It will require for 20 % of U-238 atoms in a sample of U-238 to decay.
Further Explanation:
Radioactive decay involves stabilization of unstable atomic nucleus and is accompanied by the release of energy. This emission of energy can be in form of different particles like alpha, beta and gamma particles.
Half-life is time period in which half of the radioactive species is consumed. It is denoted by .
The expression for half-life is given as follows:
[tex]\lambda = \dfrac{{0.693}}{{{t_{{\text{1/2}}}}}}[/tex] …… (1)
Where,
[tex]{t_{{\text{1/2}}}}[/tex] is half-life period
[tex]\lambda[/tex] is the decay constant.
The half-life period for decay of U-238 is [tex]4.5 \times {10^9}{\text{ yrs}}[/tex].
Substitute [tex]4.5 \times {10^9}{\text{ yrs}}[/tex] for [tex]{t_{{\text{1/2}}}}[/tex] in equation (1).
[tex]\begin{aligned}\lambda&= \dfrac{{0.693}}{{4.5 \times {{10}^9}{\text{ yrs}}}} \\&= 1.54 \times {10^{ - 10}}{\text{ yr}}{{\text{s}}^{ - 1}} \\\end{galigned}[/tex]
Since it is radioactive decay, it is first-order reaction. Therefore the expression for rate of decay of U-238is given as follows:
[tex]\lambda = \dfrac{{2.303}}{t}\log \left( {\dfrac{a}{{a - x}}} \right)[/tex]
…… (2)
Where,
[tex]\lambda[/tex] is the decay or rate constant.
t is the time taken for decay process.
a is the initial amount of sample.
x is the amount of sample that has been decayed.
Rearrange equation (2) to calculate t.
[tex]t = \dfrac{{2.303}}{\lambda }\log \left( {\dfrac{a}{{a - x}}} \right)[/tex] …… (3)
Consider 100 g to be initial amount of U-238. Since 20 % of it is decayed in radioactive process, 20 g of U-238 is decayed and therefore 80 g of the sample is left behind.
Substitute 100 g for a, 80 g for (a–x) and [tex]1.54 \times {10^{ - 10}}{\text{ yr}}{{\text{s}}^{ - 1}}[/tex] for [tex]\lambda[/tex] in equation (3).
[tex]\begin{aligned}t &= \dfrac{{2.303}}{{1.54 \times {{10}^{ - 10}}{\text{ yr}}{{\text{s}}^{ - 1}}}}\log \left( {\dfrac{{100{\text{ g}}}}{{80{\text{ g}}}}} \right)\\&= 1.449 \times {10^9}{\text{ yrs}}\\\end{aligned}[/tex]
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Grade: Senior School
Subject: Chemistry
Chapter: Radioactivity
Keywords: half-life, t, a, x, a – x, 1.449*10^9 yrs, U-238, decay constant, radioactivity, half-life period.
Ethylene gas and steam at 320°c and atmospheric pressure are fed to a reaction process as an equimolar mixture. the process produces ethanol by the reaction: c2h4(g) + h2o(g) → c2h5oh(l)c2h4(g) + h2o(g) → c2h5oh(l) the liquid ethanol exits the process at 25°c. what is the heat transfer associated with this overall process per mole of ethanol produced?
Without specific enthalpy change (ΔH) values for the reactions in question, we cannot accurately calculate the heat transfer per mole of ethanol produced. However, if these values were provided, the general formula qp = ΔH(T2-T1)/T1*T2 could be used to determine this, where qp represents heat transfer at constant pressure, T1 and T2 are the initial and final temperatures respectively.
Explanation:This question pertains to the field of thermochemistry and the role of heat transfer in chemical reactions. The process in question involves the transformation of ethylene gas and steam into ethanol, or C2H5OH. We would need the specific heat, or enthalpy change (ΔH), values for these reactions to calculate the heat transfer per mole of ethanol produced. Since these values haven't been provided, we can't provide a definite answer.
Generally, in such questions, when ΔH values and temperatures are given, one formula that can be applied is q_p = ΔH(T2-T1)/T1*T2 where T1 is initial temperature, T2 is final temperature and qp represents heat transfer at constant pressure.
This formula can be used to estimate the heat transfer associated with the process per mole of ethanol produced, once the ΔH value is known. It's also important to remember that if the reaction is exothermic (releases heat), the ΔH value would be negative, and if it's endothermic (absorbs heat), the ΔH value would be positive.
