Answer: 24.6kpa
Explanation:
This is a classic case where Boyle's Law can be applied.
The equation for Boyle's Law is given as: P1V1 = P2V2
Where P1 = initial pressure
P2 = final pressure
V1 = initial volume
V2 = final volume
From the question, P1 = ?, P2 = 98.2kpa , V1 =4L , V2 =1L
P1V1 = P2V2
P1 x 4L = 98.2 kpa x 1L
Make P1 subject of formula we then have:
P1 = 98.2kpa X 1L / 4L
= 24.6kpa
Answer:
24.6kpa
Explanation:
Which of the following is not an effect of climate change?
Rising sea levels
Coastal flooding
Increased glacial ice
Extreme weather
Answer:
coastal flooding
Explanation:
Answer:
Increased heat, drought and insect outbreaks, all linked to climate change, have increased wildfires. Declining water supplies, reduced agricultural yields, health impacts in cities due to heat, and flooding and erosion in coastal areas are additional concerns.
Explanation:
What is parthenolide
balance the equation C2H4O2 + O2 → CO2 + H2O
Answer:
Explanation:
[tex]C_{2}H_{4}O_{2}+2O_{2}[/tex] → [tex]2CO_{2}+2H_{2}O[/tex]
The balanced chemical equation is C₂H₄O₂ + (3/2)O₂ → 2CO₂ + 2H₂O.
A balanced chemical equation is an equation that represents a chemical reaction with an equal number of atoms of each element on both sides of the equation. It follows the law of conservation of mass, which states that matter cannot be created or destroyed during a chemical reaction.
In a balanced chemical equation, the coefficients are used to adjust the number of molecules or atoms involved in the reaction. Balancing chemical equations is essential for accurately representing the stoichiometry of a reaction and understanding the ratios of reactants and products involved.
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A tank contains 36.5 g of O₂ and 46.4 g of N₂ at a pressure of 3.64 atm. What is the pressure of O₂ in the tank (in atm)?
Answer:
Partial pressure O₂ = 1.49 atm
Explanation:
We can solve this, by the mole fraction concept.
Mole fraction = Moles of gas / Total moles
We can also define mole fraction as: Partial pressure of a gas / Total pressure
Total moles = Moles of O₂ + Moles of N₂
36.5 g. 1 mol / 32g + 46.4 g . 1mol /28g = 2.79 moles
1.14 moles O₂ + 1.65 moles N₂ = 2.79 moles
Partial pressure O₂ / Total pressure = Moles of O₂ / Total moles
Partial pressure O₂ = (Moles of O₂ / Total moles) . Total pressure.
We replace: (1.14 mol / 2.79 mol) . 3.64 atm = 1.49 atm
Partial pressure O₂ = 1.49 atm
24.568 cm of chlorine at 25° C will occupy what volume
at -25°C while the pressure remains constant?
Answer:
20.446cm3
Explanation:
The following were obtained from the question given:
V1 (initial volume) = 24.568 cm3
T1 (initial temperature) = 25°C = 25 + 273 = 298K
T2 (final temperature) = - 25°C = - 25 + 273 = 248K
V2 (final volume) = ?
Since the pressure is constant, the gas is obeying Charles' law. With the Charles' law equation V1/T1 = V2/T2, we can easily calculate the final volume as follow:
V1/T1 = V2/T2
24.568/298 = V2/248
Cross multiply to express in linear form as shown below:
298 x V2 = 24.568 x 248
Divide both side by 298
V2 = (24.568 x 248) /298
V2 = 20.446cm3
From the calculations made above, the volume of the gas at -25°C is 20.446cm3
The middle ear converts sound from what into vibrations that can be sent through what?
Answer:
.
Explanation:
Eardrum or tympanic membrane changes sound to vibration and pass through malleus, incus and stapes.
Cells are present
A. in only humans.
B.
in all living organisms.
C.
in only plants.
D.
in only animals.
Answer:
B
Explanation:
GOOD LUCK!!!
6) At constant temperature, the pressure of a gas is 780 mmHg and
volume of the gas is 2.0 L. Calculate the volume of the gas when
pressure decreases to 620 mmHg.
Answer:
The volume of the gas when pressure decreases to 620 mmHg is 2.52 L
Explanation:
Boyle's law is a law related to gases that establish a relationship between the pressure and the volume of a given amount of gas, without variations in temperature, that is, at constant temperature. So, this law establishes that at constant temperature, the volume of a fixed mass of gas is inversely proportional to the pressure it exerts.
