Answer:
the nucelolo is not part of the atom
Explanation:
The nucleolus is NOT a part of the atom. The nucleolus has its definition in biology as an element of the cell.
Let's remember that the atom is made up of three fundamental elements that are: electrons, protons and neutrons. Protons and neutrons are found in the nucleus while electrons are orbiting around the nucleus.
Answer:
core
Explanation:
there is no such thing as a core in an atom.
The middle is known as the nucleus
What is the standard cell potential for the spontaneous voltaic cell formed from the given half-reactions
Answer:
because it is
A(n) _____ reaction occurs when an acid and a base are present in the same solution.
Answer:
The answer is Neutralization reaction
It occurs when an acid and a base are present in the same solution and react to form salt and water only
Hope this helps you
Which of the following are meso compounds? A) trans-1,4-dimethylcyclohexane B) cis-1,3-dimethylcyclohexane C) trans-1,3-dimethylcyclohexane D) cis-1,4-dimethylcyclohexane E) trans-1,2-dimethylcyclohexane
Answer:
See explanation
Explanation:
For this question, we have to remember the definition of a meso-compound. In a meso-compound, we will have chiral carbons but we don't optical activity. This is due to the symmetry, if we have symmetry in a substance with chiral carbons the optical activity is nullified. So, if we want to find the meso-compounds we have to find symmetry planes in the molecule.
A symmetry plane is an imaginary cut that can divide the molecule in two equal parts. We have to draw the molecule first (see figure 1) and then we can try to find the symmetry planes.
With this in mind, the only compounds with symmetry planes are:
b) cis-1,3-dimethylcyclohexane
d) cis-1,4-dimethylcyclohexane
See figure 2 to more explanations
I hope it helps!
The authors state in the general procedures that the reaction was monitored by TLC. How would this be done? What would you spot in each lane? How would you know the reaction was done?
Answer:
Thin Layer Chromatography (TLC) can be used to analyze chemical reactions. During this reaction monitoring, a typical TLC plate would have three spots: the reactant lane, the reaction mixture lane, and a "co-spot" where reaction product would be spotted directly on top of reactant.
The co-spot serves as a reference point and is vital for reactions where reactant and product have similar Rfs, and many other variations of eluent tracking.
To indicate completion of the reaction, the disappearance of a spot (usually the starting reactant) is observed.
Select the oxidation reduction reactions??
Answer:
Explanation:
1 ) Cl₂ + ZnBr₂ = ZnCl₂ + Br₂
In this reaction , oxidation number of Cl decreases from 0 to -1 so it is reduced and oxidation number of Br increases from -1 to 0 so it is oxidised . Hence this reaction is oxidation - reduction reaction .
2 )
Pb( ClO₄)₂ + 2KI = PbI₂ + 2KClO₄
In this reaction oxidation number of none is changing so it is not an oxidation - reduction reaction.
3 )
CaCO₃ = CaO + CO₂
In this reaction also oxidation number of none is changing so it is not an oxidation - reduction reaction.
So only first reaction is oxidation - reduction reaction.
2nd option is correct.
For the reaction, 2SO2(g) + O2(g) <--> 2SO3(g), at 450.0 K the equilibrium constant, Kc, has a value of 4.62. A system was charged to give these initial concentrations, [SO3] = 0.254 M, [O2] = 0.00855 M, [SO2] = 0.500 M. In which direction will it go?
Answer:
To the left.
Explanation:
Step 1: Write the balanced reaction at equilibrium
2 SO₂(g) + O₂(g) ⇄ 2 SO₃(g)
Step 2: Calculate the reaction quotient (Qc)
Qc = [SO₃]² / [SO₂]² × [O₂]
Qc = 0.254² / 0.500² × 0.00855
Qc = 30.2
Step 3: Determine in which direction will proceed the system
Since Qc > Kc, the system will shift to the left to attain the equilibrium.
2) Which type movement do pivot joints allow?
