Answer:
The initial temperature of the brass sample is 90.1°C
Note: The question is incomplete. A similar but complete question is given below :
A 52.9g sample of brass, which has a specific heat capacity of 0.375·J·g−1°C−1, is put into a calorimeter (see sketch at right) that contains 100.0g of water. The temperature of the water starts off at 15.0°C. When the temperature of the water stops changing it's 18.4°C. The pressure remains constant at 1 atm. Calculate the initial temperature of the brass sample. Be sure your answer is rounded to 2 significant digits.
Explanation:
Assuming that the calorimeter is an isolated system and that no heat is lost from the calorimeter. The total heat in the system is the sum of the heat content of the brass and that of water
Total heat lost by the brass = heat gained by the water
The quantity of heat lost or gained, Q = mcΔT
Where m = mass of the substance, c = specific heat capacity of substance, ΔT = temperature change
Heat gained by water is positive while heat lost by brass is negative
mass of brass = 52.9 g, specific heat capacity of brass = 0.375·J·g−1°C−1, ΔT = (18.4 - t °C; where t is the initial temperature), mass of water = 100.0 g, specific heat capacity of water = 4.186 J/g°C, ΔT = = 18.4 - 15.0 = 3.4 °C
Heat lost by brass z= - [ 52.9 × 0.375 × (18.4 - t)] = -365.01 + 19.8375t
Heat gained by water = 100 × 4.186 × 3.4 = 1423.24
Equating heat lost by brass to heat gained by water
-365.01 + 19.8375t = 1423.24
19.8375t = 1423.24 + 365.01
19.8375t = 1788.25
t = 90.1° C
Therefore, the initial temperature of the brass sample is 90.1°C
Rank each of the following gases in order of increasing urms assuming equivalent amounts and all gases are at the same temperature and pressure where 1 has the lowest urms and 4 has the highest urms.
a. Gas 1 : H2S
b. Gas: He
c. Gas 3: NF3
d. Gas 4: H2O
The Urms refers to the root mean square speed of the gas. The order of increasing Urms for the gases shown in the question; NF3 < H2S < H2O < He.
What is the Urms?The Urms refers to the root mean square speed of the gas. This is ultimately dependent on the relative molecular mass of the gases when they are maintained at the same temperature.
Now, let us look at the order of increasing Urms for the gases shown in the question; NF3 < H2S < H2O < He.
Learnmore about Urms: https://brainly.com/question/365923
If 12.3 g of Cu is deposited at the cathode of an electrolytic cell after 5.50 h, what was the current used?
Answer:
1.88 A
Explanation:
Let's consider the reduction of copper in an electrolytic cell.
Cu²⁺ + 2 e⁻ ⇒ Cu
We can calculate the charge used to deposit 12.3 g of Cu using the following relations.
The molar mass of Cu is 63.55 g/mol.1 mole of Cu is deposited when 2 moles of electrons circulate.1 mole of electrons has a charge of 96486 C (Faraday's constant).The charge used is:
[tex]12.3 g \times \frac{1 molCu}{63.55gCu} \times \frac{2molElectron}{1molCu} \times \frac{96486C}{1molElectron} = 3.73 \times 10^{4} C[/tex]
We can convert 5.50 h to seconds using the conversion factor 1 h = 3600 s.
5.50 h × 3600 s/1 h = 1.98 × 10⁴ s
The current used is:
I = q/t = 3.73 × 10⁴ C/1.98 × 10⁴ s = 1.88 A
Linoleic acid is a polyunsaturated fatty acid found, in ester form, in many fats and oils. Its doubly allylic hydrogens are particularly susceptible to abstraction by radicals, a process that can lead to the oxidative degradation of the fat or oil.
a. True
b. Flase
Answer:
True.
