First, we must know this:
1 mole of anything = 6.022 x 10^23 6.022 formula units
Procedure:
We have 9.11 moles, so, we do this:
1 mole of Na2SO4 ------------ 6.022 x 10^23 6.022 formula units
9.11 moles of Na2SO4 ------------ x
[tex]x\text{ = }\frac{9.11\text{ moles x 6.022x}10^{23}formula\text{ units}}{1\text{ mole}}=\text{ }5.48\text{ x }10^{24}\text{ formula units}[/tex]Answer: 5.48x10^24 formula units
Perform the followingmathematical operation, andreport the answer to theappropriate number ofsignificant figures.0.00652 – 0.0003168 = [ ? ]=Enter
Let's do the operation:
0.00652 - 0.0003168 = 0.0062032
After the calculations, we choose as a reference the number with the fewest decimal places, it is 0.00652. So we need to round off the result to three significant figures.
So the answer is 0.00620.
Answer: 0.00620
Potassium chlorate is broken down into oxygen gas and potassium chloride A) write a balanced equation:B) what type of reaction is this?C) how many moles of KClO3 are needed to make 1.5 moles of oxygen gas?D) how many grams of KCl could be formed when 0.33 grams of KClO3 are used?
A) 2KClO3 -> 3O2 + 2KCl, this is the balanced equation
B) Since we have a compound being broken down into 2 new ones, we have a Decomposition Reaction
C) In the balanced equation we see that for every 2 moles of KClO3, we will end up with 3 moles of O2, therefore
2 KClO3 = 3 O2
x KClO3 = 1.5 O2
x = 1 mole of KClO3
D) First we need to see how many moles of KClO3 does 0.33g mean, we will know that by using the molar mass of the compound, which is 122.55g/mol
122.55g = 1 mol
0.33g = x moles
x = 0.0027 moles of KClO3
Now we have the number of moles, and also we need to check the reaction to know what is the mole ratio of KClO3 and KCl, and by doing that we can see that with 2 moles of KClO3 we will make 2 moles of KCl, therefore if we have 0.0027 moles of KClO3, we will make also 0.0027 moles of KCl
Now to find the mass we will use the number of moles and the molar mass, which is 74.55g/mol
74.55g = 1 mol
x grams = 0.0027 moles
x = 0.201 grams of KCl
Which of the following phrases best describes the H-O bonds present in water (H 2O)?nonpolar and ionicpolar and ionicnonpolar and covalentpolar and covalent
Answer: O-H bonding present in water molecule is best described as polar and covalent.
o Write the formulas Label the phases. Balance the equationCalcium chlorate is decomposed to produce calcium chloride and oxygen.
Answer
Ca(ClO₃)₂ → CaCl₂ + 3O₂
Procedure
Calcium chlorate decomposes to give oxygen and calcium chloride.
Calcium chlorate has a chemical formula of Ca(ClO3)₂, calcium chloride (CaCl₂), and molecular oxygen (O₂).
The reaction can be written as following
Ca(ClO₃)₂ → CaCl₂ + O₂
The equation is not balanced because the number of atoms of the oxygen element is not the same on both sides of the equation. The equation can be balanced by adding 3 as a coefficient for oxygen as follows:
Ca(ClO₃)₂ → CaCl₂ + 3O₂
How many moles are in 6.0 g of C?
Using the molar mass of Carbon, 12g/mol, we know that in 12 grams of it we have 1 mol, therefore:
12g = 1 mol
6.0g = x moles
x = 0.5 moles of Carbon in 6 grams
(Please help me) Determine the oxidation state of the metal in Cds
In this compound, we have two elements Cd (Cadmium) and S (Sulfur), and in order to determine the oxidation number or state, we need to at least know the oxidation number of the other element, since we need to counterbalance and match both values.
Even though Sulfur has some possibilities as oxidation number, the most usual is the same as Oxygen, -2 as charge, therefore Cadmium must have an oxidation number of +2, in order to match both values
Answer = +2, number 5
If you have 1.23 L of 0.347 M KNO2, 11.0 g of Al, and 1.89 L of 3.2 M NaOH, what mass of NH3 can be produced?
The mass of NH₃that can be produced is 5.49 g.
