Answer:
2 NH₃(g) + 2 O₂(g) ⇒ N₂O(g) + 3 H₂O(l) ΔH° = -684 kJ
Explanation:
Let's consider the unbalanced equation that occurs when NH₃(g) reacts with O₂(g) to form N₂O(g) and H₂O(l).
NH₃(g) + O₂(g) ⇒ N₂O(g) + H₂O(l)
We will begin balancing N atoms by multiplying NH₃(g) by 2.
2 NH₃(g) + O₂(g) ⇒ N₂O(g) + H₂O(l)
Then, we will balance H atoms by multiplying H₂O(l) by 3.
2 NH₃(g) + O₂(g) ⇒ N₂O(g) + 3 H₂O(l)
Finally, we will get the balanced equation by multiplying O₂(g) by 2.
2 NH₃(g) + 2 O₂(g) ⇒ N₂O(g) + 3 H₂O(l)
Since 342 kJ of energy are evolved for each mole of NH₃(g), and there are 2 moles of NH₃(g) in the balanced equation, the heat evolved will be2 × 342 kJ = 684 kJ. By convention, when heat is evolved, it takes a negative sign. The balanced thermochemical equation is:
2 NH₃(g) + 2 O₂(g) ⇒ N₂O(g) + 3 H₂O(l) ΔH° = -684 kJ
During the workup portion of the reaction of alkenes with HBr as described in the experiment provided, a student transferred the reaction mixture to a separatory funnel, rinsed the reaction flask with diethyl ether, and added the ether rinses to the separatory funnel. The student then added sodium bicarbonate to the separatory funnel. Extremely vigorous bubbling occurred. What did the student do wrong
Answer:
Explanation:
Because of the acid-base reaction, as sodium bicarbonate is introduced to the separatory funnel, the additional or unreacted HBr reacts vigorously to yield CO2 gas, which exits the separatory funnel together with any dissolved compound(s) in the ether layer. This is due to a wrong and incorrect selection of the solvent mixture and the addition of sodium bicarbonate to an acidic solution.
Nothing to do with it until it has leaked out of the separatory funnel. Even then, the student may separate the components from the remaining reaction mixture by washing the ether coating layer several times with brine water, then running it into a dry sodium sulfate bed and evaporating solvent ether under decreased pressure.
HELPP ASAP I WILL MARK BRAINIST
Answer:
I'm thinking ethier D or A
Explanation:
Here are the atomic masses of hypothetical elements:
X = 13.25 amu
Y = 69.23 amu
Z = 109.34 amu
3.8 moles of X2Y5Z3 is equivalent to how many grams?
Enter your answer to zero decimal places (round to the ones place). Do
not include the units of "g", just the numerical answer.
Answer:
2663 g
Explanation:
We'll begin by calculating the molar mass Of X₂Y₅Z₃. This can be obtained as follow:
Molar mass of X₂Y₅Z₃ = (13.25×2) + (69.23×5) + (109.34×3)
= 26.5 + 346.15 + 328.02
= 700.67 g/mol
Finally, we shall determine the mass of 3.8 moles of X₂Y₅Z₃. This can be obtained as follow:
Molar mass of X₂Y₅Z₃ = 700.67 g/mol
Mole of X₂Y₅Z₃ = 3.8 moles
Mass of X₂Y₅Z₃ =?
Mass = mole × molar mass
Mass of X₂Y₅Z₃ = 3.8 × 700.67
Mass of X₂Y₅Z₃ = 2663 g
Therefore, the of 3.8 moles of X₂Y₅Z₃ is
2663 g
5.0 g of copper was heated from 20°C to 80°C. How much energy was used to heat Cu?
Answer:
100 J of energy are needed to heat the copper from 20∘C to 80∘C .
8. Is specific heat an intensive or extensive property?
(will give brainlist)
Answer:
Explanation:
A specific property is the intensive property obtained by dividing an extensive property of a system by its mass. For example, heat capacity is an extensive property of a system. Dividing heat capacity, , which is an intensive property.
hope it helps :)
Explanation:
Is specific heat an intensive or extensive property?
intensive propertyHow many grams are there in 1.8055 x 10^25 molecules of sodium sulfate? Hint: Convert to moles first
Answer:
4258.82 g of Na₂SO₄
Explanation:
From the question given above, the following data were obtained;
Number of molecules of Na₂SO₄ = 1.8055x10²⁵ molecules.
Number of mole of Na₂SO₄ =?
