Answer:
option B
Explanation:
A spontaneous reaction drive favorable when enthalpy is decreasing and the entropy is increasing on the system. If that happens the reaction occurs spontaneously
Answer:
b. decreasing enthalpy and increasing entropy
Explanation:
∆H stands for enthalpy change and ∆S stands for entropy change
Spontaneity depends on the enthalpy and entropy changes of the reaction
∆G = ∆H - T∆S
When ∆H is negative and ∆S is positive
∆G will be negative
For a spontaneous reaction ∆G is negative
If ∆G = 0 then the reaction will be at equilibrium
If ∆G is positive the reaction is non spontaneous.
Decreasing enthalpy (negative) and increasing entropy (positive) will give a negative number for [tex]\Delta G[/tex]
How many moles of KBr will be produced from 10.51 moles of BaBr2?
BaBr2 + K2SO4 → KBr + BaSO4
Select one:
a. 21.0
b. 5.26
c. 17.9
d. 7.88
To find the number of moles of KBr produced from 10.51 moles of BaBr₂, we use the balanced equation to see that 1 mole of BaBr₂ gives 2 moles of KBr. Multiplying 10.51 by 2 yields 21.02 moles of KBr. The correct answer is thus 21.0 moles of KBr.
The question asks how many moles of KBr will be produced from 10.51 moles of BaBr₂ when it reacts with K₂SO₄ to form KBr and BaSO₄. The balanced chemical equation for this reaction is:
BaBr₂ + K₂SO₄ → 2 KBr + BaSO₄
From the balanced equation, we can see that 1 mole of BaBr₂ produces 2 moles of KBr. Therefore, to find the number of moles of KBr produced from 10.51 moles of BaBr2, we simply multiply the number of moles of BaBr₂ by 2.
10.51 moles of BaBr₂ x 2 moles of KBr / 1 mole of BaBr₂ = 21.02 moles of KBr
The closest answer from the given options that matches our calculation is 21.0 moles of KBr, which is option (a).
A sample of an unknown compound is vaporized at 180.°C. The gas produced has a volume of 2010.mL at a pressure of 1.00atm, and it weighs 2.73g. Assuming the gas behaves as an ideal gas under these conditions, calculate the molar mass of the compound. Round your answer to 3 significant digits.
The molar mass of the unknown compound under the given conditions, deemed to behave as an ideal gas, is approximately 176.124 g/mol. We determined this via the ideal gas law, which allowed us to find the number of moles, and subsequently the molar mass through division of the compound's mass by the number of moles.
Explanation:The problem involves calculating the molar mass of an unknown compound acting as an ideal gas under specific conditions. We begin by using the ideal gas law (PV = nRT), but we rearrange it to find the number of moles (n = PV/RT). Given the pressure (P = 1.00 atm), volume (V = 2010 mL or 2.01 L), and temperature in degrees Celsius (which must be converted to Kelvins, T = 180 + 273 = 453 K), and knowing the gas constant (R = 0.0821 L.atm/mol.K), we can calculate n.
Once we have the number of moles, we can find the molar mass by dividing the mass of the compound (given as 2.73g) by the number of moles. Thus, the molar mass of the compound is found to be approximately 176.124 g/mol.
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Aluminum metal and chlorine gas react to from aluminum chloride 2Al(s) + 3Cl2(g) --> Al2Cl6(s) How many grams of Al2Cl6 can be produced from the reaction of 2.67 mol aluminum and 3.26 mol Cl2?
Answer:
290.1 grams.
Explanation:
2 Al + 3Cl2 ---> Al2Cl6
2 moles of Al react with 3 moles of Cl2 to form 1 mole of Al2Cl6
So 2.67 moles Al reacts with 3/2 * 2.67 moles Cl2 = 4 moles of chlorine. But we only have 3.26 moles of Cl2 so this will react with only part of the available Al.
3.26 reacts with 2/3 * 3.26 = 2.173 moles of aluminium.
So the number of moles of Al2Cl6 that can be produced = 2.173 * 1/2
= 1.087 moles.
= 1.087 * (2*26.98 + 6*35.45)
= 290.1 grams.
Please answer these questions fast! Im on a time crunch right now
Answer:
1st image, 17.79 g Mg(NO3)2
2nd image 1.47 mol NH3
3rd image Molar Mass Al(OH)3 = 78 g/mol
Explanation:
1st image
Molar Mass Mg(NO3)2 = 148. 31 g/mol
0,12 mol Mg(NO3)2 x (148, 31 g Mg(NO3)2 / 1 mol Mg(NO3)2) = 17.79 gMg(NO3)2
2nd image
Molar Mass NH3 = 17g/mol
25 g NH3 x ( 1 mol NH3/ 17 g NH3) = 1.47 mol NH3
3rd image
Molar Mass Al(OH)3 = 1 Al *(27) + 3 O * (16) + 3 H * (1) = 78 g/mol
When 10.0 grams of CH4 reacts completely with 40.0 grams of O2 such that there are no reactants left over, 27.5 grams of carbon dioxide are formed. How many grams of water are formed? CH4+ 2O2 → CO2 + 2H2O
Answer:
[tex]\boxed{\text{27.4 g CO$_{2}$; 22.5 g H$_{2}$O}}}[/tex]
Explanation:
We know we will need an equation with masses and molar masses, so let’s gather all the information in one place.
