Answer:
Polysaccharides, also known as glycans, are the polymers of carbohydrate. These polymers are composed of the monosaccharide monomeric units that are joined together by glycosidic bonds.
A polysaccharide molecule composed of the same type of monosaccharide units is known as homopolysaccharide or homoglycan.
Example: Cellulose, glycogen, and cellulose.
Whereas, a polysaccharide molecule composed of more than type of monosaccharide is known as heteropolysaccharides or heteroglycans.
Examples: Hyaluronic acid and heparin
a sample of an oxide of antimony (sb) contain 19.75 g of antimony combine with 6.5 g of oxygen . what is the simplest formula for the oxide
Explanation:
The given data is as follows.
Mass of antimony = 19.75 g
Molar mass of Sb = 121.76 g/mol
Therefore, calculate number of moles of Sb as follows.
Moles of Sb = [tex]\frac{mass}{\text{molar mass}}[/tex]
= [tex]\frac{19.75 g}{121.76 g/mol}[/tex]
= 0.162 mol
Mass of oxygen given is 6.5 g and molar mass of oxygen is 16 g/mol. Hence, moles of oxygen will be calculated as follows.
Moles of oxygen = [tex]\frac{mass}{\text{molar mass}}[/tex]
= [tex]\frac{6.5 g}{16 g/mol}[/tex]
= 0.406 mol
Hence, ratio of moles of Sb and O will be as follows
Sb : O
[tex]\frac{0.162}{0.162} : \frac{0.406}{0.162}[/tex]
1 : 2.5
We multiply both the ratio by 2 in order to get a whole number. Therefore, the ratio will be 2 : 5.
Thus, we can conclude that the empirical formula of the given oxide is [tex]Sb_{2}O_{5}[/tex].
If a chemical reaction consumes reactants at a steady rate of 7.64 x 10^19 molecules per second, how long will it take for the reaction to consume 6.02 x 1017 molecules of reactant?
Answer:
time = Molecules/Rate => 0.0079 segs
Explanation:
Rate = 7.64 * 10^19 molecules/segs
Molecules = 6.02 * 10^17 molecules
time = #?
time = Molecules/Rate => 0.0079 segs
The dimensional analysis calculates the variable from the given data. A chemical reaction with a reaction rate of 7.64 X 10¹⁹ will take 0.0079 seconds to consume 6.02 x 10¹⁷ molecules of reactant.
What is the reaction rate?The reaction rate has been defined by the speed or the time taken for the product to get produced by the reactant undergoing the chemical reaction. The rate of reaction depends on the concentration of the reactants.
Given,
Rate of reaction = 7.64 x 10¹⁹ molecules per second
Molecules = 6.02 × 10¹⁷ molecules
Time is calculated by the dimensional analysis as,
Time = Molecules ÷ Rate
= 6.02 × 10¹⁷ molecules ÷ 7.64 x 10¹⁹ molecules per second
= 0.0079 seconds
Therefore, it will take 0.0079 seconds for 6.02 x 10¹⁷ molecules of reactant to yield the product.
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Why are the electronegativities of the 5d elements so high?
Explanation:
The electronegativity values of the 5d series are very high because the size of the 5d orbitals are much larger as compared to the size of the 3d and 4d orbitals. As a consequence of this, the shielding capacity of 5d elements are low or these elements are not effective at shielding the nuclear charge . Thus, this causes the increase in the effective nuclear charge which makes the electronegativity values to increase steeply from the Lutetium (1.27) to the Gold (2.54).
Trinitrotoluene (C-H5N306, 227.1 g/mol) is easily detonated. How many grams of carbon are in 57.6 grams of TNT? Avogadro's Number: 1 mole = 6.02 x 1023 species A. 403 g B. 57.6 g C. 21.3 g D. 1.78 g E. None of the above
Answer: The correct answer is Option C.
Explanation:
The chemical formula for trinitrotoluene is [tex]C_7H_5N_3O_6[/tex]
In 1 mole of TNT, 7 moles of carbon atom, 5 moles of hydrogen atom, 3 moles of nitrogen atom and 6 moles of oxygen atom are present.
We know that:
Mass of trinitrotoluene = 227.1 g/mol
Mass of carbon = 12.01 g/mol
We are given:
Mass of TNT = 57.6 grams
To calculate the mass of carbon in given amount of TNT, we apply unitary method:
In 227.1 grams of TNT, amount of carbon present is = [tex](7\times 12.01)=84.07g[/tex]
So, in 57.6 grams of TNT, the amount of carbon present is = [tex]\frac{84.07g}{227.1g}\times 57.6g=21.3g[/tex]
Hence, the correct answer is Option C.
A gaseous mixture composed of 20% CH4, 30% C2H4, 35% C2H2, and 15% C2H20. What is the average molecular weight of the mixture? a) 20 b) 9.25 c) 25 d) 6.75
Answer: The correct answer is Option d.
