Answer:
Mf = MiVi / Vf
Explanation:
To obtain the answer to the question given, let us obtain the equation for the final concentration.
This can be obtained from the dilution equation as shown below :
MiVi = MfVf
Mi is the initial concentration
Mf is the final concentration
Vi is the initial volume
Vf is the final volume.
Making Mf the subject of the above equation, we can obtain the equation which can be used to calculate the final concentration. This is illustrated below:
MiVi = MfVf
Divide both side by Vf
Mf = MiVi / Vf
Therefore, the equation which can be used to calculate the final concentration is Mf = MiVi / Vf
The first part to this question is:
What is the final volume? 150.0
the second part:
Which equation should you use to find the final concentration? 3rd option
The 3rd part:
What is the new concentration? 0.125 M NaCl
Explanation:
On edg.. Good Luck!!
At 500.K and 1.00 atm pressure, 1.5 liters of pentane gas, C5H12, is mixed with 15 liters of oxygen gas. A complete combustion results. How many liters of WATER vapor, measured at the same temperature and pressure, would be produced after the reaction?
Answer:
1.5 litre of C5H12 produce 9 litre of water vapour;
Explanation:
First balance the combustion reaction;
[tex]C_5H_12 +8O_2[/tex] → [tex]5CO_2 +6H_2O[/tex]
from the balance reacion it is clearly that,
one mole of C5H12 reacts completely with 8 mole of O2 gas;
Hence
one litre of C5H12 reacts completely with 8 litre of O2 gas;
there fore 1.5 litre C5H12 needs 12 litre oxygen gas but we have 15 litre.
so,
C5H12 is a limiing reactant and O2 is a excess reactant.
so quantity of H2O depends on limiting reactant;
one litre of C5H12 produce 6 litre of water vapour;
therefore,
1.5 litre of C5H12 produce 9 litre of water vapour;
When 3.8 g of aluminum are reacted with excess chlorine in the lab, you find you have made 17.8 g of aluminum chloride. What is the percent yield of this reaction?
Answer:
94.8% is the percentage yield of this reaction.
Explanation:
Mass of aluminium = 3.8 g
Moles of aluminium = [tex]\frac{3.8 g}{27 g/mol}=0.1407 mol[/tex]
[tex]2Al+3Cl_2\rightarrow 2AlCl_3[/tex]
According to reaction, 2 moles of aluminum gives 2 moles of aluminium chloride, then 0.1407 moles of aluminium will give:
[tex]\frac{2}{2}\times 0.1407 mol=0.01407 mol[/tex] aluminium chloride
Mass of 0.1407 moles of aluminum chloride:
= 0.1407 mol × 133.5 g/mol = 18.78 g
Theoretical yield of aluminum chloride = 18.78 g
Experimental yield of aluminum chloride = 17.8 g
The percentage yield of reaction:
[tex]=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100[/tex]
[tex]=\frac{17.8 g}{18.78 g}\times 100=94.8\%[/tex]
94.8% is the percentage yield of this reaction.
How many moles of oxygen (O2) are present in 33.6 L of the gas at 1 atm and 0°C?
1.5
O2
22.4
0 32
Mark this and retum
Answer:
The answer is 1.5 moles.
Explanation:
An ideal gas is a theoretical gas that is considered to be composed of point particles that move randomly and do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.
An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law:
P * V = n * R * T
where n is the number of moles and R is the molar constant of the gases.
In this case:
P= 1 atmV= 33.6 Ln= ?R= 0.082 [tex]\frac{atm*L}{mol K}[/tex]T= 0°C= 273 °KReplacing:
1 atm*33.6 L= n* 0.082 [tex]\frac{atm*L}{mol K}[/tex] *273 °K
Solving:
[tex]n=\frac{1 atm* 33.6 L}{0.082\frac{atm*L}{mol K}*273K }[/tex]
n= 1.5 moles
So, the answer is 1.5 moles.
Answer:
1.5 moles
Explanation:
Which image best represents a solution of distilled water at pH = 7?
Answer:
It's the middle one, B. H+ and OH-
Explanation:
I just did the question online
A 500.0 g sample of Al2(SO4)3 is reacted with 450.0 g of Ca(OH)2. A total of 596 g of CaSO4 is produced. What is the limiting reagent in this reaction, and how many moles of excess reagent are unreacted? Al2(SO4)3(aq) + 3Ca(OH)2(aq) -> 2Al(OH)3(s) + 3CaSO4(s)
Answer : The limiting reagent in this reaction is, [tex]Al_2(SO_4)_3[/tex] and number of moles of excess reagent is, 1.69 moles
Explanation : Given,
Mass of [tex]Al_2(SO_4)_3[/tex] = 500.0 g
Mass of [tex]Ca(OH)_2[/tex] = 450.0 g
Molar mass of [tex]Al_2(SO_4)_3[/tex] = 342.15 g/mol
Molar mass of [tex]Ca(OH)_2[/tex] = 74.1 g/mol
First we have to calculate the moles of [tex]Al_2(SO_4)_3[/tex] and [tex]Ca(OH)_2[/tex].
