Answer:
it will decrease
Explanation:
on test. have a nice day
Answer:
I think it is c
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.
Please answer I will give you brainliest!!
Answer:
Warm front
Explanation:
A warm front forms when a warm air mass pushes into a cooler air mass, shown in the image to the right (A). Warm fronts often bring stormy weather as the warm air mass at the surface rises above the cool air mass, making clouds and storms. Warm fronts move more slowly than cold fronts because it is more difficult for the warm air to push the cold, dense air across the Earth's surface. Warm fronts often form on the east side of low-pressure systems where warmer air from the south is pushed north.
You will often see high clouds like cirrus, cirrostratus, and middle clouds like altostratus ahead of a warm front. These clouds form in the warm air that is high above the cool air. As the front passes over an area, the clouds become lower, and rain is likely. There can be thunderstorms around the warm front if the air is unstable.
On weather maps, the surface location of a warm front is represented by a solid red line with red, filled-in semicircles along it, like in the map on the right (B). The semicircles indicate the direction that the front is moving. They are on the side of the line where the front is moving. Notice on the map that temperatures at ground level are cooler in front of the front than behind it.
t-Butyl alcohol was produced by the liquid-phase hydration (using water, W) of isobutene (I) over an Amberlyst-15 catalyst.
a. True
b. False
Answer:
True
Explanation:
This is because, The hydroboration oxidation of an alkene which is isobutene in the presence of a catalyst will result to alcohol as the product . Therefore, the OH group will attach or link itself to the carbon which is less obstructed. Thus this reaction is in accordance to Anti-Markownikoff's rule.
So isobutene on hydroboration oxidation will produce ter isobutyl alcohol.
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.
How many stoms of oxygen in 4 molecules of HNO,?
Answer: the answer is a
is scandium a transition metal?
Answer:no
Explanation:
Answer:
Scandium is a transition metal
Explanation:
Why are some chemical substances, like oil and coal, considered nonrenewable?
the process that forms them stops working after a decade
the process that forms them has not worked for millions of years
the process that forms them is very fast
the process that forms them is very slow
Answer:
The process of the formation of coal and oil is a very gradual and slow process.
Explanation:
The process of the formation of coal and oil is a very gradual process that takes up to millions of years.
How much does REAL carbon fiber cost ? lets say as big as a piece of paper
Answer:
Today, the average total production cost of “standard modulus” carbon fiber is in the range of $7-9 per pound.
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 much energy (kJ) is required to change 0.18 mole of ice (s) at 0 C to water (l) at 0 C?
Answer:25,06 kJ of energy must be added to a 75 g block of ice.
ΔHfusion(H₂O) = 6,01 kJ/mol.
T(H₂O) = 0°C.
m(H₂O) = 75 g.
n(H₂O) = m(H₂O) ÷ M(H₂O).
n(H₂O) = 75 g ÷ 18 g/mol.
n(H₂O) = 4,17 mol.
Q = ΔHfusion(H₂O) · n(H₂O)
Q = 6,01 kJ/mol · 4,17 mol
Q = 25,06 kJ.
Explanation:
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
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
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
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When NH3(g) reacts with O2(g) to form N2O(g) and H2O(l) , 342 kJ of energy are evolved for each mole of NH3(g) that reacts. Write a balanced thermochemical equation for the reaction with an energy term in kJ as part of the equation. Note that the answer box for the energy term is case sensitive. Use the SMALLEST INTEGER coefficients possible and put the energy term (including the units) in the last box on the appropriate side of the equation. If a box is not needed, leave it blank.
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
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
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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.
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:
How to pass chem!!!!!!!!!!
Answer:
refer to below
Explanation:
- read your textbook! especially the topics that are most confusing
- as you read, make sure to take notes/make an outline. write the words from the text in your own words. In chem, the textbook explanations are usually very detailed, so putting it in your own words will help you remember and understand what you write.
- make sure you really understand the concepts. don't just memorize what you read.
- draw diagrams and structures. seeing how things are structured will help you visualize what you learn
- ask questions to your teacher. if there is anything confusing, make sure to refer to your teacher as they will hopefully be able to explain it so it won't be confusing
These tips have helped me pass chem with an A. Hope all of this helped!
Answer:
Explanation:
Study material, ask for help, read textbook and do practice problems, get a tutor, beg your teacher for marks,
Rubbing alcohol and vegetable oil soluble on insoluble?
which objects could you stand near to feel the heat
Answer:
An heater, Oven, sun, and fireplaces
Chose which ever you want
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
What is the volume, in liters, of 0.350 mol of nitrogen gas at 32°C and
0.980 atm of pressure? *
A. 9.85 L
B. 8.94 L
C. 104.6 L
D. 0.94 L
Answer: The volume is 8.94 L.
Explanation:
Given: no. of moles = 0.350 mol, Pressure = 0.980 atm
Temperature = [tex]32^{o}C = (32 + 273) K = 305 K[/tex]
Formula used to calculate the volume is 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\\0.980 atm \times V = 0.350 mol \times 0.0821 L atm/mol K \times 305 K\\V = 8.94 L[/tex]
Thus, we can conclude that the volume is 8.94 L.
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:
Introduction: Reaction rates are also influenced by surface area and concentration. The surface area of a solid is a measure of how much of the solid is exposed to other substances. The concentration of a substance is a measure of how many molecules of that substance are present in a given volume. Question: How do surface area and concentration affect reaction rates
Answer:
See explanation
Explanation:
Surface area has to do with the number of solid particles that are exposed at a given time and is capable of colliding with other reactant particles. When more surface area is exposed for reaction, then it means that more particles are likely to collide with each other leading to faster chemical reaction rates. When few particles are exposed for reaction (low surface area) then less collisions occur and the rate of reaction is decreased.
Similarly, concentration refers to the amount of substance present. The greater the amount of substance present, the greater the likelihood of collision between particles and the greater the rate of reaction and vice versa.
Two dogs, one with spots and one with no spots have the same mass and
are running with different velocities. Dog 1 (Dog with no spots) has a
velocity of 1 m/s while Dog 2 (Dog with spots) has a velocity of 2 m/s.
Which dog has a larger kinetic energy and how much larger is its kinetic
energy compared to the other dog
Answer:
dog 2 because it is running 1m faster a second than dog 1. Dog 2 can run 1 meter farther in a second than the other dog.
Thorium-236 has a half life of 10.0 minutes. If I have 250 grams of Thorium-236, How many half lives will have taken place after 50.0 minutes?
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
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 100mL
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
What is the oxidation state of S in S2O32−?
A. +3 B. +2 C. –2 D. +4
Answer:
it'd be +2
Explanation:
Since 3 atoms of oxygen are present, so its combined oxidation number will be3(−2)=−6. Now this ion contains a charge of−2, where the sum of the oxidation number of all atoms in an atom is equal to−2. So the oxidation number of S in S2O2−3 should be +2.
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]
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