Answer:
80 mL
Explanation:
Step 1: Given data
Initial pressure of hydrogen (P₁): 360 mmHgInitial volume of hydrogen (V₁): 240 mLFinal pressure of hydrogen (P₂): 1080 mmHgFinal volume of hydrogen (V₂): ?Step 2: Calculate the final volume of hydrogen
If we assume ideal behavior and constant temperature, we can calculate the final volume occupied by hydrogen using Boyle's law.
P₁ × V₁ = P₂ × V₂
V₂ = P₁ × V₁/P₂
V₂ = 360 mmHg × 240 mL/1080 mmHg = 80 mL
what is the percent by mass of nitrogen in the following fertilizers? NH3
The percent by mass of nitrogen in ammonia (NH3) is approximately 82.15%
Calculating the mass of nitrogen to the total mass of the compound and then expressing the result as a percentage will allow us to determine the percent by mass of nitrogen in NH3 (ammonia).
Ammonia's molecular structure, NH3, indicates that it is made up of one nitrogen atom (N) and three hydrogen atoms (H). We must take both the molar masses of nitrogen and ammonia into account when calculating the percent by mass of nitrogen.
Nitrogen's (N) molar mass is roughly 14.01 g/mol. The molar masses of nitrogen and hydrogen are added to determine the molar mass of ammonia (NH3). Since hydrogen's molar mass is around 1.01 g/mol, ammonia's molar mass is:
(3 mol H 1.01 g/mol) + (1 mol N 14.01 g/mol) = 17.03 g/mol = NH3.
Now, we can use the following formula to get the nitrogen content of ammonia in percent by mass:
(Mass of nitrogen / Mass of ammonia) / 100% is the percentage of nitrogen by mass.
Ammonia weighs 17.03 g/mol and contains 14.01 g/mol of nitrogen by mass. By entering these values, we obtain:
(14.01 g/mol / 17.03 g/mol) 100% 82.15 % of nitrogen by mass
Ammonia (NH3) has a nitrogen content that is roughly 82.15 percent by mass.
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How many molecules of N204 are in 85.0 g of N2O4?
Answer:
5.56 x 10^23
Explanation:
Just convert and cancel out.
85 g N2O4 x 1 mol/92.01 g x 6.02 x 10^23 molecules /1 mol
A mixture of gas with a total pressure 1.47 atm is found to contain oxygen (O2), carbon monoxide (CO), and nitrogen (N2). What is the partial pressure of carbon monoxide (CO) if the partial pressure of O2is 0.82 atm and the partial pressure of N2is 0.36 atm?
Answer:
0.29 atm
Explanation: add both partial pressure and subtract from total pressure
From these four cycles which are water cycle, carbon cycle, nitrogen cycle, and phosphorus cycle which cycle has more nutrients
Answer:
Nitrogen cycle
Explanation:
The compound FeCl3 Is made of.
Answer:
iron and chlorine
Explanation:
Plants are divided into three groups based on the
Answer:Scientists have identified more than 260,000 kinds of plants. They classify plants according to whether they have body parts such as seeds, tubes, roots, stems, and leaves. The three main groups of plants are seed plants, ferns, and mosses.
Explanation:
hope i help
Use the following Balanced Equation to complete the question: 2 Al + 6 HBr → 2 AlBr3 + 3 H2
If you have 10 moles of Al how many moles of H2 can be produced?
*Will give Brainly!*
Do 5. What is the total number of atoms contained in a 1.00-mole sample of
helium?
Answer: 1 mole = 6.02 x 10^ 23
Explanation:
The number of atoms in one mole of helium is equal to 6.022×10²³ He atoms.
What is Avogadro's number?Avogadro’s constant can be demonstrated as the proportionality factor that utilizes to count the number of particles such as ions, molecules, atoms, or ions in a sample of the substance.
Avogadro's number can be defined as the approximate count of nucleons in 1 gram of substance. The value of the Avogadro number is the mass of 1 mole of a compound, in grams, and is the number of nucleons in one particle.
The value of Avogadro’s number is equal to 6.022×10²³ per mole.
Given, the number of moles of the helium = 1 mol
Helium is a monatomic gas so it contains only one atom in its one molecule.
The number of atoms of Helium in one mole = 6.022 × 10²³
Therefore, the total number of atoms contained in a 1 mole sample of
helium is 6.022 × 10²³.
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Write a short essay about life in the Han Dynasty, comparing it to life today. Make sure to include key features:
-Family
-Government
-Social Structure
-Religion
-Trade
Answer:
Life in the Han Dynasty (206 BCE - 220 CE) differed significantly from today in family, government, social structure, religion, and trade. For example, the Han Dynasty emphasized a patriarchal family structure, where the eldest male held authority, and filial piety was highly valued. In contrast, contemporary societies embrace more egalitarian family dynamics with shared decision-making.