What is the mole of the reactants present in the given solution?The moles of reactants in the given solutions are calculated using the formula given below as follows:
Moles = molar concentration * volume in litersFor KNO₂:
Volume = 1.23 Liters
molar concentration = 0.347 M
Moles = 1.23 * 0.347
Moles = 0.427 moles
For Al:
Moles = mass / molar mass
Moles = 11 / 27
moles = 0.407 moles
For NaOH:
Volume = 1.89 Liters
molar concentration = 3.2 M
Moles = 1.89 * 3.2
Moles = 6.048 moles
Equation of reaction:
NO₂⁻ (aq) + 2 Al (s) + H₂O (l) + OH⁻ (aq) → NH₃ (g) + 2 AlO₂⁻ (aq)
Al is the limiting reactant
Moles of NH₃ produced = 0.407/2 moles
Mass of NH₃ produced = 0.407/2 * 17
Mass of NH₃ produced = 5.49 g
Learn more about moles at: https://brainly.com/question/13314627
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The coefficient for the balance chemical equation below is ( note-you must FIRST balance the chemical equation)CH3CH3 + O2 ------> CO2 + H20 Group of answer choices2119134
To answer this question, we need to:
Step 1 - Balance the question
To balance the equation it is necessary to have the same number of atoms of each element on both sides of the equation (reactant side and product side). Let's count the number of atoms of each element for the unbalanced equation: C2H6 + O2 ---> CO2 + H2O
Reactant side:
C - 2
H - 6
O - 2
Product side:
C - 1
O - 3
H - 2
As you can see, carbon, oxygen and hydrogen number of atoms are not equal on both sides of the equation. Let's balance it by changing the stoichiometric coefficient.
C2H6 + O2 ---> 2 CO2 + H2O
Now we have 2 atoms of carbon on both sides.
Let's balance hydrogen:
C2H6 + O2 ---> 2 CO2 + 3 H2O
Now we have 6 atoms of hydrogen on both sides and 7 atoms of oxygen.
C2H6 + 7/2 O2 ---> 2 CO2 + 3 H2O
Now we have 7 atoms of oxygen on both sides, but the coefficient of O2 is not an integer. So let's multiply all the coefficients by 2:
2 C2H6 + 7 O2 ---> 4 CO2 + 6 H2O
Answer: 2 C2H6 + 7 O2 ---> 4 CO2 + 6 H2O
Coefficients are: 2, 7, 4, 6
The sum of them: 19
The density of an object is 10g/ml. The mass of the object is 5 grams. What is the volume of the object?Group of answer choices2 ml0.2 mi50 ml0.05 ml0.5 ml
1) List the known and unknown quantities.
Density: 10 g/mL.
Mass: 5 g.
Volume: unknown.
2) Set the equation
Equation of density
[tex]D=\frac{mass\text{ }(g)}{volume\text{ }(mL)}[/tex]3) Plug in the known quantities and solve for volume.
[tex]10\frac{g}{mL}=\frac{5\text{ }g}{volume\text{ }(mL)}[/tex][tex]Volume\text{ }(mL)=\frac{5\text{ }g}{10\text{ }\frac{g}{mL}}[/tex][tex]Volume(mL)=0.5\text{ }mL[/tex]The volume of the object is 0.5 mL.
85 degrees Fahrenheit correspond to what temperature on the kelvin scale?Group of answer choices85K90K358K302K329K
Answer:
D. 302 K.
Explanation:
Let's see the formula to convert temperature in Fahrenheit to Kelvin:
[tex]K=\frac{(\degree F-32)}{1.8}+273.[/tex]Our temperature in Fahrenheit is 85 °F, so replacing this value in the formula we obtain:
[tex]K=\frac{85-32}{1.8}+273=302\text{ K.}[/tex]85 °F on the Kelvin scale is D. 302 K.
The reaction below is at equilibrium. What would happen if more carbon wereadded?C(s) +0,92=C0 (9)A. The equilibrium position would shift to produce more products.B. The equilibrium position would remain the same.C. The equilibrium position would shift to produce more reactants.D. The equilibrium position would shift to produce more oxygen.
answer and explanation
equilibrium constant will shift to relieve any stress exerted on the reaction so when more carbon is added the equilibrium shifts to relieve this stress.
option A is correct
Not a timed or graded assignment. Please refer to requirements circled in red, quick answer = amazing review :)
For this problem, we have to use Avogadro's number which is 6.022 x 10^(23) /mol. This number is telling us that there are 6.022 x 10^(23) atoms in 1 mole of a certain compound/element. The conversion would be:
[tex]0.30molesSO_2\cdot\frac{6.022\cdot10^{23}atoms}{1\text{ mol}}=1.8\cdot10^{23}atoms.[/tex]The answer is that there are 1.8 x 10 ^(23) atoms of SO2 in 0.30 moles.
