Answer:
-376 kJ
Explanation:
The first step equation:
[tex]\mathsf{N_{2(g)} + 3H_2{(g)} \to 2NH_3{(g)} \ \ \ \Delta H = -92\ kJ}[/tex] ---- (1)
The second step equation:
[tex]\mathsf{NH_{3(g)} + 2O_2{(g)} \to HNO_3{(g)} +H_2O_{(g)} \ \ \ \Delta H = -330\ kJ}[/tex] ---- (2)
To determine the enthalpy of formation for 1 mole of HNO₃ (nitric acid), we have the following.
From the above equations; let multiply equation (1) by 1 and equation (2) by 2.
[tex]\mathsf{N_{2(g)} + 3H_2{(g)} \to 2NH_3{(g)} \ \ \ \Delta H = -92\ kJ}[/tex] ---- (3)
[tex]\mathsf{2NH_{3(g)} + 4O_2{(g)} \to 2HNO_3{(g)} +2H_2O_{(g)} \ \ \ \Delta H = 2(-330)\ kJ}[/tex] ----- (4)
adding the above two equations, we have:
[tex]\mathsf{N_{2(g)} + 3H_2{(g)}+ 2NH_{3(g)} + 4O_{2(g)} \to 2HNO_{3(g)} + 2NH_3{(g)} +2H_2O_{(g)} \ \ \ \Delta H = (-660 \ kJ -92\ kJ)}[/tex][tex]\mathsf{N_{2(g)} + 3H_2{(g)} + 4O_{2(g)} \to 2HNO_{3(g)} +2H_2O_{(g)} \ \ \ \Delta H = (-752 \ kJ)}[/tex]
Now, from the recent equation, we have:
2 moles of nitric acid = -752 kJ
∴
1 mole of nitric acid will be: = (1 mole × (-752 kJ)) ÷ 2 moles
1 mole of nitric acid will be: = -376 kJ
How many chromosomes do we not understand?
Answer:
we don't understand why humans have only 46 chromosomes
Answer:
46 chromosomes is what we don't understand
The solvent for an organic reaction is prepared by mixing 70.0 mL of acetone (C3H6O) with 75.0 mL of ethyl acetate (C4H8O2). This mixture is stored at 25.0 ∘C. The vapor pressure and the densities for the two pure components at 25.0 ∘C are given in the following table. What is the vapor pressure of the stored mixture?
Answer:
The answer is "170.9 mm Hg".
Explanation:
[tex]\text{Mass of acetone = volume} \times density[/tex]
[tex]= 70.0 \times 0.791\\\\ = 55.37\ g\\[/tex]
[tex]\text{Moles of acetone} = \frac{mass}{molar\ mass}\\\\[/tex]
[tex]=\frac{55.37}{58.08}\\\\ = 0.9533\ mol[/tex]
[tex]\text{Mass of ethyl acetate = volume} \times density[/tex]
[tex]= 73.0 \times 0.900\\\\ = 65.7\ g[/tex]
[tex]\text{Moles of ethyl acetate = mass} \times\ molar\ mass[/tex]
[tex]= \frac{65.7}{88.105} \\\\= 0.7457\ mol[/tex]
[tex]\text{Mole fraction of acetone x(acetone)} = \frac{0.9533}{(0.9533 + 0.7457)}\ = 0.5611\\\\[/tex] [tex]\text{Mole fraction of ethyl acetate x(ethyl acetate)} =\frac{0.7457}{(0.9533 + 0.7457) }= 0.4389[/tex]
Applying Raoult's law: [tex]\text{Vapor pressure = x(acetone)P(acetone) + x(ethyl acetate)P(ethyl acetate)}\\\\= 0.5611 \times 230.0 + 0.4389 \times 95.38\\\\ = 170.9\ mm \ Hg\\[/tex]
The solvent for an organic reaction is prepared by mixing 70.0 mL of acetone (C3H6O) with 75.0 mL of ethyl acetate (C4H8O2).
The vapor pressure of the stored mixture is: 170.03 mmHg
In the given information, there is some information that is still missing.
