Answer:
Constitutional Isomers
Explanation:
Constitutional isomers can be regarded as structural isomers ,these are compounds that have the same molecular formula with different structural formulas. Example is Butane and isobutane; both compound posses the same molecular formula(C4H10) with different structural formulas. compounds of Constitutional isomers are differ in term of connectivity, which
is the way the constituent atoms are been connected to another. It should be noted that Constitutional Isomers Share the same molecular formula but have different connectivity of atoms and different physical properties.
20ml of water is mixed with 40gm of fine powder. Calculate the concentration of the solution obtained.
Answer:
[tex]\%m=66.7\%[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to calculate the concentration of the solution obtained, by knowing 20 mL of water are the same to 20 g and therefore the mass of the solution is 40g+20g=60g.
Next, we apply the following equation to obtain the required concentration:
[tex]\%m=\frac{40g}{60g} *100\%\\\\\%m=66.7\%[/tex]
Regards!
This reaction was at equilibrium when 0.2 atm of iodine gas was pumped into the container, what happened to the equilibrium and the partial pressures of the gases
Answer:
Q was < K. Partial pressure of hydrogen decreased, iodine increased
Explanation:
After iodine was added the Q was [Select] K so the reaction shifted toward the Products [Select] ,The partial pressure of hydrogen [Select], Iodine [Select] |,and hydrogen iodide Decreased
Based on the equilibrium:
H2(g) + I2(g) ⇄ 2HI(g)
K of equilibrium is:
K = [HI]² / [H2] [I2]
Where [] are concentrations at equilibrium
And Q is:
Q = [HI]² / [H2] [I2]
Where [] are actual concentrations of the reactants.
When the reaction is in equilibrium, K=Q.
But as [I2] is increased, Q decreases and Q was < K
The only concentration that increases is [I2], doing partial pressure of hydrogen decreased, iodine increased
4) The initial rate of the reaction between substances P and Q was measured in a series of
experiments and the following rate equation was deduced.
rate = k[P]?[Q]
Complete the table of data below for the reaction between P and Q
*Help asap please*
Answer:
The initial rate of the reaction between substances P and Q was measured in a series of
experiments and the following rate equation was deduced.
[tex]rate = k[P]^{2} [Q][/tex]
Complete the table of data below for the reaction between P and Q
Explanation:
Given rate of the reaction is:
[tex]rate= k[P]^{2} [Q]\\=>[Q]=\frac{rate}{k.[P]^{2} } \\and \\\\\\\ [P]=\sqrt{\frac{rate}{k.[Q]} }[/tex]
Substitute the given values in this formulae to get the [P], [Q] and rate values.
From the first row,
the value of k can be calulated:
[tex]k=\frac{rate}{[P]^{2}[Q] } \\ =\frac{4.8*10^-3}{(0.2)^{2} 2. (0.30)} \\ =0.4[/tex]
Second row:
2. Rate value:
[tex]rate =0.4* (0.10)^{2} * (0.10)\\\\ =4.0*10^-3mol.dm^-3.s^-1[/tex]
3.Third row:
[tex][Q]=\frac{rate}{k.[P]^{2} } \\ =9.6*10^-3 / (0.4 *(0.40)^{2} \\ =0.15mol.dm^{-3}[/tex]
4. Fourth row:
[tex][P]=\sqrt{\frac{rate}{k.[Q]} }\\=>[P]=\sqrt{\frac{19.2*10^-3}{0.60*0.4} } \\=>[P]=0.283mol.dm^{-3}[/tex]
What is the pH of 0.6 M NaOH?
Answer:
pOH = - log[OH-]
[OH-] = 0.6M
[tex]pOH \: = - log(0.6) \\ = 0.2218487496 \\ pH \: + pOH \: = 14 \\ pH \: + 0.221848749 = 14 \\ pH = 14 - 0.221848749 \\ = 13.77815125 \\ 13.8[/tex]
38. Consider the following equilibrium:
2CO(g) + O2(g) =2CO2
Keg=4.0 x 10-10
What is the value of Key for 2CO2(g) + 2COR + O2g) ?
Answer:
[tex]Key=2.5x10^{-9}[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to calculate the equilibrium constant value for the reverse reaction:
[tex]2CO_2(g) \rightleftharpoons 2CO(g) + O_2(g)[/tex]
By knowing that the equilibrium expression is actually:
[tex]Key =\frac{[CO]^2[O_2]}{[CO_2]^2} =\frac{1}{Keg}[/tex]
Thus, we plug in and solve for the inverse of Keq to obtain Key as follows:
[tex]Key =\frac{1}{4.0x10^{-10}}\\\\Key=2.5x10^{-9}[/tex]
Regards!