Explanation:
The information presented in the question above regarding linoleic acid is true. Linoleic acid is, in fact, found in many oils and fats in the ester form. In addition, linoleic acid is considered a polyunsaturated fatty acid, due to the presence of two unsaturations in its composition. Its chemical formula is CH3-(CH2)4-CH=CH-CH2-CH=CH-(CH2)7COOH and it is an essential fatty acid for the human body, as it is essential in the composition of arachidonic acid that is responsible for building muscle, managing body fat thermogenesis, and regulating core protein synthesis.
True or false: Boron contains 2s22p1 valence electrons, so only one p orbital is needed to form molecular orbitals.
Answer:
True
Explanation:
The valence orbitals of boron are 2s2 2p1. We have to recall that all the valence orbitals whether full or empty are involved in the formation of molecular orbitals.
The number of molecular orbitals formed is equal to the number of atomic orbitals that are combined.
Since there are two valence orbitals and there is only one p orbital among the valence orbitals, it is true that only one p orbital is needed to form molecular orbitals in boron.
write any two things that should be remembered while writing chemical equation
Answer:
the product and the reactant must be balanced
if u are required to give the mechanism if the reaction it must be written
Write the number of sig. fig. in four numbers given in the sentence below. An (one) octopus has 8 legs. 13 octopi have 104 legs.
Give four answers.
A. Infinity, Infinity, Infinity, Infinity
B. 1, 1, 2, 3
C. Infinity, Infinity, 2, 3
D. No answer text provided.
Answer:
1, 1, 2, 3
Explanation:
The numbers 1 and 8 both have 1 sig. fig.
The number 13 has 2 sig. figs.
The number 104 has 3 sig. figs.
PLEASE HELP!!
How does temperature, agitation, and particle size affect solubility?
Answer:
At higher temperatures, particles move faster and collide more, increasing solubility rates.
Agitation increases solubility rates as well, by bringing fresh solvent into contact with the undissolved solute
The smaller the particle size, the higher (faster) solubility rate. Vice versa, the bigger the particle size, the lower (slower) solubility rate.
Explanation:
4.005 X 74 X 0.007 = 2.10049
Answer:
2.07459
Explanation:
this is the correct answer.
What are the uses of Sulphuric acid?
Answer:
The major use of sulfuric acid is in the production of fertilizers, e.g., superphosphate of lime and ammonium sulfate. It is widely used in the manufacture of chemicals, e.g., in making hydrochloric acid, nitric acid, sulfate salts, synthetic detergents, dyes and pigments, explosives, and drugs.
The major use of sulfuric acid is in the production of fertilizers, e.g., superphosphate of lime and ammonium sulfate. It is widely used in the manufacture of chemicals, e.g., in making hydrochloric acid, nitric acid, sulfate salts, synthetic detergents, dyes and pigments, explosives, and drugs.
4.106
Calculate the moles and the mass of solute in each of the following solutions.
(a) 150.0 mL of 0.245 M CaCl2
molarity = moles of solute / volume of solution
moles of solute = molarity × volume of solution
moles of solute = 0.245 mol/L × 0.1500 L
moles of solute = 0.03675 mol
moles of solute = 0.0368 mol
-----------------------------------------------------------
Solution: (mass of solute)Step 1: Calculate the molar mass of solute.
molar mass of solute = (40.08 g/mol × 1) + (35.45 g/mol × 2)
molar mass of solute = 110.98 g/mol
Step 2: Calculate the mass of solute.
mass of solute = moles of solute × molar mass of solute
mass of solute = 0.03675 mol × 110.98 g/mol
mass of solute = 4.08 g
Note: The volume of solution must be expressed in liters (L).
Answer:
[tex]\boxed {\sf \bold {0.0368 \ mol \ CaCl_2}}}}[/tex]
[tex]\boxed {\sf \bold {4.08 \ g \ CaCl_2}}}}}[/tex]
Explanation:
1. Moles of SoluteMolarity is a measure of concentration in moles per liter.
[tex]molarity= \frac {moles \ of \ solute}{liters \ of \ solution}[/tex]
In this solution, there are 150.0 milliliters of solution and the molarity is 0.245 M CaCl₂ or 0.245 mol CaCl₂ per liter.