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As you have seen in this lab, the density of water is near 1 g/cm3. Anything with a density lower than this will float in water, and anything with a higher density will sink. Imagine you are building a submarine. How could you be sure to have it sink and then rise?
Make sure its density is always above 1.0 g/cm3.
Make sure its density is always below 1.0 g/cm3.
Develop a way to adjust its density so that it can be above and below 1.0 g/cm3.
Answer: C) develop a way to adjust its density so that it can be above and below 1.0 gram per cubic centimeter.
Explanation: Submarine is a kind of special ship that can float above and below water. We know that an object can only float over water when its density is less than the density of water and it would sink if its density is greater than the density of water.
A submarine has tanks in which water is filled if we want to sink it and the water from the tanks is taken out with the help of water pumps to rise the submarine. These water tanks in the submarine helps to adjust its density.
The over all density is increased by filling the water tanks to sink the submarine and the density is decreased by removing the water from the tanks to rise the submarine.
So, option third or C is correct.
In an oxidation-reduction reaction, what happens to the electrons in the reduction process?
Answer:
Electrons will be gained
Write the net ionic equation (including phases) that corresponds to fe(clo4)2
The compound Fe(ClO4)2 in water would dissociate into its ions, which are Fe²+ and 2ClO4¯. A specific net ionic equation cannot be given without knowledge of the reactants.
Explanation:The net ionic equation would be the result of considering all the ions in the reaction of Fe(ClO4)2, which is iron(II) perchlorate, with all possible reactants. However, without knowing these reactants, we can't provide a specific net ionic equation. Generally, important concepts here include understanding how to balance equations and predict solubility based on common rules. In the case of Fe(ClO4)2 in water, this compound is highly soluble and would dissociate into its ions in water. Generally, you would expect it to dissociate into Fe²+ and 2ClO4¯ ions in solution.
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Provide the structure of the major product which results from 1,4-addition of br2 to the diene shown below.
A 13.0 kg iron weightlifting plate has a volume of 1650 cm3 . What is the density of the iron plate in g/cm3?
Explanation:
It is known that density is the amount of mass divided by volume.
Mathematically, Density = [tex]\frac{mass}{volume}[/tex]
It is given that mass is 13.0 kg or 13000 g (as 1 kg = 1000 g). And, volume is 1650 [tex]cm^{3}[/tex].
Therefore, calculate the density as follows.
Density = [tex]\frac{mass}{volume}[/tex]
= [tex]\frac{13000 g}{1650 cm^{3}}[/tex]
= [tex]7.87 g/cm^{3}[/tex]
Thus, we can conclude that density of the given iron plate is [tex]7.87 g/cm^{3}[/tex].
The ph of a 0.55 m aqueous solution of hypobromous acid, hbro, at 25.0 °c is 4.48. what is the value of ka for hbro?
What mass of kbr (in grams) should you use to make 300.0 ml of a 1.50 m solution of kbr?
To prepare a 300.0 mL of a 1.50 M KBr solution, you need 53.55 grams of KBr, calculated by multiplying the required moles (0.450 moles) by the molar mass of KBr (119.00 g/mol).
Explanation:Calculating the Mass of KBr for a SolutionTo find the mass of KBr needed to make a 300.0 mL (0.300 L) of a 1.50 M solution, we apply the formula:
Molarity (M) = moles of solute / liters of solution
First, calculate the moles of KBr required:
Moles of KBr = Molarity × Volume in LitersMoles of KBr = 1.50 moles/L × 0.300 LMoles of KBr = 0.450 molesNext, convert moles to grams using the molar mass of KBr (119.00 g/mol):
Mass of KBr = Moles of KBr × Molar Mass of KBrMass of KBr = 0.450 moles × 119.00 g/molMass of KBr = 53.55 gramsTherefore, you would need 53.55 grams of KBr to make a 300.0 mL of a 1.50 M KBr solution.
A gas has a volume of 590 mL at a temperature of -55°C what volume will the gas occupy at 30°C
Are there any other methods you could use to determine phosphate content in the colas
One liter of a buffer composed of 1.2 m hno2 and 0.8 m nano2 is mixed with 400 ml of 0.5 m naoh. what is the new ph? assume the pka of hno2 is 3.4.
The new pH of given solutionis [tex]\boxed{3.53}[/tex].
Further Explanation:
The aqueous solution of a weak acid and its conjugate base or a weak base and its conjugate acid is termed as buffer solution. These solutions offer strong resistance to any change in their pH on addition of small quantity of strong acid or base.
