Uranium-238
Explanations:The process by which an atomic nucleus looses energy by radiation is known as radioactivity.
An alpha particle, have an atomic number of 2 and a mass number of 4 as shown;
[tex]_2^4He[/tex]When this is combined with Th-234, it forms Uranium-238. This shows that the uranium nucleus emits alpha particle to form Th-234 according to the chemical equation below;
[tex]_{92}^{238}U\rightarrow_2^4He+_{90}^{234}Th[/tex]QUESTION 3Calculate the molarity of a 150 ml of solution that contains 25.0 g of KOH.
We are required to calculate the molarity, given:
Volume = 150 mL = 0.150 L
mass of KOH = 25.0 g
Formular for molarity:
C = n/V
Before we can get C, we need to calculate n, the number of moles of KOH.
n = m/M where m is the mass and M is the molar mass
n = 25.0 g/56,1056 g/mol
n = 0.446 mol
Now we can calculate C, the molarity
C = 0.446 mol/0.150L
C = 2.97 mol/L or 2.97 M
What is the % yield of H₂O if 144.9 g H₂O is produced from 10.04 g H₂ and excess O₂?2H₂ + O₂ → 2H₂0
ANSWER
The percentage yield of water is 160%
EXPLANATION
Given that;
The mass of water is 144.9 grams
The mass of hydrogen gas is 10.04 grams
Follow the steps below to find the percentage yield of water
Write the balanced equation of the reaction
[tex]\text{ 2H}_{2(g)}+\text{ O}_{2(g)}\text{ }\rightarrow\text{ 2H}_2O_{(g)}[/tex]Step1; Find the number of moles of hydrogen gas using the formula below
[tex]\text{ mole = }\frac{\text{ mass}}{\text{ molar mass}}[/tex]Recall, that the molar mass of H2 is 2 g/mol
[tex]\begin{gathered} \text{ mole = }\frac{\text{ 10.04}}{\text{ 2}} \\ \text{ mole = 5.02 moles} \end{gathered}[/tex]Step 2; Find the number of moles of water using a stoichiometry ratio
In the reaction, 2 moles H2 give 2 moles H2O
Therefore, the number of moles of H20 is 5.02 moles
Step 3: Find the mass of H2O
[tex]\begin{gathered} \text{ Mole = }\frac{\text{ mass}}{\text{ molar mass}} \\ \text{ cross multiply} \\ \text{ mass = mole }\times\text{ molar mass} \end{gathered}[/tex]The molar mass of water is 18.0 g/mol
[tex]\begin{gathered} \text{ mass = 18 }\times\text{ 5.02} \\ \text{ mass = 90.36 grams} \end{gathered}[/tex]Step 4; Find the percentage yield of water
[tex]\begin{gathered} \text{ \% yield = }\frac{\text{ Actual yield}}{\text{ Theoretical yield}}\times\text{ 100\%} \\ \\ \text{ \% yield = }\frac{\text{ 144.9}}{90.36}\times\text{ 100\%} \\ \\ \text{ \% yield = 1.60}\times100\text{ \%} \\ \text{ \%yield = 160\%} \end{gathered}[/tex]Therefore, the percentage yield of water is 160%
an ioc occurs when what metric exceeds its normal bounds?
An ioc occurs when KRI (key risk indicator) metric exceeds its normal bounds.
Key Risk Indicators (KRIs) are vital for predicting unfavourable events that may have a negative effect on companies. They keep an eye on changes in risk exposure levels and help provide the early warning signals that let businesses disclose concerns, avert catastrophes, and properly mitigate them. a metric that measures the maximum and lower limits of particular indications of typical network activity.
A few indicators are the total number of network logs per second, the number of unsuccessful remote login attempts, network bandwidth, and outgoing email traffic. One of these that deviates from its expected range may be a sign of compromise (IOC). For a major risk indicator to be measured and be effective over time, it must be repeatable. They give a mechanism to measure and keep track of each danger.
