Answer:
NaoH= sodium hydroxide
When a mixture of potassium chlorate and an appropriate catalyst is heated in an open test tube, potassium chloride and oxygen gas are produced. This reaction is not reversible because ______________ a the reaction does not go to completion b the system reaches equilibrium c the system is closed d the oxygen escapes
Answer:
d the oxygen escapes.
Explanation:
Hello there!
In this case, according to the given information, it turns out firstly convenient for us to write out the chemical reaction taking place:
[tex]2KClO_3\rightarrow 2KCl+3O_2[/tex]
Thus, since both potassium chlorate and chloride are solid whereas oxygen is gaseous, it tends to scape as we are working in an open test tube and that is why the products side do not have the reactants, KCl and O2, to return the reactants side, KClO3 in agreement to the concept of equilibrium reaction. This situation can be solved by working in a closed container.
Therefore the answer is d the oxygen escapes.
Regards!
please help me with b and c.
Answer:
c.sf4 b.2
Explanation:
A reaction at 5.0°C evolves 583.mmol of dinitrogen difluoride gas. Calculate the volume of dinitrogen difluoride gas that is collected. You can assume the pressure in the room is exactly 1atm . Round your answer to 3 significant digits.
Answer: The volume of dinitrogen difluoride gas collected is 13.31 L.
Explanation:
Given: Temperature = [tex]5.0^{o}C[/tex] = (5 + 273) K = 278 K
Moles = 583 mmol (1 mmol = 0.001 mol) = 0.583 mol
Pressure = 1 atm
Formula used to calculate volume is as follows.
PV = nRT
where,
P = pressure
V = volume
n = no. of moles
R = gas constant = 0.0821 L atm/mol K
T = temperature
Substitute the values into above formula as follows.
[tex]PV = nRT\\1 atm \times V = 0.583 \times 0.0821 L atm/mol K \times 278 K\\V = 13.31 L[/tex]
Thus, we can conclude that the volume of dinitrogen difluoride gas collected is 13.31 L.
What enzyme below is an exoenzyme?
A. Casease
B. Citrase
C. Catalase
D. Oxidase
A one electron species, X m, where m is the charge of the one electron species and X is the element symbol, loses its one electron from its ground state when it absorbs 3.49 x 10-17 J of energy. Using the prior information, the charge of the one electron species is:_____________
a. +8
b. +2
c. +3
d. +1
e. +4
Answer:
Option C
Explanation:
From the question we are told that:
Difference in energy [tex]\delta E =3.49 * 10^{-17} J[/tex]
The Ground state Difference in energy at n=1
[tex]\delta E_g = 2.18 * 10^{-18} × Z^2[/tex]
Generally the equation for Difference in energy is mathematically given by
[tex]\delta E=\delta E_g[/tex]
Therefore
[tex]3.49 * 10^{-17} = 2.18 * 10^{-18} * Z^2[/tex]
[tex]Z^2=16[/tex]
[tex]Z=4[/tex]
Therefore
Charge on element Z Q_Z
[tex]Q_Z= Atomic\ no\. of\ element - No.\ of\ electrons\ of\ element[/tex]
[tex]Q_Z =4-1[/tex]
[tex]Q_Z=+3[/tex]
Option C
Suppose you had a 2.4 g cracker burn down to 1.3 g, which raised the temperature of 50.1 g of water by 12.0 degrees Celsius. How many kilocalories of heat energy was released by the cracker (or absorbed by the water) per gram of cracker
Answer:
We know that the specific heat of water is:
c = 1cal/g*°C
This means, that we need 1 cal to increase the temperature of 1 gram of water by 1°C
Here, we increased the temperature of 50.1g of water by 12°C
Then the number of calories needed to do this is given by:
x = (mass of water in grams)*(how much increased the temp in °C)*1cal/g*°C
x = (50.1g*12°C)*1cal/g*°C = 601.2 cal
But we want this in Kcal, remember that:
1Kcal = 1000cal
Then:
601.2 cal = (601.2/1000) Kcal = 0.6012 Kcal
Now, for the cracker part, the energy was released by the amount of cracker that was burned.
The original mass was 2.4g
the final mass ios 1.3g
the difference is:
2.4g - 1.3g = 1.1g
This means that 1.1g was the burned mass.
The number of kilocalories of heat per gram released by the cracker is just:
n = (0.6012 Kcal)/(1.1 g) = 0.547 Kcal/g
0.547 kilocalories per gram.