[tex]\text{Moles of }Al_2(SO_4)_3=\frac{\text{Given mass }Al_2(SO_4)_3}{\text{Molar mass }Al_2(SO_4)_3}[/tex]
[tex]\text{Moles of }Al_2(SO_4)_3=\frac{500.0g}{342.15g/mol}=1.461mol[/tex]
and,
[tex]\text{Moles of }Ca(OH)_2=\frac{\text{Given mass }Ca(OH)_2}{\text{Molar mass }Ca(OH)_2}[/tex]
[tex]\text{Moles of }Ca(OH)_2=\frac{450.0g}{74.1g/mol}=6.073mol[/tex]
Now we have to calculate the limiting and excess reagent.
The given chemical reaction is:
[tex]Al_2(SO_4)_3(aq)+3Ca(OH)_2(aq)\rightarrow 2Al(OH)_3(s)+3CaSO_4(s)[/tex]
From the balanced reaction we conclude that
As, 1 mole of [tex]Al_2(SO_4)_3[/tex] react with 3 mole of [tex]Ca(OH)_2[/tex]
So, 1.461 moles of [tex]Al_2(SO_4)_3[/tex] react with [tex]1.461\times 3=4.383[/tex] moles of [tex]Ca(OH)_2[/tex]
From this we conclude that, [tex]Ca(OH)_2[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]Al_2(SO_4)_3[/tex] is a limiting reagent and it limits the formation of product.
Number of moles of excess reagent = 6.073 - 4.383 = 1.69 moles
Therefore, the limiting reagent in this reaction is, [tex]Al_2(SO_4)_3[/tex] and number of moles of excess reagent is, 1.69 moles
What are the 2 oppositional relationships?
Answer:
Ecological Relationships are of two types - Oppositional Relationships and Symbiotic Relationships. Oppositional Relationships are of two types - Predation and Competititon. Symbiotic Relationships are of four types - Mutualism, Commensalism, Amensalism & Parasitism.
Explanation:
Potassium hydroxide is very soluble in water, resulting in extremely basic solutions. A 121g sample KOH is dissolved in water at 25∘C to make up 100.0mL of solution. The molar mass of KOH is 56.11gmol. What is the pH of the solution at 25.0∘C?
Answer:
15.33
Explanation:
Given parameters:
Mass of KOH = 121g
Volume of solution = 100mL = 0.1dm³
Molar mass of KOH = 56.11g/mol
Unknown:
pH of the solution = ?
Solution:
To find the pH, we must first know the concentration of the solution obtained by mixing KOH up to this volume.
This can be done by finding the molarity of the solution.
Molarity = [tex]\frac{number of moles }{volume of solution}[/tex]
Number of moles of KOH = [tex]\frac{mass}{molar mass}[/tex] = [tex]\frac{121}{56.11}[/tex] = 2.16mole
Input parameters;
Molarity of solution = [tex]\frac{2.16}{0.1}[/tex] = 21.6moldm⁻³
KOH → OH⁻ + K⁺
21.6 21.6 21.6
In the solution we have 21.6moldm⁻³ of OH⁻ which is need to find the pH;
pOH = -log₁₀(OH⁻)
pOH = -log₁₀21.6
pOH = -1.33
Since pH + pOH = 14
pH = 14 - (-1.33) = 15.33
Diluting sulfuric acid with water is highly exothermic: (a) Use Appendix B to find for diluting 1.00 mol of H2SO4(l) (d = 1.83 g/mL) to 1 L of 1.00 M H2SO4(aq) (d = 1.060 g/mL). (b) Suppose you carry out the dilution in a calorimeter. The initial T is 25.0°C, and the specific heat capacity of the final solution is 3.50 J/g·K. What is the final T? (c) Use the ideas of density and heat capacity to explain why you should add acid to water rather than water to acid.
Answer:
[tex]H_{rxn} = -93.52 \frac{KJ}{mol}[/tex]
The final temperature [tex]T_{2} = 50.21[/tex] ° c
Explanation:
(a).
[tex]H_{rxn} = H_{products} - H_{reactents}[/tex]
[tex]H_{reactants}[/tex] = - 813.9 [tex]\frac{KJ}{mol}[/tex]
[tex]H_{Products}[/tex] = - 907.51 [tex]\frac{KJ}{mol}[/tex]
[tex]H_{rxn} = -907.51 + 813.9[/tex]
[tex]H_{rxn} = -93.52 \frac{KJ}{mol}[/tex]
(b).
[tex]H_{rxn} = -93.52 \frac{KJ}{mol}[/tex]
E = 93520 J
C = 3.5 [tex]\frac{KJ}{kg K}[/tex]
Initial temperature [tex]T_{1} = 25[/tex] ° c
Mass (m) = density × volume
m = 1060 × 1
m = 1060 gm
[tex]E = m C (T_{2} - T_{1} )[/tex]
93520 =1060 × 3.5 × ( [tex]T_{2}[/tex] - 298 )
[tex]T_{2} = 50.21[/tex] ° c
This is the final temperature.
(c).
The density of sulfuric acid is more than the water. so when water is added to acid the mixing process does not takes properly. So we have to add acid in to the water.
Adding water is exothermic process so when we add water to acid than in that case more energy produces and wasted.