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How many liters of gaseous oxygen are needed to produce 4.50 L of gaseous nitrogen dioxide, if both gases are being measured at STP?
answer and explanation
the balanced chemical reaction for the formation of nitrogen dioxie is
2NO + O₂ → 2NO₂
and so at STP the number of mols of nitrogen dioxide in 4.50L would be
PV = nRT
n = PV/RT
= (1atmx 4.50L) / (8.314 x 273.15K
= 4.50/2270.97
= 0.00204 mols
0.00204 mols of nitrogen dioxide would have been formed from the reaction.
we know from the balance reaction that 1 mol of oxygen forms 2 mols of nitrogen dioxide and so the mols of oxygen that were present would be
1/2 x0.00204 = 0.00102 mols
we know that 1 mol of a gas occupies a volume of 22.4L at STP and so
0.00102 mols of oxygen would have a volume of
0.0102 mols x 22.4 L/mol
= 0.0228L
0.0228L of oxygen would be needed
1. For the balanced equation shown below, how many moles of C₂H₂O₂F need to react toform 2 moles of H₂O?4C₂H₂O₂F+702-8CO2 + 6H₂O +2F₂
Answer:
[tex]1.33\text{ moles}[/tex]Explanation:
Here, we want to get the number of moles needed to react
From the equation of reaction:
4 moles produced 6 moles of H2O
x moles will react to form 2 moles of H2O
To get the value of x, we have it that:
[tex]\begin{gathered} 4\text{ }\times\text{ 2 = 6 }\times x \\ x\text{ = }\frac{8}{6}\text{ = }\frac{4}{3}\text{ = 1.33 moles} \end{gathered}[/tex]Use the rules for logarithms and exponents to solve for [OH-] in terms of pOH.
logarithm and power 10 are inverse operations
[tex]\lbrack\text{OH}^-\rbrack\text{ = 10}^{-\text{pOH}}[/tex]The answer is [OH-] = 10^-(pOH)
and, if pOH = 2.77
[tex]\lbrack\text{ OH}^-\rbrack\text{ = 10}^{-2.77\text{ }}\text{ = 0.0017 M}[/tex]When 1.151 grams of sucrose (Molar mass
342.3 g/mol) is burned in a bomb
calorimeter, the temperature of the
calorimeter increases from 22.41°C to
26.63°C. If the heat capacity of the
calorimeter is 4.900 kJ/°C, what is the heat
of combustion of sucrose, in kJ/mol?
The heat of combustion of the substance is obtained as -6154 kJ/mol.
What is the heat of combustion?The heat of combustion is the heat that is evolved when one mole of sucrose is burnt in oxygen. Let us now try to find the heat of combustion of the substance.
Number of moles of sucrose = 1.151 grams /342.3 g/mol = 0.00336 moles
Heat capacity of the calorimeter = 4.900 kJ/°C
Temperature change = 26.63°C - 22.41°C = 4.22 °C
Now we can find the heat of combustion of the sucrose knowing that heat is evolved.
H = -(4.900 kJ/°C) * 4.22 °C/ 0.00336 moles
H = -6154 kJ/mol
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Ex 1.4 You are trying to decide which of two automobiles to buy. The first is USA- made, which costs RM 85,000 and has the rated gasoline mileage of 22.5 miles/gal. The second car is a hybrid and of European manufacture, which costs RM 100,000, and has the rated mileage of 28.5 km/L. If the cost of petrol is RM 2.05/L and if the cars actually deliver their rated gas mileage, estimate how many miles you would have to drive for the lower fuel consumption of the second car to compensate for the higher cost of this car?
104895.10 kilometers need to drive for the lower fuel consumption of the second car to compensate for the higher cost of this car.
For calculating the number of miles we need the compare the cost equation of both cars.
Considering x be the number of kilometers, since the mileage of the USA-made car is (22.5 * 1.609) / 3.785 = 19.56-kilo meter per litre, and the cost of petrol per Litres = 2.05 .
SO for the USA made the pertrol cost is 2.05 / 9.56 = $0.214per km
and for the european made car is 2.05 / 28.5 = $0.071 per km
since the total cost is nothing like the sum of the purchase price and the fuel cost SO,
for the USA made , the total cost is = 85,000 + 0.214x and
for the eurpoean car is = 100,000 + 0.071x
By comparing the equation by considering the total cost for a car to be equal so
85,000 + 0.214x =100,000 + 0.071x
=>100,000- 85,000 = 0.214x- 0.071x
=>15000= 0.143x
=>x= 104895.10
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(e) Calculate the amount of excess reagent that remains.(1) Calculate the % yield if 6.75 g of iron (Fe) is actually formed.