Use the following key to classify each of the elements below in its elemental form:
a. Discrete atoms
b. Molecules
c. Metallic lattice
d. Covalent Network.
1. Phosporus
2. Bromine
3. Hydrogen
4. Krypton
Explanation:
Given set of elements in one column and their classification in another column.
1. Phosporus c.Metallic lattice
2. Bromine d.Covalent network.
3. Hydrogen b. Molecules
4. Krypton a. Discrete atoms
Since Krypton is an inert gas and it exists in discrete atoms.
Hydrogen exists as a diatomic gas.
Bromine exists as Br_2 liquid and is held by covalent bonds.
Phosphorus exists as P_4 molecules and it exits as a metallic lattice.
Apakah ciri-ciri ikan
Answer:
fishes are cold blooded
Classify the processes as endothermic or exothermic.
a. Ice melting
b. Water condensing on surface
c. Baking a cake
d. The chemical reaction inside an instant cold pack.
e. A car using gasoline
endothermic absorbs heat
exothermic gives heat
a. endothermic
b. exothermic
c. endothermic
d. exothermic
a. Ice melting - endothermic
b. Water condensing on the surface - exothermic
c. Baking a cake - endothermic
d. The chemical reaction inside an instant cold pack - endothermic
e. A car using gasoline - exothermic
What is an exothermic and endothermic reaction?An exothermic reaction can be described as a thermodynamic chemical reaction that emits energy from the system to its surroundings usually in the form of light, heat, or sound.
While an endothermic reaction can be described as an opposite of an exothermic reaction where the energy gains in the form of heat. In exothermic chemical reactions, the bond energy is transformed into thermal energy.
In exothermic reactions, the reaction happens the form of the kinetic energy of molecules when the energy is released. The release of energy is due to the electronic transition of electrons from one energy level to another.
The burning of gasoline, and water condensation is also an exothermic reaction in which energy is released while ice melting and baking cake is an endothermic reaction.
Learn more about the exothermic process, here:
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You should set out support, like a cork ring or clamp, before removing the glassware from a glassware kit to place the glassware in and to stop it from _________. Thoroughly check that the glasswar is________ and that it does not have any _______before using it.
Answer:
(A) Slipping and breaking
(B) Clean and dry
(C) Cracks
Explanation:
This describes the process of unpacking a glassware for use.
You should set out support like a cork ring or clamp (these are simple machines that'll hold the glassware in place) before removing the glassware from a glassware kit; to place the glassware in and to stop it from slipping and breaking.
Thoroughly check that the glassware is clean and dry and that it does not have any cracks, before using it.
When 52 grams of O2 react with excess C3H8, how many grams of CO2 would be produced?
Answer:35.1g of CO2
Explanation:
C3H8+5O2------=3CO2+4H2O
It is found that, when a dilute gas expands quasistatically from 0.40 to 5.0 L, it does 210 J of work. Assuming that the gas temperature remains constant at 300 K, how many moles of gas are present
Answer:
[tex]n=0.033mole[/tex]
Explanation:
From the question we are told that:
Initial volume [tex]V_1=0.40L[/tex]
Final Volume[tex]V_2=5.0L[/tex]
Work [tex]W=210J[/tex]
Temperature [tex]T=300k[/tex]
Generally the equation for Ideal gas is mathematically given by
[tex]W=nRTIn\frac{V_2}{V_1}[/tex]
[tex]n=\frac{W}{RTIn\frac{V_2}{V_1}}[/tex]
[tex]n=\frac{210}{8.32*300In\frac{5.0}{0.4}}[/tex]
[tex]n=0.033mole[/tex]
any two functions of crystals
Answer:
1. Participating in calcium homeostatis storage of calcium.
2. High capacity calcium (Ca) regulation and protection against herbivory
[tex]\large \boxed{\sf 2 \: functions \: of \: crystals \: are :- } [/tex]
_________________
⟹
[tex] \sf \: \underline{ Calcium \: oxalate \: (CaOx) \: crystals} \: are \: distributed \: \\\sf among \: all \: taxonomic \: levels \\ \sf\: of \: photosynthetic \: organisms \: from \\ \sf \: small \: algae \: to \: angiosperms \: and \: giant \: gymnosperms .[/tex]
__________________
⟹
[tex]\sf Bone \: is \: mostly \: made \: of \: \underline{mineral \: crystals} \: \\ \sf and \: the \: protein \: collagen. \: The \: mineral \: crystals \: bone \\ \sf\: provide \: strength \: and \: rigidity \: for \: the \: matrix \: upon \: \\ \sf \: and \: within \: which \: they \: are \: deposited.[/tex]
If a 0.320 mM solution of MnO41- has an absorbance of 0.480 at 525 nm in a 1.000 cm cell. What is the concentration of a MnO41- solution that has absorbance of 0.490 in the same cell at that wavelength?