M_r: 16.04 32.00 44.01 18.02
CH₄ + 2O₂ → CO₂ + 2H₂O
m/g: 10.0 40.0
1. Moles of CH₄
[tex]\text{Moles of CH}_{4} = \text{10.0 g CH}_{4} \times \dfrac{\text{1 mol CH}_{4}}{\text{16.04 g CH}_{4}} = \text{0.6234 mol CH}_{4}[/tex]
2. Mass of CO₂
(i) Calculate the moles of CO₂
The molar ratio is (1 mol CO₂ /1 mol CH₄)
[tex]\text{Moles of CO$_{2}$} = \text{0.6234 mol CH$_4$} \times \dfrac{\text{1 mol CO$_{2}$}} {\text{1 mol CH$_{4}$}} = \text{0.6234 mol CO$_{2}$}[/tex]
(ii) Calculate the mass of CO₂
[tex]\text{Mass of CO$_{2}$} = \text{0.6234 mol CO$_{2}$} \times \dfrac{\text{44.01 g CO$_{2}$}}{\text{1 mol CO$_{2}$}} = \textbf{27.4 g CO$_{2}$}\\\\\text{The mass of carbon dioxide formed is } \boxed{\textbf{27.4 g CO$_{2}$}}[/tex]
3. Mass of H₂O
(i) Calculate the moles of H₂O
The molar ratio is (2 mol H₂O /1 mol CH₄)
[tex]\text{Moles of H$_{2}$O}= \text{0.6234 mol CH}_{4} \times \dfrac{\text{2 mol H$_{2}$O}}{\text{1 mol CH$_{4}$}} = \text{1.247 mol H$_{2}$O}[/tex]
(ii) Calculate the mass of H₂O
[tex]\text{Mass of H$_{2}$O} = \text{1.247 mol H$_{2}$O } \times \dfrac{\text{18.02 g H$_{2}$O}}{\text{1 mol H$_{2}$O}} = \textbf{22.5 g H$_{2}$O}\\\\\text{The mass of water formed is } \boxed{\textbf{22.5 g H$_{2}$O}}[/tex]
Which substance would be found in higher concentration if the membrane were damaged? View Available Hint(s) Which substance would be found in higher concentration if the membrane were damaged? protein chloride creatinine glucose
Answer: the correct answer is protein
Explanation:
Large proteins are not normally filtered by a healthy glomerural membrane.
Be sure to answer all parts. Nitric oxide (NO) reacts with molecular oxygen as follows: 2NO(g) + O2(g) → 2NO2(g) Initially NO and O2 are separated in two different chambers connected by a valve. When the valve is opened, the reaction quickly goes to completion. Determine what gases remain at the end and calculate their partial pressures. Assume that the temperature remains constant at 25°C. Initial conditions are as follows: NO: 3.90 L, 0.500 atm O2: 2.09 L, 1.00 atm
Answer:
The remain gases are [tex]o_{2}_{(g)}[/tex] and [tex]NO_{2}_{(g)}[/tex]
Pressure of [tex]O_{2}_{(g)}[/tex] [tex]1.09 atm O_{2}_{(g)}[/tex]
Pressure of [tex]NO_{2}_{(g)}[/tex] [tex]1.09 atm NO_{2}_{(g)}[/tex]
Explanation:
We have the following reaction
[tex]2NO_{(g)} +O_{2}_{(g)}\longrightarrow 2NO_{2}_{(g)}[/tex]
Now we calculate the limit reagents, to know which of the two gases is completely depleted and which one is in excess.
Excess gas will remain in the tank when the reagent limits have run out and the reaction ends.
To calculate the limit reagent, we must calculate the mols of each substance. We use the ideal gas equation
[tex]PV= nRT[/tex]
We cleared the mols
[tex]n=\frac{PV}{RT}[/tex]
PV=nrT
replace the data for each gas
Constant of ideal gases
[tex]R= 0.082\frac{atm.l}{mol.K}[/tex]
Transform degrees celsius to kelvin
[tex]25+273=298K[/tex]
[tex]NO_{g}[/tex]
[tex]n=\frac{0.500atm.3.90l}{298k.0.082\frac{atm.l}{k.mol} } \\ \\ n=0.080mol NO_{(g)} \\[/tex]
[tex]O_{2}_{(g)[/tex]
[tex]n=\frac{1atm.2.09l}{298k.0.082\frac{atm.l}{k.mol} } \\ \\ n=0.086mol O_{2}_{(g)} \\[/tex]
Find the limit reagent by stoichiometry
[tex]2NO_{(g)} +O_{2}_{(g)}\longrightarrow 2NO_{2}_{(g)}[/tex]
[tex]0.086mol O_{2}_(g).\frac{2mol NO_{(g)} }{1mol O_{2}_{(g)} } =0.17mol NO_{(g)}[/tex]
Using [tex]O_{2}_{(g)}[/tex]as the limit reagent produces more [tex]NO_{(g)}[/tex] than I have, so oxygen is my excess reagent and will remain when the reaction is over.