The parameters that we are being given include:
The volume of acetone = 70.0 mLThe volume of ethyl acetate = 75.0 mLThe standard temperature for the mixture = 25° CThe first step we need to take is to determine the mass and number of moles of each compound (i.e. for acetone and ethyl acetate)
For us to do that:
We need the density of acetone and ethyl acetate, which is not given:
Assuming that at a standard condition of vapour pressure:
230 mmHg of acetone has a density of 0.791 g/mL95.38 mmHg of ethyl acetate has a density of 0.900 g/mLThen;
Using the relation:
[tex]\mathbf{Density = \dfrac{Mass}{volume}}[/tex]
Mass of acetone = Density of acetone × volume of acetone
Mass of acetone = 0.791 g/mL × 70.0 mL
Mass of acetone = 55.37 g
Mass of ethyl acetate = Density of ethyl acetate × volume of ethyl acetate
Mass of ethyl acetate = 0.900 g/mL × 75.0 mL
Mass of ethyl acetate = 67.5 g
At standard conditions;
For acetone, molar mass = 58.08 g/molFor ethyl acetate, molar mass = 88.11 g/molNow, using the formula for calculating the numbers of moles which can be expressed as:
[tex]\mathbf{Number \ of \ moles = \dfrac{mass}{molar \ mass}}[/tex]
For acetone:
[tex]\mathbf{Number \ of \ moles = \dfrac{55.37 \ g}{58.08 \ g/mol}}[/tex]
[tex]\mathbf{Number \ of \ moles =0.95334 \ mol}[/tex]
For ethyl acetate:
[tex]\mathbf{Number \ of \ moles = \dfrac{67.5 \ g}{88.11 \ g/mol}}[/tex]
[tex]\mathbf{Number \ of \ moles =0.76609 \ mol}[/tex]
Now, we will determine the mole fraction of each compound.
The mole fraction describes the ratio a certain constituent of a mixture to the total amount of all the constitutent in the mixture.
Using the formula:
[tex]\mathbf{mole \ fraction = \dfrac{n_A}{n_A+n_B+...n_N}}[/tex]
For Acetone:
[tex]\mathbf{mole \ fraction = \dfrac{0.95334}{0.95334+0.76609}}[/tex]
[tex]\mathbf{mole \ fraction =0.5545 }[/tex]
For ethyl acetate:
[tex]\mathbf{mole \ fraction = \dfrac{0.76609}{0.76609+0.95334}}[/tex]
[tex]\mathbf{mole \ fraction =0.4455}[/tex]
Finally, we can compute determine the vapour pressure of the stored mixture using Raoult's Law.
Raoult's Law posits that the constituent of a partial pressure in a mixture of a liquid is proportional to the mole fraction of that constituent in the mixture provided the temperature is constant.
∴ For the stored mixture = Vapor pressure of acetone + vapour pressure of ethyl acetate.
where:
Vapour pressure of the solution = (mole fraction × vapor pressure) of solventFor acetone;
Vapor pressure = 0.5545 × 230 mmHg
Vapour pressure = 127.54 mmHg
For ethyl acetate:
Vapour pressure = 0.4455 × 95.38 mmHg
Vapour pressure ==42.49 mmHg
Thus, the vapor pressure of the stored mixture is
= (127.54 + 42.49 ) mmHg
= 170.03 mmHg
Therefore, we can conclude that the vapour pressure of the stored mixture is 170.03 mmHg
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What is the formula for manganese (IV) oxide?
A. MnO4
B. Mn40
C. MNO2
D. Mn20
The formula of manganese (IV) oxide is MNO2
The formula for manganese (IV) oxide is [tex]MnO_4[/tex]. Therefore, the correct option is A.
A manganese atom is bonded to four oxygen atoms to produce an inorganic compound known as manganese(IV) oxide with the chemical formula [tex]MnO_4[/tex]. It is often referred to as manganese dioxide. The +4 oxidation state of a manganese compound is indicated by the Roman numeral "IV" in the name.
Pyrolusite, a naturally occurring mineral, is a dark brown or black solid consisting of manganese(IV) oxide. Due to its favorable properties, it is often employed in a wide variety of applications. It acts as a catalyst, especially when breaking down hydrogen peroxide. Additionally, it is used to make batteries, ceramics, glass and pigments.
Therefore, the correct option is A.
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34. 3.15 mol of an unknown solid is placed into enough water to make 150.0 mL of solution. The solution's temperature increases by 11.21°C. Calculate ∆H, in kJ/mol, for the dissolution of the unknown solid. (The specific heat of the solution is 4.184 J/g･°C and the density of the solution is 1.20 g/mL).