4. Complete the following equations:
CuCl2 + Na2CO3 → 2 NaCl +............
FeSO4 + BaCl2 →
Cu(NO3)2 + CaCO3
Answer:
2NaCl + CuCO3
FeCl2 + BaSO4
CuCO3 + Ca(NO3)2
Explanation:
Presumably this is a double replacement reaction.
A+B + C+D → A+D + C+B
It seems I may be wrong so please try to work out the problem yourself to double check, keeping in mind the charges of each compound.
To determine the freezing point depression of a LiCl solution, Toni adds 0.411 g of LiCl to the sample test tube along with 19.7 mL of distilled water. Determine the molal concentration (m) of the resulting solution. MWLiCl
Answer:
LiCl = 0.492 m
Explanation:
Molal concentration is the one that indicates the moles of solute that are contained in 1kg of solvent.
Our solute is lithium chloride, LiCl.
Our solvent is distilled water.
We do not have the mass of water, but we know the volume, so we should apply density to determine mass.
Density = mass / volume
Density . volume = mass
1 g/mL . 19.7 mL = 19.7 g
We convert g to kg → 19.7 g . 1 kg / 1000g = 0.0197 kg
Let's determine the moles of LiCl
0.411 g . 1 mol / 42.394 g = 9.69×10⁻³ moles
Molal concentration (m) = 9.69×10⁻³ mol / 0.0197 kg → 0.492 m
The molal concentration of the resulting solution is 0.492 m. The concentration number of moles of solute in one kg of solvent.
What is Molal concentration?It is the measure of the concentration number of moles of solute in one kg of solvent.
To calculate the molal concentration first calculated the mass of [tex]\bold {LiCl}[/tex],
[tex]\rm \ mass = density \times volume[/tex]
Put the values,
m = 1 g/mL x 19.7 mL
m = 19.7 g or
m = 0.0197 kg
Calculate the moles of LiCl:
[tex]n =\rm \dfrac { 0.411 \ g \times 1 mol }{ 42.394 \ g }\\\\n = 9.69\times 10^{-3} \ moles[/tex]
So, now the molal concentration,
[tex]m = \rm \dfrac { 9.69\times 10^{-3}\ mol} {0.0197 \ kg} \\\\m = 0.492[/tex]
Therefore, the molal concentration of the resulting solution is 0.492 m.
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If 4.00 moles of O2 occupies a volume of 5.0 L at a particular temperature and pressure, what volume will 3.00 moles of oxygen gas occupy under the same condition?
Answer: Volume occupied by 3.00 moles of oxygen gas under the same condition is 3.75 L.
Explanation:
Given: [tex]n_{1}[/tex] = 4.00 moles, [tex]V_{1}[/tex] = 5.0 L
[tex]n_{2}[/tex] = 3.00 moles, [tex]V_{2}[/tex] = ?
Formula used is as follows.
[tex]\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}[/tex]
Substitute the values into above formula as follows.
[tex]\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}\\\frac{5.0 L}{4.00 mol} = \frac{V_{2}}{3.00 mol}\\V_{2} = 3.75 L[/tex]
Thus, we can conclude that volume occupied by 3.00 moles of oxygen gas under the same condition is 3.75 L.
HELP!!! i will give brainliest!!
Different chemical elements have different chemical symbols, and this is determined by their atomic structure. Look at the two chemical symbols in the image. Compare and contrast the atomic symbols and the atomic structure of fluorine and oxygen.
Answer:
Explanation:
fluorine have gained one electron that is why the sign is -1. they both have different number of protons. They have different neutron numbers. F have 10 and O have 8.
hope this helps :)
what's the ph of 0.0000067 m hcl solution
Answer:
[tex]pH = - log[H {}^{ + } ] \\ = - log(0.0000067) \\ pH = 5.17[/tex]
A substance with two oxygen atoms is combined with a substance with one oxygen atom to form one product. What is true of the product?
There will be no oxygen in the product. Some of the oxygen will evaporate into the air.
Define pure substance. How is it classified on the basid of chemical properties?
Answer:
if it is pure, the substances is either an element or a compound. if a substance is not chemically pure, it is either a heterogeneous mixture or a homogeneous mixture. if its composition is uniform throughout, it is a homogeneous.