When solutions of hydrochloric acid and sodium hydroxide are mixed, a chemical reaction occurs forming aqueous sodium chloride and water. What would you expect to observe if you ran the reaction in the laboratory
Answer:
a change in temperature would be observed(ΔH is -ve)
Explanation:
Hydrochloric acid react with sodium hydroxide to give salt(sodium chloride) and water
HCl(aq) + NaOH(aq) =====> NaCl(aq) + H2O(l)
There would be no notable change since sodium chloride dissolved in water but there would be a change in temperature.
Since neutralization is exothermic(heat is evolved), therefore ΔH is negative
Do you think you could go a week without causing any chemical reactions?
yes yes yes yes
yes
yes
yes
yes
yes
The activation energy of the uncatalyzed reaction is about 3.98 times that of the catalyzed reaction with activation energy of 4.6 kJ. The uncatalyzed reaction has such a large activation energy that its rate is extremely slow. What is the activation energy for the uncatalyzed reaction
Answer:
18.308 KJ
Explanation:
From the given above, we obtained the following:
Activation energy for the catalyzed reaction = 4.6 kJ.
Activation energy for the uncatalyzed reaction =..?
Now, a careful observation of the question revealed that the activation energy of the uncatalyzed reaction is about 3.98 times that of the catalyzed reaction.
With this vital information, we can thus, calculate the activation energy of the uncatalyzed reaction as follow:
Activation energy for the uncatalyzed reaction = 3.98 times that of the catalyzed reaction.
Activation energy for the uncatalyzed reaction = 3.98 x 4.6 kJ = 18.308 KJ
Therefore, the activation energy of the uncatalyzed reaction is 18.308 KJ.
A 25.0-mL sample of 0.150 M hydrazoic acid, HN3, is titrated with a 0.150 M NaOH solution. What is the pH after 13.3 mL of base is added? The Ka of hydrazoic acid = 1.9 x 10-5.
Answer:
pH ≅ 4.80
Explanation:
Given that:
the volume of HN₃ = 25 mL = 0.025 L
Molarity of HN₃ = 0.150 M
number of moles of HN₃ = 0.025 × 0.150
number of moles of HN₃ = 0.00375 mol
Molarity of NaOH = 0.150 M
the volume of NaOH = 13.3 mL = 0.0133
number of moles of NaOH = 0.0133× 0.150
number of moles of NaOH = 0.001995 mol
The chemical equation for the reaction of this process can be written as:
[tex]HN_3 + OH- ---> N^-_{3} + H_2O[/tex]
1 mole of hydrazoic acid react with 1 mole of hydroxide to give nitride ion and water
thus the new number of moles of HN₃ = 0.00375 - 0.001995 = 0.001755 mol
Total volume used in the reaction = 0.025 + 0.0133 = 0.0383 L
Concentration of [tex]HN_3[/tex] = [tex]\dfrac{0.001755}{0.0383}[/tex] = 0.0458 M
Concentration of [tex]N^{-}_3[/tex] = [tex]\dfrac{ 0.001995 }{0.0383}[/tex] = 0.0521 M
GIven that :
Ka = [tex]1.9 x 10^{-5}[/tex]
Thus; it's pKa = 4.72
[tex]pH =4.72 + log(\dfrac{ \ 0.0521}{0.0458})[/tex]
[tex]pH =4.72 + log(1.1376)[/tex]
[tex]pH =4.72 + 0.05598[/tex]
[tex]pH =4.77598[/tex]
pH ≅ 4.80
The pH of the solution 0.150 M hydrazoic acid after 13.3 mL of NaOH base is added is 4.80.
How we calculate the pH?pH of the given solution will be used by using the following equation:
pH = pKa + log[conjugate base] / [weak acid]
Given chemical reaction will be represented as:
HN₃ + OH⁻ → N₃⁻ + H₂O
Moles will be calculated as:
n = M×V, where
M = molarity
V = volume
Moles of 0.150 M hydrazoic acid = (0.150M)(0.025L) = 0.00375 mol
Moles of 0.150 M NaOH = (0.0133)(0.150) = 0.001995 mol
From the above calculation it is clear that moles of hydrazoic acid is present in excess and it will be:
0.00375 - 0.001995 = 0.001755 mol
And 0.001995 mol of N₃⁻ is preduced by the reaction.