Vinegar is a commercial form of acetic acid, HC2H3O2 (aq). One sample vinegar has a pH value of 2.4. State the pH value of a sample that has ten times fewer hydronium ions than an equal volume of a vinegar sample with a pH value of 2.4.
3.41 is the pH value of a sample that has ten times fewer hydronium ions than an equal volume of a vinegar sample with a pH value of 2.4.
Explanation:
pH of the first sample of acetic acid = 2.4
to know the [H+] concentration in the acetic acid solution, the equation used is:
pH = -log [H+]
[H+] = [tex]10^{-pH}[/tex]
[H+] = [tex]10^{-2.4}[/tex]
[H+] = 3.98 X [tex]10^{-3}[/tex] M
The concentration of H+ ion in first case is 3.98 X [tex]10^{-3}[/tex] M, the second sample has ten times less hydronium ion so concentration of first case is divided by 10.
3.98 x [tex]10^{-4}[/tex] M
Now pH of the sample having 10 times fewer ions of acetic acid:
pH = -log [H+]
putting the values in the above equation:
pH = -log [3.98 x [tex]10^{-4}[/tex] M]
pH = 3.41
If solution has ten times less hydronium ion the pH will change to 3.41.
3.41 is the pH value of a sample that has ten times fewer hydronium ions. pH gives the concentration of Hydronium ion.
What information we have:
pH of the first sample of acetic acid = 2.4
Calculation for [H+] concentration:To know the [H+] concentration in the acetic acid solution, the equation used is:
[tex]pH = -log [H^+]\\\\H^+ = 10^{-pH}\\\\H^+ = 10^{-2.4}\\\\H^+ = 3.98 *10^{-4} M[/tex]
The concentration of H+ ion in first case is [tex]3.98 *10^{-4} M[/tex], the second sample has ten times less hydronium ion so the concentration of first case is divided by 10.
[tex]3.98 *10^{-4} M[/tex],
Now pH of the sample having 10 times fewer ions of acetic acid:
pH = -log [H+]
pH = -log [ [tex]3.98 *10^{-4} M[/tex],]
pH = 3.41
If the solution has ten times less hydronium ion the pH will change to 3.41.
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How much space would .33 moles of oxygen take up at STP?
Answer:
0.33 moles of Oxygen at stp would occupy a volume of 7.392 dm³
Explanation:
1 mole of every gas at standard temperature and pressure (stp) occupies 22.4 dm³
0.33 moles of Oxygen at stp would occupy a volume of 0.33 × 22.4 dm³ = 7.392 dm³
Hope this Helps!!!
Answer:
[tex]V = 7.391\,L[/tex]
Explanation:
Let suppose that oxygen behaves ideally. The equation of state of the ideal gas is:
[tex]P\cdot V = n\cdot R_{u}\cdot T[/tex]
The volume is now cleared:
[tex]V = \frac{n\cdot R_{u}\cdot T}{P}[/tex]
[tex]V = \frac{(0.33\,mol)\cdot \left(0.082\,\frac{atm\cdot L}{mol\cdot K} \right)\cdot (273.15\,K)}{1\,atm}[/tex]
[tex]V = 7.391\,L[/tex]
Place the capillary tube, still attached
to the ruler, in the water bath.
600 ml
x
com
COMPLETE
Temperature of gas: 44
X = 3 °C Height of the column of gas: 400
= 5.7 cm
Volume of gas (V = serh):
Temperature of gas:
cm3 DONE
Intro
Answer:
Temperature of gas: 3 °C Height of the column of gas: 5.7 cm Temperature of gas: 276k Volume of gas: .72 cm^3
Explanation:
edge 2021
The value of Temperature is 276K
and of Volume is 0.71592 [tex]cm^{3}[/tex]
How to calculate temperature?As per the formula of conversion of degree Celsius to kelvin temperature
1°C=273K
∴23 °c +273= 276 K
How to calculate volume?V=π[tex]r^{2}[/tex]h
V= volume
r = radius of capillary tube =0.2 cm
h=height of column=5.7 cm
∴V=3.14×[tex](0.2)^{2}[/tex]×5.7
∴V=0.71592[tex]cm^{3}[/tex]
Hence, the value of temperature and volume are 276K and 0.71592[tex]cm^{3}[/tex], respectively.
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What is the molarity of a HCl solution that contains 13.3 g of HCl in 447 mL of solution? Answer in units of M
Answer:
0.81M
Explanation:
Molarity, M = number of moles, n / Volume, V
number of moles of HCl, n=
mass of HCl / molar mass of HCl
Mass of HCl = 13.3g
Molar mass of HCl = 1 + 35.5 = 36.5g/mol
n = 13.3/36.5=0.364mol
V = 447mL=447/1000 L = 0.447L
Note that M is mol/L, hence, we convert 447mL to L. 1 mL = 1/1000 L
M=0.364mol/0.447L
M= 0.81M
Therefore, the molarity of the HCl solution is 0.81M
19. A sample of neon occupies a volume of 461 mL at STP.
What will be the
volume of the neon when the pressure is reduced to 93.3 kPa?
