589L volume of carbon dioxide in litres could be generated at 0.96 atm and 362. K by the combustion of 228.85 grams of oxygen gas excess pentane gas ([tex]C_5H_{12}[/tex])
What is an ideal gas equation?The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).
[tex]C_5H_{12}(g) + 8 O_2(g)[/tex] --->[tex]5 CO_2(g) + 6 H_2O(g)[/tex]
To find the volume, you need to use the Ideal Gas Law. The equation looks like this:
PV = nRT
In this equation,
P = pressure (atm)
V = volume (L)
n = number of moles
R = constant (0.0821 L*atm/mol*K)
T = temperature (K)
To find the volume of water vapour, you need to (1) convert grams C₅H₁₂ to moles C₅H₁₂ (via molar mass), then (2) convert moles C₅H₁₂ to moles H₂O (via mole-to-mole ratio from reaction coefficients), then (3) calculate volume H₂O (via Ideal Gas Law). The final answer should have 3 sig figs.
Molar Mass (C₅H₁₂) =72.151 g/mol
1 C₅H₁₂ (g) + 8 O₂ (g) -----> 5 CO₂ (g) + 6 H₂O (g)
228.85 g C₅H₁₂ 1 mole 6 moles H₂O
----------------------- x ---------------- x ----------------------- = 19.03 moles H₂O
72.151 g 1 mole C₅H₁₂
Moles of H₂O= 19.03
P =0.96 atm
R = 0.0821 L x atm/mol x K
V = ?
T = 362 K
n = 19.03 moles
PV = nRT
(0.96 atm)V = ( 19.03 moles)(0.0821 Latm/molK)(362K)
(0.96 atm)V = 565.575406
V = 589L
Hence, 589L volume of carbon dioxide in litres could be generated at 0.96 atm and 362. K by the combustion of 228.85 grams of oxygen gas excess pentane gas ([tex]C_5H_{12}[/tex]).
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For the decomposition of ammonia on a platinum surface at 856 °C
2 NH3(g)N2(g) + 3 H2(g)
the average rate of disappearance of NH3 over the time period from t = 0 s to t = 5.48×103 s is found to be 1.50×10-6 M s-1.
What is the average rate of appearance of H2 over the same time period?
M s-1.
[tex]\\ \tt\leadsto \dfrac{d[NH_3]}{dt}=1.50\times 10^{-6}[/tex]
dt remains same for reaction[tex]\\ \tt\leadsto \dfrac{d[H_2]}{dt}=\dfrac{3}{2}\dfrac{d[NH_3]}{dt}[/tex]
[tex]\\ \tt\leadsto \dfrac{d[H_2]}{dt}=\dfrac{3}{2}(1.5\times 10^{-6})[/tex]
[tex]\\ \tt\leadsto \dfrac{d[H_2]}{dt}=2.25\times 10^{-6}Ms^{-1}[/tex]
M is molarity here not metre
The average rate of appearance of H2 over the time period from t = 0 s to [tex]\rm t = 5.48\times10^3 s\ is\ 2.25\times10^{-6}[/tex] M/s.
To find the average rate of appearance of [tex]\rm H_2[/tex], we can use the stoichiometry of the reaction to relate it to the rate of disappearance of [tex]\rm NH_3[/tex].
From the balanced equation: [tex]\rm 2NH_3(g) - > N_2(g) + 3 H_2(g)[/tex]
We can see that for every 2 moles of [tex]\rm NH_3[/tex] that disappear, 3 moles of [tex]\rm H_2[/tex] appear.