The information below describes a redox reaction.
Ag+ (aq) + Al(s) — Ag(s) + Al3+ (aq)
Ag+ (aq) + -> Ag(s)
Al(s)->A3+ (aq) + 3e-
What is the coefficient of silver in the final, balanced equation for this reaction?
Answer:
Al°(s) + 3Ag⁺(aq) => Al⁺³(aq) + 3Ag(s)
Explanation:
Oxidation: Al°(s) => Al⁺³(aq) + 3e⁻
Reduction: 3Ag⁺(aq) + 3e⁻ => 3Ag°(s)
_________________________________________
Net Rxn: Al°(s) + 3Ag⁺(aq) => Al⁺³(aq) + 3Ag(s)
One mole of neutral aluminum atoms (Al°(s)) undergo oxidation delivering 3 moles of electrons to 3 moles silver ions (3Ag⁺³(aq)) that are reduced to 3 moles of neutral silver atoms (3Ag°(s)) in basic standard state 25°C; 1atm.
A balanced equation obeys the law of conservation of mass. According to the law of conservation of mass, mass can neither be created nor be destroyed. The coefficient of silver is 3.
What is a balanced equation?A balanced chemical equation can be defined as the chemical equation in which the number of reactants and products on both sides of the equation are equal. The amount of reactants and products on both sides of the equation will be equal in a balanced chemical equation.
The numbers which are used to balance the chemical equation are called the coefficients. The coefficients are the numbers which are added in front of the formula.
The balanced chemical equation for the given redox reaction is given as:
Al (s) + 3 Ag⁺ (aq) → Al³⁺ (aq) + 3Ag (s)
Thus the coefficient of silver is 3.
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Please help me, it’s my last try
Answer:
Group 1A: alkali metals, or lithium family.
Group 2A: alkaline earth metals, or beryllium family.
Group 7A: the manganese family.
Group 8A: the iron family.
Explanation:
Answer:
1A: Alkali Metals
2A: Alkaline Earth Metals
7A: Halogens
8A: Noble Gases
In the presence of excess iodide ions, the iodine formed by reaction of iodide with NBS will react further to form triiodide ions. What does the triiodide combine with to form the blue color of the endpoint
Answer:
Starch.
Explanation:
When the triiodide combine with starch, it forms dark blue colour. Amylose in starch is responsible for the occurrence of a deep blue color when the iodine is combine with the starch. The iodine molecule goes inside of the amylose coil which makes a linear triiodide ion complex that goes into the coil of the starch that leads to an intense blue-black color in the end so we can say that starch turns the colour into blue.
At a fixed volume, a four-fold increase in the temperature of a gas will lead to _______ in pressure.
Question 2 options:
A)
no change
B)
a two-fold decrease
C)
a four-fold decrease
D)
a four-fold increase
Answer:
D) a four-fold increase
Explanation:
According to Gay-Lussac's law, which states that the pressure of a given amount of gas is directly proportional to the temperature at a constant volume, the pressure increases with an increase in temperature.
According to this question, at a fixed volume, a four-fold increase in the temperature of a gas will lead to a four-fold increase in the pressure as well.
which type of chemical bond would be formed between two elements having electron configuration of 1s2 2s2 2p6 3s2 and 1s2 2s2 2p4
If we have 1.23 mol of NaOH in solution and 0.85 mol of Cl2 gas is available to react, which one is the limiting reactant? Give your reason.
Answer:
NaOH is the limiting reactant.
Explanation:
Hello there!
In this case, since the reaction taking place between sodium hydroxide and chlorine has is:
[tex]NaOH+Cl_2\rightarrow NaCl+NaClO+H_2O[/tex]
Which must be balanced according to the law of conservation of mass:
[tex]2NaOH+Cl_2\rightarrow NaCl+NaClO+H_2O[/tex]
Whereas there is a 2:1 mole ratio of NaOH to Cl2, which means that the moles of the former that are consumed by 0.85 moles of the latter are:
[tex]n_{NaOH}=0.85molCl_2*\frac{2molNaOH}{1molCl_2}\\\\n_{ NaOH}=1.7molNaOH[/tex]
Therefore, since we just have 1.23 moles out of 1.70 moles of NaOH, we infer this is the limiting reactant.
Regards!
La función de la levadura en quimica
Explanation:
las levaduras son pequeños organismos unicelulares que se alimentan de azúcares simples y los descomponen en dióxido de carbono, alcohol (etanol, específicamente), moléculas de sabor y energía. El proceso se conoce como fermentación.