Explanation:
We are given:
Mass percentage of [tex]CH_4[/tex] = 20 %
So, mole fraction of [tex]CH_4[/tex] = 0.2
Mass percentage of [tex]C_2H_4[/tex] = 30 %
So, mole fraction of [tex]C_2H_4[/tex] = 0.3
Mass percentage of [tex]C_2H_2[/tex] = 35 %
So, mole fraction of [tex]C_2H_2[/tex] = 0.35
Mass percentage of [tex]C_2H_2O[/tex] = 15 %
So, mole fraction of [tex]C_2H_2O[/tex] = 0.15
We know that:
Molar mass of [tex]CH_4[/tex] = 16 g/mol
Molar mass of [tex]C_2H_4[/tex] = 28 g/mol
Molar mass of [tex]C_2H_2[/tex] = 26 g/mol
Molar mass of [tex]C_2H_2O[/tex] = 48 g/mol
To calculate the average molecular mass of the mixture, we use the equation:
[tex]\text{Average molecular weight of mixture}=\frac{_{i=1}^n\sum{\chi_im_i}}{n_i}[/tex]
where,
[tex]\chi_i[/tex] = mole fractions of i-th species
[tex]m_i[/tex] = molar masses of i-th species
[tex]n_i[/tex] = number of observations
Putting values in above equation:
[tex]\text{Average molecular weight}=\frac{(\chi_{CH_4}\times M_{CH_4})+(\chi_{C_2H_4}\times M_{C_2H_4})+(\chi_{C_2H_2}\times M_{C_2H_2})+(\chi_{C_2H_2O}\times M_{C_2H_2O})}{4}[/tex]
[tex]\text{Average molecular weight of mixture}=\frac{(0.20\times 16)+(0.30\times 28)+(0.35\times 26)+(0.15\times 42)}{4}\\\\\text{Average molecular weight of mixture}=6.75[/tex]
Hence, the correct answer is Option d.
The average molecular weight of the gaseous mixture is approximately 27 g/mol, which is not one of the provided options.
The average molecular weight (MW) of a mixture can be calculated as the sum of the weight percentages of each component multiplied by its respective molecular weight divided by 100. For the given mixture, the molecular weights of the components are:
CH4 = 16 g/mol
C2H4 = 28 g/mol
C2H2 = 26 g/mol
C2H2O = 42 g/mol.
We multiply the weight percentages with the molecular weights of each component and then sum them up:
(20% x 16 g/mol) + (30% x 28 g/mol) + (35% x 26 g/mol) + (15% x 42 g/mol)=(0.20 x 16) + (0.30 x 28) + (0.35 x 26) + (0.15 x 42)=3.2 + 8.4 + 9.1 + 6.3=27 g/molHence, the average molecular weight of the mixture is 27 g/mol. Thus, none of the options provided (a-d) is correct.
A temperature difference of 15°C is impressed across a brick wall of a house which is 15 cm in thickness. The thermal conductivity of the brick is 1.0 W/m °C. The face of the wall is 6 meters high and 12 meters long. Compute both the heat flux and the heat transfer rate through the wall. Why aren't these numerical values the same?
Answer:
Q=7200 W
q=7200/72=100 [tex]W/m^2[/tex]
Explanation:
Given that
ΔT=15° C
Thickness ,t=15 cm
Thermal conductivity ,K=1 W/m.°C
Height,h=6 m
Length ,L=12 m
As we know that heat conduction through wall given as
[tex]Q=\dfrac{KA}{t}\Delta T[/tex]
Now by putting the values
A= 6 x 12 =72 [tex]m^2[/tex]
[tex]Q=\dfrac{KA}{t}\Delta T[/tex]
[tex]Q=\dfrac{1\times 72}{0.15}\times 15\ W[/tex]
Q=7200 W
Q is the total heat transfer.
Heat flux q
q=Q/A [tex]W/m^2[/tex]
q=7200/72=100 [tex]W/m^2[/tex]
q is the heat flux.
As w know that heat flux(q) is the heat transfer rate from per unit area and on the other hand heat transfer(Q) is the total heat transfer from the surface.
Heat flux q=Q/A
That is why these both are different.
Calculate the terminal velocity of a droplet (radius =R, density=\rho_d) when its settling in a stagnant fluid (density=\rho_f).
Answer:
[tex]V=\dfrac{2}{9\ \mu}R^2g(\rho_d-\rho_f)[/tex]
Explanation:
Given that
Radius =R
[tex]Density\ of\ droplet=\rho_d[/tex]
[tex]Density\ of\ fluid=\rho_f[/tex]
When drop let will move downward then so
[tex]F_{net}=F_{weight}-F_{b}-F_d[/tex]
Fb = Bouncy force
Fd = Drag force
We know that
[tex]F_b=\dfrac{4\pi }{3}R^3\ \times \rho_f\times g[/tex]
[tex]F_{weight}=\dfrac{4\pi }{3}R^3\ \times \rho_d\times g[/tex]
[tex]F_{d}=6\pi \mu\ R\ V[/tex]
μ=Dynamic viscosity of fluid
V= Terminal velocity
So at the equilibrium condition
[tex]F_{net}=F_{weight}-F_{b}-F_d[/tex]
[tex]0=F_{weight}-F_{b}-F_d[/tex]
[tex]F_{weight}=F_{b}+F_d[/tex]
[tex]\dfrac{4\pi }{3}R^3\ \times \rho_d\times g=\dfrac{4\pi }{3}R^3\ \times \rho_f\times g+6\pi \mu\ R\ V[/tex]
So
[tex]V=\dfrac{2}{9\ \mu}R^2g(\rho_d-\rho_f)[/tex]
This is the terminal velocity of droplet.
