Answer: 2681.81
Explanation:
What do you need to measure the amount of energy in food
Answer:
u measure how much power it has
Explanation:
for example u can power a light bulb woth it if u can it was 100eg energy
Question 11
4 pts
Using the formula 2H202 --> 2H2O + O2, if 7.30 moles of peroxide are
decomposed, how many moles of oxygen will be formed?
Answer:
3.65 mol O₂
Explanation:
Step 1: RxN
2H₂O₂ → 2H₂O + O₂
Step 2: Define
Given - 7.30 mol H₂O₂
Solve - x mol O₂
Step 3: Stoichiometry
[tex]7.30 \hspace{3} mol \hspace{3} H_2O_2(\frac{1 \hspace{3} mol \hspace{3} O_2}{2 \hspace{3} mol \hspace{3} H_2O_2} )[/tex] = 3.65 mol O₂
What can the chemical formula tell us about a compound?
Answer:
A chemical formula tells us the number of atoms of each element that is in a compound. It contains the symbols of the atoms for the elements present in the compound as well as how many there are for each element in the form of subscripts.
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An ideal gaseous reaction (which is a hypothetical gaseous reaction that conforms to the laws governing gas behavior) occurs at a constant pressure of 35.0 atm and releases 74.6 kJ of heat. Before the reaction, the volume of the system was 8.20 L . After the reaction, the volume of the system was 2.80 L . Calculate the total internal energy change, ΔE, in kilojoules.
Answer:
ΔU = −55.45 kJ
Explanation:
From first law of thermodynamics in chemistry, we have;
ΔU = Q + W
where;
ΔU is change in internal energy
Q is the net heat transfer
W is the net work done
We are given;
Q = 74.6 kJ
But Q will be negative since heat is released
Thus;
ΔU = -74.6 kJ + W
We are given;
Constant pressure; P = 35 atm = 35 × 101325 = 3546375 N/m²
Volume before reaction; Vi = 8.2 L = 0.0082 m³
Volume after reaction; V_f = 2.8 L = 0.0028 m³
Now,
W = -P(V_f - V_i)
W = - 3546375(0.0028 - 0.0082)
W = 19.15 KJ
Thus;
ΔU = Q + W
ΔU = -74.6 kJ + 19.15 KJ =
ΔU = −55.45 kJ
Calculate the pH of a solution containing a caffeine concentration of 455 mg/L . Express your answer to one decimal place.
Answer:
Explanation:
Caffeine is a weak base with pKb = 10.4
Kb = 10⁻¹⁰°⁴ = 3.98 x 10⁻¹¹
molecular weight of caffeine = 194.2
455 x 10⁻³ g / L = 455 x 10⁻³ / 194.2 moles / L
concentration of given solution a = 2.343 x 10⁻³ M
Let the caffeine be represented by B .
B + H₂O = BH + OH⁻
a - x x x
x² / ( a - x ) = Kb
x² / ( a - x ) = 3.98 x 10⁻¹¹
x is far less than a so a -x is almost equal to a
x² = 3.98 x 10⁻¹¹ x 2.343 x 10⁻³ = 9.32 x 10⁻¹⁴
x = 3.05 x 10⁻⁷
[ OH⁻ ] = 3.05 x 10⁻⁷
pOH = - log ( 3.05 x 10⁻⁷ )
= 7 - log 3.05
= 7 - 0.484 = 6.5
pH = 14 - 6.5 = 7.5
The pH of 455 mg/L of caffeine is 7.5
Using the formula;
Mass concentration = molar concentration × molar mass
Molar mass of caffeine = 194 g/mol
Mass concentration of caffeine = 455 mg/L
Molar concentration = Mass concentration/molar mass
Molar concentration = 455 × 10^-3g/L/194 g/mol
= 0.00235 M
Let Caffeine by depicted by the general formula BH
We can now set up the ICE table as follows;
:B + H2O ⇄ BH + OH^-
I 0.00235 0 0
C - x +x +x
E 0.00235 - x x x
Note that water is present in large excess
Again; pKb of caffeine =10.4
Kb = Antilog[-pKb]
Kb = Antilog [-10.4]
Kb = 3.98 × 10^-11
Kb = [BH] [OH^-]/[:B]
3.98 × 10^-11 = [x] [x]/[ 0.00235 - x ]
3.98 × 10^-11 [ 0.00235 - x ] = [x] [x]
9.4 × 10^-14 - 3.98 × 10^-11x = x^2
x^2 + 3.98 × 10^-11x - 9.4 × 10^-14 = 0
x = 3.1 × 10^-7 M
Recall [BH] = [OH^-] = 3.1 × 10^-7 M
Now;
pOH = - log [OH^-]
pOH = log [3.1 × 10^-7 M]
pOH = 6.5
But;
pH + pOH = 14
pH = 14 - pOH
pH = 14 - 6.5
pH = 7.5
The pH of 455 mg/L of caffeine is 7.5
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Missing parts
Caffeine (C8H10N4O2) is a weak base with a pKb of 10.4. Calculate the pH of a solution containing a caffeine concentration of 455 mg/L.