Boyle's law is expressed mathematically as:
Pressure * Volume = constant
o P * V = k
Having a certain volume of gas V1 that is at a pressure P1 at the beginning of the experiment, if you vary the volume of gas to a new value V2, then the pressure will change to P2, and it will be true:
P1 * V1 = P2 * V2
In this case:
P1: 780 mmHgV1: 2 LP2: 620 mmHgV2: ?Replacing:
780 mmHg* 2 L= 620 mmHg* V2
Solving:
[tex]V2=\frac{780 mmHg* 2L}{620 mmHg}[/tex]
V2= 2.52 L
The volume of the gas when pressure decreases to 620 mmHg is 2.52 L
Answer:
2.52 L
Explanation:
Step 1:
The following data were obtained from the question:
Initial pressure (P1) = 780 mmHg
Initial volume (V1) = 2.0 L.
Final pressure (P2) = 620 mmHg.
Final volume (V2) =?
Step 2:
Determination of the final volume of the gas.
From the question given, we discovered that the temperature is constant. Since the temperature is constant, the gas is simply obeying Boyle's law.
Applying the Boyle's law equation, the final volume of the gas can be obtained as follow:
P1V1 = P2V2
780 x 2 = 620 x V2
Divide both side by 620
V2 = (780 x 2) /620
V2 = 2.52 L
Therefore, the volume of the gas when the pressure decreases to 620 mmHg is 2.52 L
Consider the balanced equation for the decomposition of ozone:
2O3(g) 3O2(g)
Can the rate law for this chemical reaction be deduced from its balanced equation?
A)Yes; since the reactant and product are both oxygen gases, the reaction takes 0 seconds.
B)Yes; the coefficients of the balanced equation determine the order of the reaction.
C)No; the rate law must be determined experimentally.
D)No; the rate will change randomly.
Answer:
C)No; the rate law must be determined experimentally.
Explanation:
Rate Laws are empirical relationships defining rates of reaction. For any given reaction The Empirical Rate Law is the product of the concentration of the reactants raised to the power of their order of reaction. That is, for the hypothetical reaction aA + bB => Products its Empirical Rate Law is ...
Rate = k[A]ᵃ[B]ᵇ where k = the rate constant and a & b are orders of reaction.
Order of reaction is a 'rate trend' when one changes a rate factor such as concentration. This can only be determined by experimental observation by physically increasing or decreasing concentration and observing the change in reaction rate relative to a reference reaction of interest. Typically they are whole numbers but can be decimal fractions.
By graphing experimental outcomes of Rxn Rate vs Change in Concentration one can define the order of reaction. Observations give ...
0-order reactions => change concentration => no change in rate
1st order reactions => change concentration => proportional linear change in rate
2nd order reactions => change concentration => exponential change in rate.
Answer:
The answer is C
Explanation:
I had the question on edg
How many grams of CO2 gas would occupy 25.7 L of volume at a temperature of –22.2oC and a pressure of 997 mm Hg? (***Must convert moles into gram at the end of this problem)
Answer:
72.028g
Explanation:
Data obtained from the question. This includes:
Volume (V) = 25.7 L
Temperature (T) = –22.2°C
Pressure (P) = 997 mmHg
Mass of CO2 =..?
Mass of CO2 can be obtained as follow:
Step 1:
Conversion to appropriate unit:
It is important to express each variables in their appropriate units in order to obtain the desired result in the right unit.
For temperature:
We shall be converting from celsius to Kelvin. This is illustrated below:
Temperature (Kelvin) = temperature (celsius) + 263
temperature (celsius) = –22.2°C
Temperature (Kelvin) = –22.2°C + 263
Temperature (Kelvin) = 250.8K
For Pressure:
We shall be converting from mmHg to atm. This is illustrated below:
760mmHg = 1 atm
Therefore, 997 mmHg = 997/760 = 1.31 atm
Step 2:
Determination of the number of mole of CO2.
With the ideal gas equation, we can obtain the number of mole of CO2 as follow:
Volume (V) = 25.7 L
Temperature (T) = 250.8K
Pressure (P) = 1.31 atm
Gas constant (R) = 0.082atm.L/Kmol
Number of mole (n) =...?