First, convert the milliliters to liters. There are 1000 milliliters in 1 liter.
[tex]{150 \ mL * \frac{1 \ L}{1000 \ mL}= \frac{150}{1000} \ L = 0.150 \ L[/tex]Now, substitute the known values (molarity and liters of solution) into the formula. The moles of solution are unknown, so we can use x.
[tex]0.245 \ mol \ CaCl_2 /L= \frac{ x}{0.150 \ L}[/tex]
We are solving for x, so we must isolate this variable. It is being divided by 0.150 L. The inverse of divisions is multiplication, so we multiply both sides by 0.150 L.
[tex]0.150 \ L *0.245 \ mol \ CaCl_2 /L= \frac{ x}{0.150 \ L} * 0.150 L[/tex]
[tex]0.150 \ L *0.245 \ mol \ CaCl_2 /L=x[/tex]
The units of liters cancel.
[tex]0.150 *0.245 \ mol \ CaCl_2 =x[/tex]
[tex]0.03675 \ mol \ CaCl_2[/tex]
The original measurements have 3 significant figures, so our answer must have the same.
We should round to the ten thousandths place. The 5 to the right of this place tells us to round the 7 up to an 8.
[tex]\bold {0.0368 \ mol \ CaCl_2}[/tex]
2. Mass of the SoluteWe can convert mass to moles using the molar mass. These values are found on the Periodic Table. They are the same as the atomic masses, but the units are grams per mole (g/mol) instead of atomic mass units.
The solute is calcium chloride: CaCl₂. Look up the molar masses of the individual elements.
Ca: 40.08 g/mol Cl: 35.45 g/molNotice that chlorine has a subscript of 2. We must multiply the molar mass by 2.
Cl₂: 35.45 *2= 70.9 g/molAdd calcium's molar mass.
CaCl₂: 40.08 + 70.9 =110.98 g/molUse the molar mass as a ratio.
[tex]\frac {110.98 \ g\ CaCL_2}{ 1 \ mol \ CaCl_2}[/tex]
Multiply the moles of calcium chloride we calculated above.
[tex]0.0368 \ mol \ CaCl_2 *\frac {110.98 \ g\ CaCL_2}{ 1 \ mol \ CaCl_2}[/tex]
The units of moles of calcium chloride cancel.
[tex]0.0368 *\frac {110.98 \ g\ CaCL_2}{ 1 }[/tex]
[tex]4.084064 \ g\ CaCl_2[/tex]
Round to 3 significant figures again. For this number, it is the hundredths place. The 4 in the thousandths place tells us to leave the 8.
[tex]\bold {4.08 \ g \ CaCl_2}[/tex]
Please please help help please
En la fermentación del alcohol, la levadura convierte la glucosa en etanol y dióxido de carbono:
C6H12O6(s) → 2C2H5OH(l) + 2CO2(g)
Si reaccionan 5.97 g de glucosa y se recolectan 1.44 L de CO2 gaseoso, a 293 K y 0.984 atm, ¿cuál
es el rendimiento porcentual de la reacción
Answer:
88.9%
Explanation:
Primero convertimos 5.97 g de glucosa a moles, usando su masa molar:
5.97 g ÷ 180 g/mol = 0.0332 molDespués calculamos la cantidad máxima de moles de CO₂ que se hubieran podido producir:
0.0332 mol C₆H₁₂O₆ * [tex]\frac{2molCO_2}{1molC_6H_{12}O_6}[/tex] = 0.0664 mol CO₂Ahora calculamos los moles de CO₂ producidos, usando los datos de recolección dados y la ecuación PV=nRT:
0.984 atm * 1.44 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 293 Kn = 0.0590 molFinalmente calculamos el rendimiento porcentual:
0.0590 mol / 0.0664 mol * 100% = 88.9%Identify the possible quantitative analysis you can do using only the 28.02 g/mol as a unit factor. Select one or more:
Answer:
Calculate the moles of N2 molecules in 3.94 grams of nitrogen.