How do you know when the central atom in a lewis structure will have more than 8 electrons?
One particular example in mind is BrO3F. The structure should have 32 electrons, and it does when Br has a single bond with the other atoms and their formal charge cancels out to have a formal charge of zero.
Answer:
To know when a central atom in a lewis structure will have more than 8 electrons it is important to know where is the element at the periodic table and the orbital configuration of the atom.
Explanation:
The octet rule establishes that an atom could win, lose, or share electrons with other atoms till every atom have eight valence electrons. However exist exceptions to the octet rule. Some elements like Be, B and Al are stable with only six valence electrons because these three elements are small and with low electronegativity. On the other hand, some elements from the 3d orbital could expand their octet and still be stable, like S in SF6. This happens because elements from 3d orbital have enough space to suit more electrons.
which higher heat transfer ?
a)Fan with air
b)boiling water
c)cooled liquid
Answer:
b)boiling water
Explanation:
High heat is transferred with the thermal conduction process, also called diffusion. Always occurs from a region of higher temperatures to a region of lower temperatures. The flow of fluid may be forced into external processes. As the particles rapidly move and vibrate and the transfer of heat from one state to the other by the interaction of particles and rate of heat lost has to be the same as absorbed heat during vaporization at the same fluid pressure. A component of heat transfer can be seen as a heat sink in which transfers the heat generated within any medium. Thus evaporative cooling happens after the vapor is added to the surrounding air.What should be done in case of acid spills and mercury spills? Explain
Explanation:
Acid spills
These types of spills should be neutralized with base like sodium bicarbonate and then must be cleaned up by using paper towel or sponge. Strong base like sodium hydroxide must not be used to neutralize. Best base to use is sodium bicarbonate which has much less chance of the injury.
Mercury spills
Mercury is found commonly in the thermometers. If one have mercury spill, it must be cleaned up immediately with either commercial product like Hg Aborb ™ or by using elemental sulfur. Mercury sponges can also be purchased that form amalgam with liquid mercury and thus trapping it on surface of sponge.
Which of the following regions of the periodic table tends to prefer a –1 charge and occupies Group 17?
a. Alkaline earth metals.
b. Halogens.
c. Noble gases.
Answer:
b. halogens
Explanation:
The elements of group 17 are called halogens. These are six elements Fluorine, Chlorine, Bromine, Iodine, Astatine. Halogens are very reactive these elements cannot be found free in nature. Their chemical properties are resemble greatly with each other. As we move down the group in periodic table size of halogens increases that's way fluorine is smaller in size as compared to other halogens elements. Their boiling points also increases down the group which changes their physical states.
How many atoms are found in 3.45g of CO2?
Answer: The number of carbon and oxygen atoms in the given amount of carbon dioxide is [tex]4.72\times 10^{22}[/tex] and [tex]9.44\times 10^{22}[/tex] respectively
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass of carbon dioxide gas = 3.45 g
Molar mass of carbon dioxide gas = 44 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of carbon dioxide gas}=\frac{3.45g}{44g/mol}=0.0784mol[/tex]
1 mole of carbon dioxide gas contains 1 mole of carbon and 2 moles of oxygen atoms.
According to mole concept:
1 mole of a compound contains [tex]6.022\time 10^{23}[/tex] number of molecules
So, 0.0784 moles of carbon dioxide gas will contain [tex]1\times 0.0784\times 6.022\times 10^{23}=4.72\times 10^{22}[/tex] number of carbon atoms and [tex]2\times 0.0784\times 6.022\times 10^{23}=9.44\times 10^{22}[/tex] number of oxygen atoms
Hence, the number of carbon and oxygen atoms in the given amount of carbon dioxide is [tex]4.72\times 10^{22}[/tex] and [tex]9.44\times 10^{22}[/tex] respectively
7. For the system PCls(g) → PC13(g) + Cl2(g) Kis 26 at 300°C. In a 5.0-L flask, a gaseous mixture consists of all three gases with partial pressure as follows: Ppcis = 0.012 atm, Pc2=0.45 atm, Ppci3 -0.90 atm. a) Is the mixture at equilibrium? Explain. b) If it is not at equilibrium, which way will the system shift to establish equilibrium?
Answer:
a) Reaction is not at equilibrium
b) Reaction will move towards backward direction
Explanation:
[tex]PCl_5(g) \rightarrow PCl_3(g) + Cl_2(g)[/tex]
Equilibrium constant = 26
[tex]Reaction\ quotient (Q) = \frac{[p_{PCl_3}]\times [p_{Cl_2}]}{[p_{PCl_5}]}[/tex]
[tex][p_{PCl_5}] = 0.012 atm[/tex]
[tex][p_{PCl_3}]= 0.90 atm[/tex]
[tex][p_{Cl_2}]= 0.45 atm[/tex]
[tex]Reaction\ quotient (Q) = \frac{[p_{PCl_3}]\times [p_{Cl_2}]}{[p_{PCl_5}]}[/tex]
[tex]Reaction quotient (Q) =\frac{0.90\times 0.45} {0.012} = 33.75[/tex]
As reaction quotient (Q) is more than equilibrium constant, so reaction is not at equilibrium and reaction will move towards backward direction.