PV = nRT
n = PV/RT
n = (1.31 x 25.7)/(0.082 x 250.8)
n = 1.637 mole
Step 3:
Converting 1.637 mole of CO2 to grams. This is illustrated below:
Number of mole of CO2 = 1.637 mole
Molar Mass of CO2 = 12 + (2x16) = 12 + 32 = 44g/mol
Mass of CO2 =...?
Mass = number of mole x molar Mass
Mass of CO2 = 1.637 x 44
Mass of CO2 = 72.028g.
Air was put into an automobile tire when the tire was cold, at 5.00 C. The tire's air pressure was 3.04 atm. Later, the weather warms up. Also, the automobile is driven, further warming the tire. What will the tire's pressure be when the temperature is 25.0 C?
Answer:
[tex]P_{2} = 3.259\,atm[/tex]
Explanation:
Let suppose that air inside the tire behaves ideally. The equation of state for ideal gases is:
[tex]P\cdot V = n\cdot R_{u}\cdot T[/tex]
As tire can be modelled as a closed and rigid container, there are no changes in volume and number of moles. Hence, the following relationship is constructed:
[tex]\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}[/tex]
The final pressure is:
[tex]P_{2} = \frac{T_{2}}{T_{1}}\cdot P_{1}[/tex]
[tex]P_{2} = \frac{298.15\,K}{278.15\,K}\cdot (3.04\,atm)[/tex]
[tex]P_{2} = 3.259\,atm[/tex]
Answer:
P2=3,26 atm
Explanation:
Considering that this gas behaves like an ideal gas, the equation used in noble gases is:
PxV = nxrxT
p is pressure, v is volume, n is number of moles, r is a constant of noble gases whose value is 0.082x10exp23 as long as you use it in the equations it has the same value (that is why it is called constant) and t would be the temperature.
Because v, r and n are values that did not change throughout the reaction of the gas, we are only going to take into account the p and t that did change throughout the reaction and that is why there is an initial and final pressure and a final and initial temperature.
So:
P1 / T1 = P2 / T2 (where value 1 refers to start, and 2 to end)
Regarding this equation, since we want to know the value of P2, which is the FINAL pressure of the gas, that is, what pressure did it achieve with the reaction, the equation that we would use would be solving for P2:
P2 = (T2 / T1) xP1 = 3.26 atm
what mass of helium is in a 2.00 L balloon at STP? pls help me lol
Once upon a time there was a lovely
princess. But she had an enchantment
upon her of a fearful sort which could
only be broken by love's first kiss.
She was locked away in a castle guarded
by a terrible fire-breathing dragon.
Many brave knights had attempted to
free her from this dreadful prison,
but non prevailed. She waited in the
dragon's keep in the highest room of
the tallest tower for her true love
and true love's first kiss. (laughs)
Like that's ever gonna happen. What
a load of - (toilet flush)
Allstar - by Smashmouth begins to play. Shrek goes about his
day. While in a nearby town, the villagers get together to go
after the ogre.
Calculate the wavelength of yellow light produced by a sodium lamp if the frequency of radiation is 3.34 x 10^14 Hz
Answer : The wavelength of yellow light produced by a sodium lamp is, [tex]8.98\times 10^{-7}m[/tex]
Explanation : Given,
Frequency of radiation = [tex]3.34\times 10^{14}Hz=3.34\times 10^{14}s^{-1}[/tex]
conversion used : [tex]Hz=s^{-1}[/tex]
Formula used :
[tex]\nu=\frac{c}{\lambda}[/tex]
where,
[tex]\nu[/tex] = frequency of radiation
[tex]\lambda[/tex] = wavelength of radiation
c = speed of light = [tex]3\times 10^8m/s[/tex]
Now put all the given values in the above formula, we get:
[tex]3.34\times 10^{14}s^{-1}=\frac{3\times 10^8m/s}{\lambda}[/tex]
[tex]\lambda=8.98\times 10^{-7}m[/tex]
Therefore, the wavelength of yellow light produced by a sodium lamp is, [tex]8.98\times 10^{-7}m[/tex]
A flexible vessel contains 43L of gas where the pressure is 1.3 atm. What will the volume be when the pressure is 0.61 atm, the temperature remaining constant?
Answer:
91.64L
Explanation:
Step 1:
Data obtained from the question.
This includes the following:
Initial volume (V1) = 43L
Initial pressure (P1) = 1.3 atm
Final volume (V2) =..?
Final pressure (P2) = 0.61 atm
Step 2:
Determination of the final volume.
