Answer:
0.71 J/g°C
Explanation:
Here is the complete question
thermometer A 51.9 g sample of quartz is put into a calorimeter (see sketch at right) that contains 300.0 g of water. The quartz sample starts off at 97.8 °C and the temperature of the water starts off at 17.0 °C. When the temperature of the water stops changing it's 19.3 °C. The pressure remains constant at 1 atm. insulated container water sample Calculate the specific heat capacity of quartz according to this experiment. Be sure your answer is rounded to 2 significant digits. a calorimeter g °C
Solution
Since the temperature of the water increases from 17.0 °C to 19.3 °C, it means that it loses heat. Also, the final temperature of the quartz equals the final temperature of the water 19.3 °C. Since the quartz temperature decreases from 97.8 °C to 19.3 °C it loses heat.
So, heat lost by quartz, Q = heat gained by water, Q'
-Q = Q'
-mc(θ₂ - θ₁) = m'c'(θ₂ - θ₃) where m = mass of quartz = 51.9 g, c = specific heat capacity of quartz, θ₁ = initial temperature of quartz = 97.8 °C, θ₂ = final temperature of quartz = 19.3 °C, m' = mass of water = 300 g, c = specific heat capacity of water = 4.2 J/g °C , θ₃ = initial temperature of water = 17.0 °C, θ₂ = final temperature of water = 19.3 °C
Making c subject of the formula, we have
c = -m'c'(θ₂ - θ₃)/m(θ₂ - θ₁)
Substituting the values of the variables into the equation, we have
c = -300 g × 4.2 J/g °C(19.3 °C - 17.0 °C)/51.9 g(19.3 °C - 97.8 °C)
c = -1260 J/°C(2.3 °C)/51.9 g(-78.5 °C)
c = -2898 J/-4074.15 g°C
c = 0.711 J/g°C
c ≅ 0.71 J/g°C to 2 significant digits
^^^^Changes in state of matter are ALWAYS changes.
Answer:
physical
Explanation:
The change in the state of matter is always physical change, because it can be done with physical processes.
A 2.00-mol sample of hydrogen gas is heated at constant pressure from 294 K to 414 K. (a) Calculate the energy transferred to the gas by heat. kJ (b) Calculate the increase in its internal energy. kJ (c) Calculate the work done on the gas. kJ
Answer:
a) The energy transferred is 6.91 kJ
b) The internal energy is 4.90 kJ
c) The work done on the gas is - 2.01 kJ
Explanation:
Step 1: Data given
Number of moles of hydrogen gas = 2.00 moles
Pressure = constant
Temperature is heated from 294 K to 414 K
Molar heat capacity of hydrogen gas = 28.8 J/mol*K
Step 2: Calculate the energy transferred to the gas by heat.
Q = n* Cp * ΔT
⇒with Q =the energy transferred
⇒with n = the number of moles = 2.00 moles
⇒with Cp = the Molar heat capacity of hydrogen gas = 28.8 J/mol*K
⇒ with ΔT = Temperature 2 - Temperature 1 = 414 - 294 = 120K
Q = 2.00 * 28.8 * 120
Q = 6912 J = 6.91 kJ
Step 3: Calculate the increase in its internal energy.
ΔEint = n*Cv*ΔT
⇒with ΔEint = the increase in its internal energy.
⇒with n = the number of moles = 2.00 moles
⇒with Cv = The constant volume = 20.4 J/mol*K
⇒with ΔT = Temperature 2 - Temperature 1 = 414 - 294 = 120K
ΔEint = 2.00 * 20.4 * 120
ΔEint =4896 J = 4.90 kJ
Step 4: Calculate the work done on the gas.
Work done on the gas = -Q + ΔEint
W = -6.91 kJ + 4.90 kJ
W = -2.01 kJ
A compound with a molecular weight of about 64.47 g/mol was found to be 18.63 % of C, 1.56 % of H, 24.82 % of O, and 54.99 % of Cl by mass. Determine the molecular formula and draw the Lewis structure showing an accurate 3-D perspective. *Show your calculations
Answer:
See detailed explanation.
Explanation:
Hey there!