Given that the average rate of disappearance of [tex]\rm NH_3\ is\ 1.50\times10^{-6} M/s[/tex], we can calculate the average rate of appearance of [tex]\rm H_2[/tex] as follows:
Average rate of appearance of [tex]\rm H_2[/tex] = (3/2) x (Average rate of disappearance of [tex]\rm NH_3[/tex])
Average rate of appearance of [tex]\rm H_2[/tex] = (3/2) x [tex]\rm (1.50\times10^{-6} M/s)[/tex]
Average rate of appearance of [tex]\rm H_2[/tex] = [tex]\rm 2.25\times10^{-6} M/s[/tex]
Therefore, the average rate of appearance of [tex]\rm H_2[/tex] over the time period from t = 0 s to [tex]\rm t = 5.48\times10^3 s\ is\ 2.25\times10^{-6}[/tex] M/s.
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How much heat would need to be removed to
cool 215.1 g of water from 25.6°C to -10.7°C?
Answer:
100 KJ
Explanation:
yep yep cool is that correct
Is it possible to use subatomic particles as an energy source?
Yes, Is it possible to use subatomic particles as an energy source.
What are subatomic particles?A subatomic particle is nothing but a particle which is smaller than an atom in size. Typically, an atom can be broken down into three subatomic particles, namely: protons, electrons, and neutrons.
Yes, Is it possible to use subatomic particles as an energy source.
There are instances when the protons and electrons in an atom's outermost shells are not strongly attracted to one another. It is possible to force these electrons out of their orbits.
They may move from one atom to another by exerting force. Electricity is made up of these moving electrons.
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You have prepared an aqueous solution of a monoprotic acid (formula mass = 153.483g/mol) by dissolving 26.816g of the acid in sufficient water to make 500.0mL of solution. The pH of the resulting solution is 4.274. What is Ka for this acid?
The Ka of the solution is 8.1 * 10^-9.
What is the pH?The pH of a solution is the concentration of hydrogen ions in the solution.
Number of moles of the acid = 26.816g/153.483g/mol = 0.175 moles
Concentration of the acid = 0.175 moles/0.5 L = 0.35 M
Now we know that the acid reacts with water as follows;
HA(aq) + H2O(l) ------> H3O^+(aq) + A^-(aq)
[H3O^+] = Antilog (-4.274)
[H3O^+] = 5.32 * 10^-5 M
Since;
[H3O^+] = [A^-] = x
Ka = x^2/[HA]
Ka = (5.32 * 10^-5)^2/( 0.35)
Ka = 8.1 * 10^-9
Hence, the Ka of the solution is 8.1 * 10^-9.
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kinetic energy is proportional to what
Answer:
mass and the square of the magnitude of velocity.
Explanation:
How many grams of Al are needed to react with 63.0 g of FeO
The amount of Al required will be 15.77 grams
Stoichiometric problemFirst, the equation of the reaction:
[tex]2Al +3 FeO -- > Al_2O_3 + 3Fe[/tex]
The mole ratio is 2:3.
Mole of 63.0 g of FeO = 63/71.84 = 0.8769 moles
Equivalent moles of Al = 0.8769 x 2/3 = 0.5846 moles
Mass of 0.5846 moles Al = 0.5846 x 26.98 = 15.77 grams
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Taking into account the reaction stoichiometry, 13.28 grams of Al is required to react with 63 grams of FeO.
Reaction stoichiometryIn first place, the balanced reaction is:
2 Al + 3 FeO → 3 Fe + Al₂O₃
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
Al: 2 molesFeO: 3 molesFe: 3 molesAl₂O₃ : 1 moleThe molar mass of the compounds is:
Al: 27 g/moleFeO: 71.85 g/moleFe: 55.85 g/moleAl₂O₃ : 102 g/moleThen, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
Al: 2 moles ×27 g/mole= 54 gramsFeO: 3 moles ×71.85 g/mole= 215.55 gramsFe: 3 moles ×55.85 g/mole= 167.55 gramsAl₂O₃ : 1 mole ×102 g/mole= 102 gramsMass of Al requiredThe following rule of three can be applied: If by reaction stoichiometry 215.55 grams of FeO react with 54 grams of Al, 53 grams of FeO react with how much mass of Al?
[tex]mass of Al= \frac{53 grams of FeOx54 grams of Al}{215.55 grams of FeO}[/tex]
mass of Al= 13.28 grams
Finally, 13.28 grams of Al is required to react with 63 grams of FeO.