Total volume of the solution = 0.025 + 0.0133 = 0.0383 L
To calculate the pH after titration, first we have to calculate the concentration in terms of molarity of N₃⁻ and HN₃ as:
[N₃⁻] = 0.001995 mol / 0.0383 L = 0.0521 M
[HN₃] = 0.001755 mol / 0.0383 L = 0.0458 M
Ka for HN₃ = 1.9 × 10⁻⁵
pKa = -log( 1.9 × 10⁻⁵ ) = 4.72
On putting all these values on the above equation, we get
pH = 4.72 + log (0.0521) / (0.0458)
pH = 4.80
Hence, pH of the solution is 4.80.
To know more about pH, visit the below link:
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2.
Name the following compounds:
a. Rb20
Answer:
Rubidium oxide
Explanation:
Write the name for the following molecular compounds. Remember to use the correct prefix for each compound.
a. CS2
b. PBr3
c. NO
d. CF4
e. P2O5
Answer:
Hey there!
CS2) Carbon Disulfide.
PBr3) Phosphorus Tribromide
NO) Nitric Oxide
CF4) Carbon Tetrafluoride
P2O5) Phosphorus Pentoxide
Let me know if this helps :)
Which substance is a base? HCOOH RbOH H2CO3 NaNO3
Answer:
RbOH
Explanation:
For this question, we have to remember what is the definition of a base. A base is a compound that has the ability to produce hydroxyl ions [tex]OH^-[/tex], so:
[tex]AOH~->~A^+~+~OH^-[/tex]
With this in mind we can write the reaction for each substance:
[tex]HCOOH~->~HCOO^-~+~H^+[/tex]
[tex]RbOH~->~Rb^+~+~OH^-[/tex]
[tex]H_2CO_3~->~CO_3^-^2~+~2H^+[/tex]
[tex]NaNO_3~->~Na^+~+~NO_3^-[/tex]
The only compound that fits with the definition is [tex]RbOH[/tex], so this is our base.
I hope it helps!
RbOH or rubidium hydroxide is a base.
• RbOH or rubidium hydroxide is the inorganic compound.
• It comprises hydroxide anions and an equal number of rubidium cations.
• Rubidium hydroxide, like other strong bases is highly corrosive.
• The formation of rubidium hydroxide takes place when the metal rubidium reacts with water.
• The bases refers to the substance, which gets dissociate in an aqueous solution to produce OH- or hydroxide ions.
• Rubidium hydroxide is a base, when it is dissolved in water it give rise to an alkali, when reacts with acid it generates a rubidium salt.
Thus, RbOH is a base.
To know more about:
https://brainly.com/question/18616956
Use the following reactions and given Δh values to find standard enthalpies of reactions (in kilojoules) given below.C(s)+O2(g)→CO2(g) ΔH= -393.5 kJ 2CO(g)+O2(g)→2CO2(g) ΔH= -566.0 kJ2H2(g)+O2(g)→2H2O(g) ΔH= -483.6 kJ
Answer:
The heat of the reaction or standard enthalpy of the reaction CO(g) + H₂O(g) → CO₂(g) + H₂(g) is ΔH(rxn) = -41.2 kJ
Explanation:
The reaction whose standard enthalpy is required, as obtained from the internet is
CO(g) + H₂O(g) → CO₂(g) + H₂(g)
The formation reaction for some of the reactants and products are given as
C(s) + O₂(g) → CO₂(g) ΔH = -393.5 kJ
2CO(g) + O₂(g) → 2CO₂(g) ΔH = -566.0 kJ
2H₂(g) + O₂(g) → 2H₂O(g) ΔH = -483.6 kJ
To find the standard enthalpies of the given reaction, we need the heat of formation of each of the species involved in the reaction
ΔH (CO₂(g)) = -393.5 kJ
ΔH (H₂O(g)) = -483.6 kJ ÷ 2 = -241.8 kJ (Because 2 moles of H₂O(g) are formed in the given formation reaction)
ΔH (O₂(g)) = ΔH (H₂(g)) = 0 kJ (No heat of formation for elements)
ΔH (CO(g)) = ? (This isn't given)
But it can be calculated from the second given reaction
2CO(g) + O₂(g) → 2CO₂(g) ΔH = -566.0 kJ
Heat of reaction = ΔH(products) - ΔH(reactants)
Heat of reaction = -566.0 kJ
ΔH (products) = 2 × ΔH (CO₂(g)) = 2 × -393.5 = -787 kJ
ΔH (reactants) = [2 × ΔH (CO(g))] + [1 × ΔH (O₂(g))] = 2 × ΔH (CO₂(g))
Hence, we have