In this case, according to the given information, it turns out possible for us to solve this problem by firstly calculating the moles of each element, assuming those percentages are masses, so that we divide by their molar masses:
[tex]C=\frac{18.63}{12.01}=1.55\\\\H=\frac{1.56}{1.01} =1.55\\\\O=\frac{24.82}{16.00}=1.55\\\\Cl=\frac{54.99}{35.45}=1.55[/tex]
Then, we divide all of them by 1.55 to realize the empirical formula is:
[tex]CHOCl[/tex]
Whose molar mass is 64.47 g/mol, and therefore, since the molar mass of these two is the same, we infer the molecular formula is also CHOCl.
The Lewis structure is shown on the attached document, whereas, the central atom is C and it does complete its octet as well as both O and Cl.
Regards!
How many grams of glucose are needed to prepare 144.3 mL of a 1.4%(m/v) glucose solution?
Answer:
2.0202 grams
Explanation:
1.4% (m/v) glucose solution means: 1.4g glucose/100mL solution.
so ?g glucose = 144.3 mL soln
Now apply the conversion factor, and you have:
?g glucose = 144.3mL soln x (1.4g glucose/100mL soln).
so you have (144.3x1.4/100) g glucose= 2.0202 grams
81.5 g of metal was heated from 11 degrees Celsius to 69 degrees Celsius. If 6739 joules of heat energy were used, what is the specific heat capacity of the metal?
Answer:
the metal become red hot
Để xác định hàm lượng Cu trong hợp kim Cu-Zn người ta làm như sau: Hòa
tan hoàn toàn 2,068g mẫu hợp kim Cu-Zn trong lượng dư axit HNO3, thu được dung
dịch X. Đun đuổi axit dư, điều chỉnh tới pH 3 thu được 100mL dung dịch Y. Lấy
10mL dung dịch Y, thêm KI dư, rồi chuẩn độ dung dịch tạo thành bằng dung dịch
Na2S2O3 0,1M thì thấy hết 15,0 mL. Viết các phương trình phản ứng xảy ra. Tính
hàm lượng Cu trong mẫu hợp kim trên.
How is the compound NH3 classified?
A. As a salt
B. As a base
C. As an acid
D. As ionic
Answer:
B
Explanation:
Ammonia is considered a base as it's pH is 11
Answer from Gauthmath
The compound NH3 (Ammonia) can be classified as a weak Base. Below you can learn more about Ammonia.
What is Ammonia (NH3)?Ammonia is a chemical compound which is derived from the combination of Nitrogen and Hydrogen. It is denoted by the chemical formula NH3.
Ammonia is a base and when it reacts with acids to gives out salts. Physically, It is a colorless gas with a distinct characteristic of a pungent smell.
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Too many objects inside a laboratory fume hood can disrupt the airflow and possibly compromise you safety. Which of the following are considered best practices in the use of a laboratory fume hood?
a. Open the sash as much as possible
b. Work at least 25 cm inside the hood
c. Use fast, quick movements to limit your exposure
d. Place objects to one side—work on other side
e. Use a raised along the back of the hood
Best practices for fume hoods: work 25 cm inside, organize items to one side, use raised ledge; avoid open sash and quick movements.
Laboratory fume hoods must be used safely. Workers should operate at least 25 cm within the hood to preserve ventilation and avoid dangerous chemicals. Place things on one side of the hood to preserve ventilation and prevent clogging.
A raised ledge on the rear of the hood prevents things from falling in and impeding airflow. Avoid fully opening the sash to maintain ventilation and containment. Fast, rapid motions can interrupt airflow, so prevent them. These practises guarantee the fume hood contains harmful compounds, making the lab safer. Therefore, option (B), (D) and (E) are correct.
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Given the following reaction:
CO (g) + 2 H2(g) <==> CH3OH (g)
In an experiment, 0.42 mol of CO and 0.42 mol of H2 were placed in a 1.00-L reaction vessel. At equilibrium, there were 0.29 mol of CO remaining. Keq at the temperature of the experiment is ________.
A) 2.80
B) 0.357
C) 14.5
D) 17.5
E) none of the above
Answer:
Option D. 17.5
Explanation:
Equiibrium is: CO + 2H₂ ⇄ CH₃OH
1 mol of CO is in equibrium with 2 moles of hydrogen in order to make, methanol.
Initially we have 0.42 moles of CO and 0.42 moles of H₂
If 0.29 moles of CO remained, (0.42 - 0.29) = 0.13 moles have reacted.