How many molecules of NH3 would there be in 367.2 grams of NH3?
Number of molecules = number of moles x avogadros number
So given the above equation, we have to start by calculating NH3 number of moles.
n = m/M where m is the mass (367.2 g) of NH3 and M is the molar mass (17.031 g/mol)
n = 367.2/17.031
n = 21.56 mol
Avogadros number = 6.022x10^23 mol-1
number of molecules = 21.56 mol x 6.022x10^23 mol-1
number of molecules = 1.299x10^25
Calculate how many miles of Al are needed to react
Step 1
The equation provided:
2 Al + 3 FeO => Al2O3 + 3 Fe (completed and balanced9
------------------
Step 2
Information provided:
1.20 moles of FeO
-------------------
Step 3
By stoichiometry, procedure:
2 Al + 3 FeO => Al2O3 + 3 Fe
2 moles Al ----------- 3 moles FeO
X ---------- 1.20 moles FeO
X = 1.20 moles FeO x 2 moles Al/3 moles FeO
X = 0.80 moles
Answer: 0.80 moles
Can same element have different mass? if yes or No.
Isotopes are the same elements, with the same number of protons but different number of neutrons, which due to the rule of definition of Atomic mass, A = p + n, if we increase or decrease n, which is the number of neutrons, we will have a different mass even though we are still talking about the same element, therefore the answer is YES
For the reaction 2 Kl+Pb(NO 3 ) 2 Pbl 2 +2 KNO 3 how many grams of lead(II) nitrate, Pb(NO 3 ) 2 are needed to react completely with 71.9 g of potassium iodide, KI?
71.73grams
Explanations:Given the chemical reaction;
[tex]2KI+Pb(NO_3)_2\rightarrow PbI_2+2KNO_3[/tex]Given the following parameter
Mass of KI = 71.9grams
Determine the moles of KI
According to stoichiometry, 2moles of KI reacts with 1 mole of lead(II) nitrate,the moles of lead(II) nitrate that reacted is given as:
[tex]\begin{gathered} moles\text{ of Pb\lparen NO}_3\text{\rparen}_2=\frac{1}{2}\times0.4331 \\ moles\text{ of Pb\lparen NO}_3\text{\rparen}_2=0.2166moles \end{gathered}[/tex]Determine the mass of lead(II) nitrate
[tex]\begin{gathered} mass\text{ of Pb\lparen NO}_3\text{\rparen}_2=moles\times molar\text{ mass} \\ mass\text{ of Pb\lparen NO}_3\text{\rparen}_2=0.2166\times331.2 \\ mass\text{ of Pb\lparen NO}_3\text{\rparen}_2=71.73g \end{gathered}[/tex]Hence the mass of lead(II)nitrate needed to react completely with 71.9 g of potassium iodide is 71.73grams
What is the molarity of a solution prepared by dissolving 37.94 g of potassium hydroxide, KOH, in some water and then diluting the solution to a volume of 500.00 mL? {Ans. ~ [KOH] = 1.352 M} Write the set up of the calculations that justify the answer.
Answer:
1.352 M.
Explanation:
What is given?
Mass of KOH = 37.94 g.
Volume of KOH solution = 500.00 mL.
Molar mass of K = 39.1 g/mol.
Molar mass of O = 16 g/mol.
Molar mass of H = 1 g/mol.
Step-by-step solution:
First, let's see the formula of molarity:
[tex]Molarity\text{ \lparen M\rparen=}\frac{mole\text{s of solute}}{liter\text{s of solution}}=\frac{mol}{L}.[/tex]We have to find the moles of solute, our solute, in this case, is KOH. To convert 37.94 g of KOH to moles, we use the molar mass of KOH which is 56.10 g/mol using the given molar masses of the elements that can be found in the periodic table (39.1 g/mol + 16 g/mol + 1 g/mol = 56.1 g/mol). The conversion from grams to moles will look like this:
[tex]37.94\text{ g KOH}\cdot\frac{1\text{ mol KOH}}{56.1\text{ g KOH}}=0.6763\text{ moles KOH.}[/tex]And now, we have to find the volume in liters because we have the volume in milliliters. Remember that 1 L equals 1000 mL, so 500.0 mL in L is:
[tex]500.0mL\cdot\frac{1\text{ L}}{1000\text{ mL}}=0.5000\text{ L.}[/tex]The final step is to replace the obtained values in the molarity formula:
[tex]Molarity=\frac{0.6763\text{ moles}}{0.5000\text{ L}}=1.352\text{ M.}[/tex]The molarity of solution would be 1.352 M.