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The sugar fructose contains 40.0% C, 6.7% H, and 53.3% O by mass. A solution of 11.7 g of fructose in 325 g of ethanol has a boiling point of 78.59 °C. The boiling point of ethanol is 78.35 °C, and Kb for ethanol is 1.20 °C/m. Assuming ideal solution behavior, what is the molecular formula of fructose?
The molecular formula of fructose is C₆H₁₂O₆.
What is the molecular formula of fructose?The molecular formula is derived in the following steps:
Step 1. the molality of the solution is determined first
Molality = Change in boiling point/Kb
Molality = (78.59 - 78.35)/1.2 = 0.2 molal
Step 2. Moles of solute is determined.
Moles of solute = molality * mass of solvent (kg)
moles = 0.2 * 0.325 = 0.065
Step 3. Molar mass of fructose = mass/moles
Molar mass of fructose = 11.7/0.065 = 180 g/mol
Step 4: Mole ratio of constituent elements
Carbon = 40% * 180/12 = 6
Hydrogen = 6.7% * 180/1 = 12
Oxygen = 53.3% * 180/16 = 6
Therefore, the molecular formula of fructose is C₆H₁₂O₆.
In conclusion, the molecular formula of fructose is derived from the molality of the solution.
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The rate law for the reaction 3A --> 2B is rate = k[A] with a rate constant of 0.0447 hr-1.
What is the half-life of the reaction?
A) 0.0224 hr
B) 0.0645 hr
C) 15.5 hr
D) 22.4 hr
E) 44.7 hr
The half-life of the reaction is determined as 22.4 hr.
Half life of the reaction
The half life of the reaction is calculated as follows
rate of reaction, k = Δ[A]/3t = Δ[B]/2t
t = Δ[A]/3k = Δ[B]/2k
where;
k is rate constant = 0.0447/hrt = (3)/(3 x 0.0447) = (2)/(2 x 0.0447) = 22.4 hr
Thus, the half-life of the reaction is determined as 22.4 hr.
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the roof of a building is 0.2km^2. during a rainstorm, 5.5 cm of rain was measured to be sitting on the roof. what is the mass in kg of the water on the roof after the rainstorm? (density of rainwater = 1g/ml)
The mass in kg of the water on the roof after the rainstorm is mathematically given as
The mass of the water on the roof after a rainstorm is 1.1 *10^{10}g or 1.1*10^{7}kg
What is the mass in kg of the water on the roof after the rainstorm?Generally, the equation for the Area of the roof is mathematically given as
[tex]A=0.20 \times 10^{10} \mathrm{~cm}^{2}[/tex]
height of the rain =5.5cm
the volume of the rain on the roof =1.1 * 10^{10} CC
Generally, the equation for mass is mathematically given as
mass=volume*density
[tex]&=\left(1.1 \times 10^{10} \mathrm{cC} \times 1 \mathrm{~g} / \mathrm{cc}\right) \\&=1.1 \times 10^{10} \mathrm{~g}[/tex]
In conclusion, the Mass of the water on the roof after a rainstorm is 1.1 *10^{10} or $1.1*10^{7}
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Which most likely describes results from their experiments? tristan and arya would exert more energy to overcome friction than anoush and keran. tristan and anoush would exert more energy to overcome friction than arya and keran. tristan and keran would exert less energy to overcome friction than arya and anoush. tristan and arya would exert less energy to overcome friction than anoush and keran.
The statement that most likely describes the results from their experiments is Tristan and Arya would exert more energy to overcome friction than Anoush and Keran. Thus, the correct option is A.
What is Friction?Friction may be characterized as the force that counters the proximate motion between two surfaces of objects when they reach to contact.
According to the question, Tristan is working with static friction and Arya is working with sliding friction; static friction operates on things when they are catnapping on a surface, whereas sliding friction functions on objects when they are gliding over a surface.