From Avogadro's hypothesis,
6.02×10²³ molecules = 1 mole
Therefore,
6.02×10²³ molecules = 1 mole of Na₂SO₄
Next, we shall determine the mass of 1 mole of Na₂SO₄. This can be obtained as follow:
1 mole of Na₂SO₄ = (23×2) + 32 + (16×4)
= 46 + 32 + 64
= 142 g
Thus,
6.02×10²³ molecules = 142 g of Na₂SO₄
Finally, we shall determine the mass of Na₂SO₄ that contains 1.8055x10²⁵ molecules. This can be obtained as follow:
6.02×10²³ molecules = 142 g of Na₂SO₄
Therefore,
1.8055x10²⁵ molecules
= (1.8055x10²⁵ × 142) / 6.02×10²³
= 4258.82 g of Na₂SO₄
Thus, 4258.82 g of Na₂SO₄ contains 1.8055x10²⁵ molecules
At a certain temperature this reaction follows second-order kinetics with a rate constant of 14.1·M−1s−1 : →2SO3g+2SO2gO2g Suppose a vessel contains SO3 at a concentration of 1.44M . Calculate the concentration of SO3 in the vessel 0.240 seconds later. You may assume no other reaction is important. Round your answer to 2 significant digits.
Answer:
[tex][SO_3]=0.25M[/tex]
Explanation:
Hello there!
In this case, since the integrated rate law for a second-order reaction is:
[tex][SO_3]=\frac{[SO_3]_0}{1+kt[SO_3]_0}[/tex]
Thus, we plug in the initial concentration, rate constant and elapsed time to obtain:
[tex][SO_3]=\frac{1.44M}{1+14.1M^{-1}s^{-1}*0.240s*1.44M}\\\\[/tex]
[tex][SO_3]=0.25M[/tex]
Best regards!
Jerry is trying to classify cells by their physical characteristics. He discovers a multicellular organism containing cells that have a nucleus and a cell wall as well as the ability to conduct photosynthesis. Into which of the three domains would this organism most likely fit? A. Archaea B. bacteria C. Eukarya D. Viral
Domains eukarya would this organism most likely fit.
The domain eukarya comprised of eukaryotes or organisms whose cells contain true nucleus.
What is a domain?It is the largest of all groups in the classification of life. There are three domains:-
Archaea domainBacteria domainEukarya domainWhat is Eukarya?It is the domain of organism called eukaryotes. These are the organism who have a well defined nucleus and membrane bound organelles.
Hence, C) option is correct.
To know more about domain here
https://brainly.com/question/26344149
#SPJ2
8. Which of the following elemental gases would have properties closest to an ideal gas?
a. hydrogen
argon
b. helium
d. fluorine
Answer:
Helium
Explanation:
Please tell me if it was right
Consider the balanced equation Zn + 2HCl ZnCl2 + H2 How many
moles of ZnCl2 will be produced if 2 moles of HCl are used?
Answer:
1 mole of ZnCl₂
Explanation:
Just from the stoichiometric equation/ balanced equation:
Zn(s) + 2HCl(aq) → ZnCl₂(s) + H₂(g)
1 mole 2 moles 1 mole 1 mole
Therefore: 2 moles of 2HCl produce 1 mole of ZnCl₂
15.
A tank containing 173 grams of methane, CH (9), registers 15.1 atmospheres at 298 Kelvin. What is the volume of the tank (assuming the entire volume is available to the gas)?
A 3,410L
B. 213
C.O 175L
D. 280.
Chemistry 4/28 5454
Copyright © 2021 Illuminate Education, Inc. All Rig
Answer: The volume of tank is 17.5 L.
Explanation:
Given: Mass of methane = 173 g
Pressure = 15.1 atm
Temperature = 298 K
Molar mass of methane is 16.04 g/mol.
Therefore, moles of methane are calculated as follows.
[tex]No. of moles = \frac{mass}{molar mass}\\= \frac{173 g}{16.04 g/mol}\\= 10.78 mol[/tex]
Now, ideal gas equation is used to calculate the volume 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\\15.1 atm \times V = 10.78 mol \times 0.0821 L atm/mol K \times 298 K\\V = \frac{10.78 mol \times 0.0821 L atm/mol K \times 298 K}{15.1 atm}\\= 17.5 L[/tex]
Thus, we can conclude that volume of the tank is 17.5 L.
Which of the following is a characteristic shared by all living things?
A. They have tissues and organs,
B. They take in oxygen.
C. They make more individuals of the same kind.
D. They move.
What happens to the entropy when a solution is made?
A. The entropy increases.
B. The entropy decreases.
C. The entropy goes to zero.
D. The entropy is unaffected.
Answer:
The entropy increases
Explanation:
Just took the quiz
7. Which object would be the least dense in a tub of
water?
