH_{4}^{+}(C_{2}H_{3}O_{2})^{-}--->NH_{4}(C_{2}H_{3}O_{2})\\\\NH_{4}^{+}OH^{-} ---> NH_{4}OH[/tex]
Identify the acid, base, conjugate acid and conjugate base in the following reactions:
a. NH_3(aq) + CH_3COOH(aq) NH_4^+ (aq) + CH_3COO^-(aq)
b. HClO_4(aq) + NH_4(aq) ClO_4^- (aq) + NH_4^+ (aq)
Answer:
a. NH₃ : base
CH₃COOH (acetic acid) : acid
NH₄⁺ : conjugate acid
CH₃COO⁻ : conjugate base
b. HClO₄ (perchloric acid) : acid
NH₃ : base
ClO₄⁻ : conjugate base
NH₄⁺ : conjugate acid
Hope this helps.
. Explain why, in the sample calculations, 0.1 g of the unknown produced a GREATER freezing point depression than~e same mass of naphthalene.
Answer
Naphthalene is a non electrolyte
If the unknown compound is an electrolyte it gives 2 or more ions in solution
( NaCl >> Na+ + Cl- => 2 ions
Ca(NO3)2 >> Ca2+ + 2 NO3- => 3 ions)
the f.p. lowering is directly proportional to the molal concentration of dissolved ions in the solution )
For naphthalene
delta T = 1.86 x m
for a salt that gives 2 ions
delta T = 1.86 x m x 2
hence the lowering in freezion point of unkown is greater then napthalene
Which best describes the trends in electonegativity on the periodic table
Hey! :)
__________ ☆ ☆__________________________________
Answer:
The answer is Electronegativity increases up and to the right
Explanation:
When you move from left to right it increases ( in the periodic table )
But when you move down the table electronegativity decreases.
So “ Electronegativity increases up and to the right” describes the trends the best.
Hope this helps! :)
____________☆ ☆________________________________
By, BrainlyMember ^-^
Good luck!
Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant of nitrous acid is
Answer:
4.26 %
Explanation:
There is some info missing. I think this is the original question.
Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant of nitrous acid is 4.50 × 10 ⁻⁴.
Step 1: Given data
Initial concentration of the acid (Ca): 0.249 M
Acid dissociation constant (Ka): 4.50 × 10 ⁻⁴
Step 2: Write the ionization reaction for nitrous acid
HNO₂(aq) ⇒ H⁺(aq) + NO₂⁻(aq)
Step 3: Calculate the concentration of nitrite in the equilibrium ([A⁻])
We will use the following expression.
[tex][A^{-} ] = \sqrt{Ca \times Ka } = \sqrt{0.249 \times 4.50 \times 10^{-4} } = 0.0106 M[/tex]
Step 4: Calculate the percent ionization of nitrous acid
We will use the following expression.
[tex]\alpha = \frac{[A^{-} ]}{[HA]} \times 100\% = \frac{0.0106M}{0.249} \times 100\% = 4.26\%[/tex]