So in the equilibrium we may have:
0.29 moles of CO, and (0.42 - 0.13 . 2) = 0.16 moles of H₂
Ratio is 1:2, if 0.13 moles of CO haved reacted, (0.13 . 2) moles have reacted of hydrogen
Finally 0.13 moles of methanol, are found after the equilibrium reach the end.
Let's make expression for KC: [Methanol] / [CO] . [Hydrogen]²
0.13 / (0.29 . 0.16²)
Kc = 17.5
All of the following are characteristics of metals except: Group of answer choices good conductors of heat malleable ductile often lustrous tend to gain electrons in chemical reactions
Answer:
Hence the correct option is the last option that is tends to gain electrons in chemical reactions to become anions.
Explanation:
Metals tend to donate electrons in chemical reactions to become cations.
State what would be observed when the following pairs of reagents are mixed in a test tube.
C6H2COOH and Na2CO3(aq)
(ii) CH3CH2CH2OH and KMnO4 /H
(iii) CH3CH2OH and CH3COOH + conc. H2SO4 (iv) CH3CH = CHCH3 and Br2 /H2O
Answer:
(i). C6H2COOH and Na2CO3(aq)
observation: Bubbles of a colourless gas (carbon dioxide gas)
(ii) CH3CH2CH2OH and KMnO4 /H
observation: The orange solution turns green.
[This is because oxidation of propanol to propanoic acid occurs]
(iii) CH3CH2OH and CH3COOH + conc. H2SO4
observation: A sweet fruity smell is formed.
[This is because an ester, diethylether is formed]
(iv) CH3CH = CHCH3 and Br2 /H2O
observation: a brown solution is formed.
1.Q= {n: 7 <n<31}, list the members of the set Q
Q={x:x[tex]\epsilon[/tex]n,7<n<31}
[tex]\\ \sf\longmapsto Q=\left\{8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30\right\}[/tex]
You can write it like this too
[tex]\\ \sf\longmapsto Q=\left\{8,9......30,31\right\}[/tex]
Balance the redox reaction Al(s) + MnO4^- (aq) --> MnO2 (s) + Al(OH)4^- (aq) in aqueous basic solution
Answer:
Al + MnO4- + 2H2O → Al(OH)4- + MnO2
Explanation:
First of all, we out down the skeleton equation;
Al + MnO4- → MnO2 + Al(OH)4-
Secondly, we write the oxidation and reduction equation in basic medium;
Oxidation half equation:Al + 4H2O + 4OH- → Al(OH)4- + 4H2O + 3e-
Reduction half equation:MnO4- + 4H2O + 3e- → MnO2 + 2H2O + 4OH-
Thirdly, we add the two half reactions together to obtain:
Al + MnO4- + 8H2O + 4OH- + 3e- → Al(OH)4- + MnO2 + 6H2O + 3e- + 4OH-
Lastly, cancel out species that occur on both sides of the reaction equation;
Al + MnO4- + 8H2O→ Al(OH)4- + MnO2 + 6H2O
The simplified equation now becomes;
Al + MnO4- + 2H2O → Al(OH)4- + MnO2
Suppose a 250.mL flask is filled with 1.7mol of H2 and 0.90mol of I2. The following reaction becomes possible:
+H2gI2g 2HIg
The equilibrium constant K for this reaction is 5.51 at the temperature of the flask.
Calculate the equilibrium molarity of I2. Round your answer to two decimal places.
Explanation:
here's the answer. I just plug the expression into my calculator and find the intercept to avoid the quadratic formula
Why do we need Chemistry in Nursing?
Answer:
We need chemistry in nursing because it deals with various kinds of drugs and the reactions of these drugs on the human body as well as with each other.
Calculate the mass of isoborneol in 2.5 mmol of isoborneol and the theoretical yield (in grams) of camphor from that amount of isoborneol
isoborneol = 154.25 g mol?1
Camphor, Molar mass = 152.23 g/mol
Answer:
[tex]m_{isoborneol }=0.39g\\\\m_{Camphor}=0.38g\\[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to infer that the reaction whereby isoborneol goes to camphor occurs in a 1:1 mole ratio, that is why the theoretical yield of the latter is also 2.5 mmol (0.0025 mol) but the masses can be calculated as follows:
[tex]m_{isoborneol }=0.0025mol*\frac{154.25g}{1mol} =0.39g\\\\m_{Camphor}=0.0025mol*\frac{152.23 g}{1mol} =0.38g\\[/tex]
Because of the fact this is a rearrangement reaction whereas the number of atoms is not significantly modified.