Answer:
ΔH = 2.68kJ/mol
Explanation:
The ΔH of dissolution of a reaction is defined as the heat produced per mole of reaction. We have 3.15 moles of the solid, to find the heat produced we need to use the equation:
q = m*S*ΔT
Where q is heat of reaction in J,
m is the mass of the solution in g,
S is specific heat of the solution = 4.184J/g°C
ΔT is change in temperature = 11.21°C
The mass of the solution is obtained from the volume and the density as follows:
150.0mL * (1.20g/mL) = 180.0g
Replacing:
q = 180.0g*4.184J/g°C*11.21°C
q = 8442J
q = 8.44kJ when 3.15 moles of the solid react.
The ΔH of the reaction is:
8.44kJ/3.15 mol
= 2.68kJ/mol
For a different reaction, the plot of the reciprocal of concentration versus time in seconds was linear with a slope of 0.056 M-1 s -1 . If the initial concentration was 2.2 M, calculate the concentration after 100 seconds. Show your work.
Answer:
[tex]C_t=0.165M[/tex]
Explanation:
From the question we are told that:
Slope [tex]K=0.056 M-1 s -1[/tex]
initial Concentration [tex]C_1=2.2M[/tex]
Time [tex]t=100[/tex]
Generally the equation for Raw law is mathematically given by
[tex]\frac{1}{C}_t=kt+\frac{1}{C}_0[/tex]
[tex]\frac{1}{C}_t=0.056*100+\frac{1}{2.2}_0[/tex]
[tex]C_t=0.165M[/tex]
The second-order reaction is the reaction that depends on the reactants of the first or the second-order reaction. The concentration after 100 seconds will be 0.165 M.
What is the specific rate constant?The specific rate constant (k) of the second-order reaction is given in L/mol/s or per M per s. It is the proportionality constant that gives the relation between the concentration and the rate of the reaction.
Given,
Slope (k)= 0.056 per M per s
Initial concentration of the reactant [tex](\rm C_{1})[/tex] = 2.2 M
Time (t) = 100 seconds
The concentration of the reaction after 100 seconds can be given by,
[tex]\rm \dfrac{1}{C_{t}} = kt + \dfrac{1}{C_{1}}[/tex]
Substitute values in the above equation:
[tex]\begin{aligned} \rm \dfrac{1}{C_{t}} &= 0.056 \times 100 + \dfrac{1}{2.02}\\\\&= 0.165 \;\rm M\end{aligned}[/tex]
Therefore, after 100 seconds the concentration is 0.165 M.
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What mass of oxygen is needed for the complete combustion of 1.60-10^-3
g
of methane?
Express your answer with the appropriate units.
Answer:
6.4×10¯³ g of O₂.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
CH₄ + 2O₂ —> CO₂ + 2H₂O
Next, we shall determine the masses of CH₄ and O₂ that reacted from the balanced equation. This can be obtained as follow:
Molar mass of CH₄ = 12 + (4×1)
= 12 + 4
= 16 g/mol
Mass of CH₄ from the balanced equation = 1 × 16 = 16 g
Molar mass of O₂ = 2 × 16 = 32 g/mol
Mass of O₂ from the balanced equation = 2 × 32 = 64 g
SUMMARY:
From the balanced equation above,
16 g of CH₄ reacted with 64 g of O₂.
Finally, we shall determine the mass of O₂ needed to react with 1.6×10¯³ g of CH₄. This can be obtained as illustrated below:
From the balanced equation above,
16 g of CH₄ reacted with 64 g of O₂.
Therefore, 1.6×10¯³ g of CH₄ will react with = (1.6×10¯³ × 64) / 16 = 6.4×10¯³ g of O₂
Thus, 6.4×10¯³ g of O₂ is needed for the reaction.
What is normality in chemistry?
Answer:
a measure of concentration equal to the gram equivalent weight per liter of solution.
Explanation:
Gram equivalent weight is the measure of the reactive capacity of a molecule. The solute's role in the reaction determines the solution's normality. Normality is also known as the equivalent concentration of a solution.
hope it helped
What functional group is found in an alcohol?
A. Ester
B. Amino
C. Carbonyl
D. Hydroxyl
Answer:
an alcohol is a Hydroxyl group due to the OH~ that is associated with it's molecules
The functional group found in an alcohol is Hydroxyl . Therefore, the correct option is option D.