The government system of the Han Dynasty relied on a centralized bureaucracy and emphasized meritocracy, while modern societies often adopted democratic systems. Socially, the Han Dynasty followed a hierarchical model influenced by Confucian principles, whereas contemporary societies strive for greater equality and social mobility.
Religion in the Han Dynasty combined Confucianism, Taoism, and Buddhism, whereas modern societies exhibit diverse religious beliefs. Lastly, trade in the Han Dynasty thrived along the Silk Road, while modern trade was globally interconnected and facilitated by technological advancements. These differences highlight the evolution of society over time.
Explanation:
The precision of a method is being established, and the
following
data are obtained 22.23, 22.18 22.25, 22.09
and 22.15%, is 22.09% a valid measurement?
Answer:
No, 22.09% is not a valid measurement
Explanation:
Precision has to do with how close a given set of measured values are to each other. It is quite different from accuracy. Accuracy refers to how close a given set of values is to the true value. A given set of values may be precise but not accurate and vice versa.
If we look at the values obtained; 22.09%, 22.15%, 22.18%, 22.23%, 22.25%, the value 22.09% is too far off the other values. This implies that it does not represent a valid measurement since it is not close to all the other values obtained.
If a piece of silver specific heat .2165 j/g °C with a mass of 14.16 g and a temperature of 133.5°C is dropped into 250.0 g of fat 17.20°C what will be the final temperature of the system
Answer:
[tex]T_F=17.56\°C[/tex]
Explanation:
Hello there!
In this case, for this calorimetry problem, it is possible for us to realize that the heat lost by the hot silver is gained by the cold far whose specific heat is 3.94 J/g°c, so we can write:
[tex]-Q_{Ag}=Q_{fat}[/tex]
Which can be written in terms of mass, specific heat and temperature as shown below:
[tex]-m_{Ag}C_{Ag}(T_F-T_{Ag})=m_{fat}C_{fat}(T_F-T_{fat})[/tex]
In such a way, solving for the final temperature, we obtain:
[tex]T_F=\frac{m_{Ag}C_{Ag}T_{Ag}+m_{fat}C_{fat}T_{fat}}{m_{Ag}C_{Ag}+m_{fat}C_{fat}}}[/tex]
Then, we plug in the given data to obtain:
[tex]T_F=\frac{14.16g*0.2165J/g\°C*133.5\°C+250g*3.94J/g\°C*17.20\°C}{14.16g*0.2165J/g\°C+250g*3.94J/g\°C} \\\\T_F=17.56\°C[/tex]
Best regards!
Choose all the answers that apply.
The atmosphere:
is made mostly of nitrogen
can be used to transmit radio signals
traps heat from the sun
protects the earth from harmful radiation
is important in the water cycle
Answer:1 2 and 4
Explanation:
According to the equation below, if 2.00 g of PCl5 react completely, how many grams of HCl will be produced?
PCl5+4H2O→H3PO4+5HCl
Answer:
1.75 g HCl
Explanation:
PCl₅ + 4H₂O → H₃PO₄ + 5HClFirst we convert 2.00 g of PCl₅ into moles, using its molar mass:
2.00 g ÷ 208.24 g/mol = 0.0096 mol PCl₅Then we convert PCl₅ moles into HCl moles, using the stoichiometric coefficients of the equation:
0.0096 mol PCl₅ * [tex]\frac{5molHCl}{1molPCl_5}[/tex] = 0.048 mol HClFinally we convert HCl moles into grams, using its molar mass:
0.048 mol HCl * 36.45 g/mol = 1.75 g HClFor each of these pairs of half-reactions, write the balanced equation for the overall cell reaction and calculate the standard cell potential. Express the reaction using cell notation. You may wish to refer to Chapter 20 to review writing and balancing redox equations.
1.
Pt2+(aq)+2e-Pt(s)
Sn2+(aq)+2e-Sn(s)
2.
Co2+(aq)+2e-Co(s)
Cr3+(aq)+3e-Cr (s)
3.