[tex]NO_{(g)}[/tex]
[tex]2NO_{(g)} +O_{2}_{(g)}\longrightarrow 2NO_{2}_{(g)}\\ \\ 0.080mol NO_{(g)}.\frac{1mol O_{2}_ {(g)} }{2mol NO_{(g)} } =0.04mol O_{2}_{(g)}[/tex]
Using [tex]NO_{(g)}[/tex] as the limit reagent produces less [tex]O_{2}_{(g)}[/tex] than I have, so [tex]NO_{(g)}[/tex] is my excess reagent and will remain when the reaction is over.
Calculate the moles that are formed of [tex]NO_{2}_{(g)}[/tex]
[tex]2NO_{(g)} +O_{2}_{(g)}\longrightarrow 2NO_{2}_{(g)}\\ \\ 0.080mol NO_{(g)}.\frac{2mol NO_{2}_ {(g)} }{2mol NO_{(g)} } =0.080mol NO_{2}_{(g)}[/tex]
We know that for all [tex]NO_{(g)}[/tex] to react, 0.04 mol [tex]O_{2}_{(g)}[/tex] is consumed.
we subtract the initial amount of [tex]O_{2}_{(g)}[/tex] less than necessary to complete the reaction. And that gives us the amount of mols that do not react.
[tex]0.086-0.04= 0.046[/tex]
The remain gases are[tex]O_{2}_{(g)}[/tex] and [tex]NO_{2}_{(g)}[/tex]
calculate the volume that gases occupy
[tex]0.080 mol NO_{2}_{(g)} .\frac{22.4l NO_{2}_{(g)} }{1molNO_{2}_{(g)} }} =1.79 lNO_{2}_{(g)}[/tex]
[tex]0.046 mol O_{2}_{(g)} .\frac{22.4 l O_{2}_{(g)} }{1molO_{2}_{(g)} }} =1.03 l O_{2}_{(g)}[/tex]
Calculate partial pressures with the ideal gas equation
[tex]PV= nRT[/tex]
[tex]P=\frac{nRT}{V}[/tex]
Pressure of [tex]O_{2}_{(g)}[/tex]
[tex]P=\frac{0.046mol.0.082\frac{atm.l}{K.mol} 298K}{1.03l}= 1.09 atmO_{2}_{(g)}[/tex]
Pressure of [tex]NO_{2}_{(g)}[/tex]
[tex]P=\frac{0.080mol.0.082\frac{atm.l}{K.mol} 298K}{1.79l}= 1.09 atmNO_{2}_{(g)}[/tex]
Answer:
It will remain O₂ and NO₂, with partial pressures: pO₂ = 0.186 atm, and pNO₂ = 0.325 atm.
Explanation:
First, let's identify the initial amount of each reactant using the ideal gas law:
PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas equation (0.082 atm.L/mol.K), and T is the temperature (25°C + 273 = 298 K).
n = PV/RT
NO:
n = (0.500*3.90)/(0.082*298)
n = 0.0798 mol
O₂:
n = (1.00*2.09)/(0.082*298)
n = 0.0855 mol
By the stoichiometry of the reaction, we must found which reactant is limiting and which is in excess. The limiting reactant will be totally consumed. Thus, let's suppose that NO is the limiting reactant:
2 moles of NO ------------------ 1 mol of O₂
0.0798 mol ------------------- x
By a simple direct three rule:
2x = 0.0798
x = 0.0399 mol of O₂
The number of moles of oxygen needed is lower than the number of moles in the reaction, so O₂ is the limiting reactant, and NO will be totally consumed. The number of moles of NO₂ formed will be:
2 moles of NO --------------- 2 moles of NO₂
0.0798 mol ---------------- x
By a simple direct three rule:
x = 0.0798 mol of NO₂
And the number of moles of O₂ that remains is the initial less the total that reacts:
n = 0.0855 - 0.0399
n = 0.0456 mol of O₂
The final volume will be the total volume of the containers, V = 3.90 + 2.09 = 5.99 L, so by the ideal gas law:
PV = nRT
P = nRT/V
O₂:
P = (0.0456*0.082*298)/5.99
P = 0.186 atm
NO₂:
P = (0.0798*0.082*298)/5.99
P = 0.325 atm
A student is investigating the affect of different salts on melting points. Four patches of ice of equal size are roped off and a different type of salt is poured on each, one receives table salt (NaCl), one receives Calcium Chloride (CaCl2), one receives Potassium Carbonate (KCO3) and the fourth receives inert sand instead. Each patch receives an equal amount of salt or sand. The student measures the volume of ice remaining and subtracts it from the original volume of ice to see how much melted away. What is a control variable in this experiment?
Answer:
The inert sand
Explanation:
A control variable is the variable in which the experiment does not depend on. The outcome of the control variable does not in any way validate or invalidate an experimental procedure.