A. Golf ball
B. Glass marble
C. Rubber ball
D. Beach ball
At 47c a gas has a pressure of 140kpa. The gas is cooled until the pressure decreases to 105kpa. If the volume remains constant, what will the final temperature be in kelvin’s? In degrees Celsius
Answer:
The final temperature is equal to 240 K or -33.15°C
Explanation:
Given that,
Initial temperature of the gas, T₁ = 47°C = 320 K
Initial pressure, P₁ = 140 kpa
Final pressure, P₂ = 105 kpa
We need to find the final temperature if the volume remains constant. The relation between temperature and pressure is given by :
[tex]P\propto T[/tex]
or
[tex]\dfrac{P_1}{P_2}=\dfrac{T_1}{T_2}\\\\T_2=\dfrac{P_2T_1}{P_1}\\\\T_2=\dfrac{105\times 320}{140}\\\\T_2=240\ K\\\\T_2=-33.15^{\circ} C[/tex]
So, the final temperature is equal to 240 K or -33.15°C.
Genes influence an organism's traits by coding for:
A
Cells
Answer:
First, the protein may be a structural protein, contributing to the physical properties of cells or organisms. ... Second, the protein may be an enzyme that catalyzes one of the chemical reactions of the cell. Therefore, by coding for proteins, genes determine two important facets of biological structure and function.
Explanation:
i think it will help you
3. How many times does earth rotate on its axis in one year?
Answer:
There are approximately 366.25 sidereal days in a year so that the Earth spins 366.25 times with respect to distant stars in a year
hope this helps
have a good day :)
Explanation:
State the methods of preparing salts
Answer:
Preparation of Salts by The Action of An Acid Upon a Metal.
Preparation of Salts by Double Decomposition.
Preparation of Salts by Neutralization.
Using A Soluble Base (alkali)
(b). Using An Insoluble Base.
Preparation of Salts by the Action of An Acid on The Trioxocarbonate (IV) of A Metal.
Explanation:
How many grams of potassium carbonate are needed to make 300ml of a 4.5M solution?
Answer:
186.3g
Explanation:
4.5moles of K₂CO₃ is in 1000ml
? moles of K₂CO₃ is in 300 ml
(4.5 × 300)/ 1000 = 1.35 moles of K₂CO₃
1 mole of K₂CO₃ = (39 × 2) + 12 + (16 × 3) = 78 + 12 + 48 = 138g
1.35 moles of K₂CO₃ = ?
= (1.35 × 138)/1 = 186.3g
Which of the following is an advantage of asexual reproduction compared to sexual reproduction?
Both will produce genetically identical offspring from the parent.
Sexual reproduction will increase genetic variability within a species.
Asexual reproduction requires less energy and will produce more offspring over time.
Sexual reproduction has minimal changes of mutations compared to asexual reproduction.
Explanation:
sexual reproduction has minimal changes of mutations compared to asexual reproduction
Answer:
sexual reproduction has minimal changes of mutations compared to asexual reproduction
I hope this helps
How will an increase in wind speed affect soil erosion?
O A Soil erosion will increase.
O B. Soil erosion will decrease.
O C. Soil erosion will completely stop.
OD Soil erosion will remain the same.
Answer:
A. Soil erosion will increase.
Option A is correct. Increase in wind speed will increase the rate of erosion.
The process of eroding, transferring, and depositing tiny particles as well as nutrients from the top layer of soil is known as soil erosion by wind.
In arid and semiarid regions, wind-driven soil erosion is a serious issue that impedes the environmental sustainability of animal husbandry and agriculture and threatens ecological security.
A wind that is blowing at 30 mph will erode at a rate that is more than 3 times faster than a wind that is blowing at 20 mph. As soil moisture rises, wind erosion declines. For instance, dry soil erodes approximately 1.3 times more quickly than soil with just enough moisture to support plant life.
Learn more about soil erosion;
https://brainly.com/question/9837143
How many grams of N2 is needed to produce 2000 grams NH3?
Answer:
1644 g
Explanation:
Step 1: Write the balanced equation
N₂ + 3 H₂ ⇒ 2 NH₃
Step 2: Calculate the moles corresponding to 2000 g of NH₃
The molar mass of NH₃ is 17.03 g/mol.
2000 g × 1 mol/17.03 g = 117.4 mol
Step 3: Calculate the moles of N₂ needed to produce 117.4 moles of NH₃
The molar ratio of N₂ to NH₃ is 1:2. The moles of N₂ needed are 1/2 × 117.4 mol = 58.70 mol
Step 4: Calculate the mass corresponding to 58.70 moles of N₂
The molar mass of N₂ is 28.01 g/mol.
58.70 mol × 28.01 g/mol = 1644 g
Bears stop coming to a river ecosystem where they have been eating many fish each day. The fish the bears eat normally eat smaller fish, which eat plants along the river bottom.
What happens to the ecosystem?
Both the larger and the smaller fish populations grow quickly but then die out because the plant life is insufficient for them all to eat.
The larger fish population will drop first, and the smaller fish population will grow quickly. The plants will die off because too many of the smaller fish are eating them.