Hg2+(aq)+2e-Hg (I)
Cr2+(aq)+2e-Cr (s)
1. The standard cell potential for this reaction is 0.14 V.
2. The standard cell potential for this reaction is 0.46 V.
3. The reduction potential for Hg2+(aq) + 2e^- → Hg(l) is 0.79 V.
1. The half-reactions are:
Oxidation: Sn2+(aq) → Sn(s) + 2e^-
Reduction: Pt2+(aq) + 2e^- → Pt(s)
To balance the charges, we multiply the oxidation half-reaction by 2:
2Sn2+(aq) → 2Sn(s) + 4e^-
Now, we can combine the half-reactions to form the overall cell reaction:
2Sn2+(aq) + Pt2+(aq) → 2Sn(s) + Pt(s)
The cell notation for this reaction is:
Sn(s) | Sn2+(aq) || Pt2+(aq) | Pt(s)
To calculate the standard cell potential (E°), we can look up the reduction potentials for each half-reaction. The reduction potential for Pt2+(aq) + 2e^- → Pt(s) is typically listed as 0.00 V. The reduction potential for Sn2+(aq) + 2e^- → Sn(s) is -0.14 V. The standard cell potential is the sum of the reduction potentials:
E° = E°(reduction) - E°(oxidation)
E° = 0.00 V - (-0.14 V) = 0.14 V
2. The half-reactions are:
Oxidation: Co2+(aq) → Co(s) + 2e^-
Reduction: Cr3+(aq) + 3e^- → Cr(s)
To balance the charges, we multiply the reduction half-reaction by 2:
2Cr3+(aq) + 6e^- → 2Cr(s)
Now, we can combine the half-reactions to form the overall cell reaction:
Co2+(aq) + 2Cr3+(aq) + 6e^- → Co(s) + 2Cr(s)
The cell notation for this reaction is:
Co(s) | Co2+(aq) || Cr3+(aq) | Cr(s)
To calculate the standard cell potential (E°), we look up the reduction potentials for each half-reaction. The reduction potential for Co2+(aq) + 2e^- → Co(s) is typically listed as -0.28 V. The reduction potential for Cr3+(aq) + 3e^- → Cr(s) is -0.74 V. The standard cell potential is the sum of the reduction potentials:
E° = E°(reduction) - E°(oxidation)
E° = -0.28 V - (-0.74 V) = 0.46 V
3. The half-reactions are:
Oxidation: Cr2+(aq) → Cr(s) + 2e^-
Reduction: Hg2+(aq) + 2e^- → Hg(l)
The balanced overall cell reaction is:
Cr2+(aq) + 2Hg2+(aq) + 4e^- → Cr(s) + 2Hg(l)
The cell notation for this reaction is:
Hg(l) | Hg2+(aq) || Cr2+(aq) | Cr(s)
To calculate the standard cell potential (E°), we look up the reduction potentials for each half-reaction. The reduction potential for Cr2+(aq) + 2e^- → Cr(s) is typically listed as -0.91 V.
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What is true about renewable resources?
They are rare on Earth.
They can be replenished fast.
They can be replenished over millions of years.
They are very abundant on Earth.
They can be replenished fast.
Tarnish on silver is the compound Ag2OAg2O. A tarnished silver spoon is placed in an aluminum pan of boiling water. When enough salt is added so that the solution conducts electricity, the tarnish disappears. Imagine that the two halves of this redox reaction were separated and connected with a wire and a salt bridge.
Answer:
The answer is "2.459".
Explanation:
The cell potential or emf norm = oxidation pot. of anode -oxidation pot. of cathode
When the pot oxidizer is the opposite of a red pot size it's been transformed to either the redox reaction so that anode was its higher oxide pot material that means (AL) so,
[tex]\to emf=1.66-(-0.799) \\\\ \to emf=1.66+0.799) \\\\ \to emf =2.459[/tex]
Pls someone help me with this question pls
Answer:
So confusing but I'll try
A 10.0 mL sample of HNO3 was diluted to a
volume of 100.00 mL. Then 25 mL of that
diluted solution was needed to neutralize 50.0
mL of 0.60 M KOH. What was the
concentration of the original nitric acid?
1.2 M
12 M
none of these
O 0.12 M
0.0012M
Answer:
12 M
Explanation:
The reaction between HNO₃ and KOH is:
HNO₃ + KOH → KNO₃ + H₂OFirst we calculate how many KOH moles reacted with the diluted HNO₃ sample, using the given volume and concentration:
50.0 mL * 0.60 M = 30 mmol KOHAs 1 KOH mol reacts with 1 HNO₃ mol, in 25 mL of the diluted HNO₃ solution there are 30 HNO₃ mmoles.
With that information in mind we can calculate the HNO₃ concentration in the diluted solution:
30 mmol HNO₃ / 25 mL = 1.2 MFinally we can use the C₁V₁=C₂V₂ formula to calculate the concentration of the original solution:
C₁ * 10.0 mL = 1.2 M * 100.00 mLC₁ = 12 MWhich energy source produces less negative
environmental impacts
A)Renewable Energy Sources
B)Fossil Fuels
Help
Answer:
A)Renewable Energy Sources
Explanation:
What would you do during a zombie apocalypse
A. run and hide
B. fight back
C. save some people
D. raid survivor's homes
E. keep your family alive
Answer:
C
Explanation:
because then you'd be able to make a group to raid, run with you, and save more people.