-566 = -787 - [2 × ΔH (CO₂(g))]
2 × ΔH (CO₂(g)) = -787 + 566 = -221 kJ
ΔH (CO₂(g)) = -221 ÷ 2 = -110.5 kJ
CO(g) + H₂O(g) → CO₂(g) + H₂(g)
Heat of reaction = ΔH(products) - ΔH(reactants)
ΔH (products) = [ΔH (CO₂(g))] + [ΔH (H₂(g))]
= -393.5 + 0 = -393.5 kJ
ΔH (reactants) = [1 × ΔH (CO(g))] + [1 × ΔH (H₂O(g))] = -110.5 - 241.8 = -352.3 kJ
Heat of reaction = -393.5 - (-352.3) = -41.2 kJ
Hope this Helps!!!
What is the main side reaction that competes with elimination when a primary alkyl halide is treated with alcoholic potassium hydroxide, and why does this reaction compete with elimination of a primary alkyl halide but not a tertiary alkyl halide
Answer:
The main competing reaction when a primary alkyl halide is treated with alcoholic potassium hydroxide is SN2 substitution.
Explanation:
The relative percentage of products of the reaction between an alkyl halide and alcoholic potassium hydroxide generally depends on the structure of the primary alkylhalide. The attacking nucleophile/base in this reaction is the alkoxide ion. Substitution by SN2 mechanism is a major competing reaction in the elimination reaction intended.
A more branched alkyl halide will yield an alkene product due to steric hindrance, similarly, a good nucleophile such as the alkoxide ion may favour SN2 substitution over the intended elimination (E2) reaction.
Both SN2 and E2 are concerted reaction mechanisms. They do not depend on the formation of a carbocation intermediate. Primary alkyl halides generally experience less steric hindrance in the transition state and do not form stable carbocations hence they cannot undergo E1 or SN1 reactions.
SN2 substitution cannot occur in a tertiary alkyl halides because the stability of tertiary carbocations favours the formation of a carbocation intermediate. The formation of this carbocation intermediate will lead to an SN1 or E1 mechanism. SN2 reactions is never observed for a tertiary alkyl halide due to steric crowding of the transition state. Also, with strong bases such as the alkoxide ion, elimination becomes the main reaction of tertiary alkyl halides.
What happened to the limewater in
the experiment? What does this
prove?
Explanation:
In the reaction above, we clearly see that a chemical reaction took place but what kind of reaction you might ask?
Looks like it is Thermal decomposition of copper carbonate because decomposition reaction takes place due to the added heat. Decomposition in simple terms in chemical breakdown and here this result is obtained by adding heat.
1) What happened to the lime water?
Although not pictured above, but my assumption is that lime water turned milky or turbid because when CO2 comes in presence of limewater they react to form a percipitate of Calcium carbonate which is the milky color that you get.
2) What does this prove?
- My understanding would be that it proves that CO2 was formed, and that most metal carbonates undergo thermal decomposition into metal oxide and carbon dioxide, and also that a reaction took place since new products were made.
A critical reaction in the production of energy to do work or drive chemical reactions in biological systems is the hydrolysis of adenosine triphosphate, ATP, to adenosine diphosphate, ADP, as described by the reactionATP(aq)+ H2O(l) → ADP(aq)+ HPO4^-2 (aq)for which ΔGrxn = -30.5 kj/mol at 37.0C and pH 7.0. Required:a. Calculate the value of ΔGrxn in a biological cell in which [ATP] = 5.0 mM, [ADP] = 0.30 mM, and HPO4^-2= 5.0mMb. Is the hydrolysis of ATP spontaneous under these conditions?