Calculate the grams of N2 in 5.03 x 1020 moles of nitrogen molecules.
Explanation:
Calculate the moles of N2 molecules in 4.73 liters of nitrogen gas. FALSE. You can't make this conversion using only the conversion factor with units of g/mol. To convert liters to moles are necessaries pressure, temperature and volume of the gas to use PV = nRT
Calculate the grams of N2 in 10.58 liters of nitrogen gas. FALSE. As explained, you need, P,V and T to find the moles of the gas. With the moles you can find the mass using the conversion factor of 28.02g/mol
Calculate the moles of N2 molecules in 3.94 grams of nitrogen. TRUE. You can find the moles of N2 as follows:
3.94g N2 * (1mol/28.02g) = 0.14 moles of N2 molecules
Calculate the grams of N2 in 5.03 x 1020 moles of nitrogen molecules. TRUE. The mass in 5.03x10²⁰ moles of nitrogen molecules is:
5.03x10²⁰ moles * (28.02g/mol) = 1.4x10²²g of nitrogen.
Liquid octane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 10.3 g of octane is mixed with 23. g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to significant digits.
Answer:
9.36 g
Explanation:
The equation of the reaction is;
C8H18(g) + 25/2 O2(g) ----> 8CO2(g) + 9H2O(g)
Number of moles of octane = 10.3g/ 114 g/mol = 0.09 moles
1 mole of octane yields 9 moles of water
0.09 moles of octane yields 0.09 × 9/1 = 0.81 moles of water
Number of moles of oxygen = 23g/32g/mol = 0.72 moles
12.5 moles of oxygen yields 9 moles of water
0.72 moles of oxygen yields 0.72 × 9/12.5 = 0.52 moles of water
Hence oxygen is the limiting reactant;
Maximum mass of water produced = 0.52 moles of water × 18 g/mol = 9.36 g
Indicate how the concentration of each species in the chemical equation will change to reestablish equilibrium after reactant or product is added.
2CO(g) + O2(g) ⇌ 2CO2
Answer:
Indicate how the concentration of each species in the chemical equation will change to reestablish equilibrium after reactant or product is added.
[tex]2CO(g) + O2(g) <=> 2CO2[/tex]
Explanation:
When the reactants concentration increases, then the equilibrium will shift towards products and when the concentration of products increases, then equilibrium will shift towards reactants.
So, increases in concentration of carbon monoxide (CO) shifts the equilibrium to favor the formation of carbondioxide.
Similarly increase in concentration of oxygen also favor the formation of product carbon dioxide.
Increase in concentration of CO2 favors the formation of CO and O2.
Decrease in product concentration also favors the formation of product.
Decrease in reactant concentration favors the formation of reactants only.
If 0.250 L of a 5.90 M HNO₃ solution is diluted to 2.00 L, what is the molarity of the new solution?
Answer:
0.74 M
Explanation:
From the question given above, the following data were obtained:
Molarity of stock solution (M₁) = 5.90 M
Volume of stock solution (V₁) = 0.250 L
Volume of diluted solution (V₂) = 2 L
Molarity of diluted solution (M₂) =?
The molarity of the diluted solution can be obtained by using the dilution formula as illustrated below:
M₁V₁ = M₂V₂
5.90 × 0.250 = M₂ × 2
1.475 = M₂ × 2
Divide both side by 2
M₂ = 1.475 / 2
M₂ = 0.74 M
Thus, the molarity of the diluted solution is 0.74 M
You are asked to prepare a buffer solution with a pH of 3.50. The following solutions, all 0.100 M, are available to you: HCOOH, CH3COOH, H3PO4 , NaCHOO, NaCH3COO, and NaH2PO4. What would be the best combination to make the required buffer solution? Select one:
a. NaH2PO4 and NaCHOO
b. H3PO4 and NaH2PO4
c. NaH2PO4 and HCOOH
d. CH3COOH and NaCH3COO e. HCOOH and NaCHOO
can someone helo me with this
Answer:
e. HCOOH and NaCHOO
Explanation:
For a buffer solution, both an acid and its conjugate base are required.