1. Which statement describes a compound?
A. It contains a solute.
B. Its composition can vary.
C. Its combination of atoms never changes.
D. Its components keep separate properties.
2. Which item is NOT a type of matter?
A. force
B. mixture
C. element
D. compound
3. Which combination can be used to classify all the matter on Earth?
A. forces and energy
B. atoms and elements
C. solvents and solutes
D. substances and mixtures
Scientists are experimenting with pure samples of isotope X which is radioactive. The sample has a mass of 20. Grams. The half-life was measured to be 232 seconds. There is a second sample that weighs 80 grams. What is the half-life of the second sample
Answer:
Explanation:
Half life of radioactive materials do not depend upon the mass of the material . It only depends upon the nature of radioactive materials . The half life of 20 g is 232 seconds . That means 20 gram will be reduced to 10 gram in 232 seconds .
Half life of 80 gram is also 232 seconds . So , 80 gram will be reduced to 40 gram in 232 second .
Calcium reacts with sulfur forming calcium sulfide. What is the theoretical yield (g) of CaS(s) that could be prepared from 7.19 g of Ca(s) and 2.67 g of sulfur(s)? Enter your answer with two decimal places. Do not type units with your answer.
Answer:
The theoretical yield of CaS is 6.01 g.
Explanation:
The balanced reaction is given as:
[tex]Ca+S\rightarrow CaS[/tex]
The molar mass of Ca and S is 40.08 and 32.065 g/mol respectively.
Number of moles = [tex]\frac{Mass}{Molar Mass}[/tex]
So, 7.19 g of Ca contains [tex](\frac{7.19}{40.08})[/tex] mol of Ca or 0.179 mol of Ca
Also, 2.67 g of S contains [tex](\frac{2.67}{32.065})[/tex] mol of S or 0.0833 mol of S
According to the balanced equation:
1 mol of Ca produces 1 mol of CaS
So, 0.179 mol of Ca produces 0.179 mol of CaS
According to the balanced equation:
1 mol of S produces 1 mol of CaS
So, 0.0833 mol of S produces 0.0833 mol of CaS
As the least number of mol of CaS (product) is produced from S , therefore, S is the limiting reactant.
So, thoretically, 0.0833 mol of CaS is produced.
The molar mass of CaS is 72.143 g/mol.
So, the mass of 0.0833 mol of CaS is [tex](0.0833\times 72.143)[/tex] g or 6.01 g
Hence, the theoretical yield of CaS is 6.01 g.
a change of matter is a physical change
True or False
Answer:
true
Explanation:
hope it helps
Answer:
I'm pretty sure it's true
Explanation:
20 characters
What is the pressure if the height of a column of mercury is 0.20 m and the density of mercury is 13,600 kg/m3? (remember, gravity is 9.81 m/s2)
Answer:
[tex]p=26683.2Pa[/tex]
Explanation:
Hello,
In this case, since the pressure is computed via:
[tex]p=h*\rho*g[/tex]
Whereas h is the 0.520-m height, [tex]\rho[/tex] is the 13600-kg/m³ density and the g the 9.81-m/s² gravity. Thus, the pressure in Pa is:
[tex]p=0.20m*13,600 \frac{kg}{m^3} *9.81\frac{m}{s^2} \\\\p=26683.2\frac{kg*\frac{m}{s^2} }{m^2} =26683.2\frac{N}{m^2}\\ \\p=26683.2Pa[/tex]
Best regards.
The normal boiling point of benzene is 80.1°C. What is its enthalpy of vaporization if the vapor pressure at 26.1°C is 100 torr?