18.Ammonium hydroxide, (NH4OH), is considered a weak base because...Select one:a. it produces few hydroxide ions in aqueous solution.b. it produces few hydrogen ions in aqueous solution.c. it does not conduct electricity in aqueous solution.d. it will not take part in a neutralization reaction.
Answer
A. it produces few hydroxide ions in aqueous solution
If 18.1 grams of silicone tetrachloride reacts with 8.4 moles of hydrogen what mass of silicone is formed
Explanation:
SiCl₄ + 2 H₂ ----> Si + 4 HCl
According to the problem, 18.1 g of SiCl₄ will react with 8.4 moles of H₂ to give Si and HCl. First we have to determine which of them is the limiting reactant.
To determine the limiting reactant we can find the mass of Si that will be produced by each reactant if they were reacting with an excess of the other one.
According to the coefficients of the reaction, 1 mol of SiCl₄ will produce 1 mol of Si. Let's find the number of moles of Si produced by 18.1 g of SiCl₄
1 mol of SiCl₄ = 1 mol of Si
mass of SiCl₄ = 18.1 g
molar mass of Si = 28.09 g/mol
molar mass of Cl = 35.45 g/mol
molar mass of SiCl₄ = 1 * 28.09 g/mol + 4 * 35.45 g/mol
molar mass of SiCl₄ = 169.89 g/mol
18.1 g of SiCl₄ * (1 mol of SiCl₄/169.89 g) * (1 mol of Si/1 mol of SiCl₄) * (28.09 g of Si/1 mol of Si) = 2.99 g of Si
Now we can do something similar with the other reactant.
moles of H₂ = 8.4 moles
2 moles of H₂ = 1 mol of Si
8.4 moles of H₂ * (1 mol of Si/2 moles of H₂) * (28.09 g of Si/1 mol of Si) = 121.38 g of Si
So SiCl₄ is the limiting reactant and H₂ is in excess.
Answer: 2.99 g of silicone is formed.
LiOH(s) + CO2(g) —> LiHCO3(s)Determine the limiting reagent if you have 30.0 grams of both LiOH and CO2.
ANSWER
The limiting reagent for the reaction is CO2
EXPLANATION
Given that;
The mass of LiOH that reacted is 30.0 grams
The mass of of CO2 that reacted is 30.0 grams
Follow the steps below to find the limiting reactant of the reaction.
Step 1; Find the mole of each reactant
[tex]\text{ mole = }\frac{\text{ mass}}{\text{ molar mass}}[/tex]Recall, that the molar mass of LiOH and CO2 are 23.95 g/mol and 44.01 g/mol respectively.
For LiOH
[tex]\begin{gathered} \text{ mole = }\frac{\text{ 30}}{\text{ 23.95}} \\ \text{ mole = 1.253 moles} \end{gathered}[/tex]The number of moles of LiOH is 1.253 moles
For CO2
[tex]\begin{gathered} \text{ mole = }\frac{\text{ mass}}{\text{ molar mass}} \\ \text{ mole = }\frac{\text{ 30}}{\text{ 44.01}} \\ \text{ mole = 0.682 mol} \end{gathered}[/tex]The mole of CO2 in the reaction is 0.682 mol
In chemical reaction, the limiting reagent of the reaction always has the least number of moles.
Hence, the limiting reagent for this reaction is CO2
Would one substance have a different specific heat in different phases?
Remember that:
The specific heat capacities of the solid, liquid, and gaseous phases of the same substance are not the same, and the latent heat of fusion is not the same as the latent heat of vaporization for the same substance.
The intermolecular forces of matter in different phases are different, so different amounts of energy are needed to raise their temperature.
Therefore, the correct answer is B.
State evidence from the balanced equation for the cell with iron and copper electrodes that indicates the reaction in the cell is an oxidation-reduction reaction.
Answer
The reactions in the cells are both an oxidation-reduction
In the first reaction below
The Fe is oxidized to Fe²⁺, hence it is undergoing oxidation.
The Cu²⁺ is reduced to Cu, thus is undergoing reduction.
In the second reaction
The Zn is oxidized to Zn²⁺, therefore it is undergoing oxidation half .oxidation.
What is true about the equation below? CH4, + Cl2 - → CCI4, + HCI It is showing a physical change occuring. The equation is balanced. O It shows a chemical change occurring. O There are 2 Carbon (C) atoms in the Reactant.