To solve for the final volume, we'll apply the Boyle's law equation since the temperature is constant. The final volume is obtained as follow:
P1V1 = P2V2
1.3 x 43 = 0.61 x V2
Divide both side by 0.61
V2 = (1.3 x 43)/0.61
V2 = 91.64L
Therefore, the new volume is 91.64L
The volume will be 91.64 L when the pressure is 0.61 atm
From Boyle's law equation, we have that
P₁V₁ = P₂V₂
Where P₁ is the initial pressure
V₁ is the initial volume
P₂ is the final pressure
and V₂ is the final volume
From the given information
P₁ = 1.3 atm
V₁ = 43 L
P₂ = 0.61 atm
V₂ = ?
Putting the parameters into the formula, we get
1.3 × 43 = 0.61 × V₂
∴ [tex]V_{2} = \frac{1.3 \times 43}{0.61}[/tex]
V₂ = 91.64 L
Hence, when the pressure is 0.61 atm, the volume will be 91.64 L
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CK-12 Boyle and Charles's Laws if Mrs. Pa pe prepares 12.8 L of laughing gas at 100.0 k Pa and -108 °C and then she force s the gas into an 855 mL cylinder whe re it warms to room temperature at 22 °C, NC what is the pressure of this gas in kilopascals
Answer:
The answer to your question is P2 = 2676.6 kPa
Explanation:
Data
Volume 1 = V1 = 12.8 L Volume 2 = V2 = 855 ml
Temperature 1 = T1 = -108°C Temperature 2 = 22°C
Pressure 1 = P1 = 100 kPa Pressure 2 = P2 = ?
Process
- To solve this problem use the Combined gas law.
P1V1/T1 = P2V2/T2
-Solve for P2
P2 = P1V1T2 / T1V2
- Convert temperature to °K
T1 = -108 + 273 = 165°K
T2 = 22 + 273 = 295°K
- Convert volume 2 to liters
1000 ml -------------------- 1 l
855 ml -------------------- x
x = (855 x 1) / 1000
x = 0.855 l
-Substitution
P2 = (12.8 x 100 x 295) / (165 x 0.855)
-Simplification
P2 = 377600 / 141.075
-Result
P2 = 2676.6 kPa
Consider a mixture of two gases A and B, confined to a closed vessel. A quantity of a third gas, C, is added to the same vessel at the same temperature. How does the addition of gas C affect the total pressure in the vessel?
Answer:
Total pressure increased
Explanation:
When gas C is added in the vessel then number of mole increases and number of collision depends on the number of molecules present in the vessel and on adding gas C ,mole also increases hence number of collision increases therefore pressure also increases because number of collision increases.
Total pressure increases.
Question 2 (1 point)
A cold glass of water collects water droplets on the outside on a hot summer day.
What is the best explanation for this?
The water vapor in the air releases energy to the cold water in the glass causing
the vapor to become liquid that collects on the sides of the glass.
The water vapor in the air gains energy and collides against the sides of the cold
glass resulting in water droplets forming on the sides of the glass.
The cold water in the glass heats up and then this water collects in droplets on
the sides of the glass.
The cold water in the glass releases energy to the water vapor in the air causing
it to collect on the outside of the glass.
Answer:
The water vapor in the air releases energy to the cold water in the glass causing
the vapor to become liquid that collects on the sides of the glass.
Explanation:
Matter exist in solid, liquid and gaseous states. The energy of particles constituting each state of matter differs. In the gaseous state, the particles that compose matter are highly energetic and move at high speeds. Liquid particles possess lower energy and solid particles are the least energetic.
When water vapour in air comes in contact with a cold glass of water, heat exchange occurs. The energetic gas particles of water vapour looses energy to the cold water particles. This causes the water vapour outside the glass to condense to liquid and collects outside the glass.
The best explanation for water droplets collecting on the outside of a cold glass of water on a hot summer day is: Option (1.)The water vapor in the air releases energy to the cold water in the glass causing the vapor to become liquid that collects on the sides of the glass.
This process is called condensation.
On a hot day, the air is typically humid, meaning it contains a significant amount of water vapor. When the humid air comes into contact with the cold surface of the glass, the temperature of the air near the glass drops. If this temperature drops below the dew point (the temperature at which air becomes saturated with moisture), the water vapor in the air condenses into liquid water. This is because the cold surface of the glass absorbs heat from the water vapor, causing the vapor to lose energy and change from a gaseous state to a liquid state, forming droplets on the outside of the glass.