Answer:
Δ [tex]G_{rxn}[/tex] = −51. 4 kJ/mol
However, since Δ [tex]G_{rxn}[/tex] is negative. The hydrolysis of ATP for this reaction is said to be spontaneous
Explanation:
From the question; The equation for this reaction can be represented as :
[tex]ATP_{(aq)} + H_2O_{(l)} \to ADP_{(aq)}+ HPO_4^{2-}} _{(aq)}[/tex]
where:
[tex]\Delta G ^0 _{rxn} =[/tex]-30.5 kJ/mol
= -30.5 kJ/mol × 1000 J/ 1 kJ
= -30.5 × 10 ⁻³ J/mol
Temperature T = 37 ° C
= (37+273)
= 310 K
pH = 7.0
[ATP] = 5.0 mM
= 5.0mM × 1M/1000mM
= 0.005 M
[ADP] = 0.30 mM
= 0.30 mM × 1M/1000mM
= 0.0003 M
[tex][HPO_4^{2-}}][/tex] = 5.0 mM
= 5.0mM × 1M/1000mM
= 0.005 M
The objective is to calculate the value for Δ [tex]G_{rxn}[/tex] in the biological cell and to determine if the hydrolysis of ATP is spontaneous under these conditions.
Now;
From the equation given; the equilibrium constant [tex]K_{eq}[/tex] can be expressed as:
[tex]K_{eq} = \dfrac{[ADP][ HPO_4^{2-}]} {[ATP]}[/tex]
[tex]K_{eq} = \dfrac{(0.0003 \ M)(0.005 \ M)} {(0.005 \ M)}[/tex]
[tex]K_{eq} = 3*10^{-4}[/tex]
The Δ [tex]G_{rxn}[/tex] in the biological cell can now be calculated as:
Δ [tex]G_{rxn}[/tex] = [tex](-30.5 * 10 ^3 \ J/mol) + (8.314 \ J/mol.K)(310 K ) In ( 3*10^{-4})[/tex]
Δ [tex]G_{rxn}[/tex] = [tex](-30.5 * 10 ^3 \ J/mol) + (-20906.68126)[/tex]
Δ [tex]G_{rxn}[/tex] = −51406.68 J/mol
Δ [tex]G_{rxn}[/tex] = −51. 4 × 10³ J/mol
Δ [tex]G_{rxn}[/tex] = −51. 4 kJ/mol
Thus since Δ [tex]G_{rxn}[/tex] is negative. The hydrolysis for this reaction is said to be spontaneous
differentiate between satured and unsatured fats
Answer:
...
Explanation:
in saturated fats there is no double bond between the acids and are tightly packed and unsaturated fats arent tight and loosely packed/put together
saturated- solid at room temperature
unsaturated= liquid at room temperature
two types of unsaturated fats, Polyunsaturated fats and Monounsaturated fats
Enter an equation for the formation of CaCO3(s) from its elements in their standard states. Enter any reference to carbon as C(s). Express your answer as a chemical equation. Identify all of the phases in your answer.
Answer:
CaF2 + CO3- ----> CaCO3 + 2 F-
Explanation:
The chemical compounds found on the left side of the date are the reagents and those found on the right are the products, where calcium carbonate appears.
Calcium carbonate is a quaternary salt
Searches related to If 0.75 grams of iron (Fe) react according to the following reaction, how many grams of copper (Cu) will be produced? Fe + CuSO4 -> Cu + FeSO4
Answer:
0.83 g
Explanation:
Step 1: Write the balanced equation
Fe + CuSO₄ ⇒ Cu + FeSO₄
Step 2: Calculate the moles corresponding to 0.75 g of Fe
The molar mass of Fe is 55.85 g/mol.