For this question, we have the following reaction:
Fe2O3 + 3 CO -> 2 Fe + 3 CO2
For letter D, we have:
10 grams of Fe2O3, molar mass = 159.7g/mol
6 grams of CO, molar mass = 28g/mol
In this case, we need to find which compound is the limiting and which is the excess reactant, and it is important to have in mind the molar ratio concept, which we use the coefficients in front of the compound to determine the ratios, like for Fe2O3 and CO, the molar ratio is 1:3. Now let's check if Fe2O3 is in excess, finding the number of moles first:
159.7g = 1 mol
10g = x moles
x = 0.063 moles of Fe2O3
According to the molar ratio, we need 3 times this value for CO, therefore, 0.063 * 3 = 0.189 moles of CO, now we need to check if this is the amount that we have of CO, or if we have more than that:
28g = 1 mol
6g = x moles
x = 0.214 moles of CO, this means that we have more CO than we actually need, making it the excess reactant and Fe2O3 the limiting reactant
Now to find the mass of Fe, we will be using the number of moles of the reactant, which is Fe2O3, 0.063 moles, and according to the molar ratio, we have twice this value for Fe, therefore, 0.063 * 2 = 0.126 moles of Fe, now using the molar mass of Fe, 55.84g/mol, we can find the final mass:
55.84g = 1 mol
x grams = 0.126 moles of Fe
x = 7.03 grams of Fe are produced, or if it is possible to round up, 7 grams of Fe, this is letter D
E. In this case, we need to see how much CO remains in the reaction, and since we need only 0.189 moles of CO and we actually have 0.214 moles, we have a 0.025 moles leftover, which is, in grams:
28g = 1 mol
x grams = 0.025 moles
x = 0.7 grams of CO remaining, only 5.3 grams will react
F. The percent yield is calculated having the actual yield, given in the question and the theoretical yield, which is the value we found in letter D, the formula for it is:
%yield = (actual yield/theoretical yield)*100
%yield = (6.75g/7.03g)*100
%yield = 0.964*100
The percent yield will be 96.4% for this reaction
Calculate the number of moles in 4.5x10^23 molecules of iron. Express answer to 2 significant digits and as a decimal
Since a mol is always a constant number = 6.022*10^23, to find how many moles there are in 4.5*10^23 molecules of iron we only need to do a simple math:
6.022*10^23 = 1 mol
4.5*10^23 = x moles
x = 0.7 moles
How many grams are in 1.98x10^15 molecules of CO
The molar mass of CO = 28.01 g grams are in 1.98x [tex]10^{15}[/tex] molecules of CO is 9.20× [tex]10^{-8}[/tex] g.
Avogadro's number is a proportion that compares atomic molar mass to human physical mass.
What is the relationship between molar mass and Avogadro's number?Avogadro's number is a proportion that compares atomic molar mass to human physical mass. The number of elementary particles (molecules, atoms, compounds, etc.) per mole of a substance is defined as Avogadro's number. It has the formula 6.022 1023 mol-1 and is denoted by the symbol NAThe number of particles in one mole of anything is known as Avogadro's number. It is the number of atoms in one mole of an element in this context. Using Avogadro's number, it is simple to calculate the mass of a single atom. To get the answer in grams, divide the element's relative atomic mass by Avogadro's number.The molar mass of CO = 28.01 g
grams are in 1.98x [tex]10^{15}[/tex] molecules of CO
= 28.01x 1.98x [tex]10^{15}[/tex]/ 6.023 x 10²³
=9.20× [tex]10^{-8}[/tex] g.
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What mass of potassium nitrate is needed to generate 215.0 L of gas, composed of 111.0 L of N2 and 104.0 L of O2 at 0.920 atm and 291 K, using these two reactions?
Mass of potassium nitrate is needed to generate 215.0 L of gas, composed of 111.0 L of N₂ and 104.0 L of O₂ at 0.920 atm and 291 K, using these two reactions is 1186.75 g.