List three ways in which the octet rule can sometimes fail to be obeyed
Answer:
1. Electron-deficient molecules
2. Odd electron molecules
3. Expanded valence shell molecules
Explanation:
In what industry do fertilizers and pesticides wash off and contaminate water supplies?Construction Oil Transportation Agriculture
Answer:
The answer is agriculture.
Explanation:
Answer: Agriculture
Explanation:
I got it right on my exam
In an experiment, 135.2 grams of a mystery metal is heated to 100.0*C. The metal is then dumped into a calorimeter with 59.0 grams of water at 26.0*C. The temperature of the water in the calorimeter increases to 35.0*C after the metal is dumped into it. What is the specific heat capacity of the metal? Express your answer to 2 past the decimal. Do NOT include units
How many moles of NaOH are contained in 56.0 mL of a 2.40 M solution of 1 point
NaOH in water? (**Use only numerical answers with 3 significant figures.
The units are given in the question.)
Your answer
Answer:
1.34 mol
Explanation:
Molarity, which is the molar concentration of a solution, can be calculated by dividing the number of moles (n) by the volume (V).
That is;
Molarity (M) = n/V
According to the information provided in this question;
M = 2.40M
V = 56.0 mL = 56/1000 = 0.056 L
Since molarity = n/V
number of moles = M × V
n = 0.056 × 24
n = 1.34 mol
Show the equation you will use to calculate the volume of 1 M Cu(NO3)2 (aq) needed to prepare a set of solutions that have concentrations in the range of 1 M to 1x10-4 M in a 10-mL volumetric flask. Write the reduction half-cell reaction for the copper(II) ion. What is the standard potential for an electrochemical cell that is prepared from a copper half-cell and a zinc half-cell
Answer:
Explanation:
The equation we use to calculate the volume needed to prepare other [tex](C_1,V_1)[/tex] the solution that has a concentration [tex]C_2[/tex] and volume [tex]V_2[/tex] is:
[tex]C_1V_1 =C_2V_2[/tex]
[tex]V_1=\dfrac{C_2V_2}{C_1}[/tex]
where;
[tex]C_1[/tex]= concentration of the first solution
[tex]V_1[/tex] = volume of the first solution
[tex]C_2[/tex] = concentration of the second solution
[tex]V_2[/tex] = volume of the second solution
2) Reduction half cell reaction for the copper (II) ion is:
[tex]Cu^{2+} + 2e^- \to Cu[/tex]
3) [tex]Cu^{+2} + 2e^- \to Cu \text{ \ \ \ E = 0.3370}[/tex]
[tex]Zn^{+2} + 2 e^- \to Zn \ \ \ \ \ \ E = -0.763[/tex]
[tex]Zn \to Zn^{+2} + 2 e^- \ \ \ \ \ \ E = +0.763[/tex]
Since the reduction potential of Cu is more; it means copper will go into reduction and zinc will undergo oxidation.
Standard Potential =[tex]E^0_{left} - E^0_{right}[/tex]
[tex]= -0.763 -0.337[/tex] ( since both are reduction potential)
[tex]\mathbf{E^0_{cell} = -1.100 volt}[/tex]
What is the volume of 0.200 moles of O2 gas at STP?
Answer:
4.48 L O2
Explanation:
At STP, a mole of any gas contains 22.4 liters. Therefore, we simply have to multiply the amount of moles by 22.4
0.2mol O2 ( 22.4 L) = 4.48 L O2
20 points!
Which two structures produce energy that cells can
use?
A and B
B and C
C and D
D and A
Answer:
c and d
Explanation:
mitochondria and vacoule
Answer:
It is actually D and A
Explanation:
I took this assignment before on edge and the guy who said C and D is wrong
Many bones get their name from the bone they are
How do I solve stoichiometry problems? Please I need help with at least one of these stoichiometry problems with work and steps so I understand better. Thank you!
Answer:
2. m = 426.6 gr
4. m = 143 gr
Why is SpaceX likely to succeed in a mission to Mars?
A. It only hires expert NASA employees as its employees.
B. It takes more risks than NASA, which cannot afford them.
C. It is run by Elon Musk, who is determined to get to Mars.
D. It has fewer restrictions than NASA does.
Answer:
C
Explanation:
It is run by Elon Musk, who is determined to get to Mars.