1. The chemical equation can not be a physical change because there are changes in the composition of the compounds and substances that are involved, so this is false.
2. Let's see that in the reactants we have 1 carbon, 4 hydrogen, and 2 chlorines and in the products, we have 1 carbon, 1 hydrogen, and 4 chlorines, meaning that the equation is unbalanced, so this is false.
3. As we said before, this chemical equation represents a chemical reaction, so the chemical change indeed is occurring, so this is true.
4. In the reactants, we just have 1 carbon atom (CH4), so if we say that there are 2 carbon atoms is false.
What is the maximum number of orbitals that can be identified by the following set of quantum numbersn=5, l=1
Explanation
n = principal quantum number
l = angular momentum quantum number
-------------
The numbers provided are:
n = 5 => the electron is located on the fifth energy level
l = 1 => the electron is located in the p-subshell
------------
e.g. if n = 1 and l = 0 there is one orbital 1s
if n = 2 and l = 0 => 1 orbital 2s. If l = 1 => 3 orbitals 2p
If n = 5 there are in total 25 orbitals
answer:
If n = 5 and l = 1 we will have 3 orbitals 5p
How many moles does .365 liters of O2 gas represent
Step 1 - Understanding the molar volume
Every gas, in certain conditions known as STP conditions, occupies the same volume per mol. This is known as the molar volume, and corresponds to 22.4 L/mol. This means that each mole of a gas occupies 22.4 L (at STP).
Step 2 - Using the molar volume to solve the problem
Since we already know the relation between moles and volume in the STP, we can set the following proportion:
[tex]\begin{gathered} 1\text{ mole occupies ---- 22.4 L} \\ 0.365\text{ moles occupies --- x} \\ \\ x=\frac{22.4\times0.365}{1}=8.17\text{ L} \end{gathered}[/tex]0.365 moles of O2 gas at STP occupies thus 8.17 L.
What weight of sodium nitrate (NaNO3) must be used to prepare 506mL of a 10.5 M solution of this salt in water? Answer in units of g.
The question requires us to calculate the mass of NaNO3 necessary to prepare 506 mL of a 10.5 M solution of this salt.
To answer this question, we'll need to go through the following steps:
1) Calculate the molar mass of NaNO3, as it will be necessary during the question;
2) Calculate the number of moles necessary to prepare 506 mL of the 10.5 M solution;
3) Convert the number of moles calculated into the mass of NaNO3, using the molar mass of this salt.
Next, we'll follow the steps:
1) To calculate the molar mass of NaNO3, we need to consider the atomic masses of Na, N and O and the number of atoms of each of these elements. The atomic masses are 22.99 u, 14.01 u and 15.99 u for Na, N and O, respectively. With this information, we can calculate the molar mass:
molar mass of NaNO3 = (1 * 22.99) + (1 * 14.01) + (3 * 15.99) = 84.97 g/mol
2) Next, we use the molar concentration provided (10.5 M or 10.5 mol/L) and the volume required (506 mL = 0.506 L) to obtain the number of moles necessary to prepare this solution:
1 L of solution --------------- 10.5 mol of NaNO3
0.506 L of solution -------- x
Solving for x, we have:
[tex]x=\frac{(0.506\text{ L of solution)}\times(10.5\text{ mol of NaNO}_3)}{(1\text{ L of solution)}}=5.31\text{ mol of NaNO}_3[/tex]Therefore, the number of moles of NaNO3 necessary to prepare the solution is 5.31 moles.
3) At last, we use the molar mass obtained in step 1 (84.97 g/mol) to calculate the mass that corresponds to 5.31 moles of NaNO3:
1 mol of NaNO3 ---------------- 84.97 g
5.31 mol of NaNO3 ----------- y
Solving for y, we have:
[tex]y=\frac{(5.31\text{ mol of NaNO}_3)\times(84.97\text{ g of NaNO}_3)}{(1\text{ mol of NaNO}_3)}=451.2\text{ g of NaNO}_3[/tex]Therefore, the mass of NaNO3 necessary to prepare the solution given by the question is 451.2 g.
How many grams of lithium nitrate will be needed to make 200.0 grams of lithium sulfate, assuming that you have an adequate amount of lead (IV) sulfate to do the reaction? [a] How much lead (IV) sulfate will you actually need? [b]Pb(SO4)2 + 4 LiNO3 --> Pb(NO3)4 + 2 Li2SO4
[a] In the reaction, you can see that 4 moles of lithium nitrate (LiNO3) produce 2 moles of lithium sulfate (Li2SO4). The first thing that we need to do, is found the moles of lithium sulfate based on its mass (200.0 grams) and then, find the number of moles of lithium nitrate produced (remember that the molar mass of Li2SO4 is 110 g/mol).