Henderson-Hasselbalch equation:
This equation helps in determining the pH of buffer solution. Its mathematical form is given as follows:
[tex]{\text{pH}} = {\text{p}}{K_{\text{a}}} + {\text{log}}\dfrac{{\left[ {{{\text{A}}^ - }} \right]}}{{\left[ {{\text{HA}}} \right]}}[/tex] …… (1)
Here,
[tex]\left[ {{{\text{A}}^ - }} \right][/tex] is concentration of conjugate base.
[HA] is concentration of acid.
Given mixture is a buffer solution of [tex]{\text{HN}}{{\text{O}}_{\text{2}}}[/tex] and [tex]{\text{NaN}}{{\text{O}}_{\text{2}}}[/tex]. Therefore Henderson-Hasselbalch equation becomes as follows:
[tex]{\text{pH}} = {\text{p}}{K_{\text{a}}} + {\text{log}}\dfrac{{\left[ {{\text{NaN}}{{\text{O}}_{\text{2}}}} \right]}}{{\left[ {{\text{HN}}{{\text{O}}_{\text{2}}}} \right]}}[/tex] …… (2)
Initial moles of [tex]{\text{HN}}{{\text{O}}_{\text{2}}}[/tex] can be calculated as follows:
[tex]\begin{aligned}{\text{Moles of HN}}{{\text{O}}_2} &= \left( {1.2{\text{ M}}} \right)\left( {{\text{1 L}}} \right)\\&= 1.2{\text{ mol}} \\\end{aligned}[/tex]
Initial moles of [tex]{\text{NaN}}{{\text{O}}_{\text{2}}}[/tex] can be calculated as follows:
[tex]\begin{aligned}{\text{Moles of NaN}}{{\text{O}}_{\text{2}}} &= \left( {0.8{\text{ M}}} \right)\left( {{\text{1 L}}} \right)\\&= 0.8{\text{ mol}} \\\end{aligned}[/tex]
Moles of NaOH can be calculated as follows:
[tex]\begin{aligned}{\text{Moles of NaOH}} &= \left( {0.5{\text{ M}}} \right)\left( {{\text{400 mL}}} \right)\left( {\frac{{{{10}^{ - 3}}{\text{ L}}}}{{1{\text{ mL}}}}} \right) \\&= 0.2{\text{ mol}} \\\end{aligned}[/tex]
When addition of 0.2 moles of NaOH is done to the buffer solution, 0.2 moles of [tex]{\text{HN}}{{\text{O}}_{\text{2}}}[/tex] is neutralized while the same amount of [tex]{\text{NaN}}{{\text{O}}_{\text{2}}}[/tex] is formed. Since volumes are additive, total volume can be calculated as follows:
[tex]\begin{aligned}{\text{Total volume of solution}} &= \left( {1 + \left( {400{\text{ mL}}} \right)\left( {\frac{{{{10}^{ - 3}}{\text{ L}}}}{{1{\text{ mL}}}}} \right)} \right){\text{ L}} \\ &= {\text{1}}{\text{.4 L}} \\\end{aligned}[/tex]
Therefore concentration of [tex]{\text{HN}}{{\text{O}}_{\text{2}}}[/tex] can be calculated as follows:
[tex]\begin{aligned}\left[ {{\text{HN}}{{\text{O}}_{\text{2}}}} \right] &= \frac{{\left( {1.2 - 0.2} \right){\text{ mol}}}}{{1.4{\text{ L}}}}\\&= 0.714{\text{ M}} \\\end{aligned}[/tex]
Therefore concentration of [tex]{\text{NaN}}{{\text{O}}_{\text{2}}}[/tex] can be calculated as follows:
[tex]\begin{aligned}\left[ {{\text{NaN}}{{\text{O}}_{\text{2}}}} \right] &= \frac{{\left( {1.2 + 0.2} \right){\text{ mol}}}}{{1.4{\text{ L}}}} \\ &= 1{\text{ M}} \\\end{aligned}[/tex]
Substitute 0.714 M for [tex]\left[ {{\text{HN}}{{\text{O}}_{\text{2}}}} \right][/tex], 1 M for [tex]\left[ {{\text{NaN}}{{\text{O}}_{\text{2}}}} \right][/tex] and 3.4 for [tex]{\text{p}}{K_{\text{a}}}[/tex] in equation (2).