Regards!
What is represented by a straight line on a graph?
o the sum of the independent and dependent variables
O only the independent variable
O only the dependent variable
o the relationship between independent and dependent variable
1 2
3
4
5
Answer:
the relationship between independent and dependent variable
Explanation:
A straight line or linear graph is one of the ways to represent a given data. It shows the relationship between two given set of data; one called the independent variable is plotted on the x-axis (horizontal) while the other called the dependent variable is plotted on the y-axis (vertical).
The straighter the line is, the stronger the relationship between the two variables and vice versa. Hence, the straight line in the graph represents the relationship between independent and dependent variable.
At elevated temperatures, hydrogen iodide may decompose to form hydrogen gas and iodine gas, as follows:
2HI(g) ⇌ H2 (g) + I2 (g)
In a particular experiment, the concentrations at equilibrium were measured to be [HI] = 0.85 mol/L, [I2] = 0.60 mol/L, and [H2] = 0.27 mol/L. What is Kc for the above reaction?
Explanation:
Since Kc is
[tex]k = \frac{(products)}{(reactants)} [/tex]
You can insert the Hydrogen and Iodine gas on top, and Hydrogen Iodide in the denominator.
Note: you can only include gases and aqueous species in an equilibrium expression, and all the species in this reaction are gaseous so you're good.
Inserting their molarity at equilibrium into their places, and you can solve. Don't forget to make the coefficient of HI turn into a power.
Avogradro's number is the number of particles in one gram of carbon- 12 atom true or false?
Answer:
True
Explanation:
The value of the mole is equal to the number of atoms in exactly 12 grams of pure carbon-12. 12.00 g C-12 = 1 mol C-12 atoms = 6.022 × 1023 atoms • The number of particles in 1 mole is called Avogadro's Number (6.0221421 x 1023).
A 1 liter solution contains 0.370 M hypochlorous acid and 0.493 M sodium hypochlorite. Addition of 0.092 moles of barium hydroxide will: (Assume that the volume does not change upon the addition of barium hydroxide.)
In the original solution you have the mixture of a weak acid (Hypochlorous acid) and its conjugate base (Sodium hypochlorite). That is a buffer.
The barium hydroxide will react with hypochlorous acid. If this reaction cause the complete reaction of hypochlorous acid, the buffer break its capacity and the pH change in several units. In this case:
The addition of barium hydroxide will raise the pH slightly because the buffer still working.
The initial moles of those species are:
Hypochlorous acid:
[tex]1L * \frac{0.370mol}{1L} = 0.370 moles[/tex]
Sodium hypochlorite:
[tex]1L * \frac{0.493mol}{1L} = 0.493 moles[/tex]
Now, a strong acid as barium hydroxide (Ba(OH)₂) reacts with a weak acid as hypochlorous acid (HClO) as follows:
Ba(OH)₂ + 2HClO → Ba(ClO)₂ + 2H₂O
For a complete reaction of 0.092 moles of barium hydroxide are required:
[tex]0.092 moles Ba(OH)_2*\frac{2mol HClO}{1molBa(OH)_2} = 0.184 moles HClO[/tex]
As there are 0.370 moles, the moles of HClO after the reaction are:
0.370 moles - 0.184 moles = 0.186 moles of HClO will remain
As you still have hypochlorite and hypochlorous acid you still have a buffer.
Thus, the pH will raise slightly because the amount of acid is decreasing and slightly because the buffer can keep the pH.
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e. Which of the following is a mixture? i. Water ii. Hydrogen iii. Air iv. Iron
water is known as the mixture
Answer:
iv. Iron
water is not a mixture
hydrogen is the simplest element
air is pure
When 1 mole of CO(g) reacts with H2O(l) to form CO2(g) and H2(g) according to the following equation, 2.80 kJ of energy are absorbed. CO(g) + H2O(l)CO2(g) + H2(g) Is this reaction endothermic or exothermic? _________ What is the value of q? kJ
why does D2O
have a higher boiling point than H2O
Answer:
deuterium is heavier isotope of hydrogen need
high weight of atom will lead to higher boiling point
Which of the following is a reduction half-reaction?