Answer:
Hence the concentration of a MnO41- solution that has absorbance of 0.490 in the same cell at that wavelength is 0.3266.
Explanation:
Now A = el, el=const
Then,
[tex]A2 / A1 = C2/ C1\\\\0.49/ 0.48 = C2 / 0.32\\\\C2 = 0.3266[/tex]
10. Hydrogen peroxide (H2O2: M = 34 g mol-1) is a powerful oxidising agent that is used in concentrated solution in rocket fuel systems and in dilute solution as in hair bleach. An aqueous solution of H2O2 is 30 % by mass and has a density of 1.11 g cm-3. Express the concentration of the solution in terms of: (i) Molality
Answer:
The molality of solution=12.605 m
Explanation:
We are given that
Molar mass of Hydrogen peroxide, M=34 g/mol
Density of solution, [tex]\rho=1.11gcm^{-3}[/tex]
30% Means mass of solute (Hydrogen peroxide)=30 g
Mass of solvent =100-30=70 g
Total mass of solution, m=100 g
Number of moles of solute=[tex]\frac{given\;mass}{molar\;mass}[/tex]
Using the formula
Number of moles of hydrogen peroxide=[tex]\frac{30}{34}[/tex]
Now, molality of solution
[tex]m=\frac{number\;of\;moles\;of\;solute}{mass\;of\;solvent}\times 1000[/tex]
[tex]m=\frac{30}{70\times 34}\times 1000[/tex]
[tex]m=12.605 m[/tex]
Hence, the molality of solution=12.605 m
differences between expansion of solid and liquid
Sometimes in lab we collect the gas formed by a chemical reaction over water . This makes it easy to isolate and measure the amount of gas produced.
Suppose the CO, gas evolved by a certain chemical reaction taking place at 50.0°C is collected over water, using an apparatus something like that in the sketch, and the final volume of gas in the collection tube is measured to be 132. mL. Calculate the mass of CO, that is in the collection tube. Round your answer to 2 significant digits.
Answer:
0.17 g
Explanation:
Since the volume of gas collected is 132 mL, we need to find the number of moles of gas present in 132 mL.
So, number of moles, n = volume of gas, v/molar volume, V
n = v/V where v = 132 mL = 0.132 L and V = 22.4 L
So, substituting the values of the variables into the equation, we have
n = v/V
n = 0.132 L/22.4 L
n = 0.005893 mol
We then need to calculate the molar mass of CO, M = atomic mass of carbon + atomic mass of oxygen = 12 g/mol + 16 g/mol = 28 g/mol
Also, number of moles of gas, n = m/M where m = mass of CO and M = molar mass of CO
m = nM
m = 0.005893 mol × 28 g/mol
m = 0.165004 g
m ≅ 0.17 g to 2 significant digits
The decomposition of ethyl amine, C2H5NH2, occurs according to the reaction: C2H5NH2(g)⟶C2H4(g)+NH3(g) At 85∘C, the rate constant for the reaction is 2.5 x 10-1 s-1. What is the half-life (in sec) of this reaction?
Answer:
2.772 seconds
Explanation:
Given that;
t1/2 = 0.693/k
Where;
t1/2 = half life of the reaction
k= rate constant
Note that decomposition is a first order reaction since the rate of reaction depends on the concentration of one reactant
t1/2 = 0.693/2.5 x 10-1 s-1
t1/2= 2.772 seconds
how many of the electrons in a molecule of ethane are not involved in bondind
Ethane consists of 6C−H bonds and 1C−C bond. Total number of bonds is 7. Each bond is made up of two electrons
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A system receives 425 J of heat from and delivers 425 J of work to its surroundings. What is the change in internal energy of the system (in J)?
Answer:
0 J
Explanation:
Applying,
ΔE = q+w................ Equation 1
Where ΔE = change in internal energy of the system, q = Heat of the system, w = work of the system.
Note: q is positive, while w is negative
From the question,
Given: q = 425 J, w = -425 J
Substitute these values into equation 1
ΔE = 425-425
ΔE = 0 J
Hence the change in internal energy of the system is 0 J
The homework question reads:
"A sample of gas in a cylinder of volume 3.42 L at 298 K
and 2.57 atm expands to 7.39 L by two different pathways.