Answer:
500.65mL
Explanation:
The following information were obtained from the question:
V1 (initial volume) = 461 mL
P1 (initial pressure) = stp = 101325Pa
P2 (final pressure) = 93.3 kPa
Recall: 1KPa = 1000Pa
Therefore, 93.3 kPa = 93.3x1000 = 93300Pa
V2 (final volume) =?
Using the Boyle's law equation P1V1 = P2V2, the final volume of the gas can be obtained as follow:
P1V1 = P2V2
461 x 101325 = 93300 x V2
Divide both side by 93300
V2 = (461 x 101325)/93300
V2 = 500.65mL
Therefore, the volume of Neon at 93.3 kPa is 500.65mL
an aqueous solution is 65% (v/v) rubbing alcohol. How many millilitres of water are in a 95-ml sample of this solution?
Answer:
The amount of water in a 95 mL sample of the solution is 33.25 mL
Explanation:
65% (v/v) rubbing alcohol is a solution containing 65 mL rubbing alcohol in 100 mL solution of the rubbing alcohol and water
Therefore, 95 mL sample of the solution will contain x mL of rubbing alcohol;
x = 65% × 95 = 0.65 × 95 = 61.75 mL rubbing alcohol and therefore, the amount of water contained = 95 - 61.75 = 33.25 mL
The amount of water in a 95 mL sample of the solution = 33.25 mL.
The volume of water present in 95 ml of 65 % rubbing alcohol solution has been 33.25 ml.
The v/v has been the concentration representation of the volume of solute present in the total volume of the solution.
The given aqueous solution has 65 % v/v concentration of the rubbing alcohol. This states that there has been a presence of 65 ml of rubbing alcohol in 100 ml of the solution.
Since, in 100 ml sample there has been a presence of 65ml alcohol. The volume of water in 100 ml sample has been the remaining portion of the solution.
[tex]\rm Volume \;of \;water \;= \;Total \;volume - volume \;of \;alcohol\\Volume \;of\; water = 100 \;ml - 65 \;ml\\Volume \;of \;water = 35 \;ml[/tex]
The aqueous solution of rubbing alcohol has been consisted of 35 ml water per 100 ml solution.
Since, 100 ml solution = 35 ml water
[tex]\rm 95\;ml\;solution\;=\;\dfrac{35}{100}\;\times\;95\;ml\;water\\95\;ml\;solution=33.25\;ml\;water[/tex]
The volume of water present in 95 ml of 65 % rubbing alcohol solution has been 33.25 ml.
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A galvanic (voltaic) cell consists of an electrode composed of nickel in a 1.0 M nickel(II) ion solution and another electrode composed of copper in a 1.0 M copper(I) ion solution, connected by a salt bridge. Calculate the standard potential for this cell at 25 °C . Refer to the list of standard reduction potentials.
The standard potential for a galvanic cell consisting of nickel and copper electrodes at 25 °C is calculated using the standard reduction potentials for each electrode, resulting in a cell potential of +0.77 V.
To calculate the standard potential for the galvanic cell at 25 °C, you need to know the standard reduction potentials for the nickel and copper electrodes. These potentials can be found in a table of standard reduction potentials. The cell potential (Ecell) is the difference between the reduction potential of the cathode and the anode.
In our case, for a cell composed of nickel and copper electrodes, the standard reduction potential for the nickel(II) ion to nickel reaction is approximately -0.25 V, and for the copper(I) ion to copper reaction, it is +0.52 V.
Considering that the copper electrode undergoes reduction (acts as cathode) and the nickel electrode undergoes oxidation (acts as anode), we use the following formula:
[tex]E_{cell} = E_{cathode} - E_{anode[/tex]
Therefore: Ecell = (+0.52 V) - (-0.25 V) = +0.52 V + 0.25 V = +0.77 V
The standard potential for this galvanic cell at 25 °C is +0.77 V.
You computed the specific heat for each of the four metals using the formula below:
Check all items that you used, either directly or indirectly, each time you did this computation.
Answer: All of them are right there’s no wrong answer
Explanation:
Masses for the metal and the water in the calorimeter, temperature changes for the water and the metal, and the known specific heat of the water.
First, the masses of both the metal and the water within the calorimeter are crucial, as they determine the amount of substance undergoing the temperature change. Second, the temperature changes (Tw, final −Tw, initial and T metal, final −T metal, initial ) are necessary to quantify the heat exchange during the experiment. Lastly, the known specific heat of water (C water ) plays a role in the calculation, as it is a fundamental constant representing the amount of heat needed to raise the temperature of water.
The first law of thermodynamics, which states that the heat lost by the metal equals the heat gained by the water, and the ability of heat to flow from a hot object to a cooler one are overarching principles guiding the experimental setup and the interpretation of results. However, these principles are not directly inputted into the formula but are fundamental concepts in calorimetry and thermodynamics that inform the experimental design and the interpretation of the calculated specific heat values.
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Why are fatty acids generally composed of an even number of carbons? Why are fatty acids generally composed of an even number of carbons? Fatty acids have an even number of carbons because they are the most stable. Fatty acids have an even number of carbons because they are synthesized in basic media. Fatty acids have an even number of carbons because they are synthesized from a starting material (acetyl-CoACoA) that has an even number of carbons. Fatty acids have an even number of carbons because they are synthesized in acidic media.