What is functional group?A functional group in organic chemistry is a substituent and moiety inside a molecule that triggers the molecule's distinctive chemical processes. No matter how the rest of a molecule is made up, the very same functional group would experience the same or a similar set of chemical events.
This permits the design of synthetic chemistry as well as the methodical forecasting of chemical reactions as well as the behaviour of chemical molecules. Other functional groups close by can affect a functional group's reactivity. Retrosynthetic analysis can be used to design organic synthesis by using functional group interconversion. The functional group found in an alcohol is Hydroxyl .
Therefore, the correct option is option D.
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A system fitted with a piston expands when it absorbs 53.1 ) of heat from the surroundings. The piston is working against a pressure of 0.677 atm. The final volume is 63.2 L. What was the initial volume of the system if the internal energy of the system decreased by 108.3 J?
a. 65.6 L
b. 64.0 L
c. 70.8 L
d. 60.8 L
e. 54.4L
Answer:
d. 60.8 L
Explanation:
Step 1: Given data
Heat absorbed (Q): 53.1 JExternal pressure (P): 0.677 atmFinal volume (V2): 63.2 LChange in the internal energy (ΔU): -108.3 JStep 2: Calculate the work (W) done by the system
We will use the following expression.
ΔU = Q + W
W = ΔU - Q
W = -108.3 J - 53.1 J = -161.4 J
Step 3: Convert W to atm.L
We will use the conversion factor 1 atm.L = 101.325 J.
-161.4 J × 1 atm.L/101.325 J = -1.593 atm.L
Step 4: Calculate the initial volume
First, we will use the following expression.
W = - P × ΔV
ΔV = - W / P
ΔV = - 1.593 atm.L / 0.677 atm = 2.35 L
The initial volume is:
V2 = V1 + ΔV
V1 = V2 - ΔV
V1 = 63.2 L - 2.35 L = 60.8 L
Classify these bonds as ionic, polar covalent, or nonpolar covalent. You are currently in a sorting module.
Ionic Polar Covalent Nonpolar covalent
C-O , Mg-F , Cl-Cl
Answer:
C-O: polar covalent
Mg-F: ionic
Cl-Cl: nonpolar covalent
Explanation:
Ionic bonds are formed between an atom of a metallic element and another atom of a non-metallic element. Thus, Mg-F is an ionic bond, in which Mg is the metal and F is the nonmetal.
Covalent bonds are formed between two non-metallic elements. So, C-O and Cl-Cl are covalent bonds, because C, O, and Cl are nonmetals.
In C-O, the atom of oxygen (O) has more electronegativity than the atom of carbon (C). Thus, O will attract the electrons with more strength and a difference in charge will be established between the two bonded atoms. So, this covalent bond is polar.
In Cl-Cl, both atoms have the same electronegativity because they are from the same chemical element (Cl). Thus, this bond is nonpolar.
how is the molecule of substance formed
Answer:
When atoms approach one another closely, the electron clouds interact with each other and with the nuclei. If this interaction is such that the total energy of the system is lowered, then the atoms bond together to form a molecule.
Explanation:
convert 36.52 mg to ?___ g
Answer:
0.0365 is your answer
Explanation:
hope it will help uAnswer:
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11 Explain how you would obtain solid lead carbonate from a mixture of lead carbonate and sodium chloride
Explanation:
Add water, Na2CO3 dissolves, filter, PbCO3 stays in the paper and dissolved Na2CO3 goes through as the solution. Dry the PbCO3 and you have the dry solid.
OR
Add water to dissolve then filter to obtain PbCo3 as you're residue and Na2Co3 as the filtrate. Dry the insoluble PbCo3 between filter papers and you obtain solid PbCo3
What is the volume of 6 moles of hydrogen gas, H2 (g)
Answer:
Molar volume, or volume of one mole of gas , depends on pressure and temperature, and is 22.4 liters - at 0 °C (273.15 K) and 1 atm (101325 Pa), or STP (Standard Temperature and Pressure), for every gas which behaves similarly to an ideal gas. The ideal gas molar volume increases to 24.0 liters as the temperature increases to 20 °C (at 1 atm).
How many oxygen atoms are there in 0.25 mole of CO2?
Answer:
It is known that 1 mol of a molecule contains 6.023×1023 6.023 × 10 23 number of molecules. So, 0.25 moles of CO2 C O 2.
The correct option for the given question about Mole Concept is Na / 2 atom of oxygen.