The student seeks to investigate the effect of different salts on melting point. He used NaCl, CaCl₂, K₂CO₃ and inert sand. The first three are salts. Sand is made up of silica, SiO₂ and it is not a salt. Therefore, the control variable here is the inert sand. Sand is highly unreactive and would not in any significant way help the investigation.
Answer: Fourth one which received inert sand.
Explanation:
A control variable is the one which remains unchanged in an experiment. It does not receive any treatment. It is considered as a benchmark or standard for comparison of changes occurring in the dependent or experimental variable.
According to the given situation, the fourth one is the correct option as the inert sand will not have any effect over the melting of ice. This can be useful for comparison of the effect of salts on the ice.
Compared with its corresponding unsaturated fatty acid, a saturated fatty acid has _____.
more hydrogen
less hydrogen
more oxygen
less oxygen
Answer: more hydrogen
Explanation:
Saturated fatty acids are defined as the fatty acids in which a single bond is present between carbon and carbon atoms and hydrocarbon chain is attached to carboxylic group. Example: [tex]C_{18}H_{34}O_2[/tex]
Unsaturated fatty acids are defined as the fatty acids which have double or triple covalent C-C bonds and hydrocarbon chain is attached to carboxylic group. Example: [tex]C_{18}H_{32}O_2[/tex]
Thus a saturated fatty acid has more hydrogens than unsaturated fatty acid.
Calculate the pH of 1.00 L of the buffer 0.95 M CH3COONa/0.92 M CH3COOH before and after the addition of the following species. (Assume there is no change in volume.) (a) pH of starting buffer: (b) pH after addition of 0.040 mol NaOH: (c) pH after further addition of 0.100 mol HCl:
Answer: (a) pH = 4.774, (b) pH = 4.811 and (c) pH = 4.681
Explanation: (a) pH of the buffer solution is calculated using Handerson equation:
[tex]pH=pKa+log(\frac{base}{acid})[/tex]
pKa for acetic acid is 4.76. concentration of base and acid are given as 0.95M and 0.92M. Let's plug in the values in the equation and calculate the pH of starting buffer.
[tex]pH=4.76+log(\frac{0.95}{0.92})[/tex]
pH = 4.76 + 0.014
pH = 4.774
(b) When 0.040 mol of NaOH (strong base) are added to the buffer then it reacts with 0.040 mol of acetic acid and form 0.040 mol of sodium acetate.
Original buffer volume is 1.00 L. So, the original moles of sodium acetate will be 0.95 and acetic acid will be 0.92.
moles of acetic acid after addition of NaOH = 0.92 - 0.040 = 0.88
moles of sodium acetate after addition of NaOH = 0.95 + 0.040 = 0.99
Let's again plug in the values in the Handerson equation:
[tex]pH=4.76+log(\frac{0.99}{0.88})[/tex]
pH = 4.76 + 0.051
pH = 4.811
(c) When 0.100 mol of HCl are added then it reacts with exactly 0.100 moles of sodium acetate(base) and form 0.100 moles of acetic acid(acid).
so, new moles of acetic acid = 0.92 + 0.100 = 1.02
new moles of sodium acetate = 0.95 - 0.100 = 0.85
Let's plug in the values in the equation:
[tex]pH=4.76+log(\frac{0.85}{1.02})[/tex]
pH = 4.76 - 0.079
pH = 4.681
The pH of the buffer can be calculated at each step using the Henderson-Hasselbalch equation, which requires the pKa of the acid and the ratio of the base to acid concentrations. When NaOH is added, it reacts with the acetic acid, changing this ratio and thus the pH. When HCl is added, it reacts with the acetate, again changing the ratio and pH.
Explanation:The initial pH of the buffer can be calculated using the Henderson-Hasselbalch equation. This equation requires the pKa, which for acetic acid is 4.74, and the ratio of the concentrations of the base (CH3COONa, or acetate) and acid (CH3COOH, or acetic acid). So we have pH = pKa + log([base]/[acid]) = 4.74 + log(0.95/0.92) ≈ 4.75. That's the starting pH of the buffer.
Now, when we add 0.040 mol NaOH, it will react with the acetic acid, converting it to acetate and thus increasing the [base]/[acid] ratio. The change in pH can again be calculated via the Henderson-Hasselbalch equation: pH = 4.74 + log((0.95 + 0.040)/(0.92 - 0.040)). Then, when we add 0.100 mol HCl, it will react with the acetate, converting it back to acetic acid and thus decreasing the [base]/[acid] ratio, again resulting in a pH shift that you can calculate the same way.
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how would you use a grignard reaction of an aldehyde or ketone to synthesize 2-pentanol? draw the grignard reagent and the aldehyde or ketone you would use below.
You start from butanal which will react with methyl magnesium bromide (Gringard reagent) forming a compound with a new carbon-carbon bond. After that using an acid media the magnesium bromide salt is eliminated with formation of the 2-pentanol.
Members of the order Testudines have an anapsid-like skull with one opening. However, molecular studies indicate that turtles descended from a diapsid ancestor. Why might this be the case?