Answer:
Fatty acids have an even number of carbons because they are synthesized from a starting material (acetyl-CoACoA) that has an even number of carbons.
Explanation:
Fatty acids are carboxylic acids and are known to have even number of carbon atoms because they are synthesized from a 2 carbon atom acetyl Co-A molecules which are assembled together. It contains a carboxylate group covalently joined with an hydrophobic head of CH3-(CH2)n and may have an unsatireated CH=CH group within its CH2 chain. The synthesis of fatty acids from acetyly Co-A involves the activation reaction of an enzyme acetyl Co-A synthetase with an acid. It involves dehydrogebation, hydration, oxidation and thiolysis. Fatty acids vary in length of its chain, the number of carbon-carbon double bonds attached to its CH2 chain nd also the location of those double bonds in the carbon chain. Saturated fatty acids are those without carbon-carbon double bond in its chain while unsaturated have carbon- carbon double bonds. Monounsaturated fatty acids have one carbon-carbon double bonds and polyunsaturated fatty acids have two or more carbon-carbon double bonds.
Fatty acids are the building blocks of fat molecules. The fat when broken down into simpler forms, yields fatty acids and glycerol.
The correct answer is:
Option C. Fatty acids have an even number of carbons because they are synthesized from a starting material (acetyl-CoACoA) that has an even number of carbons.
The fatty acids are generally even numbers because of the even precursor.
The metabolism of fatty acids starts with the two molecules of the acetyl coenzyme A, which on assembling yields the even-numbered of fatty acids.
The acetyl coenzyme A is an even-numbered two-carbon molecule, which undergoes enzyme-catalyzed reactions to yield an even number of fatty acids.
The even number of fatty acids can be saturated (made up of single bonds) or can be unsaturated (presence of double or triple bonds).
Therefore, Option C is correct.
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The volume of 62.0 mL of gas 175 K is increased to 82.9 mL at constant pressure. What is the final temperature of the gas?
The final temperature of the gas is 234K. As the volume of the increases to the given value, the temperature of the gas also increases.
What is Charle's law?Charle's law states that the volume of an ideal gas is directly proportional to the absolute temperature provided pressure is kept at constant.
It is expressed as;
V₁/T₁ = V₂/T₂
Given the data in the question;
Initial volume V₁ = 62.0mL = 0.062L
Initial temperature T₁ = 175K
Final volume V₂ = 82.9mL = 0.0829L
Final temperature T₂ = ?
To calculate the final temperature, we subtsitute our given values into the expression above.
V₁/T₁ = V₂/T₂
V₁T₂ = V₂T₁
T₂ = V₂T₁ / V₁
T₂ = ( 0.0829L × 175K ) / 0.062L
T₂ = 14.5075LK / 0.062L
T₂ = 234K
The final temperature of the gas is 234K. As the volume of the increases to the given value, the temperature of the gas also increases.
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A 30.5-g sample of an alloy at 93.0°C is placed into 50.0 g of water at 22.0°C in an insulated coffee-cup calorimeter with a heat capacity of 9.2 J/K. If the final temperature of the system is 31.1°C, what is the specific heat capacity of the alloy?
Answer:
The specific heat of alloy [tex]C_{alloy} = 1.007 \frac{KJ}{Kg K}[/tex]
Explanation:
Mass of the alloy = 30.5 gm = 0.0305 kg
Initial temperature = 93 °c = 366 K
Mass of water = 50 gm = 0.05 kg
Initial temperature = 22 °c = 295 K
Final temperature of the mixture = 31.1 °c = 304.1 K
From the energy conservation principal the heat lost by the alloy is equal to heat gain by the water.
Heat lost by alloy
[tex]Q_{alloy} = m C (T_{f}- T_{i} )[/tex]
[tex]Q_{alloy} = (0.0305) C_{alloy} (366-304.1)[/tex]
[tex]Q_{alloy} = (1.88795) C_{alloy}[/tex] ------- (1)
Heat gain by water
[tex]Q_{w} = (0.05) (4.18) (304.1 - 295)[/tex]
[tex]Q_{w} = 1.9019 \frac{KJ}{kg K}[/tex] ------- (2)
Equation (1) = Equation (2)
[tex](1.88795) C_{alloy} = 1.9019[/tex]
[tex]C_{alloy} = 1.007 \frac{KJ}{Kg K}[/tex]
This is the specific heat of alloy.
Taking into account the definition of calorimetry, the specific heat capacity of the alloy is 1.063 [tex]\frac{kJ }{kgK}[/tex]
In first place, calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.
In this way, between heat and temperature there is a direct proportional relationship.
The constant of proportionality depends on the substance that constitutes the body and its mass, and is the product of the specific heat by the mass of the body.
So, the equation that allows to calculate heat exchanges is:
Q = C× m× ΔT
where Q is the heat exchanged by a body of mass m, made up of a specific heat substance C and where ΔT is the temperature variation.