A chemist measures the amount of hydrogen gas produced during an experiment. She finds that 264. g of hydrogen gas is produced. Calculate the number of moles of hydrogen gas produced. Round your answer to 3 significant digits. W mol x 6 ?
Answer:
The answer is 130.953 g of hydrogen gas.
Explanation:
Hydrogen gas is formed by two atoms of hydrogen (H), so its molecular formula is H₂. We can calculate is molecular weight as the product of the molar mass of H (1.008 g/mol):
Molecular weight H₂= molar mass of H x 2= 1.008 g/mol x 2= 2.01568 g
Finally, we obtain the number of mol of H₂ there is in the produced gas mass (264 g) by using the molecular weight as follows:
mass= 264 g x 1 mol H₂/2.01568 g= 130.9731703 g
The final mass rounded to 3 significant digits is 130.973 g
To find the number of moles of hydrogen gas produced from 264 grams, divide the mass by the molar mass of hydrogen (2.02 g/mol), resulting in approximately 130.7 moles of hydrogen gas to three significant digits.
Explanation:To calculate the number of moles of hydrogen gas (H₂) produced from 264 grams of hydrogen gas, you would use the molar mass of H₂ which is approximately 2.02 g/mol (1 mole of H₂ = 2.02 grams). Using the formula:
number of moles = mass of substance (g) / molar mass (g/mol)
We find the number of moles of hydrogen gas to be:
number of moles = 264 g / 2.02 g/mol
After performing the division, this gives us approximately 130.7 moles of H₂.
This result is to three significant digits, aligned with the precision provided by the initial mass of the hydrogen gas.
Calculate the freezing point of a solution made from 220g of octane (C Hua), molar mass = 114,0 gmol dissolved in 1480 g of benzene. Benzene freezes at 5.50"C and its Kvalue is 5.12C/m. -1.16°C 0.98°C 666"C 12 2°C 5.49°C 10 12 AM A A 2019 Backspace yuo Pill но кL
Answer: Freezing point of a solution will be [tex]-1.16^0C[/tex]
Explanation:
Depression in freezing point is given by:
[tex]\Delta T_f=i\times K_f\times m[/tex]
[tex]\Delta T_f=T_f^0-T_f=(5.50-T_f)^0C[/tex] = Depression in freezing point
i= vant hoff factor = 1 (for non electrolyte)
[tex]K_f[/tex] = freezing point constant = [tex]5.12^0C/m[/tex]
m= molality
[tex]\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}[/tex]
Weight of solvent (benzene)= 1480 g =1.48 kg
Molar mass of solute (octane) = 114.0 g/mol
Mass of solute (octane) = 220 g
[tex](5.50-T_f)^0C=1\times 5.12\times \frac{220g}{114.0 g/mol\times 1.48kg}[/tex]
[tex](5.50-T_f)^0C=6.68[/tex]
[tex]T_f=-1.16^0C[/tex]
Thus the freezing point of a solution will be [tex]-1.16^0C[/tex]
For which of these is there an increase in entropy? KCI(aq)+AgNO3(aq)KNO3(aq)+AgCI(s) NaCl(s)NaCl(aq) 2NaOH(aq)+CO2(g)Na2CO3(aq)+H20(1) C2H5OH(g)C2H5OH(I)
Answer: NaCl (s) → NaCl (aq)
Explanation:
Entropy is often associated with the disorder or randomness of a system. Therefore, in each reaction, it is necessary to evaluate if the disorder increases or decreases to understand what happens to the entropy:
1) KCl (aq) + AgNO₃ (aq) → KNO₃ (aq) + AgCl (s) - In this reaction, we have only aqueous reactants in the beginning and in the product we have a precipitate. The solid state is more organised than the liquid, consequently, the entropy decreases.
2) NaCl (s) → NaCl (aq) - In this case, oposite to the first one, we go from a solid state to an aqueous state. The solvation of the ions Na⁺ and Cl⁻ is random while the solid state is very organised. Therefore, in this reaction the entropy increases.
3) 2NaOH (aq) + CO₂ (g) → Na₂CO₃ (aq) + H₂O (l) - In this reaction, the reactants have higher entropy because of the gas CO₂. Therefore, the entropy decreases.
4) C₂H₅OH (g) → C₂H₅OH (l) - In this reaction, the reactant is a gas and the product a liquid. Therefore, the entropy decreases.
What is the pH of a solution containing 0.049 M of formic acid and 0.055 M of sodium formate?
Answer:
pH of solution is 3.80
Explanation:
Formic acid an weak acid and formate is conjugate base of formic acidHence solution containing formic acid and formate acts as a buffer.According to Henderson-Hasselbalch equation for a buffer consist of an weak acid (formic acid) and it's conjugate base (formate)-[tex]pH=pK_{a}(formic acid)+log(\frac{C_{formate}}{C_{formic acid}})[/tex]
where, C stands for concentration
[tex]pK_{a}[/tex] of formic acid 3.75So, [tex]pH=3.75+log(\frac{0.055}{0.049})=3.80[/tex]
Blood substitute. As noted in this chapter, blood contains a total concentration of phosphate of approximately 1 mM and typically has a pH of 7.4. You wish to make 100 liters of phosphate buffer with a pH of 7.4 from NaH 2 PO 4 (molecular weight, 119. 98 g mol 1) and Na 2 HPO 4 (molecular weight, 141. 96 g mol 1). How much of each (in grams) do you need? Berg, Jeremy M.. Biochemistry (p. 25). W. H. Freeman. Kindle Edition.
Answer:
Mass of NaH₂PO₄ = 4.707 gMass of Na₂HPO₄ = 8.627 gExplanation:
The equilibrium relevant for this problem is:
H₂PO₄⁻ ↔ HPO₄⁻² + H⁺
The Henderson–Hasselbalch (H-H) equation is needed to solve this problem:
pH= pka + [tex]log\frac{[A^{-} ]}{[HA]}[/tex]
In this case, [A⁻] = [HPO₄⁻²], [HA] = [H₂PO₄⁻], pH = 7.4; from literature we know that pka=7.21.
We use the H-H equation to describe [HPO₄⁻²] in terms of [H₂PO₄⁻]:
[tex]7.4=7.21+log\frac{[HPO4^{-2} ]}{[H2PO4^{-} ]} \\0.19=log\frac{[HPO4^{-2} ]}{[H2PO4^{-} ]} \\10^{0.19}= \frac{[HPO4^{-2} ]}{[H2PO4^{-} ]} \\1.549*[H2PO4^{-} ]=[HPO4^{-2} ][/tex]
The problem tells us that the concentration of phosphate is 1 mM, which means:
[HPO₄⁻²] + [H₂PO₄⁻] = 1 mM = 0.001 M
In this equation we can replace [HPO₄⁻²] with the term expressed in the H-H eq:
1.549 * [H₂PO₄⁻] + [H₂PO₄⁻] = 0.001 M
2.549 * [H₂PO₄⁻] = 0.001 M
[H₂PO₄⁻] = 3.923 * 10⁻⁴ M
With the value of [H₂PO₄⁻] we can calculate [HPO₄⁻²]:
[HPO₄⁻²] + 3.923 * 10⁻⁴ M = 0.001 M
[HPO₄⁻²] = 6.077 * 10⁻⁴ M
With the concentrations, the molecular weight, and the volume, we calculate the mass of each reagent:
Mass of NaH₂PO₄ = 3.923 * 10⁻⁴ M * 100 L * 119.98 g/mol = 4.707 gMass of Na₂HPO₄ = 6.077 * 10⁻⁴ M * 100 L * 141.96 g/mol = 8.627 gVarious members of a class of compounds, alkenes, react with hydrogen to produce a corresponding alkane. Termed hydrogenation, this type of reaction is used to produce products such as margarine. A typical hydrogenation reaction is C10H20() + H2(g) → C10H22(5) Decene Decane How much decane can be produced in a reaction of excess decene with 2.45 g hydrogen? Give your answer in scientific notation. O *10 g decane
Answer: The mass of decane produced is [tex]1.743\times 10^2g[/tex]
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)
Mass of hydrogen gas = 2.45 g
Molar mass of hydrogen gas = 2 g/mol
Putting values in equation 1:, we get:
[tex]\text{Moles of }H_2=\frac{2.45g}{2g/mol}=1.225mol[/tex]
The chemical equation for the hydrogenation of decene follows:
[tex]C_{10}H_{20}(l)+H_2(g)\rightarrow C_{10}H_{22}(s)[/tex]
As, decene is present in excess. So, it is considered as an excess reagent.
Thus, hydrogen gas is a limiting reagent because it limits the formation of products.
By Stoichiometry of the reaction:
1 mole of hydrogen gas produces 1 mole of decane.
So, 1.225 moles of hydrogen gas will produce = [tex]\frac{1}{1}\times 1.225=1.225mol[/tex] of decane
Now, calculating the mass of decane by using equation 1, we get:
Moles of decane = 1.225 mol
Molar mass of decane = 142.30 g/mol
Putting values in equation 1, we get:
[tex]1.225mol=\frac{\text{Mass of decane}}{142.30g/mol}\\\\\text{Mass of carbon dioxide}=(1.225mol\times 142.30g/mol)=174.3g=1.743\times 10^2g[/tex]
Hence, the mass of decane produced is [tex]1.743\times 10^2g[/tex]
The weight of the body in the air is .... the weight of the submerged body
a) equal
b) more than
c) lower than
d) Not related
Answer:
the correct answer is option 'b': More than
Explanation:
The 2 situations are represented in the attached figures below
When an object is placed in air it is acted upon by force of gravity of earth which is measured as weight of the object.
While as when any object is submerged partially or completely in any fluid the fluid exerts a force in upward direction and this force is known as force of buoyancy and it's magnitude is given by Archimedes law as equal to the weight of the fluid that the body displaces, hence the effective force in the downward direction direction thus the apparent weight of the object in water decreases.
Which of the following gives the net ionic reaction for the reaction used in this experiment?
H+(aq) + OH-(aq) → H2O(l)
no net reaction
Ba2+(aq) + SO42-(aq) → BaSO4(s)
Ba2+(aq) + 2 OH-(aq) + 2 H+(aq) + SO42-(aq) → BaSO4(s) + 2 H2O(l)
Answer: Option (d) is the correct answer.
Explanation:
The given reaction will be as follows.
[tex]Ba(OH)_{2}(aq) + H_{2}SO_{4}(aq) \rightarrow BaSO_{4}(s) + 2H_{2}O(l)[/tex]
In the ionic form, the equation will be as follows.
[tex]Ba^{2+}(aq) + SO^{2-}_{4}(aq) \rightarrow BaSO_{4}(s)[/tex] ........ (1)
[tex]H^{+}(aq) + OH^{-}(aq) \rightarrow H_{2}O(l)[/tex] ............ (2)
Hence, for the net ionic equation we need to add both equation (1) and (2). Therefore, the net ionic equation will be as follows.
[tex]Ba^{2+}(aq) + SO^{2-}_{4}(aq) + H^{+}(aq) + OH^{-}(aq) \rightarrow BaSO_{4}(s) + H_{2}O(l)[/tex]
Now, balancing the atoms on both the sides we get the net ionic equation as follows.
[tex]Ba^{2+}(aq) + SO^{2-}_{4}(aq) + 2H^{+}(aq) + 2OH^{-}(aq) \rightarrow BaSO_{4}(s) + 2H_{2}O(l)[/tex]
The correct net ionic reaction for the experiment is H+(aq) + OH-(aq) → H2O(l), an example of an acid-base neutralization reaction. The other options included the formation of a precipitate, which is not part of the net ionic reaction.
Explanation:Based on the provided options, the correct net ionic reaction for the experiment seems to be H+(aq) + OH-(aq) → H2O(l). This reaction is a classic example of an acid-base neutralization reaction, where an acid (H+) and a base (OH-) react to form water. The other two reactions involve the formation of a precipitate (BaSO4), but the full reaction is simplified to leave out the precipitate ions on either side. This does not occur in the first reaction. Hence, the first reaction is the correct net ionic reaction for this experiment.
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Use the References to access important values if needed for this question. The radius of a potassium atom is 231 pm. How many potassium atoms would have to be laid side by side to span a distance of 2.91 mm? atoms Submit Answer Try Another Version 1 item attempt remaining
Answer:
6298702 potassium atoms would have to be laid side by side to span a distance of 2,91 mm
Explanation:
If the radius of a potassium atom is 231 pm, then the diameter would be:
d = 2r = 2*(231) = 462 pm
So each potassium atom occupies a space of 462 pm, we can express this relationship as follows:
[tex]\frac{1 potassium atom}{462 pm}[/tex]
To solve this problem, we'll use the following conversion factors:
1 pm = 1×E−12 m
1000 mm = 1 m
We begin to accommodate all our relationships starting from that numerical expression that is not written as a relationship (usually the one in the question), and in such a way that the units are eliminated between them.
[tex]2, 91 mm * \frac{1 m}{1000 mm}*\frac{ 1 pm}{1*10^{-12}m }*\frac{1 potassium atom}{462 pm}= 6298701.3[/tex] potassium atoms
So, we'll need 6298702 potassium atoms to span a distance of 2,91 mm
QUESTION 1 1.041 points If a 50.00 ml aliquot of a 0.12 M NaCl solution is added to 30.00 mL of a 0.18 M CaCl solution, what is the concentration of calcium ion in the mixture? 0.10M 0.086 M 0,068 M 0.36 M 0.090 M
Answer:
[Ca²⁺] = 0.068 M
Explanation:
The concentration of the calcium ion will be equal to the amount of calcium in CaCl₂, divided by the total volume:
C = n/V
When CaCl₂ dissociates in water, one mole of calcium ion is produced for every mole of CaCl₂, so the molar ratio of CaCl₂ to Ca²⁺ is 1:1. The moles of Ca²⁺ are calculated as follows:
(0.18 mol/L)(30.00 mL) = 5.4 mmol CaCl₂ = 5.4 mmol Ca²⁺
The total volume is (50.00 mL + 30.00 mL) = 80.00 mL
Thus, the concentration of Ca²⁺ is:
C = n/V = (5.4 mmol)/(80.00 mL) = 0.068 M
Define ""point source pollution"" and ""non-point pollution"".
Answer:
Point source pollution:
If pollution comes from a fix source then is called point source pollution.It have a specific location where pollution comes.
Ex: Air pollution ,water pollution.
Non point source pollution:
If pollution comes from number of sources then is called point source pollution.This pollution does not have a specific point of source.
Ex: Spills ,leaks ,Sewage over flow etc.
Answer:
Defined as under.
Explanation:
A point source pollution is a single identifiable source of pollutants like air, water, thermal and light. And has a negligible extent, distinguishes from other geometric or area sources of nonpoint, etc. While nonpoint source pollution is derived from various sources. Such as results of land runoff. atmospheric drainage and seepage they may affect a body of water such as from the agricultural areas draining to a river. Various pathogens and viruses found in the body of water, also the presence of poorly managed livestock, pets, and other these problems is also elated to the urban and rural asphalt and concrete, highway runoff and mining activities.
The decomposition of hydrogen peroxide, H2O2, has been used to provide thrust in the control jets of various space vehicles. Using the Supplemental Data, determine how much heat (in kJ) is produced by the decomposition of 1.05 mol of H2O2 under standard conditions.
2 H2O2(l) → 2 H2O(g) + O2(g)
Answer:
The heat released is 56.7 kJ.
Explanation:
To solve this problem, first we need to find out the standard enthalpy of reaction, that is, the energy released at constant pressure in standard conditions (P=1bar, T=298.15K). We can find it using the expression:
ΔH°r = Σn(p).ΔH°f(p) - Σn(r).ΔH°f(r)
where,
n refers to the number of moles of reactants and products in the balanced equation
ΔH°f refers to standard enthalpies of formation (which can be found in tables).
Given the equation:
2 H₂O₂(l) → 2 H₂O(g) + O₂(g)
We can replace with the proper data in the equation:
ΔH°r = Σn(p).ΔH°f(p) - Σn(r).ΔH°f(r)
ΔH°r= [2 mol . ΔH°f H₂O(g) + 1 mol . ΔH°f O₂(g)] - [2 mol . ΔH°f H₂O₂(l)]
ΔH°r= [2 mol . (-241.8 kJ/mol) + 1 mol . 0 kJ/mol] - [2 mol . (-187.8 kJ/mol)]
ΔH°r = -108.0 kJ
Since enthalpy is an extensive property, it depends on the amount of reagents. In this case, 108.0 kJ of heat are released every 2 moles of H₂O₂(l) decomposed. Then, for 1.05 mol of H₂O₂(l):
[tex]1.05 mol.\frac{-108.0kJ}{2mol} =-56.7kJ[/tex]
By convention, the negative sign means that heat is released.
You make 1 Liter of an aqueous solution containing 9.20 ml of 57.8 mM acetic acid and 56.2 mg of sodium acetate (MW = 82.0 g/mole). a. How many moles of acetic acid did you add? b. How many moles of sodium acetate did you add? c. What is the appropriate formula for calculating the pH of the above solution? d. What is the pH of the above solution? Is it acidic or basic? Slightly or strongly so? (Use the same scale as on HW 1)
Answer:
a) 5,3176x10⁻⁴ moles
b) 6,85x10⁻⁴ moles
c) The appropriate formula to calculate is Henderson-Hasselbalch.
d) pH = 4,86. Acidic solution but slighty
Explanation:
a) moles of acetic acid:
9,20x10⁻³L × 57,8x10⁻³M = 5,3176x10⁻⁴ moles
b) moles of sodium acetate:
56,2x10⁻³g ÷ 82,0 g/mole = 6,85x10⁻⁴ moles
c) The appropriate formula to calculate is Henderson-Hasselbalch:
pH= pka + log₁₀ [tex]\frac{[A^-]}{[HA]}[/tex]
d) pH= 4,75 + log₁₀ [tex]\frac{[6,85x10_{-4}]}{[5,3176x10_{-4}]}[/tex]
pH = 4,86
3 < pH < 7→ Acidic solution but slighty
I hope it helps!
Be sure to answer all parts. The thermal decomposition of phosphine (PH3) into phosphorus and molecular hydrogen is a first-order reaction: 4PH3(g) → P4(g) + 6H2(g) The half-life of the reaction is 35.0 s at 680°C. Calculate the first-order rate constant for the reaction: s−1 Calculate the time required for 77.0 percent of the phosphine to decompose: s
Answer:
k = 0.0198 s⁻¹
t = 74.25 seconds
Explanation:
Given that:
Half life = 35.0 s
[tex]t_{1/2}=\frac {ln\ 2}{k}[/tex]
Where, k is rate constant
So,
[tex]k=\frac {ln\ 2}{t_{1/2}}[/tex]
[tex]k=\frac {ln\ 2}{35.0}\ s^{-1}[/tex]
The rate constant, k = 0.0198 s⁻¹
Using integrated rate law for first order kinetics as:
[tex][A_t]=[A_0]e^{-kt}[/tex]
Where,
[tex][A_t][/tex] is the concentration at time t
[tex][A_0][/tex] is the initial concentration
Given:
77.0 % is decomposed which means that 0.77 of [tex][A_0][/tex] is decomposed. So,
[tex]\frac {[A_t]}{[A_0]}[/tex] = 1 - 0.77 = 0.23
t = ?
[tex]\frac {[A_t]}{[A_0]}=e^{-k\times t}[/tex]
[tex]0.23=e^{-0.0198\times t}[/tex]
t = 74.25 seconds
When in a reaction heat is a reactant then the reaction is called thermal decomposition reaction. The first-order rate constant is 0.0198 per second and the time required is 74.25 seconds.
What is a rate constant?The rate constant is the specific rate that is the proportionality constant in a reaction and depicts the relation between the chemical reaction rate and the concentration of the reactants.
Rate constant can be given as,
[tex]\rm t\dfrac{1}{2} = \dfrac{ln 2}{k}[/tex]
The half-life of the reaction is 35 seconds.
Substituting values in the above equation:
[tex]\begin{aligned}\rm k &= \rm \dfrac{ln\;2}{t\frac{1}{2}}\\\\&= \dfrac{\rm ln\2}{35}\\\\&= 0.0198 \;\rm s^{-1}\end{aligned}[/tex]
The time taken can be calculated by the rate law for first order:
[tex]\rm [A_{t}] = [A_{o}]e^{-kt}[/tex]
Here,
Concentration at time t = [tex]\rm [A_{t}][/tex]The initial concentration = [tex]\rm [A_{o}][/tex]Solving for time (t):
[tex]\begin{aligned}\rm \dfrac {[A_{t}]}{[A_{o}]} &= 1 - 0.77\\\\0.23 &= \rm e ^{-0.0198 \times t}\\\\\rm t &= 74.25\;\rm seconds\end{aligned}[/tex]
Therefore, the rate constant is 0.0198 per second and the time taken is 74.25 seconds.
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What is the pH of a 0.18 M CH3NH3+Cl– aqueous solution? The pKb ofCH3NH2 is 3.44
Answer:
5.65
Explanation:
Given that:
[tex]pK_{b}\ of\ CH_3NH_2=3.44[/tex]
[tex]K_{b}\ of\ CH_3NH_2=10^{-3.44}=3.6308\times 10^{-4}[/tex]
[tex]K_a\ of\ CH_3NH_3^+Cl^-=\frac {K_w}{K_b}=\frac {10^{-14}}{3.6308\times 10^{-4}}=2.7542\times 10^{-11}[/tex]
Concentration = 0.18 M
Consider the ICE take for the dissociation as:
[tex]CH_3NH_3^+[/tex] ⇄ H⁺ + [tex]CH_3NH_2[/tex]
At t=0 0.18 - -
At t =equilibrium (0.18-x) x x
The expression for dissociation constant of acetic acid is:
[tex]K_{a}=\frac {\left [ H^{+} \right ]\left [CH_3NH_2 \right ]}{[CH_3NH_3^+]}[/tex]
[tex]2.7542\times 10^{-11}=\frac {x^2}{0.18-x}[/tex]
x is very small, so (0.18 - x) ≅ 0.18
Solving for x, we get:
x = 0.2227×10⁻⁵ M
pH = -log[H⁺] = -log(0.2227×10⁻⁵) = 5.65
A gram of gasoline produces 45.0kJ of energy when burned. Gasoline has a density of 0.77/gmL. How would you calculate the amount of energy produced by burning 5.2L of gasoline? Set the math up. But don't do any of it. Just leave your answer as a math expression. Also, be sure your answer includes all the correct unit symbols.
Answer:
Math expression: [tex]=\frac{0.77\ g/ml*5200\ ml}{1\ g} *45.0\ kJ[/tex]
Explanation:
Given:
Energy produced per gram of gasoline = 45.0 kJ
Density of gasoline = 0.77 g/ml
Volume of gasoline = 5.2 L=5200 ml
To determine:
The amount of energy produced by burning 5.2 L gasoline
Calculation set-up:
1. Calculate the mass (m) of gasoline given the density (d) and volume (v)
[tex]m = d*v\\\\m = 0.77 g/ml*5200 ml[/tex]
2. Calculate the amount of energy produced
[tex]=\frac{0.77\ g/ml*5200\ ml}{1\ g} *45.0\ kJ=180180 kJ[/tex]
Not only due trees "fix" carbon but so do green vegetables. Photosynthesis in spinach leaves produces glucose via the Calvin cycle which involves the fixation of CO2 with ribulose 1-5 bisphosphate to form 3-phosphoglycerate via C3H8P2011(aq) + H2O(aq) + CO2(g) → 2 CzH4PO3(aq) + 2 H+(aq) If 15.0 g of 3-phosphoglycerate is formed by this reaction at T = 298 K and P = 1.00 atm what volume of CO2 is fixed? [1.00 L]
Answer : The volume of [tex]CO_2[/tex] gas is 1.00 L
Explanation :
First we have to determine the moles of [tex]C_3H_4PO_7^{3-}[/tex].
Molar mass of [tex]C_3H_4PO_7^{3-}[/tex] = 182.9 g/mole
[tex]\text{ Moles of }C_3H_4PO_7^{3-}=\frac{\text{ Mass of }C_3H_4PO_7^{3-}}{\text{ Molar mass of }C_3H_4PO_7^{3-}}=\frac{15.0g}{182.9g/mole}=0.0820moles[/tex]
Now we have to calculate the moles of [tex]CO_2[/tex].
The given balanced chemical reaction is:
[tex]C_5H_8P_2O_{11}^{4-}(aq)+H_2O(aq)+CO_2(g)\rightarrow 2C_3H_4PO_7^{3-}(aq)+2H^+(aq)[/tex]
From the reaction we conclude that,
As, 2 moles of [tex]C_3H_4PO_7^{3-}[/tex] produce from 1 mole of [tex]CO_2[/tex]
So, 0.0820 moles of [tex]C_3H_4PO_7^{3-}[/tex] produce from [tex]\frac{0.0820}{2}=0.041moles[/tex] of [tex]CO_2[/tex]
Now we have to calculate the volume of [tex]CO_2[/tex] gas.
Using ideal gas equation:
[tex]PV=nRT[/tex]
where,
P = pressure of gas = 1.00 atm
V = volume of gas = ?
T = temperature of gas = 298 K
n = number of moles of gas = 0.041 mole
R = gas constant = 0.0821 L.atm/mole.K
Now put all the given values in the ideal gas equation, we get:
[tex](1.00atm)\times V=(0.041mole)\times (0.0821L.atmK^{-1}mol^{-1})\times (298K)[/tex]
[tex]V=1.00L[/tex]
Therefore, the volume of [tex]CO_2[/tex] gas is 1.00 L