[tex]200.0gLi_sSO_4\cdot\frac{1molLi_2SO_4}{110gLi_2SO_4}=1.82molesLi_2SO_4[/tex]Now, doing a rule of three to find the number of moles of LiNO3, that would be:
[tex]\begin{gathered} 4molesLiNO_3\to2molesLi_2SO_4 \\ \text{?molesLiNO}_3\to1.82molesLi_2SO_4. \end{gathered}[/tex]The calculation to find the number of moles of LiNO3 would be:
[tex]1.82\text{ moles}Li_2SO_4\cdot\frac{4molesLiNO_3}{2\text{ moles }Li_2SO_4}=3.64molesLiNO_3.[/tex]Using this value and the molar mass of LiNO3 which is 69 g/mol, we're going to obtain the mass in grams of LiNO3, like this:
[tex]3.64molesLiNO_3\cdot\frac{69gLiNO_3}{1molLiNO_3}=251.16gLiNO_3.[/tex]We will need 251.16 grams of lithium nitrate to make 200.0 grams of lithium sulfate.
[b] You can see that we're producing 3.64 moles of LiNO3. In the reaction, 4 moles of LiNO3 are reacting with 1 mole of Pb(SO4)2 and we will take into account this information because LiNO3 is the reactant that runs out first; the calculation to find the number of moles of lead (VI) sulfate would be:
[tex]3.64molesLiNO_3\cdot\frac{1molPb(SO_4)_2}{4molesLiNO_3}=0.91molesPb(SO_4)_2.[/tex]Using this value and the molar mass of Pb(SO4)2 which is 399.1 g/mol, we're going to obtain the mass of Pb(SO4)2:
[tex]0.91molesPb(SO_4)_2\cdot\frac{399.1\text{ g }Pb(SO_4)_2}{1\text{ mol }Pb(SO_4)_2}=363.18\text{ g }Pb(SO_4)_2.[/tex]We will need 363.18 grams of lead (IV) sulfate.
___Al(NO3)3 + ___FeCl2 ___Fe(NO3)2 + ___AlCl31. Write the names or formulas of the reactants in this equation.2. Write the names or formulas of the products in this equation
Answer:
Explanation:
'Here, we want to get the reactants and the products in the given reaction
The reactants are on the left side, while the products are on the right side
The reactants are the compounds before the arrow and the products are the compounds after the arrow
Thus, we have it that:
[tex]\begin{gathered} \text{ Reactants} \\ Al(NO_3)_3 \\ \text{FeCl}_2 \\ \text{Products} \\ Fe(NO_3)_2 \\ \text{AlCl}_3 \end{gathered}[/tex]A sample of hydrogen (H2) gas at 122 ∘C and has a pressure of 1.10 atmAt what final temperature, in degrees Celsius, will the pressure of the H2 decrease to 0.26 atm, if volume and amount of gas do not change?Express your answer using three significant figures.
Explanantion:
We are given: initial temperature of H2 = 122 degrees Celsius
: initial pressure of H2 = 1.10 atm
: final pressure of 0.26 atm
Using Gay-Lussac's Law:
[tex]\begin{gathered} \frac{P_1}{T_1}\text{ = }\frac{P_2}{T_2} \\ \\ \therefore T_2\text{ = }\frac{P_2T_1}{P_1}\text{ = }\frac{0.26\times122}{1.10}\text{ = 28.9 }\degree C \end{gathered}[/tex]Answer:
At 28.9 degrees Celsius
4.The molar enthalpy of combustion of glucose, C6H12O6(s) is –2538.7 kJ/mol.a.Calculate the amount of energy produced by 2.35 mol of glucose.
Answer:
-5965.9 kJ.
Explanation:
What is given?
Molar enthalpy of combustion of glucose (C6H12O6) = -2538.7 kJ/mol.
Moles of glucose = 2.35 moles.
Steb-by-step solution:
Based on the molar enthalpy and the moles, we can do a dimensional analysis to find the amount of energy. Remember that kJ is the unit for energy, so the dimensional analysis will look like this:
[tex]-2538.7\text{ }\frac{kJ}{mol}\cdot2.35\text{ mol=-5965.9 kJ.}[/tex]The amount of energy produced by 2.35 mol of glucose is -5965.9 kJ.