With the information above in mind, we can discard options a) and c), as those combinations are not of an acid and its conjugate base.
Now it is a matter of comparing the pKa (found in literature tables) of the acids of the remaining three acids:
H₃PO₄ pKa = 2.12CH₃COOH pKa = 2.8HCOOH pKa = 3.74The acid with the pKa closest to the desired pH is HCOOH, so the correct answer is e. HCOOH and NaCHOO
What is the difference between conjugate acid-base pair?
a. a H atom. c. a mole water
b. a H+ ion d. a OH– ion
Answer:
b. a H+ ion
Explanation:
The concept of conjugate acid-base pair is related to Bronsted-Lowry acid-base theory and according to this theory, acid is a proton acceptor.
In short,
conjugate base is formed when an acid donates a proton.
conjugate acid is formed when a base accepts a proton.
A quantity of 1.435 g of naphthalene , was burned in a constant-volume bomb calorimeter. Consequently, the temperature of the water rose from 20.28oC to 25.95oC If the heat capacity of the bomb plus water was , calculate the heat of combustion of naphthalene on a molar basis; that is, find the molar heat of combustion.
Answer:
molar heat of combustion = -5156 *10³ kJ/mol
Explanation:
A quantity of 1.435 g of naphthalene , was burned in a constant-volume bomb calorimeter. Consequently, the temperature of the water rose from 20.28oC to 25.95oC If the heat capacity of the bomb plus water was 10.17 kJ/°C, calculate the heat of combustion of naphthalene on a molar basis; that is, find the molar heat of combustion.
Step 1: Data given
Mass of naphthalene = 1.435 grams
Initial temperature of water = 20.28 °C
Final temperature of water = 25.95 °C
heat capacity of the bomb plus water was 10.17 kJ/°C
Molar mass naphtalene = 128.2 g/mol
Step 2:
Qcal = Ccal * ΔT
⇒with Qcal =the heat of combustion
⇒with Ccal = heat capacity of the bomb plus water = 10.17 kJ/°C
⇒with ΔT = the difference in temperature = T2 - T1 = 25.95 - 20.28 = 5.67°C
Qcal = 10.17 kJ/°C * 5.67 °C
Qcal = 57.7 kJ
Step 3: Calculate moles
Moles naphthalene = 1.435 grams / 128.2 g/mol
Moles naphthalene = 0.01119 moles
Step 4: Calculate the molar heat of combustion
molar heat of combustion = Qcal/ moles
molar heat of combustion = -57.7 kJ/ 0.01119 moles
molar heat of combustion = -5156 *10³ kJ/mol
Suppose you ran this reaction without triethylamine and simply used an excess of reactant 1. At the end of the reaction, your methylene chloride solution would contain mostly reactant 1 and the product. What would you do to remove reactant 1 from the solution
ummm is that chemistry?
Answer:
is this chem
Explanation:
#6 and #7. How many carbon atoms are in a mixture of 7.00 mol c2F2 and 0.400 mol carbon dioxide and also #7
Answer:
#6 8.67x10²⁴ atoms
#7
1. Atom
2. Formula unit
3. Molecule
4. Ion
Explanation:
#6 First we calculate how many carbon moles are there in 7.00 moles of C₂F₂, keeping in mind that there are 2 C moles per C₂F₂ mol:
7.00 mol C₂F₂ * 2 = 14.00 mol CAs for carbon dioxide, there are 0.400 C moles in 0.400 moles of CO₂.