Sliding friction is much more fragile and weaker than static friction and is comparatively uncomplicated and utilizes less energy to drag the objects.
The complete question is as follows:
Four students are doing different friction experiments with a cylinder.
Tristan is working with static friction,Arya is working with sliding friction,Anoush is working with rolling friction,Keran is working with fluid friction.Therefore, the correct option for this question is A, i.e. Tristan and Arya would exert more energy to overcome friction than Anoush and Keran.
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I NEED HELP! A rigid cylinder contains 3.00 liters of gas at a temperature of 25°C. If the pressure of this gas is changed from
0.500 atmospheres to 1.50 atmospheres, what will be the new temperature of the gas? (R= 0.0821 Latm/mol K)
O 612 °C
894 °C
99.3 °C
885 °C
O 75.0 °C
Gay Lussac's Law
P₁/T₁=P₂/T₂
25°C = 25+273 = 298 K
0.5/298 = 1.5/T₂
T₂ = 894 K = 621 °C
A gas law known as Gay-Lussac's law asserts that a gas's pressure (when kept at a constant volume and mass) varies directly with its absolute temperature. The new temperature is 612 °C and the correct option is A.
When the mass is fixed and the volume is constant, the pressure a gas exerts is proportional to the temperature of the gas. In the year 1808, French scientist Joseph Gay-Lussac created this law. Gay-Lussac's law can be expressed mathematically as follows:
P₁ / T₁ = P₂ / T₂
25°C = 25 + 273 = 298 K
0.5 / 298 = 1.5 / T₂
T₂ = 894 K
T₂ = 894-273
T₂ = 621 °C
Thus the correct option is A.
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For this same exercise, if 2.5 liters of this solution (2,500 ml) are administered in a 24-hour interval, what amount in grams of glucose did the patient receive in one day? the glucose to dissolve is 50 grams
Answer:
j
Explanation:
becaisdbyhe wrl is so not goood
Please help with Question 4 :(
Answer:
cf4, ch20,vhjuyu8hc hi l bro tap
Which three words best describe the composition of the inner planets?solid, smooth, giantrocky, solid, densegiant, dense, gaseousgaseous, smooth, small
The three words which best describe the composition of the inner planets are rocky, solid, and dense.
What is an inner planet?A planet classified as an inner planet is one that is rocky, solid, and dense by nature, such as Mercury, Venus, Earth, and Mars. Inner planets are those that are nearest to the Sun.
There are two types of planets:
Based on astronomical information and records, there are two main types of planets :
Outer planetInner planetThe moons of the inner planets are either absent or limited to one (Earth) or two (Mars). The inner planets don't have any rings. The inner planets have shorter orbits around the Sun than the outer planets, but they all rotate more slowly.
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Answer:rocky soild and dense
Explanation:
If a sample of Cl2 gas in a 1000 mL container is exerting a pressure of 14.2 atm, how many grams of chlorine are present? The temperature of the container is 256 K.
Taking into account the ideal gas law, the mass of chlorine present is 47.9284 grams.
Ideal gas lawAn ideal gas is a theoretical gas that is considered to be composed of randomly moving point particles that do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.
The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:
P×V = n×R×T
where
P is the gas pressure.V is the volume that occupies.T is its temperature.R is the ideal gas constant. The universal constant of ideal gases R has the same value for all gaseous substances. n is the number of moles of the gas. Mass of chlorine presentIn this case, you know:
P= 14.2 atmV= 1000 mL= 1 LT= 256 KR= 0.082 [tex]\frac{atmL}{molK}[/tex]n= ?Replacing in the ideal gas law:
14.2 atm× 1 L = n× 0.082 [tex]\frac{atmL}{molK}[/tex]× 256 K
Solving:
(14.2 atm× 1 L)÷ (0.082 [tex]\frac{atmL}{molK}[/tex]× 256 K)= n
0.676 moles= n
Being the molar mass of chlorine Cl₂ 70.9 g/mole, the mass of the compound can be calculated as:
0.676 moles× 70.9 g/mole= 47.9284 grams.