In this case, you know:
for alloy:
Calloy= ?m= 30.5 g= 0.0305 kg (being 1000 g= 1 kg)ΔT=Tfinal - Tinitial= 93 C - 31.1 C= 61.9 C= 61.9 K because it is a temperature difference, it has the same value in both unitsfor water:
Cwater= 4.18 [tex]\frac{kJ}{kgK}[/tex]m= 50 g= 0.050 kg (being 1000 g= 1 kg)ΔT=Tfinal - Tinitial= 31.1 C - 22 C= 9.2 C= 9.2 KReplacing in the expression to calculate heat exchanges:
for alloy:
Qalloy= Calloy× 0.0305 kg× 61.9 Kfor water:
Qwater= 4.18 [tex]\frac{kJ}{kgK}[/tex]× 0.050 kg× 9.2 KOn the other hand, the heat of the calorimeter can be expressed as:
Qcalorimeter= Ccalorimeter×ΔT
Being:
Ccalorimeter= 9.2 [tex]\frac{J}{K}[/tex]= 0.0092 [tex]\frac{kJ}{K}[/tex]ΔT=Tfinal - Tinitial= 31.1 C - 22 C= 9.2 C= 9.2 Kand replacing you get:
Qcalorimeter= 0.0092 [tex]\frac{kJ}{K}[/tex]× 9.2 K
It should be taken into account that a system at different temperatures evolves spontaneously towards a state of equilibrium in which all bodies have the same temperature. Then, mixing two quantities of liquids at different temperatures generates an energy transfer in the form of heat from the hottest to the coldest. Said energy transit is held until temperatures equalize, when it is said to have reached thermal equilibrium.
So the heat released by the sample is absorbed by the calorimeter and the water.
Qalloy= Qcalorimeter + Qwater
Replacing the corresponding expressions and solving:
Calloy× 0.0305 kg× 61.9 K= 0.0092 [tex]\frac{kJ}{K}[/tex]× 9.2 K + 4.18 [tex]\frac{kJ}{kgK}[/tex]× 0.050 kg× 9.2 K
Calloy× 1.88795 kg× K= 0.08464 kJ + 1.9228 kJ
Calloy× 1.88795 kg× K= 2.00744 kJ
[tex]Calloy=\frac{2.00744 kJ }{1.88795 kgK}[/tex]
Calloy= 1.063 [tex]\frac{kJ }{kgK}[/tex]
Finally, the specific heat capacity of the alloy is 1.063
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brainly.com/question/11586486?referrer=searchResultsbrainly.com/question/24724338?referrer=searchResultsWhich is not a characteristic of most nonmetals?
Their atoms lose electrons when bonding with metal atoms.
Their atoms gain electrons when bonding with metal atoms.
Their atoms share electrons with other nonmetal atoms.
They readily form compounds with other elements.
Most nonmetals do not lose electrons when bonding with metal atoms; instead, they tend to gain electrons, forming anions. They share electrons in covalent bonds with other nonmetals.
Explanation:The characteristic that is not typical of most nonmetals is their atoms lose electrons when bonding with metal atoms. When bonding occurs, nonmetals usually do the opposite; they tend to gain electrons to achieve more stable electron configurations. This is often seen in ionic bonds, where metals lose electrons to become positively charged cations, and nonmetals gain electrons to become negatively charged anions. However, in the case of bonding with other nonmetals, they typically share electrons, resulting in covalent bonds. Furthermore, nonmetals do indeed readily form compounds with other elements, either through ionic or covalent bonding.
Final answer:
Most nonmetals gain electrons when bonding with metal atoms to achieve stability, and they can also share electrons with other nonmetal atoms.
Explanation:
The characteristic that is not a characteristic of most nonmetals is that their atoms lose electrons when bonding with metal atoms. Nonmetals tend to gain electrons when bonding with metal atoms in order to achieve a stable electron configuration. They can also share electrons with other nonmetal atoms to form covalent bonds. Additionally, nonmetals readily form compounds with other elements.
The final solution had a volume of 1.0L and a molarity of 0.925. How many moles of Nickel (II) chloride were present in the solution? Show your work.
Answer:
0.925 mole
Explanation:
From the question given, the following were obtained:
Volume = 1.0L
Molarity = 0.925 M
Number of mole of Nickel (II) chloride =?
Molarity is simply defined as the mole of solute per unit litre of the solution.
It is represented mathematically as:
Molarity = mole /Volume
With the above equation, we can easily find the mole of Nickel (II) chloride present in the solution as follow:
Molarity = mole /Volume
0.925 = mole / 1
Mole = 0.925 x 1
Mole of Nickel (II) chloride = 0.925 mole
A container has a total pressure of 1439 mmHg. There are 3 gases in this container, one has a pressure of 523.3 mm Hg, and another has a pressure of 509.8 mm Hg. What is the pressure of the 3rd gas?
Answer:
P₃ = 594.1 mmHg
Explanation:
Given data:
Total pressure of container = 1439 mmHg
Partial pressure of 1st gas = 523.3 mmHg
Partial pressure of 2nd gas = 509.8 mmHg
Partial pressure of 3rd gas = ?
Solution:
According to Dalton law of partial pressure,
The total pressure inside container is equal to the sum of partial pressures of individual gases present in container.
Mathematical expression:
P(total) = P₁ + P₂ + P₃+ ............+Pₙ
Now we will solve this problem by using this law.