[tex]\begin{aligned} {\text{pH}} &= 3.4 + {\text{log}}\left( {\frac{{{\text{1 M}}}}{{0.714{\text{ M}}}}} \right) \\&= 3.54 \\\end{aligned}[/tex]
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Grade: High School
Subject: Chemistry
Chapter: Acid, base and salts
Keywords: pH, buffer, pKa, NaNO2, HNO2, 3.4, 1 M, 0.714 M, concentration, total volume of solution, 1.2 mol, 0.8 mol, 0.2 mol, 3.54.
A solution is prepared by dissolving 17.75 g sulfuric acid, h2so4, in enough water to make 100.0 ml of solution. if the density of the solution is 1.1094 g/ml, what is the molality?
Answer: The molality of solution is 1.94 m.
Explanation:
We are given:
Mass of sulfuric acid = 17.75 grams
Volume of solution = 100 mL
Density of solution = 1.1094 g/mL
To calculate the mass of solution, we use the equation:
[tex]Density=\frac{Mass}{Volume}[/tex]
Putting values in above equation, we get:
[tex]1.1094g/mL=\frac{\text{Mass of solution}}{100mL}[/tex]
[tex]\text{Mass of solution}=110.94g[/tex]
Mass of solvent = Mass of solution - Mass of solute
Mass of solvent = 110.94 - 17.75 = 93.19 g
To calculate the molality of solution, we use the equation:
[tex]Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}[/tex]
Where,
[tex]m_{solute}[/tex] = Given mass of solute [tex](H_2SO_4)[/tex] = 17.75 g
[tex]M_{solute}[/tex] = Molar mass of solute [tex](H_2SO_4)[/tex] = 98 g/mol
[tex]W_{solvent}[/tex] = Mass of solvent = 93.19 g
Putting values in above equation, we get:
[tex]\text{Molality of }H_2SO_4=\frac{17.75\times 1000}{98\times 93.19}[/tex]
[tex]\text{Molality of }H_2SO_4=1.94m[/tex]
Hence, the molality of solution is 1.94 m.
To calculate the molality of the sulfuric acid solution, we convert the volume to mass, find the number of moles of sulfuric acid, and divide by the mass of the solvent in kilograms.
Explanation:To calculate the molality of a solution, we need to find the number of moles of solute and the mass of the solvent.
In this case, we are given the mass of sulfuric acid (17.75 g) and the volume of the solution (100.0 ml). First, we need to convert the volume to mass by multiplying it by the density of the solution (1.1094 g/ml). 100.0 ml x 1.1094 g/ml = 110.94 g.
To find the number of moles of sulfuric acid, we divide the mass by the molar mass of sulfuric acid (98.09 g/mol). 17.75 g / 98.09 g/mol ≈ 0.1808 mol.
Finally, we can calculate the molality by dividing the moles of solute by the mass of the solvent in kilograms. 0.1808 mol / 0.11094 kg = 1.63 mol/kg.
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A 6 l volume of ideal neon gas (monatomic) is at a pressure of 3.2 atmospheres and a temperature of 310 k. the atomic mass of neon is 20.2 g/mol. in this situation, the temperature of the gas is increased to 410 and the volume is increased to 8.0 l. the final pressure of the gas, in atmospheres, is closest to:
Answer is: the final pressure of the gas is closest to 3,17 atm.
p₁ =
3,2 atm.
T₁ =
310 K.
V₁ =
6 L.
p₂ = ?
T₂ =
410K.
V₂ = 8,0 L.
Use combined gas law - the volume of amount of gas is proportional to the ratio of
its Kelvin temperature and its pressure.
p₁V₁/T₁ =
p₂V₂/T₂.
3,2 atm · 6,0 L ÷ 310 K = p₂ · 8,0 L ÷ 410 K.
0,0619 = 0,0195p₂.
p₂ =
3,17 atm.
Using the periodic table, choose the more reactive metal. (hint: reactivity of Ga > Al: reactivity of Zn > Ga) Pt or Ag
Answer:
Silver (Ag) is more reactive than platinum (Pt) :) hope this helps!!!
Explanation:
100 POINTS Which energy source produces electricity at a hydroelectric dam? A. solar panels B. wind turning turbines C. falling water D. pressure from water on the dam
100 POINTS
The answer is C falling water hope this helps
Electricity at a hydroelectric dam is produced by falling water, which turns turbines to generate electricity. The ultimate source of this energy is solar energy that drives the hydrological cycle.