The reactions are :
2KNO₃ ---> 2KNO₂ + O₂
4KNO₂ ----> 2K₂O + 3O₂ + 2N₂
Therefore,
4KNO₃ ----> 2K₂O + 5O₂ + 2N₂
using the formula:
PV = nRT
moles of N₂ = PV / RT
P = 0.920 atm
T =291 K
R = 0.082 Latm/mol/k
moles of N₂ = 4.279 mol
moles of O₂ = (0.92 × 104) / 0.082 × 291
= 4.00 mol
from the equation .
moles of potassium nitrate to produce nitrogen = 2 × 4.279
= 8.55 mol
moles of nitrate tp produce oxygen = (4 / 5) × 4
= 3.2 mol
total moles of nitrate = 8.55 mol + 3.2 mol
= 11.75 mol
mass of potassium nitrate = number of moles × molar mass
= 11.75 × 101
= 1186.75 g
Thus, Mass of potassium nitrate is needed to generate 215.0 L of gas, composed of 111.0 L of N₂ and 104.0 L of O₂ at 0.920 atm and 291 K, using these two reactions is 1186.75 g.
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in a reaction, if you start with 2g of carbon and 15g of oxygen, how many grams of carbon dioxide should you have in the system, if the reactants completely reacted?
Writing a balanced equation:
[tex]C_{(s)}+O_{2(g)}\rightarrow CO_{2(g)}[/tex]Determine the moles of C = mass/ molar mass
mole of C= 2g / 12.01 g = 0.17 mol
mole of oxygen = 15g/ 32g = 0.47 mol
Mole ratio between carbon and carbon dioxide is 1:1. Therefore 0.17 mol Carbon reacts with 0.17 carbon dioxide.
mass of carbon dioxide = moles x molar mass
mass of carbon dioxide = 0.17 x 44.01
mass of carbon dioxide = 7.48 g
6. How is the pressure of a gas affected by temperature changes? Assume no change in volume or mass in your explanation.
Gaseous butane (CH₂(CH₂), CH₂) will react with gaseous oxygen (O₂) to produce gaseous carbon dioxide (CO₂) and gaseous water (H₂O). Suppose 26. g
of butane is mixed with 168. g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Round your answer
to 2 significant digits.
When 26. g of butane is mixed with 168. g of oxygen. Then, the maximum mass of carbon dioxide that could be produced by the chemical reaction is 77.44g.
Chemical reaction2Ch3CH₂CH₂CH3 + 13O₂ ------- 8CO₂ + 10H₂O
From the above chemical reaction we get to know that, 2 moles of butane react with 13 moles of oxygen.
1 moles of butane react with 6.5 moles of oxygen to produce 4 moles. of carbon dioxide.
Given,
mass of oxygen = 168 g
Mass of butane = 26g
Molar mass. of oxygen = 32 g
Molar mass of butane = 58 g
We will calculate the number of moles
As we know that,
Mole = given mass/ molar mass
Moles of oxygen = 168/32 = 5.2 mol
Moles of butane = 26/ 58 = 0.44 moles
So, we can say that 0.4 moles of butane react with 6.5 × 0.44 moles of oxygen to produce 4 × 0.44 moles. of carbon dioxide.
Or 0.4 moles of butane react with 2.86 moles of oxygen to produce 1.76 moles of carbon dioxide.
1.6 = mass / molas mass
Molar mass of carbon dioxide = 44
Mass = 1.76 × 44
77.44g
Thus, we calculated that when 26. g of butane is mixed with 168. g of oxygen. Then, the maximum mass of carbon dioxide that could be produced by the chemical reaction is 77.44g.
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The hydrogen ion (H+) concentration of a solution is 1.2 x 10-2 M. What is the pH of the solution?
Based on the calculations, the pH of this solution is equal to 1.9208.
What is pH?pH is literally the power of hydrogen ions and it can be defined as a measure of the molar concentration of hydrogen ions in a given solution.
How to calculate the pH of this solution?Mathematically, the pH of this solution can be calculated by using this formula;
pH = -log[H⁺]
Where:
H⁺ represents the hydrogen ion concentration of a solution.
Given the following data:
Concentration of hydrogen ion (H⁺) = 1.2 × 10⁻² M.
Substituting the given parameters into the formula, we have;
pH = -log(1.2 × 10⁻²).
pH = -(-1.9208)
pH = 1.9208
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What is an excess reactant?The reactant that is depleted first in a chemical reaction.The calculated amount of product formed in a chemical reaction.The reactant that is left over in a chemical reaction.The actual amount of product formed in a chemical reaction.
Explanation:
The excess reactant is the reactant that is left over once the limiting reactant or reagent is consumed. We can also say that it is a reactant present in an amount in excess of that require to combine with all of the limiting reactant.
Answer: The reactant that is left over in a chemical reaction.
All other things equal, if you had used twice the mass of water, what would you expect to happen to the observed value?