What
was the initial volume of the hydrogen in cm3?
Answer:
255.51cm3
Explanation:
Data obtained from the question include:
V1 (initial volume) =?
T1 (initial temperature) = 50°C = 50 + 273 = 323K
T2 (final temperature) = - 5°C = - 5 + 237 = 268K
V2 (final volume) = 212cm3
Using the Charles' law equation V1/T1 = V2/T2, the initial volume of the gas can be obtained as follow:
V1/T1 = V2/T2
V1/323 = 212/268
Cross multiply to express in linear form
V1 x 268 = 323 x 212
Divide both side by 268
V1 = (323 x 212)/268
V1 = 255.51cm3
Therefore, the initial volume of the gas is 255.51cm3
Answer:
Explanation:
To get the initial volume of the hydrogen in cm³, we apply the equation generated from Charles' law.
V1/T1=V2/T2
Where;
V1=initial volume of gas
T1=initial temperature of gas
V2=final volume of gas
T2=final temperature of gas
We make V1 subject of formula
V1=(V2×T1)/T2
Given;
V2=212cm³
T1=50°C=(50+273)K=323K
T2=-5.0°C=(-5.0+273)K=268K
V1=(212×323)/268
V1=255.5cm³
Therefore, the initial volume of the hydrogen was 255.5cm³
2. What is the percentage by mass of sodium (Na) in a formula unit of sodium
hydrogen carbonate (NaHCO3)?
The percentage by mass of sodium in sodium hydrogen carbonate (NaHCO3) is 27.4%, calculated by dividing the atomic mass of sodium by the molar mass of NaHCO3 and multiplying by 100%.
Explanation:
To calculate the percentage by mass of sodium (Na) in a unit of sodium hydrogen carbonate (NaHCO3), you first need to know the atomic masses of the elements involved. The atomic mass of Na is approximately 23 g/mol, H is about 1 g/mol, C is approximately 12 g/mol and O is approximately 16 g/mol.
Using these numbers, the molar mass of NaHCO3 is about 84 g/mol (23 for Na, 1 for H, 12 for C, and 48 for the three O atoms). To find the percentage of Na in NaHCO3, divide the atomic mass of Na by the molar mass of NaHCO3 and multiply by 100%:
(23 g/mol ÷ 84 g/mol) x 100% = 27.4%.
Therefore, the percentage by mass of sodium in a formula unit of sodium hydrogen carbonate is 27.4%.
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The mass percentage of sodium (Na) in a formula unit of sodium hydrogen carbonate (NaHCO₃) is 27.36%. The calculated molar mass of NaHCO₃ is 84.01 g/mol.
To determine the percentage by mass of sodium (Na) in a formula unit of sodium hydrogen carbonate (NaHCO₃), we need to follow these steps:
1. Find the atomic masses of each element:
Sodium (Na) = 22.99 g/mol
Hydrogen (H) = 1.01 g/mol
Carbon (C) = 12.01 g/mol
Oxygen (O) = 16.00 g/mol
2. Calculate the molar mass of NaHCO₃:
NaHCO₃ = 22.99 g/mol (Na) + 1.01 g/mol (H) + 12.01 g/mol (C) + 3 x 16.00 g/mol (O)
NaHCO₃ = 22.99 + 1.01 + 12.01 + 48.00 = 84.01 g/mol
3. Determine the mass percentage of sodium (Na):
Mass % Na = (22.99 g / 84.01 g) x 100
Mass % Na = 27.36%
Therefore, the mass percentage of sodium (Na) in sodium hydrogen carbonate (NaHCO₃) is 27.36%.
For the vaporization of a liquid at a given pressure: A.ΔG is positive at all temperatures. B.ΔG is negative at all temperatures. C.ΔG is positive at low temperatures, but negative at high temperatures (and zero at some temperature). D.ΔG is negative at low temperatures, but positive at high temperatures (and zero at some temperature). E.None of these.
Answer:
C. ΔG is positive at low temperatures, but negative at high temperatures (and zero at some temperature).
Explanation:
Since we need to give energy in the form of heat to vaporize a liquid, the enthalpy is positive. In a gas, molecules are more separated than in a liquid, therefore the entropy is positive as well.