Solution : An oxidation reduction (redox) reaction is a type of chemical reaction that involves a transfer of electrons between tow species an oxidaion reductin reaction is any chemical reaction in which the oxidation number of a molecule atom or ion changes by gaining or losing an electron
Arrange the following in order of increasing melting point: NaCl, H2O, CH4, C6H4(OH)2.
a. NaCl < H2O < CH4 < C6H4(OH)2
b. CH4 < H2O < NaCl < C6H4(OH)2
c. CH4 < H2O < C6H4(OH)2 < NaCl
d. CH4 < C6H4(OH)2 < H2O < NaCl
e. CH4 < NaCl < C6H4(OH)2 < H2O
Explanation:
one thing to know is that higher surface area = higher boiling point.
NaCl has the smallest surface area, so it's the first one.
H2O has less surface area than methane, so it's second.
Methane has more surface area than H20, so it's third.
The big molecule has the most surface area, so it's last
Temperature measures the average kinetic energy of particles of the substances. Melting point is directly proportional to surface area. Therefore, the correct option is option C.
What is temperature?Temperature is used to measure degree or intensity of heat of a particular substance. Temperature is measured by an instrument called thermometer.
Temperature can be measured in degree Celsius °c, Kelvin k or in Fahrenheit. Temperature is a physical quantity. Heat always flow from higher temperature source to lower temperature source.
We can convert these units of temperature into one another. The relationship between degree Celsius and Fahrenheit can be expressed as:
°C={5(°F-32)}÷9
Melting point is directly proportional to surface area. NaCl has the smallest surface area. Water has less surface area than methane. Methane has more surface area than H[tex]_2[/tex]O.
Therefore, the correct option is option C.
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Given the equation: 2C6H10(l) 17 O2(g) ---> 12 CO2(g) 10 H2O(g) MM( g/mol): 82 32 44 18 If 115 g of C6H10 reacts with 199 g of O2 and 49 g of H2O are formed, what is the percent yield of the reaction
Answer:
74%
Explanation:
Step 1: Write the balanced equation
2 C₆H₁₀(l) + 17 O₂(g) ⇒ 12 CO₂(g) + 10 H₂O(g)
Step 2: Determine the limiting reactant
The theoretical mass ratio (TMR) of C₆H₁₀ to O₂ is 164:544 = 0.301:1.
The experimental mass ratio (EMR) of C₆H₁₀ to O₂ is 115:199 = 0.578:1.
Since EMR > TMR, the limiting reactant is O₂.
Step 3: Calculate the theoretical yield of H₂O
The theoretical mass ratio of O₂ to H₂O 544:180.
199 g O₂ × 180 g H₂O/544 g O₂ = 65.8 g H₂O
Step 4: Calculate the percent yield of H₂O
%yield = (experimental yield/theoretical yield) × 100%
%yield = (49 g/65.8 g) × 100% = 74%
Answer:
Percentage yield of H₂O = 74.24%
Explanation:
The balanced equation for the reaction is given below:
2C₆H₁₀ + 17O₂ —> 12CO₂ + 10H₂O
Next, we shall determine the masses of C₆H₁₀ and O₂ that reacted and the mass of H₂O produced from the balanced equation. This is can be obtained as follow:
Molar mass of C₆H₁₀ = 82 g/mol
Mass of C₆H₁₀ from the balanced equation = 2 × 82 = 164 g
Molar mass of O₂ = 32 g/mol
Mass of O₂ from the balanced equation = 17 × 32 = 544 g
Molar mass of H₂O = 18 g/mol
Mass of H₂O from the balanced equation = 10 × 18 = 180 g
SUMMARY:
From the balanced equation above,
164 g of C₆H₁₀ reacted with 544 g of O₂ to produce 180 g of H₂O.
Next, we shall determine the limiting reactant. This can be obtained as follow:
From the balanced equation above,
164 g of C₆H₁₀ reacted with 544 g of O₂.
Therefore, 115 g of C₆H₁₀ will react to produce = (115 × 544)/164 = 381 g of O₂.
From the calculations made above, we can see that a higher mass (i.e 381 g) of O₂ than what was given (i.e 199 g) is needed to react with 115 g of C₆H₁₀.