Answer:
Explanation:because turtles will eventually get old and they will leave behind their family like the next generation
How long does helium last in latex balloons
Answer:
it will last I'll say about 12-19 or 20 hours
Explanation:
my explanation is it all depends on the time and the degrees because it can last for weeks (5-7) but it won't
float
hope that helps
Helium typically lasts about 12 to 24 hours in a latex balloon. This is due to helium's high effusion rate, meaning it escapes quickly through small pores in the latex material.
Explanation:In a latex balloon, helium typically lasts about 12 to 24 hours. This is mainly due to the effusion rate of helium, which is quite high because helium is a very light gas. A gas's effusion rate relates to how quickly it escapes through small pores or openings.
The balloon material, which in this case is latex, can only contain the gas for so long before it starts to leak out. As shown in the given photographs, a helium-filled balloon noticeably deflated after just 12 hours, whereas the argon-filled balloon remained inflated due to argon's lower effusion rate.
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You are performing an experiment in a lab to attempt a new method of producing pure elements from compounds. The only problem is that you do not know what element will form. By your previous calculations you know that you will have 3 moles of product. When it is complete, you weigh it and determine you have 209.1 grams. What element have you produced?
You need to calculate the atomic weight of that element.
number of moles = mass / atomic weight
atomic weight = mass / number of moles
atomic weight = 209.1 / 3 = 69.7 g/mol
The element produced will be gallium, Ga.
Answer:
The element produced is gallium, Ga
Explanation:
Given:
Moles of the product = 3 moles
Weight of the product = 209.1 g
To determine:
The identity of the product formed i.e. the unknown element
Calculation:
The identity of element can be deduce from its atomic weight and comparing the calculated weight to that of the elements in the periodic table.
[tex]Number\ of\ moles = \frac{mass\ of\ the\ element}{atomic\ weight}[/tex]
[tex]Atomic\ weight = \frac{mass\ of\ the\ element}{number\ of\ moles}[/tex]
In this case:
[tex]Atomic\ weight = \frac{209.1\ g}{3\ moles} = 69.7\ g/mol[/tex]
From the periodic table, the element with an atomic mass = 69.7 g/mol is gallium, Ga
At a certain temperature, Kc = 0.0500 and ∆H = +39.0 kJ for the reaction below, 2 MgCl2(s) + O2(g) → 2MgO(s) + Cl2(g) Calculate Kc for the reaction, MgO(s) + ½ Cl2(g) → MgCl2(s) + ½ O2(g) and indicate whether the value of Kc will be larger or smaller at a lower temperature.
Explanation:
Since, it is shown that the reaction has been reversed. Therefore, value of [tex]K_{c}[/tex] will become [tex]\frac{1}{K_{c}}[/tex].
Hence, new [tex]K_{c'} = \frac{1}{K_{c}}[/tex]
= [tex]\frac{1}{0.0500}[/tex]
= 20
Also, the number of moles of each reactant has been halved. So, [tex]K_{c''}[/tex] for the reaction [tex]MgO(s) + \frac{1}{2}Cl2(g) → MgCl_{2}(s) + \frac{1}{2} O2(g)[/tex] will also get halved.
Therefore, [tex]K_{c''}[/tex] = [tex]K_{c'}[/tex] = [tex](20)^{0.5}[/tex]
= 4.47
As the value of [tex]\Delta H[/tex] is given as +39.0 kJ. So, it means that the reaction is endothermic in nature. So, energy of reactants will be more than the products. Hence, according to Le Chatelier's principle reaction will move in the forward direction.
As a result, [tex]K_{c}[/tex] will also increase with increase in temperature.
The value of Kc for the reaction MgO(s) + ½ Cl2(g) → MgCl2(s) + ½ O2(g) is 1.00. The value of Kc will be smaller at a lower temperature for an exothermic reaction.
Explanation:To calculate Kc for the reaction MgO(s) + ½ Cl2(g) → MgCl2(s) + ½ O2(g), we need to use the equation Kc = K1*K2, where K1 is the equilibrium constant for the forward reaction and K2 is the equilibrium constant for the reverse reaction. In this case, K1 is equal to 0.0500, as given, and K2 can be found using the equation K2 = 1/K1. So, K2 = 1/0.0500 = 20. Therefore, Kc = 0.0500 * 20 = 1.00.
To determine whether the value of Kc will be larger or smaller at a lower temperature, we need to consider the effect of temperature on the equilibrium constant. In general, for exothermic reactions, the equilibrium constant decreases as the temperature decreases. Since the reaction given is exothermic (∆H = +39.0 kJ), the value of Kc will be smaller at a lower temperature.
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Consider the combustion reaction for octane (C8H18), which is a primary component of gasoline. 2C8H18+25O2⟶16CO2+18H2O How many moles of CO2 are emitted into the atmosphere when 28.6 g C8H18 is burned?
To find out the number of moles of CO2 produced from 28.6 g of octane, convert the mass of octane to moles and then apply the stoichiometry of the balanced equation. This results in 2 moles of CO2 being emitted.