Considering the Gibbs free energy equation:
ΔG= ΔH - TΔS
+ +
When both the enthalpy and entropy are positive, the reaction proceeds spontaneously (ΔG is negative) at high temperatures. At low temperatures, the reaction is spontaneous in the reverse direction (ΔG is positive).
Why is stoichiometry is used in cooking?
Answer:
Explanation:
stoichiometry is used in cooking because it helps you determine the amount or proportion of compounds you will need in a chemical reaction. Stoichiometry is present in daily life, even in the cooking recipes we make at home. The reactions depend on the compounds involved and how much of each compound is needed to determine the product that will result.
Michelle has a cup of hot coffee. The liquid coffee weighs 236 grams. Michelle adds a few teaspoons sugar and 25 grams of milk to the coffee. Michelle stirs the mixture until everything is combined. The mixture now weighs 271 grams. How many grams of sugar did Michelle add to the coffee?
Answer:
10grams
Explanation:
So it weighs 236 grams added with 25 grams. So it now weighs 261 grams so 10 grams of sugar remains in it.
Final answer:
Michelle added 10 grams of sugar to her coffee. The calculation involved subtracting the initial coffee weight and the weight of added milk from the final mixture weight.
Explanation:
To determine how many grams of sugar Michelle added to her coffee, we need to analyze the before and after weights of the mixture. The initial weight of the coffee was 236 grams, and after adding milk, the total weight became 271 grams. Given that Michelle also added 25 grams of milk, we can calculate the weight of the sugar.
The calculation to find the weight of the sugar added is as follows:
Subtract the initial weight of the coffee (236 grams) from the final weight of the mixture (271 grams) to account for the added ingredients: 271 grams - 236 grams = 35 grams.
Subtract the weight of the milk (25 grams) from the 35 grams to find the weight of the sugar: 35 grams - 25 grams = 10 grams.
Therefore, Michelle added 10 grams of sugar to her coffee.
A balloon of air occupies 10.0L(V2)at25.0°C(T2)and1.00atm(P2). What temperature (T1) was it initially, if it occupied 9.40 L (V1) and was in a freezer with a pressure of 0.939 atm (P1)?
The initial temperature (T₁) was 263.0145 K.
Given Conditions:
Final volume (V₂) = 10.0 L
Final temperature (T₂) = 25.0°C = 298.15 K
Final pressure (P₂) = 1.00 atm
Initial volume (V₁) = 9.40 L
Initial pressure (P₁) = 0.939 atm
Using the Combined Gas Law:
[tex]\[ \dfrac{P_1 \times V_1}{T_1} = \dfrac{P_2 \times V_2}{T_2} \][/tex]
Rearrange to Solve for T₁:
[tex]\[ T_1 = \dfrac{P_1 \times V_1 \times T_2}{P_2 \times V_2} \][/tex]
Plugging in the Values:
[tex]\[ T_1 = \dfrac{0.939 \, \text{atm} \times 9.40 \, \text{L} \times 298.15 \, \text{K}}{1.00 \, \text{atm} \times 10.0 \, \text{L}} \][/tex]
Calculate T₁:
[tex]\[ T_1= \dfrac{2630.145}{10} = 263.0145 \, \text{K} \][/tex]
Nitrifiers are bacteria
a. capable of living in symbiosis in the roots of some plants.
b. capable of oxidizing NH3 using NO.
c. that convert N2 to NH3.
d. that generate nitrites or nitrates.
e. capable of using NH3 as an electron donor.
Answer:
(D.) Nitrifiers are bacteria that generate nitrites or nitrates.
Explanation:
In the nitrogen cycle which occurs in nature, ammonia and ammonium compounds in the soil from organic sources and are converted to nitrites and nitrates by aerobic microorganisms.
Nitrifiers, as the name implies, are these such aerobic bacteria which oxidize inorganic constituents in the soil to generate energy. Examples of these nitrifiers are nitrobacter and nitrosomonas.
When 189.6 g of ethylene (C2H4) burns in oxygen to give carbon dioxide and water, how many grams of CO2 are formed? C2H4(g) + O2(g) → CO2(g) + H2O(g) (unbalanced)
Answer:
596 g of CO₂ is the mass formed
Explanation:
Combustion reaction:
C₂H₄(g) + 3O₂(g) → 2CO₂(g) + 2H₂O(g)
We determine moles of ethylene that has reacted:
189.6 g . 1mol / 28g = 6.77 moles
We assume the oxygen is in excess so the limiting reagent will be the ethylene.