What is mole?In chemistry, a mole is a unit. A mole of a substance is the mass of the substance containing precisely 12 gram of the same number of atoms as 12C.The mole is the SI unit used to measure a substance's quantity. It has the symbol mol.
How to solve this question?
In 1 mole number of molecules of CO₂ = Na (Avagadro Number)
In 0.25 mole number of molecules of CO₂ = 0.25 × Na molecules
In 1 molecule number of oxygen atom = 2 atom
In 0.25 × Na molecules number of oxygen atom = 2 × 0.25 × Na atom
In 0.25 × Na molecules number of oxygen atom = 0.5 × Na atom
So in 0.25 mole of CO₂ number of oxygen atom = Na / 2 atom
Thus we can conclude that in 0.25 mole of CO₂ number of oxygen atom will be Na / 2, where Na is Avagadro Number.
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When should a line graph be used
Answer:
Line graphs are used to track changes over short and long periods of time. When smaller changes exist, line graphs are better to use than bar graphs. Line graphs can also be used to compare changes over the same period of time for more than one group.
Many important analgesic compounds are derived from simple aromatic starting materials.
a. True
b. False
WHAT WOULD THE RIGHT OPTION??
how many resonance structures of benzene are known?
A) 3
B) 4
C) 5
D) 6
Answer:
C) 5
Step-by-step explaination:
Benzene has 5 resonance structures.
During electrophilic aromatic substitution, a resonance-stabilized cation intermediate is formed. Groups, already present on the benzene ring, that direct ortho/para further stabilize this intermediate by participating in the resonance delocalization of the positive charge. Assume that the following group is present on a benzene ring at position 1 and that you are brominating the ring at positon 4. Draw the structure of the resonance contributor that shows this group actively participating in the charge delocalization.
-----OCH3
Answer:
See explanation and image attached
Explanation:
Aromatic compounds undergo electrophilic aromatic substitution reactions in which the aromatic ring is maintained.
Substituted benzenes may be more or less reactive towards electrophilic aromatic substitution than benzene depending on the nature of the substituent present in the ring.
Substituents that activate the ring towards electrophilic substitution such as -OCH3 are ortho-para directing.
The major products of the bromination of anisole are p-bromoanisole and o-bromoanisole. The resonance structures leading to these products are shown in the image attached.
Low-density polyethylene is formed because _______ polymerization is very unpredictable and difficult to control.
dehydration-condensation
anionic-initiated
radical-initiated
esterification
Answer:
radical-initiated
Explanation:
Radical-initiated polymerization is unpredictable and difficult to control. The reaction proceeds indiscriminately and produces shortened chains, loops, and branches that create holes in the polymer. This reduces its mass to volume ratio.
Which of these is an organic compound?
CaO
H2SO4
C3H8
H2O2
Answer:
which class are you please mention
Select True or False: The equilibrium constant for the chemical equation 2NO(g) O2(g) 2NO2(g) is two times the equilibrium constant for the chemical equation NO(g) 1/2O2(g) NO2(g).
Answer:
False
Explanation:
The first reaction is;
NO(g) + 1/2O2(g) ---->NO2(g)
K= [NO2]/[NO] [ O2]^1/2
The second reaction is;
2NO(g) + O2(g) ---->2NO2(g)
K'= [NO2]^2/[NO]^2 [O2]
It now follows that;
K'= K^2
Hence the statement in the question is false
Which type of organic compound is shown below?
A. Carboxylic acid
B. Ester
C. Amine
D. Alcohol
Answer:
I think its A maybe am not sure
According to the ideal gas law, a 9.998 mol sample of argon gas in a 0.8311 L container at 502.7 K should exert a pressure of 496.2
atm. What is the percent difference between the pressure calculated using the van der Waals' equation and the ideal pressure? For Ar
gas, a = 1.345 L’atm/mol? and b = 3.219x10-2 L/mol.
Pideal – Puan der Waals |
Percent difference
x 100
Answer:
[tex]\%diff=24.0\%[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out firstly necessary for us to set up the van der Waals' equation as shown below:
[tex]p=\frac{RT}{v-b}-\frac{a}{v^2}[/tex]
Thus, we secondly calculate the molar volume as:
[tex]v=\frac{0.8311L}{9.998mol} =0.083L/mol[/tex]
Then, we plug in the entire variables in the vdW equation to get such pressure:
[tex]p=\frac{0.08206\frac{atm*L}{mol*K}*502.7K}{0.08313L/mol-0.03219L/mol}-\frac{1.345L*atm/mol}{(0.08313L/mol)^2}\\\\p=615.2atm[/tex]
And the ideal gas pressure:
[tex]p=\frac{0.08206\frac{atm*L}{mol*K}*502.7K}{0.08313L/mol}\\\\p=496.2atm[/tex]
Finally, the percent difference:
[tex]\%diff=\frac{|496.2atm-615.2atm|}{496.2atm} *100\%\\\\\%diff=24.0\%[/tex]
Regards!