[tex]0.75g \times \frac{1mol}{55.85g} = 0.013 mol[/tex]
Step 3: Calculate the moles of Cu produced from 0.013 moles of Fe
The molar ratio of Fe to Cu is 1:1. The moles of Cu produced are 1/1 × 0.013 mol = 0.013 mol.
Step 4: Calculate the mass corresponding to 0.013 moles of Cu
The molar mass of Cu is 63.55 g/mol.
[tex]0.013mol \times \frac{63.55g}{mol} = 0.83 g[/tex]
Answer:
If 0.75 grams of iron (Fe) react, 0.85 grams of copper (Cu) will be produced.
Explanation:
You know the following balanced reaction:
Fe + CuSO₄ ⇒ Cu + FeSO₄
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following quantities react and are produced:
Fe: 1 moleCuSO₄: 1 moleCu: 1 moleFeSO₄: 1 moleBeing:
Fe: 55.85 g/moleCu: 63.54 g/moleS: 32 g/moleO: 16 g/molethe molar mass of the compounds participating in the reaction is:
Fe: 55.85 g/moleCuSO₄: 63.54 g/mole + 32 g/mole+ 4* 16 g/mole= 159.54 g/moleCu: 63.54 g/moleFeSO₄: 55.85 g/mole + 32 g/mole+ 4* 16 g/mole= 151.85 g/moleThen, by stoichiometry of the reaction, the amounts of reagent and product that participate in the reaction are:
Fe: 1 mole*55.85 g/mole= 55.85 gCuSO₄: 1 mole* 159.54 g/mole= 159.54 gCu: 1mole* 63.54 g/mole= 63.54 gFeSO₄: 1 mole* 151.85 g/mole= 151.85 gThen you can apply a rule of three as follows: if 55.85 grams of Fe produces 63.54 grams of Cu, 0.75 grams of Fe how much mass of Cu does it produce?
[tex]mass of Cu=\frac{0.75 grams of Fe*63.54 grams of Cu}{55.85 grams of Fe}[/tex]
mass of Cu= 0.85 grams
If 0.75 grams of iron (Fe) react, 0.85 grams of copper (Cu) will be produced.
plllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllls help me
Answer:
Explanation:
equilibrium constant
Kc = [ C ]² / [ A ] [ B ]
= .5² / .2 x 3
= .4167
Let moles of A to be added be n
concentration of A unreacted becomes .2 + n M
increase of product C by .2 M will require use of A and B be .1 M
So unreacted A = .2 + n - .1 = n + .1
Kc = [ C ]² / [ A ] [ B ]
.4167 = .7² / ( n + .1 ) ( 3 - .1 )
n + .1 = .4
n = . 3 moles .
So .3 moles of A to be added .
What is the pH of a solution prepared by dissolving 0.140 g of potassium hydroxide in sufficient pure water to prepare 250.0 ml of solution
Answer:
pH= 12
Explanation:
Potassium hydroxide (KOH) is a strong base, so it dissociates completely in water by giving OH⁻ anions as follows:
KOH⇒ K⁺ + OH⁻
Since dissociation is complete, it is assumed that the concentration of OH⁻ is equal to the initial concentration of KOH:
[OH⁻]= [KOH]
In order to find the initial concentration of KOH, we have to divide the mass (0.140 g) into the molecular weight of KOH (Mw):
Mw (KOH)= K + O + H = 39 g/mol + 16 g/mol + 1 g/mol = 56 g/mol
moles KOH: mass/Mw= 0.140 g/(56 g/mol) = 2.5 x 10⁻³ moles
The molality of the solution is the number of moles of KOH per liter of solution:
V= 250.0 ml x 1 L/1000 ml= 0.250 L
M = (2.5 x 10⁻³moles)/(0.250 L)= 0.01 M
Now, we calculate pOH:
pOH = -log [OH⁻]= - log [KOH]= -log (0.01) = 2
Finally, we calculate pH from pOH:
pH + pOH = 14
⇒pH = 14 - pOH= 14 -2 = 12
The constant pressure molar heat capacity of argon, C_{p,m}C
p,m
, is
20.79\text{ J K}^{-1}\text{ mol}^{-1}20.79 J K
−1
mol
−1
at 298\text{ K}298 K. What
will be the value of the constant volume molar heat capacity of argon,
C_{V,m}C
V,m
, at this temperature?