Answer:
the density is halved
Explanation:
doubling the volume results in the density being halved
A coffee cup calorimeter with a heat capacity of 5.80 J/∘ C was used to measure the change in enthalpy of a precipitation reaction. A 50.0 mL solution of 0.360 M AgNO3 was mixed with 50.0 mL of 0.460 M KI. After mixing, the temperature was observed to increase by 4.71∘C. Calculate the enthalpy of reaction, Δrxn, per mole of precipitate formed (AgI). Assume the specific heat of the product solution is 4.18 J / (g⋅∘C) and that the density of both the reactant solutions is 1.00 g/mL.
Calculate the theoretical moles of precipitate formed from AgNO3 and KI.
moles of precipitate formed from AgNO3:
moles of precipitate formed from KI:
Calculate the heat change experienced by the calorimeter contents, contents.
contents=
Calculate the heat change experienced by the calorimeter contents, cal.
cal=
Calculate the heat change produced by the solution process, solution.
solution=
Calculate Δsolution for one mole of precipitate formed.
Δsolution=
The heat change experienced by the calorimeter is 5.1339J.
Total mass of solution = 50 + 50 = 100ml
Density = 1 g/ml
mass = 100 gm
Q = (msΔt)solution + (sΔt) mixture
= (100×4.18×4.71) + (4.70 ×4.71)
= 1980.78 + 22.137
=2002.9 J
mole of AgNo₃ = 50×0.360/1000
= 0.0018
a mole of Kcl = 50×0.200/1000
= 0.010
The limiting reagent in a chemical reaction is a reactant that is definitely eaten up whilst the chemical reaction is completed. the amount of product fashioned is limited via this reagent, for the reason, that response can not hold without it.
The limiting reagent in a chemical reaction is the reactant with a view to be consumed completely. as soon as there is no greater of that reactant, the response can not proceed. Therefore it limits the response from continuing. The extra reagent is the reactant that would maintain reacting if the other had now not been consumed.
The limiting reagent is KCl so, moles of precipitate formed 0.010
ΔH exothermic = - 0,6568/0.01 = -65.68
Hence ΔH ext = - 65.68 kJ/mol
Theoretical moles of precipitate formed = 0.01 moles
Heat change by combustion = 100×4.70×4.18
= 1964.6 J
Heat change experienced by calorimeter is = 4.70 × 1.09
= 5.1339J
The heat of solution = 1964.6 J + 5.1339J
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Put the following liquids in order of least dense to most dense: vegetable oil, water, and rubbing alcohol
Explanation
First, let's list the density of each substance:
vegetable oil - approximately 0.92 g/mL
water - approximately 0.99 g/mL
rubbing alcohol - approximately 0.79 g/mL
So the least dense is rubbing alcohol, then vegetable oil and then water.
Answer: rubbing alcohol < vegetable oil < water.
Part B
What are the strengths and weaknesses of the cell model?
BIUX2² X₂ 15px
Characters used: 0/ 15000
AVEE
Carbon, hydrogen, nitrogen, oxygen, phosphorus, and sulfur are the strengths of the cell. The first four are the most significant since they are utilized to build the molecules required to create live cells.
What are the strengths and weaknesses of the cell model?The models draw attention to characteristics that are less obvious in actual cells. Color is used by the models to highlight various components. example of a weakness The models are static, but real cells feature moving elements. While the models depict cross-sections in two dimensions, real cells are three dimensional.
The inadequacy of cellular models to accurately represent the function of complicated decision-making by human beings is one of their shortcomings. Although transition rules are handled probabilistically in Markov models and may be affected by temporal lags in cell response, cellular frameworks can be coupled with these models.
Carbon, hydrogen, nitrogen, oxygen, phosphorus, and sulfur are the strengths of the cell. The first four are the most significant since they are utilized to build the molecules required to create live cells.
The complete question is:
What are the strengths and weaknesses of the cell model?
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The pressure of a gas which occupies 500cm3 at27°C is 900 mm Hy. what is the pressure of thegasat -48°c if thethe volume is reduced to250cm3 ?