Acetylide ions react with aldehydes and ketones to give alcohol addition products.
a. True
b. False
Answer:
a
Explanation:
Calculate the number of milliliters of 0.587 M NaOH required to precipitate all of the Fe3 ions in 197 mL of 0.654 M FeCl3 solution as Fe(OH)3. The equation for the reaction is: FeCl3(aq) 3NaOH(aq) Fe(OH)3(s) 3NaCl(aq)
Answer: The number of milliliters of 654 mL for 0.587 M NaOH required to precipitate all of the [tex]Fe^{3+}[/tex] ions in 197 mL of 0.654 M [tex]FeCl_{3}[/tex] solution as [tex]Fe(OH)_{3}[/tex].
Explanation:
The reaction equation is as follows.
[tex]FeCl_{3}(aq) + 3NaOH(aq) \rightarrow Fe(OH)_{3}(s) + 3NaCl(aq)[/tex]
Therefore, moles of [tex]Fe(OH)_{3}[/tex] are calculated as follows.
Moles = Molarity of [tex]Fe(OH)_{3}[/tex] [tex]\times[/tex] Volume (in L)
= 0.654 M [tex]\times[/tex] 0.197 L
= 0.128 mol
Now, according to the given balanced equation 1 mole of [tex]FeCl_{3}(aq)[/tex] reacts with 3 moles of NaOH(aq). Hence, moles of [tex]Fe(OH)_{3}[/tex] reacted are calculated as follows.
3 [tex]\times[/tex] 0.128 mol = 0.384 moles of NaOH
As moles of NaOH present are as follows.
Moles of NaOH = Molarity of NaOH [tex]\times[/tex] Volume (in L)
0.384 mol = 0.587 M [tex]\times[/tex] Volume (in L)
Volume (in L) = 0.654 L (1 L = 1000 mL) = 654 mL
Thus, we can conclude that the number of milliliters of 654 mL for 0.587 M NaOH required to precipitate all of the [tex]Fe^{3+}[/tex] ions in 197 mL of 0.654 M [tex]FeCl_{3}[/tex] solution as [tex]Fe(OH)_{3}[/tex].
A student sets up the following equation to convert a measurement.
(The stands for a number the student is going to calculate.)
Fill in the missing part of this equation.
Answer:
–0.13 Pa.m²
Explanation:
From the question given above, the following data were obtained:
Measurement (Pa.mm²) = –1.3×10⁵ Pa.mm²
Measurement (Pa.m²) =?
We can convert from Pa.mm² to Pa.m² by doing the following:
1 Pa.mm² = 1×10¯⁶ Pa.m²
Therefore,
–1.3×10⁵ Pa.mm² = –1.3×10⁵ Pa.mm² × 1×10¯⁶ Pa.m² / 1 Pa.mm²
–1.3×10⁵ Pa.mm² = –0.13 Pa.m²
Thus, –1.3×10⁵ Pa.mm² is equivalent to –0.13 Pa.m².
The complete equation will be:
[tex](-1.3\times 10^5 Pa.mm^2)\times 10^{-6}=(-0.13) Pa.m^2[/tex]
Explanation:
Given:
The equation to convert a measurement:
[tex](-1.3\times 10^5 Pa.mm^2)\times ? = ? Pa.m^2[/tex]
To find:
The missing part of the equation.
Solution:
[tex](-1.3\times 10^5 Pa.mm^2)\times ? = ? Pa.m^2[/tex]
On LHS the unit is in [tex]Pa. mm^2[/tex] and RHS the unit is in [tex]Pa.m^2[/tex] which means that we have to convert [tex]mm^2[/tex] to [tex]m^2[/tex]
In 1 millimeter there are 0.001 meters.
[tex]1 mm = 0.001 m\\1 mm^2=0.000001 m^2=10^{-6} m^2[/tex]
So, the complete equation will be:
[tex](-1.3\times 10^5 Pa.mm^2)\times 10^{-6}=(-0.13) Pa.m^2[/tex]
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For the reaction...
N2 + O2 <=> 2NO: AH = +182 kJ mol-1.
If the temperature is increased the equilibrium position will shift
Your answer:
a) to the left
b) to the right
c) to the left and right
d) neither left nor
right
Answer:
B
Explanation:
AH is positive so the forward reaction is endothermic. Thus, increasing temperature would cause equilibrium to shift to the right as endothermic reaction favors higher temperature. This increases the yield of NO.