Finally, the mass of chlorine present is 47.9284 grams.
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Which of the following ions is formed when an acid is dissolved in a solution?
OH+
0-
OH-
OSO4²+
Answer:
The ion that is formed when an acid is dissolved in a solution is OH-.
Explanation:
An acid is a molecule that donates hydrogen ions, or protons, to another molecule. In solution, the acid dissociates, or breaks apart, to form hydrogen ions and the negatively-charged anion of the acid. The anion of a typical acid is the molecule minus the hydrogen ions, so it is OH- for hydrochloric acid, HCl.
A sample of He gas (2.35 mol) occupies 57.9 L at 300.0 K and 1.00 atm. The volume of this sample is __________ L at 423 K and 1.00 atm.
Considering the combined law, the volume of the sample is 81.639 L at 423 K and 1.00 atm.
Charles's lawCharles's law establishes the relationship between the temperature and the volume of a gas when the pressure is constant. This law says that the volume is directly proportional to the temperature of the gas: if the temperature increases, the volume of the gas increases while if the temperature of the gas decreases, the volume decreases.
Mathematically, Charles's law states:
[tex]\frac{V}{T}=k[/tex]
Boyle's lawBoyle's law states that the pressure of a gas in a closed container is inversely proportional to the volume of the container, when the temperature is constant. That is, if the pressure increases, the volume decreases while if the pressure decreases, the volume increases.
Mathematically, this law is expressed as:
P×V= k
Gay-Lussac's lawGay-Lussac's law establishes the relationship between the temperature and pressure of a gas when the volume is constant. This law says that the pressure of a gas is directly proportional to its temperature: if the temperature increases, the pressure will increase, while if the temperature decreases, the pressure will decrease.
Mathematically, Gay-Lussac's law states:
[tex]\frac{P}{T}=k[/tex]
Combined law equationCombined law equation is the combination of three gas laws called Boyle's, Charlie's and Gay-Lusac's law:
[tex]\frac{VxP}{T}=k[/tex]
Considering an initial state 1 and a final state 2, it is fulfilled:
[tex]\frac{V1xP1}{T1}=\frac{V2xP2}{T2}[/tex]
Volume of the sampleIn this case, you know:
V1= 57.9 LP1= 1 atmT1= 300 KV2= ?P2= 1 atmT2= 423 KReplacing in combined law equation:
[tex]\frac{57.9 Lx1 atm}{300 K}=\frac{V2x1 atm}{423 K}[/tex]
Solving:
[tex]V2= \frac{423 K}{1 atm} \frac{57.9 Lx1 atm}{300 K}[/tex]
V2= 81.639 L
Finally, the volume of the sample is 81.639 L at 423 K and 1.00 atm.
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If you know that lithium reacts strongly with oxygen in a 2:1 ratio, which of
the following properties would you predict about cesium
Answer:
Same as lithium
Explanation:
Lithium and Ceaseum both are group one elements i.e alkali metals
So they have same chemical properties almost
Hence Ceaseum would react with oxygen in 2:1 ratio
Show all work:
A saturated solution of the Ca(OH)2 is prepared and allowed to equilibrate at 25°C. A 32.40ml
sample of the solution is titrated with 0.050M HCL solution. 14.90mL of the HCI solution are
required to fully react with the sample.
Write the balanced equation for the reaction of the salt with HCl
The equation of the reaction is;
Ca(OH)2(aq) + 2HCl(aq) ------> CaCl2(aq) + 2H2O(l)
What is neutralization?The term neutralization has to do with the process in which an acid reacts with a base to form salt and water only. A neutralization reaction is known by the fact that water is obtained by the reaction of hydrogen and hydroxide ions.
The equation of the reaction is;
Ca(OH)2(aq) + 2HCl(aq) ------> CaCl2(aq) + 2H2O(l)
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What temperature would constitute a fever for a child? What effects on dental treatment would a fever pose?