ANSWER
The final pressure of the gas is 1350.22 mmHg
STEP-BY-STEP EXPLANATION:
Given information
[tex]\begin{gathered} \text{The initial volume of the gas = 500 cm}^3 \\ \text{ Initial temperature = 27}\degree C \\ \text{ Initial pressure = 900 mmHg} \\ \text{ Final temperature = -48}\degree C \\ \text{ Final volume = }250cm^3 \end{gathered}[/tex]From the question provided, you were asked to find the final pressure of the gas, hence, we assume that x represents the final pressure of the gas
To find the final pressure of the gas, we need to apply the general gas law
[tex]\frac{P1V1}{T1}\text{ = }\frac{P2V2}{T2}[/tex]Where,
P1 = initial pressure
V1 = initial volume
T1 = initial temperature
P2 = final pressure
V2 = final volume
T2 = final temperature
The next process is to convert the final and initial temperature from degree Celcius to degree kelvin
[tex]\begin{gathered} T\degree K\text{ = T}\degree C\text{ + 273.15} \\ T1\text{ = 27 + 273.15} \\ T1\text{ = 300.15K} \\ T2\text{ = -48 + 273.15} \\ T2\text{ = 225.15K} \end{gathered}[/tex]The next thing is to substitute the given data into the above formula
[tex]\begin{gathered} \frac{P1V1}{T1}\text{ = }\frac{P2V2}{T2} \\ \\ \frac{900\cdot\text{ 500}}{300.15}\text{ = }\frac{x\cdot\text{ 250}}{225.15} \\ \frac{450000}{300.15}\text{ = }\frac{250x}{225.15} \\ Cross\text{ multiply} \\ 300.15\cdot\text{ 250x = 450000 }\cdot\text{ 225.15} \\ 75037.5x\text{ = 101317500} \\ \text{Divide both sides 75027.5} \\ \frac{75037.5x}{75037.5}\text{ = }\frac{101317500}{75037.5} \\ x\text{ = 1350.22 mmHg} \end{gathered}[/tex]Therefore, the final pressure of the gas is 1350.22 mmHg
Which of the following compounds is a base?Select one:a. HNO3b. HClc. Al(OH)3d. H2SO4
ANSWER
option C
EXPLANATION
A base is a substance that reacts with an acid to produce salt and water only.
A base will also give hydroxide ion (OH-) when dissolved in water
Hence, the only compound that is a base in the given options is Al(OH)3
Therefore, the correct is option C
Why do ALL of these examples of popcorn - stove-top, microwave, camp fire, orair popped - exemplify heat transfer? Describe in detail, using vocab termswhere appropriate.
The grains are heated by microwave radiation which heats the grains and gives off more heat to the surrounding grains.
Popcorn pops using conduction convection and radiation. Students will use the particle states of matter to model three types of heat transfer. Radiant energy is converted to heat energy when the popcorn kernels absorb microwaves. As a result, the grains heat up and burst. You can also make loose corn kernels in the microwave if you have the right popcorn supply.
Convection heating is a great way to make popcorn. The reason popcorn pops is that the wick is filled with water. When the core is heated hot enough, this water turns into steam. Due to the hard, almost non-porous shell steam has nowhere to escape and pressure builds up within the core.
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Question 8 options:
An adult human lungs can only hold a volume of up to 6.0 L of air when fully expanded. Suppose that at sea level where the pressure is 1.00 atm, a sky diver traps 3.0 L of air in her lungs and holds her breath before rapidly ascending to an altitude where the pressure drops to 0.46 atm. If she doesn't exhale, determine if her lungs will rupture at this high altitude, then complete the sentence:
An adult human lungs can only hold a volume of up to 6.0 L of air when fully expanded. Suppose that at sea level where the pressure is 1.00 atm, a sky diver traps 3.0 L of air in her lungs and holds her breath before rapidly ascending to an altitude where the pressure drops to 0.46 atm. If she doesn't exhale, determine if her lungs will rupture at this high altitude, then complete the sentence:
At the higher altitudes, the gas volume will be increases, on higher altitudes pressure will decreases.
Thus , If she doesn't exhale, determine if her lungs will rupture at this high altitude, then complete the sentence: At the higher altitudes, the gas volume will be increases, on higher altitudes pressure will decreases.
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What is the concentration in mol/L of sodium cations when 2.88 g of Na3PO4 is dissolved in 25.0 mL of water?
The concentration in mol/L of sodium cations when 2.88 g of Na3PO4 is dissolved in 25.0 mL of water is 0.0303 mol/L Na+ ion
What is concentration ?