Path A is an isothermal, reversible expansion. Path B has two
steps. In the fi rst step, the gas is cooled at constant volume to
1.19 atm. In the second step, the gas is heated and allowed to
expand against a constant external pressure of 1.19 atm until
the final volume is 7.39 L. Calculate the work for each path.
Answer:
Explanation:
this guy on brainly already did it:
Alleei
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Answer : The work done for path A and path B is -685.3 J and -478.1 J respectively.
Explanation :
To calculate the work done for path A :
First we have to calculate the moles of the gas.
where,
= initial pressure of gas = 2.57 atm
= initial volume of gas = 3.42 L
n = moles of gas = ?
R = gas constant = 0.0821 atm.L/mol.K
T = temperature of gas = 298 K
Now put all the given values in the above formula, we get:
According to the question, this is the case of isothermal reversible expansion of gas.
As per first law of thermodynamic,
where,
= internal energy
q = heat
w = work done
As we know that, the term internal energy is the depend on the temperature and the process is isothermal that means at constant temperature.
So, at constant temperature the internal energy is equal to zero.
The expression used for work done will be,
where,
w = work done on the system = ?
n = number of moles of gas = 0.359 mole
R = gas constant = 8.314 J/mole K
T = temperature of gas = 298 K
= initial volume of gas = 3.42 L
= final volume of gas = 7.39 L
Now put all the given values in the above formula, we get :
Thus, the work done of path A is, -685.3 J
To calculate the work done for path B :
The formula used for isothermally irreversible expansion is :
where,
w = work done
= external pressure = 1.19 atm
= initial volume of gas = 3.42 L
= final volume of gas = 7.39 L
Now put all the given values in the above formula, we get :
Thus, the work done of path B is, -478.1 J
Viết các đồng phân cấu tạo mạch hở của C4H6O2 cùng nhóm chức axit
Answer:
+ axit
CH2=CH-CH2-COOH,
CH3-CH=CH-COOH (tính cả đồng phân hình học)
CH2=C(CH3)-COOH.
+ este
HCOOCH=CH-CH3 (tính cả đồng phân hình học)
HCOO-CH2-CH=CH2,
HCOOC(CH3)=CH2.
CH3COOCH=CH2
CH2=CH-COOCH3
When taking pH measurements of solutions, why is it important to stir thoroughly after adding each reagent? a. Mixing helps keep the solution components from contaminating the pH meter. b. A pH meter can only measure ions that are flowing past the meter at a fast speed. c. Mixing helps prevent precipitation of insoluble salts out of solution. d. Mixing helps ensure that the heasured pH is reflective of the entire solution.
Answer:
d. Mixing helps ensure that the measured pH is reflective of the entire solution.
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to reason that the answer is d. Mixing helps ensure that the measured pH is reflective of the entire solution because mixing help us to move the ions all around the solution so it undergoes homogenization and the measured pH at any point of the solution will be the same.
Moreover, we need to keep in mind that the incomplete stirring leads to regions with more concentration of the acid or base, or what we know as a gradient of concentration, which may lead to a incorrect measurement.
Regards!
1. Most of the chemicals included in your General Chemistry Lab kit can be discarded down a drain. Describe a situation in which you would need to neutralize a chemical before discarding down a drain.
Answer: Chemicals like acids and bases are harmful and must be neutralized before draining.
Explanation:
A strong acid or strong base is required to be diluted or neutralized before it is discarded in the drain as if is discarded without diluting and neutralization it can spill and splash from sink or drain and can harm people in chemistry lab, moreover the fumes of the discarded chemical on spilling can cause respiratory tract burning and can even cause fire hazard so it must be converted into less harmful form and then must be drained.
42 Organic compound may have names ending in -ane, -ene, -ol or -oic acid. How many of these endings indicate the compounds contain double bonds in their molecules? * (1 Point)
Answer: Organic compounds ending with the name (-ene) indicate that the compounds contain double bonds in their molecules.
Explanation:
Organic compounds are those molecules that contains carbon atoms (as their main element), hydrogen and oxygen which are usually present. The presence of numerous organic compounds is due to the following properties of carbon:
--> the exceptional ability of carbon atoms to catenate, that is, to combine with one another to form straight chains, branched chains or ring compounds containing many carbon atoms.