1 mol of ethylene produce 2 moles of CO₂ then,
6.77 moles will produce the double of CO₂, 13.5 moles.
We convert the moles to mass: 13.5 mol . 44 g /1mol = 596 g
17. A gas tank holds 2785 L of propane, C3H8, at 830. mm
Hg. What is the
volume of the propane at standard pressure?
Answer:
3041.5L
Explanation:
Data obtained from the question include:
V1 (initial volume) = 2785 L
P1 (initial pressure) = 830mmHg
P2 (final pressure) = stp = 760mmHg
V2 (final volume) =?
Applying the Boyle's law equation P1V1 = P2V2, the final volume propane can be achieved by as illustrated below:
P1V1 = P2V2
830 x 2785 = 760 x V2
Divide both side by 760
V2 = (830 x 2785)/760
V2 = 3041.5L
Therefore, the volume of the propane at standard pressure is 3041.5L
By applying Boyle's Law and using the given initial conditions, the volume of propane at standard pressure is calculated to be 3041.71 L.
The question involves applying the gas law to determine the volume of propane at standard pressure. The given conditions are 2785 L of propane at 830 mmHg, and we need to find the volume at standard pressure (760 mmHg). Using Boyle's Law, which states that P1V1 = P2V2 where P is pressure and V is volume, we can calculate the new volume at standard pressure.
First, convert the initial pressure to the same units as the standard pressure, which is already in mmHg. Then, set up the equation with P1 = 830 mmHg, V1 = 2785 L, P2 = 760 mmHg, and solve for V2:
V2 = (P1 × V1) / P2 = (830 mmHg × 2785 L) / 760 mmHg
V2 = 3041.71 L
So, the volume of propane at standard pressure is 3041.71 L.
Which of the following is an example of maintaining homeostasis? A. Learning B. Jumping C. Shivering D. Smiling
Answer:
Shivering.
Explanation:
Homeostasis is the bodies need to keep things at a normal. When you are cold you shiver to produce body heat and get back to normal
Which of these did you include in your response?
A strong electrolyte dissociates or ionizes completely in water.
A weak electrolyte dissociates or ionizes partially in water.
A nonelectrolyte does not dissociate or ionize in water.
The more ions in solution, the better the solution will conduct electricity.
To test each solution, use a conductivity apparatus. If the bulb lights, the solution is an electrolyte. The brighter the bulb, the stronger the electrolyte.
Answer:
Check all of them
Explanation:
All are correct.
- stong electrolytes dissolve completely
- weak disolve partially
- nonelectrolytes do not dissolve
Answer:
Check allExplanation:
Will mark brainliest, please help!
Answer:
Ammonia
Explanation:
Answer:
ammonia NH3
Explanation:
pls mark brainliest
What do methanol (CH2O), ethanoic acid (C2H4O2), and glucose (C3H6O3) all have in common?
a. They are all carbohydrates
b. They all have the same empirical formula
c. They all have the same % composition
d. All of the above are true
Answer:
d
Explanation:
Carbohydrates are compounds containing carbon, hydrogen, and oxygen. Therefore, a is true.
An empirical formula is the simplest ratio of atoms present in a compound. Therefore, C2H4O2 and C3H6O3, (if you simplified them like you would a fraction) would be CH2O. Therefore b is correct,
They also have the same % composition, with a ratio of 1 carbon : 2 hydrogen : 1 oxygen. Therefore, c is correct.
Since a, b and c are all correct, the answer is d, all of the above are true.
Final answer:
Methanol, ethanoic acid, and glucose all share the same empirical formula CH2O, despite having different molecular structures and functions.
Explanation:
What methanol (CH2O), ethanoic acid (C2H4O2), and glucose (C6H12O6) all have in common is that they all have the same empirical formula, which is CH2O. This simplified empirical formula indicates a molar ratio between carbon, hydrogen, and oxygen, which is 1:2:1. However, these compounds have different molecular formulas, meaning they contain different numbers of atoms, but when simplified to the lowest whole number ratio, they share the same empirical formula.
Methanol, a simple alcohol with the molecular formula CH4O, ethanoic acid (also known as acetic acid, found in vinegar), with a molecular formula of C2H4O2, and glucose, a simple sugar with the molecular formula C6H12O6, while all different compounds, share the empirical formula CH2O. This showcases an interesting aspect of organic chemistry where different compounds can have identical empirical formulas but vary greatly in structure and function.