Explanation:The question asks how many moles of CO2 are produced when 28.6 g of C8H18 is burned according to the balanced chemical equation 2 C8H18(l) + 25 O2(g) → 16 CO2(g) + 18 H2O(g). The first step in this problem is to convert the given mass of octane to moles using its molar mass (114.23 g/mol for C8H18). With 28.6g of C8H18, we calculate the moles of octane:
moles of C8H18 = 28.6 g / 114.23 g/mol = 0.25 moles of C8H18
Next, we use the stoichiometric ratio from the balanced equation, which shows that 2 moles of C8H18 produce 16 moles of CO2. Therefore, for 0.25 moles of octane, the moles of CO2 produced are calculated as:
moles of CO2 = (0.25 moles C8H18/1) × (16 moles CO2/2 moles C8H18) = 2 moles CO2
So, 28.6 g of octane will produce 2 moles of CO2 when completely burned.
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A mixture of gases A2 and B2 are introduced to a slender metal cylinder that has one end closed and the other fitted with a piston that makes a gas-tight seal so that the gases are a closed system. The cylinder is submerged in a large beaker of water whose temperature is 25∘C, and a spark is used to trigger a reaction in the cylinder. At the completion of the reaction, the piston has moved downward, and the temperature of the water bath has increased to 28∘C. If we define the system as the gases inside the cylinder, which of the following best describes the signs of q, w, and ΔE for this reaction?
(a) q < 0, w < 0, ΔE < 0
(b) q < 0, w > 0, ΔE < 0
(c) q < 0, w > 0, the sign of ΔE cannot be determined from the information given
(d) q > 0, w > 0, ΔE > 0
(e) q > 0, w < 0, the sign of ΔE cannot be determined from the information given
Answer:
q < 0, w > 0, the sign of ΔE cannot be determined from the information given
Explanation:
Determination of sign of q
Temperature of the water bath before the reaction = 25 °C
Temperature of the water bath after the completion of the reaction = 28 °C
After the completion of the reaction, temperature of the water bath is increased that means heat is released during the reaction and flows out of the system.
If heat is absorbed by the system, then q is indicated by positive sign and if heat is released by the system, then q is indicated by negative sign.
As in the given case, heat is released by the system, so sign of q is negative, or q < 0
Determination of sign of w
After the completion of the reaction, piston moved downward, that means volume of the system decreases or compression occur. During the compression, work is done on the system.
if work is done on the system, sign of w is positive.
If work is done by the system, sign of w is negative.
In the given case, work is done on the system, therefore sign of w is positive, or w > 0
Determination of sign of ΔE
Relationship between ΔE, q and w is given by first law of thermodynamics:
ΔE = q + w
In this case, q is positive and w is negative, so the sign of ΔE depends of magnitude of q and w. As magnitude of w and q cannot be determined in this case, thus, the given information is insufficient for the determination of sign of ΔE.
So, among the given option, option c is correct.
q < 0, w > 0, the sign of ΔE cannot be determined from the information given
The best option that describes the signs of q, w, and ΔE for this reaction is q < 0, w > 0, the sign of ΔE cannot be determined from the information given.
Option C is correct.
From the given information, if we examine the mixture of the gases A2 and B2 and take them to be the system, and the slender metal cylinder as well as the water bath to be the surroundings.
Then, we can infer that:
if there is a movement of heat flow(q) into the system, it signifies a positive signif there is an outflow of heat (q) out of the system, it signifies a negative signHence, when the reaction is said to be completed, we are being informed from the question that the temperature increased. This implies that there is an outflow and release of heat away from the system. As such, when heat flows away from the system, the sign (q) will be negative.
For the work done (w):
suppose work is done on a system show that the sign will be positive;and suppose the work is done by the system, then it will be negative.When the reaction goes into completion, the movement of the piston downward indicates that there is a decrease in the volume of the system.
As such, work is done on the system, hence, the work done (w) is positive.
For the change in the internal energy ΔE. If we look at the first law of thermodynamics, we know that:
ΔE = q + w
here,
we know q to be -ve and w to be +ve;However, the sign of ΔE largely depends on the magnitude of q and w. We can infer that there is little information given for the determination of ΔE
Therefore, we can conclude that q < 0 since it is negative, w > 0 since it is positive and the sign of ΔE cannot be determined from the information given.
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Write balanced molecular equation for the reaction between nitric acid and calcium hydroxide.
Answer:
2HNO3 + Ca(OH)2 → 2H2O + Ca(NO3)2
Explanation:
HNO3 + Ca(OH)2 → ?
This is an acid-base reaction. It is also a double displacement reaction, in which the positive ions change partners.
Thus, H pairs with OH and Ca pairs with NO3.
HNO3 + Ca(OH)2 → HOH + Ca(NO3)2
HOH is our old friend, H2O (water) so, after balancing atoms, the equation becomes:
2HNO3 + Ca(OH)2 → 2H2O + Ca(NO3)2
Which one of the following substances has three unpaired d electrons? Which one of the following substances has three unpaired d electrons? [Cu(NH3)4]2+ [V(H2O)6]4+ [Ag(NH3)2]+ [Zn(NH3)4]2+ [Cr(CN)6]3-
Answer:
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Explanation:
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Need help fast 20 PIONTS
The main component of smog is NO2(g). Smog is a product that is formed through the following series of intermediate chemical
reactions.