Therefore, O₂ is the limiting reactant and C₆H₁₀ is the excess reactant.
Next, we shall determine the theoretical yield of H₂O. This can be obtained by using the limiting reactant as shown below:
From the balanced equation above,
544 g of O₂ reacted to produce 180 g of H₂O.
Therefore, 199 g of O₂ will react to produce = (199 × 180)/544 = 66 g of H₂O.
Thus, the theoretical yield of H₂O is 66 g.
Finally, we shall determine the percentage yield. This can be obtained as follow:
Actual yield of H₂O = 49 g
Theoretical yield of H₂O = 66 g
Percentage yield of H₂O =?
Percentage yield = Actual yield /Theoretical yield × 100
Percentage yield of H₂O = 49/66 × 100
Percentage yield of H₂O = 74.24%
what is the complete ionic equation between the reaction of sodium carbonate and calcium hydroxide to form a precipitate
Answer:
CaCl2(aq) + Na2CO3(aq) → CaCO3(s) + 2NaCl(aq)
Reference:
H.W. Hanna and A. Dittmar; Laboratory Manual for General Chemistry, 4th ed.; Morton Publishing Company, Denver, Colorado
A mixture of methane and carbon dioxide gases contains methane at a partial pressure of 431 mm Hg and carbon dioxide at a
partial pressure of 504 mm Hg. What is the mole fraction of each gas in the mixture?
XCHA
Xc02
Answer:
XCH₄ = 0.461
XCO₂ = 0.539
Explanation:
Step 1: Given data
Partial pressure of methane (pCH₄): 431 mmHgPartial pressure of carbon dioxide (pCO₂): 504 mmHgStep 2: Calculate the total pressure in the container
We will sum both partial pressures.
P = pCH₄ + pCO₂
P = 431 mmHg + 504 mmHg = 935 mmHg
Step 3: Calculate the mole fraction of each gas
We will use the following expression.
Xi = pi / P
XCH₄ = pCH₄/P = 431 mmHg/935 mmHg = 0.461
XCO₂ = pCO₂/P = 504 mmHg/935 mmHg = 0.539
Select all that are true for both a voltaic cell and an electrolytic cell. Group of answer choices There must be a complete circuit to allow electrons to flow. The redox reaction is spontaneous. Oxidation happens at the anode. Reduction happens at the cathode. The redox reaction is non-spontaneous. g
Answer:
There must be a complete circuit to allow electrons to flow. Oxidation happens at the anode. Reduction happens at the cathode.
Explanation:
Select all that are true for both a voltaic cell and an electrolytic cell.
There must be a complete circuit to allow electrons to flow. YES. This is true for both cells.
The redox reaction is spontaneous. NO. This is only true for the voltaic cell.
Oxidation happens at the anode. YES. This is true for both cells.
Reduction happens at the cathode. YES. This is true for both cells.
The redox reaction is non-spontaneous. NO. This is only true for the electrolytic cell.
The statements that have been true for both voltaic and electrolytic cells are:
Option A: There must be a complete circuit to allow electrons to flow.
Option C: Oxidation happens at the anode.
Option D: Reduction happens at the cathode.
The electrolytic cells are the one that converts electrical energy into chemical energy. Voltaic cells convert chemical energy into electrical energy.
From the given options, the best-suited option for both electrolytic and voltaic cells are:
For any of the cells to perform the current flow, the circuit must be complete. Option A stating that there must be a complete circuit to allow electrons to flow is correct.The redox reaction has been a spontaneous reaction and results in the chemical conversion to form energy. Since the chemical energy has been converted into electrical energy, statement B has been true for only Voltic cells. Thus option B is incorrect.Oxidation is the reaction to the loss of electrons. In both cells, oxidation happens at the anode. Thus, option C is correct.Since oxidation takes place at anode, cathode acts as the reduction end in both the cells. Thus option D is correct.The redox reaction has been non-spontaneous only in the electrolytic cell. Thus option E is incorrect.The statements that have been true for both voltaic and electrolytic cells are, Option A, C, and D.
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Determine the rate of reaction based on the data below:
75-
rate of reaction = y
hydrogen produced (cm)
х
y
25
50
75
100
125
time (seconds)
Answer:
75
Explanation:
is the answer for your question