P(total) = P₁ + P₂ + P₃
439 mmHg = 523.3 mmHg + 509.8 mmHg + P₃
439 mmHg = 1033.1 mmHg + P₃
P₃ = 1033.1 mmHg -439 mmHg
P₃ = 594.1 mmHg
A solution is saturated at 25°C. It is then slowly cooled to 20°C with no change to the appearance of
the liquid. What term would be associated with this solution?
A.saturated
B.unsaturated
C.supersaturated
D.oversaturated
The term associated with the solution is saturated.
Explanation:The term associated with a solution that is saturated at 25°C and then slowly cooled to 20°C with no change to its appearance is saturated.
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On Venus, the atmospheric pressure is 2700 mmHg and nitrogen gas makes up 3.0% of the atmosphere. What is the partial pressure of nitrogen on Venus?
___ mmHg
Answer:
81mmHg
Explanation:
Since the total pressure is 2700mmHg and we are told that nitrogen occupies 3% of that pressure in the Venus atmosphere. We have to find the pressure of nitrogen using its percentage.
It should be noted that the total pressure of the atmosphere is the sum total of the partial pressures of all the respective gases in the atmosphere.
Hence partial pressure of nitrogen= 3.0%× 2700mmHg= 3.0/100× 2700
Partial pressure of nitrogen= 81mmHg
Which statement about a hypothesis is true
It is written in an “if…then” format.
It is an educated guess based on background research.
It includes a prediction regarding the dependent variable.
All these statements are true.
Answer:
All these statements are true.
Explanation:
Many describe it as an "educated guess," based on prior knowledge and observation. While this is true, the definition can be expanded. A hypothesis also includes an explanation of why the guess may be correct.
hope this helps :)
The true statement about hypothesis is: All these statements are true. The option (4) is correct.
A hypothesis is a testable statement that is formulated based on observations and background research.
It is typically written in an "if... then" form, which clearly states the relationship between two variables: the independent variable (the one at that the researcher manipulates) and the dependent variable (the one that the researcher measures).
A well-constructed hypothesis will also include a prediction regarding the dependent variable, which is an educated guess about the outcome of the experiment or study. Therefore, all the given statements about a hypothesis are true.
1. "It is written in an 'if...then' format." - This is true because a hypothesis often takes the form of ""If [independent variable], then [dependent variable]"" to clearly indicate the cause-and-effect relationship being tested.
2. "It is an educated guess based on background research." - This is true because a hypothesis is not just a random guess; it is formulated after conducting a thorough literature review and observing patterns or phenomena that suggest a certain relationship or outcome.
3. "It includes a prediction regarding the dependent variable." - This is true because a hypothesis predicts the effect that changing the independent variable will have on the dependent variable.
Since all three statements accurately describe the characteristics of a hypothesis, the correct answer is that all these statements are true.
The complete question is:
Which statement about a hypothesis is true?
1) It is written in an “if…then” format.
2) It is an educated guess based on background research.
3) It includes a prediction regarding the dependent variable.
4) All these statements are true.
How many milliliters of 1.0 M HCl are needed to exactly neutralize 15 milliliters of 2.0 M Ba(OH)2?
Answer:
OK, I don't know if this is right but this is my thinking of it. I hope this helps you out.
Let's do problem "a"
Step 1. Write a balanced equation.
2 HCl + Ba(OH)2 --------------> BaCl2 + 2 HOH
Step 2. Underneath the HCl and Ba(OH)2 in the equation write what you are given and what you need to find out.
2 HCl + Ba(OH)2 --------------> BaCl2 + 2 HOH
C = 0.130 mol/L v = 56 mL
V = ? C = 0.109 mol/L
Step 3. They ALWAYS provide a way to calculate the number of moles of one of the substances. In this case it is Ba(OH)2 because you have C and V
n = C X V n = 0.109 x 56 = 6.104 millimoles
Step 4. Use the equation to figure out how many moles of the unknown (HCl) you will need.
the equation says you need TWICE as many moles as you have of Ba(OH)2. So you will need 6.104 millimoles x 2 or 12.208 milimoles.
Step 5. Now that yo have moles and concentration of HCl you can now calculate volume
C = n / V, so V = n / C
12.208 millimoles / 0.130 mol/L = 93.9 mL
Use this method for all these kind of problems.
In B) you will have to convert he g of NaOH into moles by 0.240 / 40.00 g/mol and then following trhe rest of the procedure.
C) is done the same way. convert the g of Na2SO4 into moles first.
d) is almost the same as a) except this time when you find the moles of Ca(OH)2 you just have to convert the moles into mass by multiplying moles x Molar mass of Ca(OH)2.
Best of luck.
Explanation:
To neutralize 15 mL of 2.0 M Ba(OH)₂, one would need 60 mL of 1.0 M HCl. The calculation is based on the mole ratio from the balanced chemical equation, with the final step being the conversion from liters to milliliters.