The energy source that produces electricity at a hydroelectric dam is C. falling water. This process involves the conversion of potential energy stored in the water behind the dam into kinetic energy as the water flows down. The kinetic energy turns the blades of turbines, which then drive a generator to produce electricity. The ultimate source of the electrical energy produced by a hydroelectric plant is solar energy, which drives the hydrological cycle that lifts water to high elevations through evaporation and precipitation.
What volume of a 0.25 m phosphoric acid solution is required to react completely with 1.0 l of 0.35 m sodium hydroxide?
describe the placement of the crucible lid on the crucible when heating the magnesium. Why is it important that this be done correctly?
Set the crucible's lid slightly off-center to allow air to enter while keeping the magnesium oxide from escaping.
What is a crucible lid?A crucible is a cup-shaped piece of laboratory equipment used to keep chemical compounds contained while they are heated to extremely high temperatures.
Crucibles come in a variety of sizes and are usually packaged with a crucible cover (or lid).
Keep the lid on the crucible while cooling to prevent moisture from the atmosphere from interacting with the anhydrous salt, especially if the lab is humid. As a result, the mass of water will be too low.
The most important apparatus because it will be used to obtain the final precipitate, which will tell us how much salt is in the solution.
The lid is used to cover the crucible so that the heated precipitates do not oxidize when they come into contact with air.
Thus, it is important that the lid should be kept correctly.
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Which element is reduced in this reaction? 2cr(oh)3+3ocl−+4oh−→2cro4−2+3cl−+5h2o enter the chemical symbol of the element?
A 42 kg sample of water absorbs 347 kJ of heat. If the water was initially at 23.2 ∘C, what is its final temperature?
Major groups of minerals include _____. oxides and carbonates ions and isotopes inorganics and halides silicates and magnetics
oxides and carbonates, grad point..
If the calcium oxide were to be obtained by the heating of calcium hydroxide, how much hydroxide would be needed to obtain the 15.0 g?
A reaction in which a , b , and c react to form products is zero order in a , one-half order in b , and second order in
c. by what factor does the reaction rate change if [b] is doubled (and the other reactant concentrations are held constant)? -g
When the concentration of b is doubled in a chemical reaction that is zero order in a, one-half order in b, and second order in c, the reaction rate will increase by a factor of √2 (approximately 1.414), with other reactant concentrations held constant.
Explanation:The question you're asking relates to the rate of a chemical reaction and how it changes with varying concentrations of reactants. Specifically, there is a reaction where the rate is zero order in a, one-half order in b, and second order in c. According to the given reaction orders, the rate expression would be:
Rate = k [a]0[b]1/2[c]2
Since the reaction is zero order in a, changing the concentration of a does not affect the rate. However, since it is one-half order in b, if the concentration of b is doubled, the rate will increase by a factor of the square root of 2. This is because:
New Rate = k [a]0(2[b])1/2[c]2 = k [a]0[b]1/2 × √2 [c]2
= Rate × √2
Therefore, when the concentration of b is doubled, and a and c remain constant, the reaction rate will increase by a factor of √2 (approximately 1.414).
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A ball-and-stick model of a molecule provides much the same information as a
Both the ball-and-stick and space-filling models are 3D molecular representations, with the former displaying the bonds and the latter showing the relative atom sizes and general shape.
A ball-and-stick model of a molecule provides much the same information as a space-filling model. Both are used to represent the three-dimensional structure of a molecule in different ways. The ball-and-stick model shows atoms as spheres and bonds as sticks, highlighting the bond angles and lengths between atoms.
In contrast, the space-filling model represents atoms by spheres that are sized proportionally to the atoms' radii, depicting the actual volume occupied by the atoms without showing the bonds explicitly.
While the ball-and-stick model may exaggerate the space between the atoms, it is beneficial for understanding the geometry of the chemical bonds. The space-filling model, however, allows for visualization of the molecule's general shape and the relative sizes of the atoms. Both models are simplifications and do not fully capture subtle variations in bond lengths and angles.
how can an atom that has seven valence electrons complete its outermost level
Answer: The element needs to react with other element by gaining an electron to complete its valence electron.
Explanation: The element which exhibit 7 valence electrons are halogens. They readily react with other element, for example: Hydrogen and Sodium, in order to gain an electron to complete their outermost shell.