The heat of vaporization of benzene is required.
The heat of vaporization of benzene is 33009 J/kg.
[tex]T_0[/tex] = Normal boiling point = 80.1+273.15 K
[tex]T_B[/tex] = Boiling point at given pressure = 26.1+273.15 K
[tex]R[/tex] = Gas constant = 8.314 J/mol K
[tex]P[/tex] = Pressure at given [tex]T_B[/tex] = 100 torr
[tex]\Delta H[/tex] = Heat of vaporization
From the Clausius–Clapeyron equation
[tex]\dfrac{1}{T_B}=\dfrac{1}{T_0}-\dfrac{R\ln(\dfrac{P}{P_0})}{\Delta H}\\\Rightarrow \Delta H=\dfrac{R\ln\dfrac{P}{P_0}}{\dfrac{1}{T_0}-\dfrac{1}{T_B}}\\\Rightarrow \Delta H=\dfrac{8.314\times \ln\left(\frac{100}{760}\right)}{\frac{1}{80.1+273.15}-\frac{1}{26.1+273.15}}\\\Rightarrow \Delta H=33008.99\ \text{J/kg}[/tex]
The heat of vaporization of benzene is 33009 J/kg.
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help me please loves
Answer:
K and ba
Explanation:
Answer: K and Ba
Explanation:
convert 575.1 mmHg to atm
Answer:
= .7567105263
Explanation:
1 atm = 760 mmHg
575.1 mmHg (1 atm/760mmHg) = .7567105263 atm
why are copper pipes used in place of old lead pipes for plumbing systems?
A sample of an ideal gas has a volume of 2.23 L at 289 K and 1.05 atm. Calculate the pressure when the volume is 1.08 L and the temperature is 304 K. P= atm
Answer:
2.28 atm
Explanation:
V₁ = 2.33L, V₂ = 1.08L
T₁ = 289K, T₂ = 304K
P₁ = 1.05 atm, P₂ = ?
Where V₁ and V₂ are initial and final volume respectively
T₁ and T₂ are initial and final temperature respectively
P₁ and P₂ are initial and final pressure respectively
The formula to be used here is the general gas equation:
P₁V₁/T₁=P₂V₂/T₂
1.05 × 2.23/289 = P₂ × 1.08/304
P₂ × 1.08 × 289 = 1.05 × 2.23 × 304
P₂ = (1.05 × 2.23 ×304) ÷ (1.08 × 289)
P₂ = 711.82 ÷ 312.12
P₂ = 2.28 atm
How many moles of precipitate will be formed when 100.0 mL of 0.200 M NaBr is reacted with excess Pb(NO₃)₂ in the following chemical reaction?
2 NaBr (aq) + Pb(NO₃)₂ (aq) → PbBr₂ (s) + 2 NaNO₃ (aq)
Answer : The number of moles of precipitate, [tex]PbBr_2[/tex] formed will be 0.01 moles.
Explanation : Given,
Concentration of NaBr = 0.200 M
Volume of solution = 100.0 mL = 0.1 L (1 L = 1000 mL)
First we have to calculate the moles of NaBr.
[tex]\text{Moles of NaBr}=\text{Concentration of NaBr}\times \text{Volume of solution in L}[/tex]
[tex]\text{Moles of NaBr}=0.200M\times 0.1L=0.02mol[/tex]
Now we have to calculate the moles of precipitate, [tex]PbBr_2[/tex] formed.
The balanced chemical reaction is:
[tex]2NaBr(aq)+Pb(NO_3)_2(aq)\rightarrow PbBr_2(s)+2NaNO_3(aq)[/tex]
From the balanced chemical reaction we conclude that:
As, 2 moles of NaBr react to give 1 mole of [tex]PbBr_2[/tex]
So, 0.02 moles of NaBr react to give [tex]\frac{0.02}{2}=0.01[/tex] mole of [tex]PbBr_2[/tex]
Therefore, the number of moles of precipitate, [tex]PbBr_2[/tex] formed will be 0.01 moles.