N2(g) + O2(g)—>2NO(g)
2NO(g) + O2(g)—>2NO2(g)
NON(g) light_» NO(g) + O(g)
O2(g)+O(g)—>Os(9)
NO(g) + O2(g)—>NO2(g) + O2(g)
What is the overall chemical equation for smog after the above intermediate reactions are combined?
N2(g)+302(9)+203(g)+20(g) NO(g)+ 4NO2(g) +203(9)
N2(g) + 302(g) + O2(g)+O() — 9NO(g)+3NO2(g) + Os(g)
16N2(g) + 302(g) → 3N02(0)
N2(g) +202(g) → 2NO2(g)
Answer:
N₂(g) + 2O₂(g) → 2NO₂(g).
Explanation;
Photo-chemical smog is considered to be a new type of air pollution that is basically a mixture of pollutants that are formed when volatile compounds of organic nature (Hydrocarbons) react and Nitrogen oxides (NOx) react with solar radiations of sun and lead to the formation of photochemical smog that looks like a brown haze in the skies of cities.
pounds (VOCs) react to sunlight, creating a brown haze above cities. It tends to occur more often in summer, because that is when we have the most sunlight.
As mentioned in the question, the formation of smog involves a series of reactions:
N2(g) + O2(g)—>2NO(g)
2NO(g) + O2(g)—>2NO2(g)
NON(g) light_» NO(g) + O(g)
O2(g)+O(g)—>Os(9)
NO(g) + O2(g)—>NO2(g) + O2(g)
If we combine all these chemical equations, it gives an answer as:
N₂(g) + 2O₂(g) → 2NO₂(g).
This is because main components involve in the formation of smog is initial N2 and 2 molecules of O2.
Hope it helps!
Answer:
D
Explanation:
Edge 2020
In the equation for the formation of magnesium sulfide shown, which atom gains electrons?
Mg + S → MgS
S + 2e⁻ → S²⁻ (sulfur will gain electrons)
Mg - 2e⁻ → Mg²⁺ (magnesium will lose electrons)
Magnesium is a metal which tends to give electrons which are received by the sulfur, which is a nonmetal and tends to gain electrons.
Answer: The atom which gains electrons is sulfur
Explanation:
Oxidation reaction is defined as the chemical reaction in which an atom looses its electrons. The oxidation number of the atom gets increased during this reaction.
[tex]X\rightarrow X^{n+}+ne^-[/tex]
Reduction reaction is defined as the chemical reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.
[tex]X^{n+}+ne^-\rightarrow X[/tex]
For the given chemical equation:
[tex]Mg+S\rightarrow MgS[/tex]
Oxidation half reaction: [tex]Mg\rightarrow Mg^{2+}+2e^-[/tex]
Reduction half reaction: [tex]S+2e6-\rightarrow S^{2-}[/tex]
Hence, the atom which gains electrons is sulfur
A mineral is observed to become soft and crumbly and develop a reddish color as it undergoes weathering, what process has most likely taken place?
Answer:
Corrosion
Explanation:
The corrosion is a natural process that occurs because of the influence of the water and wind. It is part of the processes of erosion. It manages to soften the minerals, make them less stable and more crumbly. It can easily be noticed as it gives the mineral reddish, or rather dark orange color. This process occurs where the humidity is high, as the water plays a crucial role for this process to occur. Apart from occurring in nature, this process is also a big problem with the refined metals, as it gradually destroys them, and taking in consideration that the metals are included in all sorts of objects, it can make a lot of damage.
Rank the following elements that make up the majority of Earth’s crust according to increasing ionization energy Mg Ca Si O
The order of increasing ionization energy of the elements are; Ca < Mg < Si < O.
Ionization energy is a periodic trend that decreases down the group but increases across the period.
Ionization energy decreases down the group due to addition of more shells and increased screening of outermost electrons by the inner electrons.
Ionization energy increases across the period due to increase in size of the nuclear charge.
Hence, the order of increasing ionization energy of the elements are; Ca < Mg < Si < O.
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Horizontal rows of the periodic table are called
Horizontal rows of the periodic table are called periods.
Periods are the horizontal rows of the periodic table that correspond to successive layers of electron shells surrounding the atomic nucleus. The atomic number rises as one moves over a period from left to right, adding more protons and electrons in the process.
The elements' characteristics gradually change as a result of this process, changing in reactivity and with an increase in atomic radius. While all elements in the same period have the same number of electron shells, the amounts of valence electrons in each element cause the elements' chemical characteristics to vary.
How many grams of aluminum will be required to produce 4.5 g of copper metal?
Answer:
Mass of aluminum required is 1.277 grams
Explanation:
the reaction between copper and aluminium is a redox reaction and this can be written as:
[tex]3Cu^{+2}+2Al(s) ---> 2Al^{+3} +3Cu(s)[/tex]
Thus we need two moles of aluminum to reduce or to produce three moles of copper metal.