Kemmi pipets 25.00 ml of pure 1-propanol (c3h7oh, a liquid organic alcohol) into a 100.0 ml volumetric flask. she dilutes it with di water. help her calculate the strength of her diluted solution. volumes: 25.00 ml volumetric pipet, 100.00 ml volumetric flask. molar mass of pure 1-propanol: 60.09 g/mol. density of pure 1-propanol: 0.803 g/ml calculate the number of moles in exactly 25.00 ml of 1-propanol. (3 significant digits; units of mol) [x] calculate the molarity of the diluted solution. (3 significant digits; units of m or mol/l) [y]
The number of moles of 1-propanol in 25.00 mL is 0.334 (to three significant digits). The molarity of the diluted 1-propanol solution in a 100.0 mL flask is 3.34 M (to three significant digits).
To calculate the number of moles of 1-propanol in 25.00 mL, we use the density of 1-propanol (0.803 g/mL) to find the mass:
Mass = Volume × Density = 25.00 mL 0.803 g/mL = 20.075 g
Next, we convert this mass to moles using the molar mass of 1-propanol (60.09 g/mol):
Moles of 1-propanol = Mass / Molar Mass = 20.075 g / 60.09 g/mol
This calculation yields 0.334 moles of 1-propanol (to three significant digits).
For the molarity of the diluted solution, we take into account that the total volume is now 100.0 mL. Since molarity is moles per liter (mol/L), we must first convert the volume from mL to L:
Volume in L = 100.0 mL (1 L / 1000 mL) = 0.100 L
Molarity = Moles of 1-propanol / Volume in L = 0.334 mol / 0.100 L
The molarity of the solution is 3.34 M (to three significant digits).
In two or more complete sentences, explain the law of conservation of mass and how it relates to this experiment?
Answer:
The mass can not be neither created nor destroyed but transformed by chemical reactions or physical transformations in an isolated recipient.
Explanation:
Hello,
In case, no matter the carried out experiment, the law of conservation of mass always leads the same: the mass can not be neither created nor destroyed but transformed by chemical reactions or physical transformations in an isolated recipient.
In such a way, we must consider that any system closed to every form of transport of matter, will show a no change in its mass as time goes by, since the system's mass cannot change neither by additions nor withdrawals. Therefore, the quantity of mass is conserved over time.
Best regards.
Think about a carbon atom that is released into the atmosphere from the burning of wood in a campfire. If it were to go through the whole carbon cycle,
Step 1 A caterpillar gets the carbon by eating the tree's leaves.
Step 2 The bird decomposes and the carbon is added to the atmosphere.
Step 3 A bird gets the carbon by eating the caterpillar.
Step 4 The bird flies into a building and dies instantly. It falls to the ground
Step 5 A tree absorbs the carbon from the atmosphere into its leaves for photosynthesis.
plz help me by telling me what order this go's in?
Answer: the actual answer is
step 1: A tree absorbs the carbon from the atmosphere into its leaves for photosynthesis
step 2: A caterpillar gets the carbon by eating the tree's leaves
step 3: A bird gets the carbon by eating the caterpillar
step 4: The bird flies into a building and dies instantly. It falls to the ground
step 5: The bird decomposes and the carbon is added to the atmosphere
Explanation:
What do you need to know describe the velocity of an object
Answer:
You need to know the speed, direction (displacement) and time to describe the velocity of an object.
Identify element 3. EXPLAIN YOUR REASONING
A gas sample has a volume of 178 mL at 0.00oC.The temperature is raised at constant pressure until the volume reaches 211 mL. What is the temperature of that gas sample at this volume?
The temperature of the gas will be 324K or [tex]\rm \bold{51^\cdot C}[/tex] when volume reaches 211 mL.
Charle's law stated that the volume occupied by fixed amount of gas is directly proportional to the temperature,while pressure is constant.
[tex]\rm \bold{\frac{V_1}{T_1} =\frac{V_2}{T_2}}[/tex]
Where,
[tex]\rm \bold V_1[/tex] is initial volume = 178mL
[tex]\rm \bold {V_2}[/tex] is final volume = 211mL
[tex]\rm \bold{ T_1}[/tex] is initial temperature = [tex]\rm \bold{0.00^\cdot C}[/tex] = 273K
[tex]\rm \bold {T_2}[/tex] is final temperature = ?
Put the values,
[tex]\rm \bold{\frac{178}{273} =\frac{211}{T_2}}\\\\\rm \bold{T_2=324 K}[/tex]
Hence, we can conclude that the temperature of the gas will be 324K or [tex]\rm \bold{51^\cdot C}[/tex] when volume reaches 211 mL.
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