The number of mole of the precipitate (i.e PbBr₂) formed when 100 mL of 0.2 M NaBr react with excess Pb(NO₃)₂ is 0.01 mole
We'll begin by calculating the number of mole of NaBr in 100 mL of 0.2 M NaBr solution. This can be obtained as follow:Volume = 100 mL = 100 / 1000 = 0.1 L
Molarity of NaBr = 0.2 M
Mole of NaBr =?Mole = Molarity x Volume
Mole of NaBr = 0.2 × 0.1
Mole of NaBr = 0.02 mole Finally, we shall determine the number of mole of the precipitate (i.e PbBr₂) produced from the reaction. This can be obtained as follow:2NaBr(aq) + Pb(NO₃)₂(aq) → PbBr₂(s) + 2NaNO₃ (aq)
From the balanced equation above,
2 moles of NaBr reacted to produce 1 mole of PbBr₂.
Therefore,
0.02 mole of NaBr will react to produce = [tex]\frac{0.02}{2} \\\\[/tex] = 0.01 mole of PbBr₂.
Thus, the number of mole of the precipitate (i.e PbBr₂) produced from the reaction is 0.01 mole
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Which of the following is an Elementary compound?
A. CO2
B. N2
C. SO2
D. H2S
heeeeeeeeeelp please please please
Answer:
Explanation:
In my opinion the answer should be SO2
Answer:
a should be answer i think.
Which of the following is NOT a strong electrolyte?
:
Answers:
Na2SO4
KI
CaCl2
LiOH
C6H1206
Answer:
C6H1206
Explanation:
C6H12O6 is a monomer of carbohydrates also known as glucose, so it is not an electrolyte at all.
Which of the following is a good definition of matter?
O A. Anything that is made up of light and gravity
O B. Anything that has mass and takes up space
O C. Anything that produces heat and mass
O D. Anything that has energy and creates heat
Answer:
B
Explanation:
I did the question before and got it right.
Based on the visible cell structure, which of the following statements is true?
Answer:I think it would be the third one.
Explanation:
True or False: Particles that are moving faster have a higher temperature
Answer:
true
Explanation:
I'm not sure why cause I dont know how to explain but it's TRUE
Answer:
True
Explanation:
The particles moving faster in a substance the hotter it gets.
What occurs after cytokinesis is completed at the end of meiosis I?
O Four haploid cells are formed.
O Two diploid cells are formed.
OTwo haploid cells are formed.
O Four diploid cells are formed.
Answer. After cytokinesis is completed at end of meiosis - I two haploid cells are formed.on:
Answer:
C. TWO HAPLOID CELLS ARE FORMED
Explanation:
I TOOK THE EDGUNITY TEST AND I GOT IT CORRECT
What happens when the elements in group 2 react with water?
Answer:
The Group 2 metals become more reactive towards the water as you go down the Group.
Explanation:
These all react with cold water with increasing vigour to give the metal hydroxide and hydrogen. ... You get less precipitate as you go down the Group because more of the hydroxide dissolves in the water. Summary of the trend in reactivity.
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________ and ________ fibres are obtained from animals
Answer:
Explanation:
Silk and wool
Answer:
silk and wool fibers are obtained from animals
Explanation:
good luck :)
Determine the value of the equilibrium constant (report your answer to three significant figures) for the following reaction if an equilibrium mixture contains 0.010 mol of solid PbBr2, and is 0.0100 M in Pb2+ ions and 0.0250 M in Br1- ions. Use the notation 4.31e-5 to indicate a number such as 4.31 x 10-5.
Answer:
6.25e-6 is the value of the equilibrium constant
Explanation:
we have this equation
[tex]PbBr(s) ----- Pb^{2+}(aq) + 2Br(aq)[/tex]
When at a state of equilibrium,
we have the concentration of Pb^2+ to be 0.01
we have the concentration of Br^- to be 0.025
the equilibrium constant concentration of both pure solids and liquid s are said to be equal to 1
[PbBR2] = 1
such tht
Keq = [Pb^2+] x [Br-]^2
we already know the values of these from the above.
0.01x0.025^2
= 0.01 x 0.000625
= 0.00000625
= 6.25 x 10^-6
= 6.25e^-6
A sample of an unknown gas weighs 0.419 grams and produced 5.00L of gas at 1.00atm (unknown gas only) and 298.15K, what is the molar mass (g/mole) of this unknown gas
Answer:
molar mass of unknown gas = 1.987 g/mol
Explanation:
First, the number of moles of the unknown gas is found
Using the ideal gas equation: PV = nRT
P = 1.00 atm, V = 5.00 L, T = 298.15 K, R = 0.082 L.atm.mol⁻¹K⁻¹
n = PV/RT
n = (1.00 atm * 5.00 L)/(298.15 K *0.082 L.atm.mol⁻¹K⁻¹)
n = 0.2109 moles
Molar mass = mass/ number of moles
molar mass = 0.419 g/ 0.2109 mols
molar mass of unknown gas = 1.987 g/mol
The molar mass of unknown gas by using ideal gas equation = 1.987 g/mol.