Answer:
38
Explanation:
that I believe is correct
Cellular Respiration is the process of what? A. Producing oxygen at the cellular level. B. The release of energy from food. C. Breathing. D. The creation of food, powered by the energy of the sun.
Cellular Respiration is the process of the release of energy from food. Hence, option B is correct.
What is cellular respiration?Cellular respiration is the process by which organisms use oxygen to break down food molecules to get chemical energy for cell functions.
Through the process of cellular respiration, the energy in food is converted into energy that can be used by the body's cells. During cellular respiration, glucose and oxygen are converted into carbon dioxide and water, and the energy is transferred to ATP.
Hence, option B is correct.
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Question 1 of 25 Hydrogen gas can be produced using this equilibrium reaction. Which change would drive this process toward the product side? C + H₂O CO + H₂ + energy A. Removing the H₂O as it forms B. Increasing the temperature C. Removing the C as it forms D. Removing the CO as it forms SUBMIT
What we need to do to drive the equilibrium to the products side is removing the CO as it forms.
What is the equilibrium position?The equilibrium position is dependent in the conditions of the reaction. We know that the direction to which the equilibrium is shifted can be altered by manipulating the parameters of the reaction.
Now the reaction is; C + H₂O -----> CO + H₂ + energy. The case in point here is to drive the equilibrium position towards the products side.
What we need to do to drive the equilibrium to the products side is removing the CO as it forms.
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The table describes how some substances were formed.
Substance
Description
P
Formed by boiling pure water
Q
Formed by combining three hydrogen atoms to every nitrogen atom
R
Formed by adding 5 g of sugar to 1 L of water
S
Formed by compressing carbon under high pressure
Based on the given descriptions, which substance is most likely a mixture?
P
Q
R
S
R is the mixture formed by adding 5 g of sugar to 1 L of water.
Which is a mixture?A mixture is an impure substance made up of two or more substances which are miscible or immiscible in each other.
Dissolving the sugar in the water makes a homogenous mixture and once dissolved a solution is made by the combination of a solute (sugar) and solvent (water).
Hence, R is the mixture formed by adding 5 g of sugar to 1 L of water.
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Which group on the periodic table has the most reactive metals?
A.
group 1
B.
group 3
C.
group 18
D.
group 17
Answer:
A group 1
Explanation:
Group 1
reactivity decreases as you go from L to R or downward on the periodic table
How much heat will be absorbed by a 66.3 g
piece of aluminum (specific heat = 0.930 J/
g. °C) as it changes temperature from 23.0°C
to 67.0°C?
Q=mcΔt
Q=66.3 x 0.93 x (67-23)
Q=2712.996 J
[tex]\quad \huge \quad \quad \boxed{ \tt \:Answer }[/tex]
[tex]\qquad \tt \rightarrow \:Q = 2.713 \:\:kJ [/tex]
____________________________________
[tex] \large \tt Solution \: : [/tex]
[tex]\qquad \tt \rightarrow \: Q = ms\Delta T[/tex]
[tex] \textsf{Q = heat evolved/absorbed = ??} [/tex]
[tex] \textsf{m = mass in gram = 66.3 g} [/tex]
[tex] \textsf{s = specific heat = 0.930 J/g °C} [/tex]
[tex] \textsf{ΔT = change in temp = 67 - 23 =44°C} [/tex]
[tex] \large\textsf{Find Q : } [/tex]
[tex]\qquad \tt \rightarrow \:Q=( 66.3) \sdot (0.93) \sdot(44)[/tex]
[tex]\qquad \tt \rightarrow \:Q = 2713 \: \: joules[/tex]
[tex]\qquad \tt \rightarrow \:Q= 2.713 \: \: kj[/tex]
Answered by : ❝ AǫᴜᴀWɪᴢ ❞
It requires 334 kJ of heat to melt 1 kg of ice. The largest known iceberg had a volume of about 3.1 x 1013 m³. How much heat was required
to melt the iceberg if 1 m³ of ice has a mass of 917 kg?
The amount of heat required to melt the iceberg at 9.58 * 10¹⁸ kJ.
What is the amount of heat required to melt an iceberg that has a volume of about 3.1 x 1013 m³?The heat required to melt a unit mass of ice is known as the latent heat of fusion.
The latent heat of fusion of ice of 1 kg= 334 kJ of heat
Mass of iceberg = 3.1 x 10¹³ m³ * 917 kg = 2.8427 * 10¹⁶ kg
Amount of heat required = 2.8427 * 10¹⁶ * 334 = 9.58 * 10¹⁸ kJ.
Therefore, the amount of heat required to melt the iceberg at 9.58 * 10¹⁸ kJ.
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What volume in milliliters of 0.0150 M Ba(OH)2 is required to neutralize 50.0 mL of 0.0500 M HBr?
The volume of [tex]Ba(OH)_2[/tex] required will be 83.33 mL
Stoichiometric reactionsThe equation of the reaction is as follows:
[tex]Ba(OH)_2 + 2HBr --- > 2H_2O + BaBr_2[/tex]
Mole ratio of reactants = 1:2
Mole of 50 mL, 0.05 M HBr = 0.05 x 0.05 = 0.0025 moles
Equivalent mole of [tex]Ba(OH)_2[/tex] = 0.0025/2 = 0.00125 moles
Volume of 0.015 M, 0.00125 moles [tex]Ba(OH)_2[/tex] = 0.00125/0.015 = 0.0833 L
0.0833 L = 83.33 mL
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which of these is/are conserved during a chemical reaction ?
a)Mass only
b) Charge only
c)both mass and charge
d)neither
Answer:
C- both!
Explanation:
In a chemical reaction, the total mass and total charge has to be balanced on both sides of the "equation", therefore both are conserved
The _______ effect means water can "climb" up thin tubes and water plants. This is due to the ______ nature of water.
The Capillary effect means water can "climb" up thin tubes and water plants. This is due to the surface tension nature of water.
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The capillary effect means water can "climb" up thin tubes and water plants. This is due to the adhesive nature of water.
The capillary effect is a phenomenon where liquids, like water, can rise against gravity in narrow tubes or spaces, such as the stems of plants. This effect is possible due to the adhesive nature of water, which allows it to stick to the walls of the tube or surface it comes into contact with.
The cohesive forces between water molecules create a sort of "pull" that allows the water to climb up the tube, defying gravity. This property is essential for plants, as it helps them transport water and nutrients from their roots to their leaves and other parts of the plant, supporting their growth and survival.
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What is the half-life in minutes of a compound if 75.0 percent of a given sample decomposes in 40.0 minutes? Assume first-order kinetics.
Answer:
See below
Explanation:
.75 = 1/2^(40/h)
log .75 / ( log 1/2) = 40 / h
h = half life = 96.37683 min
To solve this type of question we must know the concept of rate law. The half-life of a compound is 23 years for 75% of a given sample of the compound decomposes in 40 minutes.
What is the expression for rate law for first order kinetics?There are two kinds of rate law in chemical kinetics one is differential rate law and other is integrated rate law
The rate law for first order kinetics is
[tex]k=\frac{2.303}{t}log\frac{a}{a-x}[/tex]
Where
k - rate constant
t - time passed by the sample
a - initial amount of the reactant
a-x - amount left after the decay process
The expression for the half-life of the compound is
[tex]t_{1/2} = \frac{0.693}{K}[/tex]
Where k is the rate constant
t = 40 minutes
a = 100 g
a-x = 100 g - 75g = 25 g
(Since it is given that 75% of the given compound decomposes)
Then,
[tex]k=\frac{2.303}{40}log\frac{100}{25}[/tex]
=0.03years⁻¹
[tex]t_{1/2} = \frac{0.693}{0.03}[/tex]
=23 years
Therefore, the half-life of the given compound is 23years.
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