We calculate the total number of C moles:
14.00 mol + 0.400 mol = 14.4 mol CFinally we calculate the number of atoms in 14.4 C moles, using Avogadro's number:
14.4 mol * 6.023x10²³ atoms/mol = 8.67x10²⁴ atoms#7
1. Radon - Atom (Ra)2. Formula unit (It is a crystalline solid, BaBr₂)3. Molecule (NH₃)4. Ion (It has a formal charge, +2)1.rain pours from the sky
2.leaves of the plant dried
3.fluffy clouds form in the sky
4.bathing suit dries after swim
5.water puddles disappear
A.Evaporation
B.Condensation
C.Precipitation
D.Transpiration
Yan po pag pipilian
Answer:
1.Precipitation
2.Transpiration
3.Condensation
4.Evaporation
5.Evaporation
3.Condensation
Explanation:
Rain pours from the sky occurs due to the process of precipitation, leaves of the plant dried due to the process of transpiration in which the water is evaporated from the body of plant, fluffy clouds form in the sky occurs in the process of condensation, bathing suit dries after swim is due to evaporation in which water is removed and goes into the atmosphere and water puddles disappear due to the process of evaporation. Evaporation is the removal of water from the any surface whereas transpiration is the removal of water from plant body parts.
what is the machine used to check melting point called?
Answer:
Melting-point apparatus
Which equation obeys the law of conservation of
mass?
Answer:2C4H10+2C12+12O2 4CO2+CC14+H20
bio-chemisty of protain
Answer:
Protein biochemistry is the study of proteins. Protein biochemistry is a scientific field dedicated to the study of proteins, complex chains of amino acids which make up the building blocks of all living organisms.
Explanation:
I hope that helped
Copy and Pasted!
Answer:
Listen to what guy said on top.
Explanation:
polypeptide structures consisting of one or more long chains of amino acids residue.....
or my answer
A scientific hypothesis is
ANSWER:
predictive.
testable.
explanatory.
all of the above.
Answer:
All of the above.
Explanation:
For a scientific hypothesis to be considered a hypothesis, it has to be testable. When conducting a lab experiment, it also allows the tester to predict what might occur during and after the experimentation. They are also explanatory. For example, theories are hypotheses that have been verified and can explain why something in nature takes place.
Que es la actividad física y en qué mejora
Given the following list of densities, which materials would float in a molten vat of lead provided that they do not themselves melt? Densities (g/mL): lead = 11.4, glass = 2.6, gold = 19.3, charcoal = 0.57, platinum = 21.4.
a. gold and platinum
b. glass and charcoal
c. gold, platinum, glass and coal
d. gold and charcoal
e. None of these
Answer:
b. glass and charcoal
Explanation:
Step 1: Given data
Density of Pb: 11.4 g/mLDensity of Glass: 2.6 g/mLDensity of Au: 19.3 g/mLDensity of charcoal: 0.57 g/mLDensity of platinum: 21.4 g/mLStep 2: Determine which material will float in molten lead
Density is an intrinsic property of matter. Less dense materials float in more dense materials. The materials whose density is lower than that of lead and will therefore float on it are glass and charcoal.
once a recrystallization is completed and filtered, what solvent would be suitable for transferring the leftover solids to filtration funnel
Answer:
To transfer leftover solids to the filtration funnel and wash out crystals after recrystallization, ice cold methanol should be used (the mother liquor used for recrystallization).
Explanation:
Hope this helped
Draw structures corresponding to the following IUPAC names:(a) (Z)-2-Ethyl-2-buten-1-ol (b) 3-Cyclohexen-1-ol(c) trans-3-Chlorocycloheptanol (d) 1,4-Pentanediol(e) 2,6-Dimethylphenol (f ) o-(2-Hydroxyethyl)phenol
Answer:
Draw structures corresponding to the following IUPAC names:(a) (Z)-2-Ethyl-2-buten-1-ol (b) 3-Cyclohexen-1-ol(c) trans-3-Chlorocycloheptanol (d) 1,4-Pentanediol(e) 2,6-Dimethylphenol (f ) o-(2-Hydroxyethyl)phenol
Explanation:
According to IUPAC rules, the name of a compound is:
Prefix+root word+suffix
1) Select the longest carbon chain and it gives the root word.
2) The substituents give the prefix.
3) The functional group gives the secondary suffix and the type of carbon chain gives the primary suffix.
The structure of the given compounds are shown below:
