Each triglyceride is different from the others on the basis of presence of fatty acids in it.
What are triglycerides?Triglycerides is a kind of fat and derivative of ester which is formed by the combination of glycerol and three fatty acids.
So in the triglyceride molecule three sub divided parts are present due to the presence of three fatty acids groups and these fatty acids will make difference in each triglyceride molecules.
Hence of fatty acids in triglyceride molecule makes it different from other.
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Answer:
It has different fatty acids.
Explanation:
This is a signature of triglycerides
Balance the chemical equation below using the smallest possible whole number stoichiometric coefficients.
P4(s) + NaOH(aq) + H2O(l) -> Ph3(g) + Na2HPO3(aq)
Answer:
I used a,b c, d in the equation as substituted coefficients to find the unknown for each element of P, Na, O, H, and I got
P4(s) + 4NaOH(aq) + 2H20(l)---->2Ph3 +2Na2HPO3(aq).
which I think should be the answer.
what is the machine used to check melting point called?
Answer:
Melting-point apparatus
How many liters of hydrogen can be produced at a pressure of 2 atm and a temperature of 298 K
Answer:
1.17 L of H₂
Explanation:
We'll begin by calculating the number of mole in 2.3 g of Mg. This can be obtained as follow:
Mass of Mg = 2.3 g
Molar mass of Mg = 24 g/mol
Mole of Mg =?
Mole = mass /molar mass
Mole of Mg = 2.3 / 24
Mole of Mg = 0.096 mole
Next, we shall determine the number of mole of H₂ produced by the reaction of 2.3 g (i.e 0.096 mole) of Mg. This can be obtained as follow:
Mg + 2HCl —> MgCl₂ + H₂
From the balanced equation above,
1 mole of Mg reacted to 1 mole of H₂.
Therefore, 0.096 mole of Mg will also react to produce 0.096 mole of H₂.
Finally, we shall determine volume of H₂ produced from the reaction. This can be obtained as follow:
Number of mole (n) of H₂ = 0.096 mole
Pressure (P) = 2 atm
Temperature (T) = 298 K
Gas constant (R) = 0.0821 atm.L/Kmol
Volume (V) of H₂ =?
PV = nRT
2 × V = 0.096 × 0.0821 × 298
Divide both side by 2
V = (0.096 × 0.0821 × 298) /2
V = 1.17 L
Therefore, 1.17 L of H₂ were obtained from the reaction.
Compound X has a molar mass of 316.25 g*mol^-1 and the following composition:
element & mass %
phosphorus & 39.18%
sulfur & 60.82%
Write the molecular formula of X.
Answer:
Compound X has a molar mass of 316.25 g*mol^-1 and the following composition:
element & mass %
phosphorus & 39.18%
sulfur & 60.82%
Write the molecular formula of X.
Explanation:
The given molecule of phosphorus and sulfur has molar mass --- 316.25 g.
Empirical formula calculation:
element: phosphorus sulfur
co9mposition: 39.185% 60.82%
divide with
atomic mass: 39.185/31.0 g/mol 60.82/32.0g/mol
=1.26mol 1.90mol
smallest mole ratio: 1.26mol/1.26mol =1 1.90mol/1.26 mol =1.50
multiply with 2: 2 3
Hence, the empirical formula is:
P2S3.
Mass of empirical formula is:
158.0g/mol
Given, molecule has molar mass --- 316.25 g/mol
Hence, the ratio is:
316.25g/mol/158.0 =2
Hence, the molecular formula of the compound is :
2 x (P2S3)
=[tex]P_4S_6[/tex]
1. What is the equivalent pressure measurement in mmHg of 2.50 atm?
Answer:
Atmosphere to mmHg Conversion Example. Task: Convert 8 atmospheres to mmHg (show work) Formula: atm x 760 = mmHg Calculations: 8 atm x 760 = 6,080 mmHg Result: 8 atm is equal to 6,080 mmHg.
Explanation:
This answer is helpfull for you I nowwhy might the melting point of the crystals obtained in this experiment be close to but below one of the reference melting points and melt slowly over several degrees
Answer: hello the experiment related to your question is missing but I will provide a more general answer within the scope of your question
answer :
presence of Impurities
Explanation:
The melting point of the crystals as obtained in the experiment will be close to but below reference melting points and will also melt slower because of the presence of impurities in the compound
Impurities alter the melting and freezing points from ideal freezing and melting points of compounds
6ooo kg into quintal
Answer:
60 q
Explanation:
The conversion factor is 100; so 1 quintal = 100 kilograms. In other words, the value in q multiply by 100 to get a value in kg.
g A piece of solid Zn metal is put into an aqueous solution of Cu(NO3)2. Write the net ionic equation for any single-replacement redox reaction. (Use the lowest possible coefficients for the reaction. Be sure to specify states such as (aq) or (s). If a box is not needed, leave it blank. If no reaction occurs, leave all boxes blank and click on Submit.)
Answer:
Zn(s) + Cu²⁺(aq) ⇒ Zn²⁺(aq) + Cu(s)
Explanation:
Let's consider the molecular single displacement equation between Zn and Cu(NO₃)₂
Zn(s) + Cu(NO₃)₂(aq) ⇒ Zn(NO₃)₂(aq) + Cu(s)
The complete ionic equation includes all the ions and insoluble species.
Zn(s) + Cu²⁺(aq) + 2 NO₃⁻(aq) ⇒ Zn²⁺(aq) + 2 NO₃⁻(aq) + Cu(s)
The net ionic equation includes only the ions that participate in the reaction and insoluble species.
Zn(s) + Cu²⁺(aq) ⇒ Zn²⁺(aq) + Cu(s)
Silver has two naturally occurring isotopes with the following isotopic masses: 10747Ag 10947Ag 106.90509 108.9047 The average atomic mass of silver is 107.8682 amu. The fractional abundance of the lighter of the two isotopes is ________.
Answer: The fractional abundance of lighter isotope is 0.518
Explanation:
Average atomic weight is the sum of the masses of the individual isotopes each multiplied by its fractional abundance. The equation used is:[tex]\text{Average atomic weight}=\sum_{i=1}^{n}\text{(Atomic mass of isotope)}_i\times \text{(Fractional abundance)}_i[/tex] ......(1)
Let the fractional abundance of Ag-107 isotope be 'x'
For Ag-107 isotope:Atomic mass = 106.90509 amu
Fractional abundance = x
For Ag-109 isotope:Atomic mass = 108.9047 amu
Fractional abundance = (1 - x)
Average atomic mass of silver = 107.8682 amu
Plugging values in equation 1:
[tex]107.8682=(106.90509 \times x) + (108.9047 \times (1-x))\\\\107.8682=106.90509x+108.9047-108.9047x\\\\1.99961x=1.0365\\\\x=0.518[/tex]
Fractional abundance of Ag-107 isotope (lighter) = x = 0.518
Hence, the fractional abundance of lighter isotope is 0.518
Aqueous hydrobromic acid (HBr) will react with solid sodium hydroxide (NaOH) to produce aqueous sodium bromide (NaBr) and liquid water (H2O). Suppose 5.7 g of hydrobromic acid is mixed with 0.980 g of sodium hydroxide. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to significant digits.
Answer:
The maximum mass of water that could be produced by the chemical reaction=0.441g
Explanation:
We are given that
Given mass of HBr=5.7 g
Given mass of sodium hydroxide=0.980 g
Molar mass of HBr=80.9 g/ Mole
Molar mass of NaOH=40 g/mole
Molar mass of H2O=18 g/mole
Reaction
[tex]HBr+NaOH\rightarrow H_2O+NaBr[/tex]
Number of moles=[tex]\frac{given\;mass}{molar\;mass}[/tex]
Using the formula
Number of moles of HBr=[tex]\frac{5.7}{80.9}=0.0705 moles[/tex]
Number of moles of NaOH=[tex]\frac{0.980}{40}=0.0245moles[/tex]
Hydrogen bromide is in a great excess and the amount of water produced.
Therefore,
Number of moles of water, n(H2O)=Number of moles of NaOH=0.0245moles
Now,
Mass of water=[tex]n(H_2O)\times Molar\;mass\;of\;water[/tex]
Mass of water=[tex]0.0245moles\times 18=0.441g[/tex]
Hence, the maximum mass of water that could be produced by the chemical reaction=0.441g
Consider the balanced chemical equation below.
3 A ⟶ C + 4 D
How many moles of C would be produced if 7 moles of A were used?
Answer:
2.33 mol C
Explanation:
Step 1: Write the balanced generic chemical equation
3 A ⟶ C + 4 D
Step 2: Establish the appropriate molar ratio
According to the balanced equation, the molar ratio of A to C is 3:1.
Step 3: Calculate the number of moles of C produced from 7 moles of A
We will use the previously established molar ratio.
7 mol A × 1 mol C/3 mol A = 2.33 mol C
1. Show that heat flows spontaneously from high temperature to low temperature in any isolated system (hint: use entropy change that occurs during the process for your proof).
2. Work out the entropy change for the decomposition of mercuric oxide using mathematical and graphical arguments.
Answer:
1 ) Δs ( entropy change for hot block ) = - Q / th ( -ve shows heat lost to cold block )
Δs ( entropy change for cold block ) = Q / tc
∴ Total Δs = ΔSc + ΔSh
= Q/tc - Q/th
2) ΔSdecomposition = Δh / Temp = ( 181.6 * 10^3 / 773 ) = 234.928 J/k
Explanation:
1) To show that heat flows spontaneously from high temperature to low temperature
example :
Pick two(2) solid metal blocks with varying temperatures ( i.e. one solid block is hot and the other solid block is cold )
Place both blocks for time (t ) in an insulated system to reduce heat loss or gain to or from the environment
Check the temperature of both blocks after time ( t ) it will be observed that both blocks will have same temperature after time t ( first law of thermodynamics )
Δs ( entropy change for hot block ) = - Q / th ( -ve shows heat lost to cold block )
Δs ( entropy change for cold block ) = Q / tc
∴ Total Δs = ΔSc + ΔSh
= Q/tc - Q/th
2) Entropy change for Decomposition of mercuric oxide
2HgO (s) → 2Hg(l) + O₂ (g)
Δs = positive
there is transition from solid to liquid and the melting point of mercury ( the point at which reaction will take place ) = 500⁰C
hence ΔSdecomposition = S⁻ Hg - S⁻ HgO =
Δh of reaction = 181.6 KJ
Temp = 500 + 273 = 773 k
hence ΔSdecomposition = Δh / Temp = ( 181.6 * 10^3 / 773 ) = 234.928 J/k
Calculate the osmotic pressure of 5.0g of sucrose ssolution in 1L. Answer should be in Torr
Answer: The osmotic pressure of 5.0g of sucrose solution in 1 L is 271.32 torr.
Explanation:
Given: Mass = 5.0 g
Volume = 1 L
Molar mass of sucrose = 342.3 g/mol
Moles are the mass of a substance divided by its molar mass. So, moles of sucrose are calculated as follows.
[tex]Moles = \frac{mass}{molarmass}\\= \frac{5.0 g}{342.3 g/mol}\\= 0.0146 mol[/tex]
Hence, concentration of sucrose is calculated as follows.
[tex]Concentration = \frac{moles}{Volume (in L)}\\= \frac{0.0146 mol}{1 L}\\= 0.0146 M[/tex]
Formula used to calculate osmotic pressure is as follows.
[tex]\pi = CRT[/tex]
where,
[tex]\pi[/tex] = osmotic pressure
C = concentration
R = gas constant = 0.0821 L atm/mol K
T = temperature
Substitute the values into above formula as follows.
[tex]\pi = CRT\\= 0.0146 \times 0.0821 L atm/mol K \times 298 K\\= 0.357 atm (1 atm = 760 torr)\\= 271.32 torr[/tex]
Thus, we can conclude that the osmotic pressure of 5.0g of sucrose solution in 1 L is 271.32 torr.
If a hydrogen atom and a helium atom have the same kinetic energy:________
a. the wavelength of the hydrogen atom will be about 4 times longer than the wavelength of the helium atom.
b. the wavelength of the hydrogen atom will be about 2 times longer than the wavelength of the helium.
c. the wavelength of the hydrogen atom will be roughly equal to the wavelength of the helium atom.
d. the wavelength of the helium atom will be about 2 times longer than the wavelength of the hydrogen atom.
e. the wavelength of the helium atom will be about 4 times longer than the wavelength of the hydrogen atom.
Answer: If a hydrogen atom and a helium atom have the same kinetic energy then the wavelength of the hydrogen atom will be roughly equal to the wavelength of the helium atom.
Explanation:
The relation between energy and wavelength is as follows.
[tex]E = \frac{hc}{\lambda}\\[/tex]
This means that energy is inversely proportional to wavelength.
As it is given that energy of a hydrogen atom and a helium atom is same.
Let us assume that [tex]E_{hydrogen} = E_{helium} = E'[/tex]. Hence, relation between their wavelengths will be calculated as follows.
[tex]E_{hydrogen} = \frac{hc}{\lambda_{hydrogen}}[/tex] ... (1)
[tex]E_{helium} = \frac{hc}{\lambda_{helium}}[/tex] ... (2)
Equating the equations (1) and (2) as follows.
[tex]E_{hydrogen} = E_{helium} = E'\\\frac{hc}{\lambda_{hydrogen}} = \frac{hc}{\lambda_{helium}} = E'\\\lambda_{helium} = \lambda_{hydrogen} = E'[/tex]
Thus, we can conclude that if a hydrogen atom and a helium atom have the same kinetic energy then the wavelength of the hydrogen atom will be roughly equal to the wavelength of the helium atom.
when a polar bond is formed between 2 atoms which atom receives a partial positive charge
Answer:
The more electronegative atom in a covalent bond
determine the budget for tge fitness event.consider the attendance (should tou ask for registration fee.give shirts etc)
Choose the correct statement. A) The cathode is the electrode where the oxidation takes place. B) The cathode is the electrode where the reduction takes place. C) Both oxidation and reduction may take place at the cathode, depending on the cell. D) The cathode is always positive
can anybody pls help me with chemistry how to balance an equation of class 7 ??? pls
really I don't like chemistry..
I need only the step(s)..
Answer:
[tex] This\:may\: help[/tex]
Answer:
(1) Write down the chemical reaction in the form of word equation,keeping reactants on left hand side and products on right hand side.
(2) Write symbol and formula of all reactants and products in word equation. (3) Balance the equation by multiplying the symbols and formula by smallest possible figures.
The mass of a single tantalum atom is 3.01×10-22 grams. How many tantalum atoms would there be in 37.1 milligrams of tantalum?
Answer: There are [tex]1.23 \times 10^{22}[/tex] atoms present in 37.1 mg of tantalum.
Explanation:
Given: Mass of single tantalum atom = [tex]3.01 \times 10^{-22} g[/tex]
Mass of tantalum atoms = 37.1 mg (1 mg = 0.001 g) = 0.0371 g
Therefore, number of tantalum atoms present in 0.0371 grams is calculated as follows.
[tex]No. of atoms = \frac{0.0371 g}{3.01 \times 10^{-22}}\\= 1.23 \times 10^{20}[/tex]
Thus, we can conclude that there are [tex]1.23 \times 10^{22}[/tex] atoms present in 37.1 mg of tantalum.
There are [tex]1.23\times 10^{20}[/tex] atoms of tantalum in 37.1 mg of tantalum.
Explanation:
Given:
Mass of single atom of tantalum =[tex]3.01\times 10^{-22} g[/tex]
To find:
The number of atoms of tantalum in 37.1 milligrams.
Solution:
Mass of tantalum = 37.1 mg
[tex]1 mg = 0.001 g\\37.1 mg=37.1\times 0.001 g=0.0371 g[/tex]
The number of atoms in 0.0371 grams of tantalum = N
Mass of a single atom of tantalum = [tex]3.01\times 10^{-22} g[/tex]
Then a mass of N atoms of tantalum will be:
[tex]0.0371 g=N\times 3.01\times 10^{-22} g\\N=\frac{0.0371 g}{ 3.01\times 10^{-22} g}\\=1.23\times 10^{20}[/tex]
There are [tex]1.23\times 10^{20}[/tex] atoms of tantalum in 37.1 mg of tantalum.
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Give your familiarity for following terms
1. roasting 2. smelting 4. zone refining 5. polling
Answer:
The roasting process is a delicate combination of art and science . Roasters are familiar with how the beans look and the smells Well, familiarity with the machine makes things much easier to predict, but the best way is to do many different tests .Well, familiarity with the machine makes things much easier to predict, but the best way is to do many different tests.Smelting is a process of applying heat to ore in order to extract a base metal. It is a form of extractive metallurgy. It is used to extract many metals from their ores, including silver, iron, copper, and other base metals.In zone refining, solutes are segregated at one end of the ingot in order to purify the remainder, or to concentrate the impurities. For example, in the preparation of a transistor or diode semiconductor, an ingot of germanium is first purified by zone refining. In zone refining, solutes are segregated at one end of the ingot in order to purify the remainder, or to concentrate the impurities. ... For example, in the preparation of a transistor or diode semiconductor, an ingot of germanium is first purified by zone refining.Polling is the process where the computer or controlling device waits for an external device to check for its readiness or state, often with low-level hardware. For example, when a printer is connected via a parallel port, the computer waits until the printer has received the next character.Explanation:
hope it heloed
Zn-64 = 48.63%
Zn-66 = 27.90%
Zn-67 = 4.10%
Zn-68 = 18.75%
Zn-70 = .62%
Calculate the average atomic mass/given their percent abundance
Answer:
A = 65.46 u
Explanation:
Given that,
The composition of zinc is as follows :
Zn-64 = 48.63%
Zn-66 = 27.90%
Zn-67 = 4.10%
Zn-68 = 18.75%
Zn-70 = .62%
We need to find the average atomic mass of the given element. It can be solved as follows :
[tex]A=\dfrac{48.63\times 64+27.90\times 66+4.1\times 67+18.75\times 68+0.62\times 70}{100}\\A=65.46\ u[/tex]
So, the average atomic mass of zinc is 65.46 u.
the pressure of a sample of gas at constant volume is 942 kPa at 338K. what will the pressure be 293k?
Explanation:
4 tctcgcgcgctctchvvyctctc
Which equation obeys the law of conservation of
mass?
Answer:2C4H10+2C12+12O2 4CO2+CC14+H20
Which is the primary type of radiation from the sun that is absorbed by the ozone layer?
A. infrared radiatin
B. UV-B
C. X-rays
D. UV-C
E. UV-A
the answer to the question is B.UV-B
Question 16(Multiple Choice Worth 5 points)
(04.01 LC) Which statement is true about the total mass of the reactants during a chemical change?
O It is destroyed during chemical reaction.
O It is less than the total mass of the products. O It is equal to the total mass of the products.
O It is greater than the total mass of the products.
Answer:
It is equal to the total mass of the products.
Explanation:
Hope this helps :)
The chemical formula is different from the empirical formula in
Answer:be careful and relax
Explanation:
Answer:
Hahaha be careful and relax
Calculate the displacement (the total volume of the cylinder through which the piston move) of a 5.70L automobile engine in cubic inches, (1inch=2.54cm)
Answer:
348 inches³
Explanation:
From our previous knowledge of units conversion:
We know that 1000 cm³ makes 1 Liter.
Thus, for a 5.70 L automobile engine in cubic meters will be:
= 5.70 × 1000 cm³
= 5700 cm³
Now, the displacement of the automobile in cubic inches provided that 1 inch = 2.534 cm is:
⇒ 5700× (1/ (2.54)³) in³
= 5700×0.0610 in³
= 347.7 in³
≅ 348 inches³
a. Draw 2,3-dichloro octane.
b. Write the lewis structure for H20 molecule.
Answer:
a.draw 2,3 dicholoro octane
Explanation:
mag isip ka kung paano hehe
Which Group is in the second column of the periodic table?
O A. Noble gases
O B. Alkaline earth metals
O C. Alkali metals
O D. Halogens
Answer:
O B. Alkaline earth metals
Explanation:
Noble gases → 8th column.
Alkali metal → first column.
Halogen → 7th
Answer:
B. Alkaline earth metals
Explanation:
Alkaline-earth metals: The alkaline-earth metals make up Group 2 of the periodic table, from beryllium (Be) through radium (Ra). Each of these elements has two electrons in its outermost energy level, which makes the alkaline earths reactive enough that they're rarely found alone in nature. But they're not as reactive as the alkali metals. Their chemical reactions typically occur more slowly and produce less heat compared to the alkali metals.
Consider the following equilibrium:
2H2(g)+S2(g)⇌2H2S(g)Kc=1.08×107 at 700 ∘C.
What is Kp?
Answer:
Consider the following equilibrium:
2H2(g)+S2(g)⇌2H2S(g)Kc=1.08×107 at 700 ∘C.
What is Kp?
Explanation:
Given,
[tex]Kc=1.08 * 10^7[/tex]
The relation between Kp and Kc is:
[tex]Kp=Kc * (RT)^d^e^l^t^a^(^n^)[/tex]
Where delta n represents the change in the number of moles.
For the given equation,
The Delta n = Number of moles of products - number of moles of reactants
(2-(2+1))
=-1.
Hence,
Kp=Kc/RT.
Thus,
[tex]Kp=1.08 * 10^7 / 8.314 J.K6-1.mol^-^1 x 973 K\\Kp=1335.06[/tex]
The answer is Kp=1335.06
The value of [tex]K_p[/tex] is [tex]1.35\times 10^5[/tex].
Explanation:
The relation between [tex]K_p \& K_c[/tex] is given by:
[tex]K_p=K_c(RT)^{\Delta n_g}[/tex]
Where:
[tex]K_c[/tex] = The equilibrium constant of reaction in terms of concentration
[tex]K_p[/tex] = The equilibrium constant of reaction in terms of partial pressure
R= The universal gas constant
T = The temperature of the equilibrium
[tex]n_g[/tex]= Change in gaseus moles
Given:
An equilibrium reaction, 700°C:
[tex]2H_2(g)+S_2(g)\rightleftharpoons 2H_2S(g),K_c=1.08\times 10^7[/tex]
To find:
The equilibrium constant in terms of partial pressure, [tex]K_p[/tex].
Solution:
The equilibrium constant of reaction in terms of concentration= [tex]K_c[/tex]
[tex]K_c=1.08\times 10^7[/tex]
The equilibrium constant of reaction in terms of partial pressure =[tex]K_p=?[/tex]
The gaseous moles of reactant side = [tex]n_r= 3[/tex]
The gaseous moles of product side = [tex]n_p= 2[/tex]
The temperature at which equilibrium is given = T
[tex]T = 700^oC+273.15 K=973.15K[/tex]
The change in gaseous mole = [tex]n_g=n_p-n_r=2-3 = -1[/tex]
[tex]K_p=1.08\times 10^7\times (0.0821 atm L/mol K\times 973.15 K)^{-1}\\K_p=1.35\times 10^5[/tex]
The value of [tex]K_p[/tex] is [tex]1.35\times 10^5[/tex].
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