Answer:
Constant-volume molar heat capacity of argon is 12.47 J K ⁻¹mol⁻¹
Explanation:
Argon is a monoatomic gas that behaves as an ideal gas at 298K.
Using the first law of thermodinamics you can obtain:
Work, Q, for constant pressure molar heat capacity,CP:
CP = (5/2)R
For constant-volume molar heat capacity,CV:
CV = (3/2)R
That means:
2CP/5 = 2CV/3
3/5 = CV / CP
As CP of Argon is 20.79 J K ⁻¹mol⁻¹, CV will be:
3/5 = CV / CP
3/5 = CV / 20.79 J K ⁻¹mol⁻¹
12.47 J K ⁻¹mol⁻¹ = CV
Constant-volume molar heat capacity of argon is 12.47 J K ⁻¹mol⁻¹Draw the structure 2 butylbutane
Answer:
please look at the picture below.
Explanation:
a piece of copper weighing 850 grams is placed in a cup with 450 ml of water at 21 C and the Cp of the cup is 47 J/K, how many grams of gasoline would it take to heat the entire system to 110 C?
Answer:
4.2g of gasoline
Explanation:
In the problem, you need to give energy to the cup from the combustion of gasoline. The energy you need to give is:
Qcup + QWater + QCopper
As you need to increase (110ºC - 21ºC = 89º = Increase 89K) 89K, the Qcup is:
Qcup = 89K × (47J/K) = 4183J.
You can find Qwater using its specific heat, C (4.18Jg⁻¹K⁻¹), its mass (450mL = 450g) and the change of temperature, 89K:
QWater = CₓmₓΔT
QWater = 4.184Jg⁻¹K⁻¹ ₓ 450g×89K
QWater = 167569J
And Q of Copper, QCu, could be obtained in the same way (Specific heat Cu: 0.387 J/g⁻¹K⁻¹:
QCu = CₓmₓΔT
QCu = 0.387 J/g⁻¹K⁻¹ₓ850gₓ89K
QCu = 29277J
Thus, total heat you need is:
Q = Qcup + QWater + QCopper
Q = 4183J + 167569J + 29277J
Q = 201029J = 201kJ
The combustion of gasoline (Octane) produce 47.8kJ/g (Its heat of combustion). that means to produce 201kJ of energy you require:
201kJ × (1g / 47.8kJ) =
4.2g of octane = Gasoline you requireSuppose that you add 26.7 g of an unknown molecular compound to 0.250 kg of benzene, which has a K f of 5.12 oC/m. With the added solute, you find that there is a freezing point depression of 2.74 oC compared to pure benzene. What is the molar mass of the unknown compound
Answer: The molar mass of the unknown compound is 200 g/mol
Explanation:
Depression in freezing point is given by:
[tex]\Delta T_f=i\times K_f\times m[/tex]
[tex]\Delta T_f=2.74^0C[/tex] = Depression in freezing point
i= vant hoff factor = 1 (for molecular compound)
[tex]K_f[/tex] = freezing point constant = [tex]5.12^0C/m[/tex]
m= molality
[tex]\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}[/tex]
Weight of solvent (benzene)= 0.250 kg
Molar mass of solute = M g/mol
Mass of solute = 26.7 g
[tex]2.74^0C=1\times 5.12\times \frac{26.7g}{Mg/mol\times 0.250kg}[/tex]
[tex]M=200g/mol[/tex]
Thus the molar mass of the unknown compound is 200 g/mol
The molar mass of an unknown solute compound in the solution has been 199.626 g/mol.
