Answer:
1. Electron-deficient molecules
2. Odd electron molecules
3. Expanded valence shell molecules
Explanation:
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:
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 HClHow 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
Pls someone help me with this question pls
Answer:
So confusing but I'll try
Help it’s about Binary compounds and transition metals
Answer:
Cr 6+ & SO4 2-
Explanation:
Sulfate is a polyatomic ion that has a charge of 2-. There are three of them in the chemical formula so it equates to a 6- total charge. Thus, chromium must have a 6+ charge to give the compound a neutral charge.
Which energy source produces less negative
environmental impacts
A)Renewable Energy Sources
B)Fossil Fuels
Help
Answer:
A)Renewable Energy Sources
Explanation:
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
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
2. Gas A has twice as much mass as Gas B. Compared to one mole of gas A, one mole of
gas B contains:
a) one-half as many particles
b) the same number of particles
c) twice as many particles
d) 22.4 times as many particles
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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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 MChoose 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:
state two differences between palmitic acid, oleic acid and linoleic acid in full sentences
Answer:
Linoleic acid is omega-6 fatty acid, Oleic acid is omega-9 fatty acid and Palmitic acid is omega-7 fatty acid
Explanation:
Linoleic acid is omega-6 fatty acid which is not naturally produced by the body while Oleic acid is omega-9 fatty acid which occurs naturally in various animal and vegetable fats and oils.
As compared to oleic acid, Linoleic acid is lightweight and thin.
Palmitic acid is omega-7 monounsaturated fatty acid which occurs naturally and is found in plant and marine sources.
Palmitic acid is lighter than both the oleic acid and Linoleic acid
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.
Many bones get their name from the bone they are
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
A student investigated whether the number of bees he saw was related to the temperature outside. His results are on the graph below. Explain the relationship between temperature and the number of bees he saw. Use evidence from the graph to support your answer
Answer:
OOH a lot of points! too bad you scammed me :) its what you get i gueess
Explanation:
Answer:
The number of bees increased as the heat increased.
Explanation:
The line is on a steady increase.
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.
Use the following Balanced Equation to complete the question: 2 Al + 6 HBr → 2 AlBr3 + 3 H2
If you have 12 moles of HBr how many moles of H2 can be produced?
*Will give Brainly!*
what is the mass concentration in ppm of NaCl of 0.01% mass/mass
A-10
B-100
C-10^3
D-10^4
E-10^5
Answer:
B-100
Explanation:
ppm is an unit of concentration that could be defined as the mass in mg of solute (In this case, NaCl) per kg of solution.
Now, a solution of NaCl that is 0.01% by mass, contains 0.01g of NaCl in 100g of solution.
To solve this question, we must convert the mass of NaCl to mg and the mass of solution to kg:
Mass NaCl:
0.01g * (1000mg / 1g) = 10mg
Mass Solution:
100g * (1kg / 1000g) = 0.10kg
The ppm are:
10mg / 0.10kg =
100ppm
Right answer is:
B-100If 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!
For 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 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
The compound FeCl3 Is made of.
Answer:
iron and chlorine
Explanation:
calculate the difference in slope of the chemical potential against temperature on either side of the normal freezing point of water and the normal boiling point of water the molar entropy change accompanying fussion is 22.0 and that accompanying evaporation
Answer:
(a) The normal freezing point of water (J·K−1·mol−1) is [tex]-22Jmole^-^1k^-^1[/tex]
(b) The normal boiling point of water (J·K−1·mol−1) is [tex]-109Jmole^-^1K^-^1[/tex]
(c) the chemical potential of water supercooled to −5.0°C exceed that of ice at that temperature is 109J/mole
Explanation:
Lets calculate
(a) - General equation -
[tex](\frac{d\mu(\beta )}{dt})p-(\frac{d\mu(\alpha) }{dt})_p[/tex] = [tex]-5_m(\beta )+5_m(\alpha )[/tex] = [tex]-\frac{\Delta H}{T}[/tex]
[tex]\alpha ,\beta[/tex] → phases
ΔH → enthalpy of transition
T → temperature transition
[tex](\frac{d\mu(l)}{dT})_p -(\frac{d\mu(s)}{dT})_p[/tex] =[tex]= -\frac{\Delta_fH}{T_f}[/tex]
= [tex]\frac{-6.008kJ/mole}{273.15K}[/tex] ( [tex]\Delta_fH[/tex] is the enthalpy of fusion of water)
= [tex]-22Jmole^-^1k^-^1[/tex]
(b) [tex](\frac{d\mu(g)}{dT})_p-(\frac{d\mu(l)}{dT})_p= -\frac{\Delta_v_a_p_o_u_rH}{T_v_a_p_o_u_r}[/tex]
= [tex]\frac{40.656kJ/mole}{373.15K}[/tex] ([tex]\Delta_v_a_p_o_u_rH[/tex] is the enthalpy of vaporization)
= [tex]-109Jmole^-^1K^-^1[/tex]
(c) [tex]\Delta\mu =\Delta\mu(l)-\Delta\mu(s)[/tex] =[tex]-S_m\DeltaT[/tex]
[tex][\mu(l-5[/tex]°[tex]C)-\mu(l,0[/tex]°[tex]C)][/tex] = [tex][\mu(s-5[/tex]°[tex]C)-\mu(s,0[/tex]°[tex]C)][/tex][tex]=-S_m[/tex]ΔT
[tex]\mu(l,-5[/tex]°[tex]C)-\mu(s,-5[/tex]°[tex]C)=-Sm\DeltaT [\mu(l,0[/tex]
[tex]\Delta\mu=(21.995Jmole^-^1K^-^1)\times (-5K)[/tex]
= 109J/mole
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
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
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.
From these four cycles which are water cycle, carbon cycle, nitrogen cycle, and phosphorus cycle which cycle has more nutrients
Answer:
Nitrogen cycle
Explanation:
A voltaic cell is constructed in which the anode is a Fe|Fe2 half cell and the cathode is a Cd|Cd2 half cell. The half-cell compartments are connected by a salt bridge. (Use the lowest possible coefficients. Be sure to specify states such as (aq) or (s). If a box is not needed, leave it blank.) The anode reaction is:
Answer: The Anode reaction is [tex]Fe(s)\rightarrow Fe^{2+}(aq)+2e^-[/tex]
Explanation:
Oxidation-reduction reaction or redox reaction is defined as the reaction in which oxidation and reduction reactions occur simultaneously.
Oxidation reaction is defined as the reaction in which a substance looses its electrons.Oxidation occurs at anode.
Anode : [tex]Fe(s)\rightarrow Fe^{2+}(aq)+2e^-[/tex]
Reduction reaction is defined as the reaction in which a substance gains electrons. Reduction occurs at cathode.
Cathode : [tex]Cd^{2+}(aq)+2e^-\rightarrow Cd(s)[/tex]