--> The ease with which carbon combines with hydrogen, oxygen, Nitrogen and halogens
--> The ability of carbon atoms to form single, DOUBLE or triple bonds.
The organic compound that has the name ending with -ene are known as the alkenes. The members of the alkene series are formed from the alkanes by the removal of two hydrogen atoms and the introduction of a DOUBLE BOND in the carbon chain. They are named after the corresponding alkanes by changing the -ane ending to -ene.
Note: the systematic name of a compound is formed from the root hydrocarbon by adding a suffix and prefixes to denote the substitution of the hydrogen atoms.
5.60g of glyceraldehydes was dissolved in 10ml of a solvent and placed in a 50mm cell if the rotation is 1.74 calculate the specific rotation?
Answer:
6.214 degrees-mL/gdm
Explanation:
The specific rotation α' = α/LC where α = observed rotation, L = length of tube and C = concentration of solution.
Given that α = 1.74, L = length of cell = 50 mm = 0.50 dm and C = m/V where m = mass of glyceraldehyde = 5.60 g and V = volume = 10 ml
So, C = m/V = 5.60 g/10 ml = 0.560 g/ml
Since α' = α/LC
substituting the values of the variables into the equation, we have
α' = α/LC
α' = 1.74/(0.50 dm × 0.560 g/ml)
α' = 1.74/(0.28 gdm/l)
α' = 0.006214 °mL/gdm
α' = 6.214 °mL/gdm
α' = 6.214 degrees-mL/gdm
Use the Ka values for weak acids to identify the best components for preparing buffer solutions with the given pH values.
Name Formula Ka
Phosphoric acid H3PO4 7.5 x 10^-3
Acetic acid CH3COOH 1.8 x 10^-5
Formic acid HCOOH 1.8 x 10^-4
pH 1.9 =_________
pH 5.0 = ________
pH 3.9= ________
Answer:
pH= 1.9 then [tex]H_{3} PO_{4}[/tex]
pH = 5.0 , [tex]CH_{3} COOH[/tex]
pH = 3.9 , HCOOH
As we know range left [tex]pH= pKa+/- 1[/tex]
Which functional group is used in other functional groups?
A. Ester
B. Carbonyl
c. Hydroxyl
D. Amino
q9
Answer:
The answer is B. Carbonyl
Carbonyl is the functional group is used in other functional groups. Therefore, option (B) is correct.
What do you mean by carbonyl functional group?A functional group with the formula C=O that is composed of a carbon atom double-bonded to an oxygen atom and is divalent at the C atom is known as a carbonyl group in organic chemistry.
A carbonyl gathering is a synthetically natural utilitarian gathering made out of a carbon iota twofold clung to an oxygen molecule - - > [C=O] The most straightforward carbonyl gatherings are aldehydes and ketones typically connected to another carbon compound.
A functional group with a carbon double bonded to an oxygen is called a carbonyl group. They have unsurprising properties, like extremity, instability, and reactivity.
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If 52.5 mL of lead(II) nitrate solution reacts completely with excess sodium iodide solution to yield 0.248 g of precipitate, what is the molarity of lead(II) ion in the original solution
Explanation:
The volume of given lead nitrate solution is:
52.5 mL.
The amount of lead iodide formed is ---0.248 g.
To get the molarity of lead (II) ion follow the below-shown procedure:
The number of moles of lead iodide formed is:
[tex]number of moles of lead iodide(n)=mass of lead iodide/its molecular mass\\n=0.248 g/461.01g/mol\\n=0.000537mol[/tex]
0.000537 mole of lead iodide contains --- 0.000537 moles of lead (II) ion.
Thus, the number of moles are there, volume is there, and to get the molarity of lead (II) ion use the formula:
[tex]Molarity=\frac{number of moles}{volume in L.} \\M=0.000537 mol / 0.0525 L.\\M=0.0102mol/L[/tex]
Molarity of lead iodide is --- 0.0102 M.
The compound IF5 contains Question 16 options: polar covalent bonds with partial negative charges on the F atoms. ionic bonds. polar covalent bonds with partial negative charges on the I atoms. nonpolar covalent bonds.
Answer:
See explanation
Explanation:
The molecule IF5 possesses five I-F polar bonds. However, the presence of polar bonds does not automatically imply that the molecule will be polar.
The geometry of the molecule is very important in determining the polarity of a compound. Since IF5 has a lone pair of electrons, the molecule is bent and as such there is a permanent dipole moment created in the molecule thereby making IF5 polar in nature.