With the addition of the solute to the solution, there has been a depression in the freezing point. The depression in the freezing point can be expressed as:
Depression in freezing point = Van't Hoff factor × Freezing point constant × molality
The molality can be defined as the moles of solute per kg solvent
Molaity = [tex]\rm \dfrac{Mass\;of\;solute\;(g)}{Molecular\;mass\;of\;solute}\;\times\;\dfrac{1}{Mass\;of\;solvent\;(kg)}[/tex]
The depression in freezing point can be given as:
Depression in freezing point = Van't Hoff factor × Freezing point constant × [tex]\rm \dfrac{Mass\;of\;solute\;(g)}{Molecular\;mass\;of\;solute}\;\times\;\dfrac{1}{Mass\;of\;solvent\;(kg)}[/tex] ......(i)
Given, the depression in freezing point = 2.74 [tex]\rm ^\circ C[/tex]
Van't Hoff factor = 1 (Molecular compound)
Freezing point constant (Kf) = 5.12 [tex]\rm ^\circ C[/tex]/m
Mass of solute = 26.7 g
Mass of solvent = 0.250 kg
Substituting the values in equation (i):
2.74 [tex]\rm ^\circ C[/tex] = 1 × 5.12
[tex]\rm \dfrac{2.74}{5.12}[/tex] = [tex]\rm \dfrac{1}{Molecular\;mass\;of\;solute}\;\times\;\dfrac{26.7}{0.250\;kg}[/tex]
0.535 = [tex]\rm \dfrac{1}{Molecular\;mass\;of\;solute}\;\times\;106.8[/tex]
Molecular mass of solute = [tex]\rm \dfrac{106.8}{0.535}[/tex] g/mol
Molecular mass of solute = 199.626 g/mol
The molar mass of an unknown solute compound in the solution has been 199.626 g/mol.
For more information about the molality of the compound, refer to the link:
https://brainly.com/question/7229194
Determine the the mass of one molecule of hydrogen sulfide gas.
Answer:
the molecular mass of hydrogen sulphide, which contains two atoms of hydrogen and one atom of sulphur is = 2 — 1 + 1 — 32 = 34 a.m.u.
An aqueous solution of glucose (C6H12O6), called D5W, is used for intravenous injection. D5W contains 54.30 g of glucose per liter of solution. What is the molar concentration of glucose in D5W
Answer:
The correct answer is 0.30 M
Explanation:
The molar concentration or molarity of a solution is defined as moles of solute per liter of solution. We found the moles of solute (glucose) by dividing the mass (54.30 g) into the molecular weight (MW) of glucose (C₆H₁₂O₆):
MW(C₆H₁₂O₆)= (12 g/mol x 6) + (1 g/mol x 12) + (16 g/mol x 6) = 180 g/mol
Moles of glucose= mass/MW= 54.30 g/(180 g/mol)= 0.30 mol
There is 0.30 mol of solute per liter of solution, thus the molarity is:
M= moles solute/L solution= 0.30 mol/1 L = 0.30 M
The activation energy for the decomposition of HI is 183 kJ/mol. At 573 K, the rate constant was measured to be 2.91 x 10^{-6} M/s. At what temperature in Kelvin does the reaction have a rate constant of 0.0760 M/s
Answer:
[tex]T_2=453.05K[/tex]
Explanation:
Hello,
In this case, the temperature-variable Arrhenius equation is written as:
[tex]\frac{k(T_2)}{k(T_1)}=exp(\frac{Ea}{R}(\frac{1}{T_2}-\frac{1}{T_1} ))[/tex]
Now, for us to solve for the temperature by which the reaction rate constant is 0.0760M/s we proceed as shown below:
[tex]ln(\frac{k(T_2)}{k(T_1)})=\frac{Ea}{R}(\frac{1}{T_2}-\frac{1}{T_1} )\\ln(\frac{0.0760M/s}{0.00000291M/s} )=\frac{183000J/mol}{8.314J/(mol*K)} *(\frac{1}{T_2} -\frac{1}{573K} )\\\frac{1}{T_2} -\frac{1}{573K} =\frac{10.17}{22011.06K^{-1}} \\\\\frac{1}{T_2}=4.62x10^{-4}K^{-1}+\frac{1}{573K}\\\\\frac{1}{T_2}=2.21x10^{-3}K^{-1}\\\\T_2=453.05K[/tex]
Regards.