The moles of copper need to produce
= [tex]\frac{mass}{atomicmass}=\frac{4.5}{63.5}=0.071mol[/tex]
These will produced from = [tex]\frac{2X0.071}{3}=0.0473mol[/tex] of aluminum
The mass of aluminum required = moles X atomic mass = 0.0473X27=1.277g
To produce 4.5 g of copper metal, approximately 1.28 g of aluminum will be required, based on stoichiometry calculations using a typical chemical reaction.
Explanation:To determine the amount of aluminum needed to produce 4.5 g of copper metal, we need to use stoichiometry. Initially, we'll need to know the balanced chemical equation for the reaction in which aluminum is used to produce copper.
Typically, this equation might be something like: 2 Al(s) + 3 CuSO4(aq) -> Al2(SO4)3(aq) + 3 Cu(s). From this equation, we see that 2 moles of aluminum will produce 3 moles of copper. In terms of mass, 54 g (2 moles) of aluminum will produce about 190 g (3 moles) of copper (calculated using atomic masses of Al = 27 g/mol and Cu = 63.5 g/mol).
Therefore, to find the amount of aluminum required to produce 4.5 g of copper, we can use a simple ratio: (54 g Al / 190 g Cu = x g Al / 4.5 g Cu). Solving for 'x' gives us approximately 1.28 g of aluminum.
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One disadvantage of vacuum distillation is it is inherently less efficient than fractional distillation at atmospheric pressure.
a. True
b. False
Answer:
One disadvantage of vacuum distillation is it is inherently less efficient than fractional distillation at atmospheric pressure. :True -a.
Approximately when did the first plants and arthropods appear on land? A. 444 Ma B. 488 Ma C. 299 Ma D. 416 Ma
Answer:
A.
Explanation:
444 Ma
Answer:
444 ma is right
Explanation:
A balloon with a volume of 38.2 L at 91 psi is allowed to expand into a larger can with a total volume of 77.4 L. What is the pressure in psi of the gas in the larger container?
Answer:
45 psiExplanation:
1) Data:
a) V₁ = 38.2 liter
b) P₁ = 91 psi
c) V₂ = 77.4 liter
d) P₂ = ?
2) Formula:
According to Boyle's law, at constant temperature, the pressure and volume of a fixed amount of gas are inversely related:
PV = constant ⇒P₁V₁ = P₂V₂3) Solution:
Solve for the unknown: P₂ = P₁V₁ /v₂Substitute the values: V₂ = 91 psi × 38.2 liter / 77.4 liter = 44.9 9si ≈ 45 psi.Hydrogen fuel cells burn with _______ and produce _______.
Hydrogen fuel cells burn with Oxygen and produces Water.
It is commonly used to produce electricity.
Answer:
with oxygen to form water
Explanation:
In a flame of pure hydrogen gas, burning in air, the hydrogen (H2) reacts with oxygen (O2) to form water (H2O) and releases heat. If carried out in atmospheric air instead of pure oxygen (as is usually the case), hydrogen combustion may yield small amounts of nitrogen oxides, along with the water vapor.
How is the enthalpy of reaction related to the enthalpies of formation for the products and reactants?
Answer:
ΔH rxn = ∑ enthalpy of formation for the products - ∑enthalpy of formation for the reactantsExplanation:
The enthalpy of reaction is the change of enthalpy of the chemical reaction. It is equal to the heat released or absorbed during the chemical reaction.
The enthalpy of reaction is equal to the difference of the enthalpy at the end of the reaction and the enthalpy at the begining of the reaction.
Since the substances at the end are the products, and the substances at the begining are the reactants, this is equivalent to say that the enthalpy of reaction is equal to the enthalpy of formation for the products less the enthalpy of formation for the reactants.
In the form of equation that is:
ΔH rxn = ∑ enthalpy of formation for the products - ∑enthalpy of formation for the reactants.The enthalpy of a reaction (ΔH_rxn) is calculated by subtracting the sum of the enthalpies of formation of the reactants from the sum of the enthalpies of formation of the products. This calculation uses Hess's Law and accounts for the stoichiometric coefficients in the balanced equation. As such, the enthalpy change can indicate whether a reaction is exothermic or endothermic.
The enthalpy of a reaction (ΔH_rxn) is determined using the enthalpies of formation (ΔH_f) for the reactants and products. According to Hess's Law, the enthalpy change of any chemical reaction is equal to the sum of the enthalpies of formation of the products minus the sum of the enthalpies of formation of the reactants. Mathematically, this is expressed as:
ΔH_rxn = ∑ΔH_f(products) - ∑ΔH_f(reactants)
The enthalpies of formation are multiplied by the respective number of moles of each substance (as given by their coefficients in the balanced chemical equation) before summing up. For example, in the combustion of methane:
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)The reaction releases 890.5 kJ at 25°C because the energy needed to break the chemical bonds in the reactants differs from the energy released when forming the bonds in the products, reflecting an exothermic process.