Explanation:To determine how many milliliters of 1.0 M HCl are needed to neutralize 15 milliliters of 2.0 M Ba(OH)₂, you need to use the concept of mole ratios from the balanced chemical equation. The reaction between hydrochloric acid (HCl) and barium hydroxide (Ba(OH)₂) is:
2 HCl + Ba(OH)₂ ightarrow BaCl2 + 2 H₂O
The balanced equation shows that two moles of HCl react with one mole of Ba(OH)2. First, calculate the moles of Ba(OH)2:
moles Ba(OH)2 = volume (L) times molarity (M) = 0.015 L times 2.0 M = 0.03 moles
Since the ratio of HCl to Ba(OH)₂ is 2:1, you will need twice as many moles of HCl to neutralize Ba(OH)₂. Therefore, you'll need:
moles HCl needed = 0.03 moles Ba(OH)₂ times 2 = 0.06 moles HCl
Now, convert moles of HCl to volume:
volume HCl (L) = moles HCl / molarity HCl = 0.06 moles / 1.0 M = 0.06 L
Finally, convert liters to milliliters:
volume HCl (mL) = volume HCl (L) times 1000 mL/L = 0.06 L times 1000 mL/L = 60 mL
Therefore, 60 mL of 1.0 M HCl are needed to exactly neutralize 15 mL of 2.0 M Ba(OH)₂.
Calculate the energy absorbed when 6.20 g of aluminum is heated from 10.0°C to 65.0°C if the specific heat of aluminum is 0 .900 J/g°C
Answer:
The energy absorbed when 6.20 grams of aluminum is heated from 10.0°C to 65.0°C is 306.9 Joules.
Explanation:
[tex]Q=mc\Delta T=mc\times (T_2-T_1)[/tex]
Where:
Q = heat absorbed or heat lost
c = specific heat of substance
m = Mass of the substance
ΔT = change in temperature of the substance
[tex]T_1,T_2[/tex] : Initial and final temperature of the substance
We have mass of aluminium = m = 6.20 g
Specific heat of aluminium= c = 0.900 J/g°C
Initial and final temperature of the aluminium [tex]T_1=10.0^oC [/tex]
Final temperature of the aluminium = [tex]T_2=65.0^oC[/tex]
Heat absorbed by the aluminium:
[tex]Q=6.20 g\times 0.900J/g^oC\times (65.0^oC-10.0^oC)=306.9 J[/tex]
The energy absorbed when 6.20 grams of aluminum is heated from 10.0°C to 65.0°C is 306.9 Joules.
Calculate the net change in enthalpy for the formation of one mole of lead(ii) sulfate from lead, lead(iv) oxide, and sulfuric acid from these reactions. round your answer to the nearest .
Answer:
ΔHr = -275 kj
Explanation:
It is possible to obtain the net change in enthalpy for the formation of one mole of lead(II) sulfate from lead, lead(IV) oxide, and sulfuric acid using the reactions:
(1) H₂SO₄(l) → SO₃(g) + H₂O (l) ΔH=+113kJ
(2) Pb(s) + PbO₂(s) + 2SO₃(g) → 2PbSO₄(s) ΔH=−775kJ
If you sum (1) + ¹/₂(2) you will obtain:
H₂SO₄(l) + ¹/₂Pb(s) + ¹/₂PbO₂(s) → PbSO₄(s) + H₂O(l)
Using Hess's law, the net change in enthalpy for this reaction could be obtained as:
ΔHr = ΔH(1) + ¹/₂ΔH(2)
ΔHr = +113kJ + ¹/₂ -775kJ
ΔHr = -275 kJ
According to the reaction below, what mass of oxygen gas is required to produce 22 g of carbon dioxide?
C(s) + O2(g) --> CO2(g)
A.) 16 g
B.) 4.0 g
C.) 8.0 g
D.) 32 g
E.) 12 g
Final answer:
To produce 22 g of CO₂, we calculate the moles of CO₂ and use the 1:1 molar ratio with O₂ to find that 16 g of oxygen gas is required, according to the balanced chemical equation for the reaction.
Explanation:
Calculating Mass of Oxygen Required to Produce Carbon Dioxide
To determine the mass of oxygen gas required to produce 22 g of carbon dioxide in the reaction C(s) + O₂(g)
ightarrow CO₂(g), we will first consider the molar masses of the substances involved. The molar mass of carbon dioxide (CO₂) is 44.009 g/mol, which means 1 mole of CO₂ weighs 44.009 grams. Since carbon is a pure element in its solid form, its molar mass is equivalent to its atomic mass, which is 12.0 g/mol. The molar mass of oxygen (O₂) is 32.0 g/mol because one mole of O₂ consists of 2 atoms of oxygen and the atomic mass of oxygen is 16.0 g/mol per atom.
Using the stoichiometry from the balanced chemical equation, we know that 1 mole of C will react with 1 mole of O₂ to produce 1 mole of CO₂. Therefore, if we have 22 g of CO₂, we can calculate the moles of CO₂ and then the moles of O₂ required:
Calculate moles of CO₂ produced: (22 g CO₂) / (44.009 g/mol) = 0.5 mol CO₂.
We need 0.5 mol of O2 since the mole ratio is 1:1.
Calculate mass of O₂ required: (0.5 mol O₂) times (32.0 g/mol) = 16 g O₂.
So, the mass of oxygen gas required to produce 22 g of carbon dioxide is 16 g.
In cats, the allele B is for black fur and allele b is for brown fur. Identify the phenotypes of the following individuals who have different genotypes.
A. Bb
B. BB
C. bb
Answer:
Explanation:
M