Which is the electronic configuration for oxygen?
When water reaches the metastable state, the phase of water will be?
Answer:
water, when the metastable state is reached, is cooled below the zero temperature. It freezes abruptly. this is called metastable. They are not at equilibrium per se; as at negative temperatures the only equilibrium state of water is ice.
Explanation:
Predict whether reactants or products will be favored at equilibrium for the below reaction.
Kp= 2.2 x 10^6 at 298K
2COF2 (g) + ⇌ CO2(g) + CF4(g)
Answer:
The products will be favored at equilibrium.
Explanation:
The balanced chemical equation for the reaction is the following:
2 COF₂ (g) + ⇌ CO₂(g) + CF₄(g)
The reactant is COF₂ (left side) and the products are CO₂ and CF₄ (right side).
The equilibrium constant is given by the ratio between the partial pressures (P) of products and reactants, because they are in the gas phase. Thus, the expression of the equilibrium constant is the following:
[tex]Kp = \frac{P(CO_{2}) P(CF_{4}) }{P(COF_{2} )^{2} } = 2.2 x 10^{6}[/tex]
Since Kp>>>>1 ⇒ (P(CO₂) x P(CF₄)) > (P(COF₂))²
So, the partial pressures of the products (CO₂ and CF₄) are higher than the partial pressure of the reactant (COF₂).
Therefore, products will be favored at equilibrium at 298 K.
A nuclease enzyme breaks the covalent bond originally connecting the phosphate to the 5' carbon in a nucleic acid. After allowing this enzyme to completely digest the nucleic acid down to monomers, you perform tests to determine where the phosphate is attached to each monomer. Where do you expect to find this phosphate
Answer:
The phosphate will remain attached to the 5' carbon of the deoxy or the ribose sugar in the nucleic acid monomers.
Explanation:
The structure of nucleic acid polymers is built up from monomers of nucleotides.
A nucleotide consists of a sugar backbone which is either a ribose or deoxyribose sugar, a nitogenous base which is either a purine or pyrimidine, and a phosphate group. The nitrogenous base is attached to the carbon number 1 or C-1 of the sugar backbone by a covalent bond. The phosphate group on the other hand is covalently attached to the carbon number 5 or 5' carbon of the sugar backbone.
When polymers of nucleic acids are formed, the phosphate at the 5' carbon of the sugar backbone is covalently linked in a phosphodiester bond to the 3' carbon of the sugar backbone in another nucleotide molecule, thus extending the strands of the nucleic acid molecule.
Nucleases are enzymes that break down the phosphodiseter bonds in nucleic acids resulting in nucleotide monomers. After complete digestion ofmthe nucleic acid polymer by nucleases, the phosphate will remain attached to the 5' carbon of the deoxy or the ribose sugar in the nucleic acid monomers.
A metal (C = 0.2158 cal/g· °C) is removed from a hot (350. °F) oven in which it had achieved thermal equilibrium. The metal is placed into 200. mL acetic acid. The temperature of the acid increases to 90.3 °C from 24.3 °C. What is the mass of the metal? (dacetic acid = 1.04 g/cm3; Cs, acetic acid = 2.055 J/g·°C) Group of answer choices 120. g 362 g 1452 g 347 g 281 g
Answer:
362g
Explanation:
heat lost by metal= heat gained by acetic acid
tfs are the same so you cando delta T
convert Cal/gc to J/gc
thectgod ig follow
GIVING BRAINLIEST
Which equations are used to calculate the velocity of a wave?
O velocity = distance ~ time
velocity = wavelength x frequency
velocity = distance/time
velocity = wavelength/frequency
velocity = distance/time
velocity = wavelength x frequency
velocity = distance ~ time
velocity = wavelength/frequency
Answer:
velocity = distance/time
velocity = wavelength × frequency
Both of these are commonly known equations to calculate velocity with different variables.