Ideal gas equationThis equation gives the relation between pressure, volume, temperature as given below:
[tex]PV = nRT[/tex]
P = 1.00 atm, V = 5.00 L, T = 298.15 K, R = 0.082 L.atm.mol⁻¹K⁻¹
Substitute the above values in the above equation as follows:
n = (1.00 atm * 5.00 L)/(298.15 K *0.082 L.atm.mol⁻¹K⁻¹)
n = 0.2109 moles
Formula for molar mass[tex]Molar mass = mass/ number of moles[/tex]
Calculate molar mass by using the above equation,
molar mass = 0.419 g/ 0.2109 moles
The molar mass of unknown gas = 1.987 g/mol
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Label the parts of the electric circuit that best correspond to the heart, arteries, veins, and cells.
Answer:
1 ➡️ Cells
2 ➡️ Arteries
3 ➡️ Veins
4 ➡️ Heart
Explanation:
The parts of the electric circuit that best correspond to the heart, arteries, veins, and cells have been properly labeled.
The circulatory system involves the transportation of nutrients, oxygen and water by blood to other the parts of the body.
From the electric circuit, we see that arteries transport blood away from the heart to the other cells in the body. The veins actually return the blood back to the heart from the cells. The heart pumps the blood
The electric circuity diagram has the label 1 bulb analogous to cell, label 2 analogous to arteries, label 3 analogous to veins, and label 4 cell analogous to heart.
What is an electric circuit?The electric circuit has been given as the power source and the conducting wires that allows the flow of the current in the circuit.
In the human body, the heart has been transported the oxygenated blood through the arteries to the cell and carried the deoxygenated blood from the cells back to the heart via veins.
In the circuit, the battery has been the source of the power/blood. The current has been carried from the heart to the cell/bulb through the arteries labeled, 2, and transported back to the battery via veins labeled 3.
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The smallest form of matter that still retains the properties of an element
Answer:
atom
Explanation:
the atom is the smallest form.
What is the volume of a substance that has a mass of 59 g and a density of 1.98 g/mL?
(show all work)
Answer:
29.8
Explanation:
The formula for volume is mass/ density, so 59/1.98. 29.8 is the answer.
A chemical engineer must calculate the maximum safe operating temperature of a high-pressure gas reaction vessel. The vessel is a stainless-steel cylinder that measures 41.0cm wide and 49.2cm high. The maximum safe pressure inside the vessel has been measured to be 3.70MPa. For a certain reaction the vessel may contain up to 2.50kg of dinitrogen difluoride gas.
Required:
Calculate the maximum safe operating temperature the engineer should recommend for this reaction. Write your answer in degrees Celsius. Be sure your answer has the correct number of significant digits.
Answer:
[tex]T=2.78x10^3 \°C[/tex]
Explanation:
Hello,
In this case, considering that the safe temperature may be computed via the ideal gas law as we now the pressure, mass and volume via the dimensions:
[tex]V=\pi r^2 h=\pi *(41.0cm)^2*49.2cm=2.60x10^5cm^3*\frac{1L}{1000cm^3} =260L[/tex]
The pressure in atm is:
[tex]P=3.70MPa*\frac{1x10^6Pa}{1MPa} \frac{1atm}{101325Pa} =36.5atm[/tex]
And the moles considering the mass and molar mass (66 g/mol) of dinitrogen difluoride (N₂F₂):
[tex]n_{N_2F_2}=2.50kg*\frac{1000g}{1kg}*\frac{1mol}{66g} =37.9mol[/tex]
In sich a way, by applying the ideal gas equation, which is not the best assumption but could work as an approximation due to the high temperature, the temperature, with three significant figures, will be:
[tex]T=\frac{PV}{nR}=\frac{36.5Pa*260L}{37.9mol*0.082\frac{atm*L}{mol*K} }\\ \\T=3053.6K-273.15\\\\T=2.78x10^3 \°C[/tex]
Best regards.