Concentration in chemistry is calculated by dividing a constituent's abundance by the mixture's total volume. Mass concentration, molar concentration, number concentration, and volume concentration are four different categories of mathematical description. [1] Any type of chemical mixture can be referred to by the term "concentration," however solutes and solvents in solutions are most usually mentioned. There are different types of molar (quantity) concentration, including normal concentration and osmotic concentration.
Na3PO4 <=====> 3Na+ + PO4^-3
Molar mass of Na3PO4 = 3*23+31+4*16 = 164g
164g Na3PO4 in 1000ml water contains 69g Na^+ ion
2.88g Na3PO4 in 1000ml water contains 69*2.88/164g Na^+ ion
2.88g Na3PO4 in 25ml water contains 1.212*25/1000Na^+ ion
= 0.0303 mol/L Na+ ion
The concentration in mol/L of sodium cations when 2.88 g of Na3PO4 is dissolved in 25.0 mL of water is 0.0303 mol/L Na+ ion
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The energy diagram above is for the reaction A + B -> C. The product has more energy than the reactants and the reaction s is best classified as A)exothermic B)endothermic C)electrostatic D)neutral
The energy diagram given is for endothermic reaction process
This is characterized by the system to require more energy to form products.
In this question, specie A and B are the reactant while C is the product.
What happens in this type of reaction is that for products to be formed, the reaction has to attain activation energy which is used to break bonds and form products and this energy has to be gotten from outside the system in order for the reaction to proceed.
The flame produced by the burner of a gas (propane) grill is a pale blue color when enough air mixes with the propane (C3H8) to burn it completely. For every gram of propane that flows through the burner, what volume of air is needed to burn it completely? Assume that the temperature of the burner is 195.0°C, the pressure is 1.10 atm, and the mole fraction of O2 in air is 0.210.
The volume of the air that we require is 18.85 liters air.
What is the volume required?The term combustion has to do with a situation where we are burning a substance in oxygen. By implication, combustion is actually a type of oxidation reaction and it does occur with hydrocarbons to produce energy.
The combustion of the butane occurs as follows:
C₃H₈ + 5O₂ = 4 H₂O + 3CO₂
MW 44 5x32 4x18 3x44
The number of moles of one gram of butane is; 1/44 = 0.02272 mol propane
The amount of oxygen that is required is; 5 x 0.02272 = 0.1136 mol oxygen
Given that 0.21 is the mol fraction of oxygen in air; 0.1136 / 0.21 = 0.54 mol air is required for the combustion of propane.
From the ideal gas law;
PV=nRT, or V = nRT/P
P = 1.1 x 101325 Pa = 111457 PaT = 195°C + 273 = 468 KR = 8.314n= 0.54 molThe volume in liters = 0.54 x 8.314 x 468 / 111457 = 0.0188 m³ = 18.85 liters air.
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in the following reactions,decide which reactant is oxidized and which is reduced. Designate the oxidzing agent and the reducing agent
In the reaction that is shown in the question;
1) Dichromate ion is reduced while tin II ion is oxidized
2) Fe II is oxidized while nitrate ion is reduced.
What is oxidation and reduction?We know that oxidation has to do with the loss of electrons. Thus we say that a specie has been oxidized if the specie looses electrons and we say that a specie has been reduced if the specie gains electrons. The species that is oxidized in a reaction is the oxidizing agent while the specie that is reduced in the reaction is a reducing agent.
Let us note that the procedures of oxidation and reduction does occur at the same time and when this happens, we have a specie whose oxidation number is increased and we say that it has been oxidized while the other specie has received some new electrons and has been said to be reduced.
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1 quart = 946 mL
This is an example of an
1 quart = 946 mL is an example of unit conversion in volume.
What is unit conversion?The unit conversion of a substance is the process through which the unit of the substance is converted from one form to another.
The expression 100 cm/1 m is called a conversion factor, and it is used to formally change the unit of a quantity into another unit.
For example, converting the volume of a substance from one unit to another, we make use of conversion factor such as 1 L/1000 mL, 1 m³/1,000,000 cm³, etc.
There different units in the volume of a substance or liquid is measured and they include the following;
quart cubic centimeter (cm³)cubic meter (m³)cubic feet (ft³)liters (L)milliliters (mL)The unit of volume can be converted from one unit to another as shown in the given example.
1 quart = 946 mL
Thus, we can conclude that 1 quart = 946 mL is an example of unit conversion in volume, from quart to milliliters.
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