The larger fish population explodes at first, and the smaller fish population begins to drop. Eventually, the river runs out of smaller fish so larger fish die out, and the plant population grows.
The smaller fish population begins to eat more plants and to grow. The larger fish have more food to eat so their population is able to grow, too.
Answer:
The larger fish population explodes at first, and the smaller fish population begins to drop. Eventually, the river runs out of smaller fish so larger fish die out, and the plant population grows.
Explanation:
If a substance has a density of 0.123 g/mL, 10.0 dL would weigh in g?
Answer:
123 g
Explanation:
First, we convert 10.0 dL into mL, keeping in mind that:
1 dL = 100 mL; then10.0 dL * 100 = 1000 mLNow we can multiply the density by the volume in order to calculate the mass:
Density = mass / volumeDensity * volume = mass0.123 g/mL * 1000 mL = 123 g10.0 dL of a substance with a density of 0.123 g/mL would weigh 123 grams.
What's you're favorite year?
Answer:
2020 cause of the lockdown
Explanation:
PLZ HELP *NO LINKS*
1) How many moles of gaseous arsine (AsH3) occupy 0.372 L at STP?
In sig fig 4
2) What is the density of gaseous arsine?
In sig fig 4
Thanks!
Answer: (1). There are 0.0165 moles of gaseous arsine (AsH3) occupy 0.372 L at STP.
(2). The density of gaseous arsine is 3.45 g/L.
Explanation:
1). At STP the pressure is 1 atm and temperature is 273.15 K. So, using the ideal gas equation number of moles are calculated as follows.
PV = nRT
where,
P = pressure
V = volume
n = number 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 0.372 L = n \times 0.0821 L atm/mol K \times 273.15 K\\n = 0.0165 mol[/tex]
2). As number of moles are also equal to mass of a substance divided by its molar mass.
So, number of moles of Arsine [tex](AsH_{3})[/tex] (molar mass = 77.95 g/mol) is as follows.
[tex]No. of moles = \frac{mass}{molar mass}\\0.0165 mol = \frac{mass}{77.95 g/mol}\\mass = 1.286 g[/tex]
Density is the mass of substance divided by its volume. Hence, density of arsine is calculated as follows.
[tex]Density = \frac{mass}{volume}\\= \frac{1.286 g}{0.372 L}\\= 3.45 g/L[/tex]
Thus, we can conclude that 0.0165 moles of gaseous arsine (AsH3) occupy 0.372 L at STP and the density of gaseous arsine is 3.45 g/L.
a) How do you prepare %3 (w/v) Na2CO3 solution from Na2CO3⸱2H2O? (15p) Na2CO3 MW=106 g/mol
Answer:
4.02g of Na2CO3⸱2H2O must be added completing the volume of the solution to 100mL
Explanation:
A 3%(w/v) solution contains 3g of solute (In this case, Na2CO3) in 100mL of solution.
Assuming we require 100mL of solution we must add 3g of Na2CO3. The reactant that is available is its dihydrate, with molar mass:
106g/mol + 2*MW H2O
106g/mol + 2*18g/mol = 142g/mol
That means the mass of Na2CO3.2H2O that must be added to prepare the solution is:
3g Na2CO3 * (142g/mol Na2CO3.2H2O / 106g/mol Na2CO3) =
4.02g of Na2CO3⸱2H2O must be added completing the volume of the solution to 100mL131.39 g/mol C2HCl3
find the molecular formulas
Answer:
Find the mass of 1 mole.
mass of
1
mole of
131.39
g
m
o
l
C
2
H
C
I
⋅
3
=
37.04000067
g
Explanation:
hope it helps make brainlliest ty
What volume will 28 grams of nitrogen gas occupy at 27 Celsius and a
pressure of 785 mm Hg?
Answer: [tex]2.49\ m^3[/tex]
Explanation:
Given
Mass of nitrogen present [tex]m=28\ g[/tex]
Temperature [tex]T=27^{\circ}C\equiv 300\ K[/tex]
Pressure [tex]P=785\ mm\ \text{of}\ Hg\ \text{or}\ 1.032\ atm[/tex]
The molar mass of Nitrogen [tex]M=28\ g/mol[/tex]
No of moles of nitrogen present
[tex]n=\dfrac{m}{M}\\\\n=\dfrac{28}{28}\\\\n=1[/tex]
Using [tex]PV=nRT[/tex]
[tex]\Rightarrow 1.032\times V=1\times 8.314\times 300\\\\\Rightarrow V=\dfrac{2494.2}{1.032}\\\\\Rightarrow V=2494.2\ L\ \text{or}\ 2.49\ m^3[/tex]
How can a cation be formed?
Add 1 clectron
Take away 1 clectron
Add 1 proton
Take away 1 prekon