Which of the following is not a polymer
A. Glucose.
B. Starch.
C. Cellulose.
D. DNA.
Answer:
A. Glucose
Explanation:
Glucose is a monomer and not a polymer. So, option (A) is not a polymer.
Glucose is not a polymer because it is a kind of molecule while Starch , cellulose and DNA are polymers.
The correct answer is option A. Glucose.
a sample of cobalt, A, with a mass of 5.00g, is initially at 25 C. When this sample gains 6.70 J of heat, the temperature rises to 27.9 C. Another sample of cobalt, B, with a mass of 7.00 g, iw initially at 25 C. If sample B gains 5.00 J of heat, what is the final temperature of sample B
Answer:
26.5°C
Explanation:
We can solve this question using the equation:
q = m*S*ΔT
Where Q is heat gained in joules, m is the mass of the sample, S is specific heat and ΔT change in temperature.
With the sample A we can find specific heat of cobalt in order to find, in sample B, the ΔT and the final temperature:
Sample A:
q = m*S*ΔT
6.70J = 5.00g*S*(27.9°C-25.0°C)
0.462J/g°C = Specific heat of cobalt
Sample B:
q = m*S*ΔT
5.00J = 7.00g*0.462J/g°C*ΔT
1.5°C = ΔT
As the initial temperature of sample B is 25°C, final temperature is:
25°C + 1.5°C = 26.5°C
Star
Planet
*
As the planet makes one completer revolution around the star, starting at the position shown the gravitational attraction between the star
and the planet will
A Continually decrease
3 Decrease, then increase
increase then decrease
Romain the same
RI
12.20 AM
618/2001
Answer:
according to the path shown in the figure it will start decreasing then again it will start increasing when the path will be nearer to the star.
Reason is gravitation force is indirectly proportional to the distance.
So, option B. decrease then increase is correct
1) 7.269 moles of oxygen gas are used in combusting butane (C H..). How many moles of carbon dioxide
gas are produced? You must start with a balanced chemical equation. Start with a balanced equation
Explanation:
C4H10 + 13/2O2 ---------> 4CO2 + 5H2O
so u can work out the amount of moles by doing
moles=mass/mr
mr of C4H10 is 12 × 4 + 10 =58
=7.269/58
= 0.125moles
Then u can use the molar ratio which is
6.5:4
0.125 ÷6.5 × 4 = 0.0769moles
hope this helps:)
An elephant walks north from a watering hole for 1 mile. The elephant
encounters a hill and turns east. The elephant continues to walk east until it
reaches a tree. It stays near the tree for 1 hour before continuing on to a
grassy field. Which point of reference should be used to describe the
elephant's motion?
A. The grassy area
B. The hill
STA
Tel
C. The tree
D. The watering hole
Answer:
i think its c because thats when the direction changes and you dont know where he went.
Explanation:
A pressure cooker contains 5.68 L of air at a temperature of 390 4K if the absolute pressure of the air in the pressure cooker is 205 Pa how many moles of air are in the cooker
Answer:
3.59x10⁻⁴ mol
Explanation:
Assuming ideal behaviour we can solve this problem by using the PV=nRT formula, where:
P = 205 PaV = 5.68 Ln = ?R = 8314.46 Pa·L·mol⁻¹·K⁻¹T = 390.4 KWe input the data given by the problem:
205 Pa * 5.68 L = n * 8314.46 Pa·L·mol⁻¹·K⁻¹ * 390.4 KAnd solve for n:
n = 3.59x10⁻⁴ mola) If we have a 4.5 L container of CH 10 gas at a temperature of 178 K and a pressure of 0.50 atm, then how many moles of CaHio do
we have?
b) How many grams of C4H1o do we have?
Answer:
a) 0.15 mol.
b) 8.95 g.
Explanation:
Hello there!
In this case, according to the given information, it is possible for us to infer this problem is solved by using the ideal gas equation:
[tex]PV=nRT[/tex]
And proceed as follows:
a) Here, we solve for the moles, n, as follows:
[tex]n=\frac{PV}{RT} \\\\n=\frac{0.50atm*4.5L}{0.08206\frac{atm*L}{mol*K}*178K} \\\\n=0.15mol[/tex]
b) for the calculation of the mass, we recall the molar mass of butane, 58.12 g/mol, to obtain:
[tex]0.15mol*\frac{58.12g}{1mol} =8.95g[/tex]
Regards!