3rd Law of Thermodynamics
The correct answer is the third law of thermodynamics.
"The entropy of a perfect crystal is zero when the temperature of the crystal is equal to absolute zero (0 K).”
The temperature scale they refer to is the zero in kelvin degrees, this is what we call absolute zero.
If the entropy is zero, all physic processes stop, and the entropy of the system is minimum and constant.
Final answer:
The concept described refers to the Third Law of Thermodynamics, which states that the entropy of a perfect crystalline substance is zero at absolute zero temperature. This law is often used to compute standard entropy values and predict entropy changes during phase transitions and chemical reactions. Option b.
Explanation:
The statement describing a pure crystalline substance at absolute zero temperature having no movement refers to the Third Law of Thermodynamics. This law states that the entropy of a perfectly ordered, crystalline substance at absolute zero temperature (0 K) is zero. Entropy, often denoted by S, is a measure of the disorder or randomness in a system, and it increases with temperature as molecular motion increases.
At absolute zero, all molecular motion ceases, meaning a perfectly crystalline substance has only a single microstate available to it (W = 1). Since there is just one possible arrangement for the particles, the entropy is zero as per the Boltzmann equation. This is part of the Third Law of Thermodynamics, which can be used to calculate entropy changes for phase transitions and chemical reactions under standard conditions.
The area of a telescope lens is 6.676 x 10 mm. (a) What is the area in square feet (ft)? Enter your answer in scientific notation. x 10ft (b) If it takes a technician 51.7 s to polish 1.46 x 10mm', how long does it take her to polish the entire lens?
Answer:
a) A = 7.186 E-4 ft²
b) t = 236.403 s
Explanation:
A = 6.676 E1 mm²
a) A = 6.676 E1 mm² * ( ft / 304.8 mm )²
⇒ A = 7.186 E-4 ft²
b) t = 51.7 s → A = 1.46 E1 mm²
⇒ relation (r) = 6.676 E1 mm² / 1.46 E1 mm² = 4.573
⇒ t = 51.7 s * 4.573 = 236.403 s
How many U.S. gallons are there in a cubic mile? The total proven oil reserves of the U.S. are roughly 30 x 10°bbl. How many cubic miles is this?
Answer:
1 cubic mile = 1.101 * 10^12 US gallons
1 US bbl oil = 42 US gallons = 3.8143*10^ -11 cubic miles
Explanation:
The number of the exponent of the oil reserve is not very well shown in the question so, I provide you the conversion of bbl oil into cubic mile, the only thing you have to do is multiply the number of bbls of the reserve for the conversion in cubic miles and you'll have the answer.
Calculate the number of moles in 369 grams of CaoCl2? a) 1 b) 2 c) 3 d) 4
Answer:
First of all, take account the molar mass of the CaOCl2 (Calcium hypochlorite), which is 126.97 g/mol. If we have 126.97 g in a mol, 369g should be in aproximately 3 moles. Try to the think the rule of 3.
Explanation:
Write 450.0000345 in Scientific Notation with 4 significant figures.
Answer:
[tex]4.500\times 10^{2}[/tex]
Explanation:
Scientific notation is the way of writing numbers which are either large or small. The number is written in the scientific notation when the number is between 1 and 10 and then multiplied by the power of 10.
The given number:
450.0000345 can be written as [tex]4.500000345\times 10^{2}[/tex]
Answer upto 4 significant digits = [tex]4.500\times 10^{2}[/tex]
Which statement below matches the correct response with the proper reasoning when comparing the volatility of CH2Cl2 with CH2Br2? a. CH2Cl2 is more volatile than CH2Br2 because the dipole-dipole interactions in CH2Cl2 are greater than the dipole-dipole interactions in CH2Br2. b. CH2Cl2 is more volatile than CH2Br2 because of the large dispersion forces in CH2Br2. c. CH2Br2 is more volatile than CH2Cl2because because the dipole-dipole interactions in CH2Br2 are greater than the dipole-dipole interactions in CH2Cl2d. CH2Br2 is more volatile than CH2Cl2 because of the large dispersion forces in CH2Br2
Answer:
b. CH₂Cl₂ is more volatile than CH₂Br₂ because of the large dispersion forces in CH₂Br₂
Explanation:
CH₂Cl₂ is more volatile than CH₂Br₂ (b.p of CH₂Cl₂ = 39,6 °C; b.p of CH₂Br₂ = 96,95°C). Thus, c. and d. are FALSE
Dipole-dipole interactions in CH₂Cl₂ are greater than the dipole-dipole interactions in CH₂Br₂ because Cl is more electronegative that Br (Cl = 3,16; Br = 2,96). But this mean CH₂Cl₂ is less volatile than CH₂Br₂ but it is false.
There are large dispersion forces in CH₂Br₂ because Br has more electrons and protons than Cl. Large disperson forces mean CH₂Br₂ is less volatile than CH₂Cl₂ and it is true.
I hope it helps!
(c) Draw the structure of any two aminoacids which bears a heterocyclic unit as all or part of the R side chain of the aminoacid, and highlight and name the particular heterocyclic unit present in the aminoacid.
Answer:
Here I show you tryptophan and tyrosine
Explanation:
of the 21 amino acids, there is five amino acid that have a heterocyclic group as part of the R side chain: Histidine, proline, Phenylalanine, tyrosine, and tryptophan. Each one has a unique nature and the heterocyclic group, mainly in tyrosine, and tryptophan allows to absorb the UV light (280 nm)
Which is more stable, a trans-1,4-disubstitutedcyclohexane or
its cis isomer?
Answer:
If the substituents are the same, the cis distribution is more stable than trans distribution.
Explanation:
A cis cyclohexane is one in which both substituents are oriented towards the same face of the ring regardless of the conformation.
A trans cyclohexane has substituents on opposite sides of the ring.
In trans-1,4-disubstituted cyclohexane, the conformation with the two substituent groups in equatorial is more stable than the constitution that has the axial groups.
In cis-1,4-disubstituted cyclohexane, both conformations have the same stability as they have an axial and an equatorial substituent, so they have the same energy and there is no equilibrium shift.
Arrange the following aqueous solutions in order of increasing boiling points 0.050 m Mg(NO3): 0.100 m ethanol 0.090 m NaCl Ethanol
Answer: The order of increasing boiling points follows:
[tex]\text{Ethanol }<Mg(NO_3)_2<\text{ NaCl}[/tex]
Explanation:
The expression of elevation in boiling point is given as:
[tex]\Delta T_b=i\times k_b\times m[/tex]
where,
[tex]\Delta T_b[/tex] = Elevation in boiling point
i = Van't Hoff factor
[tex]T_b[/tex] = change in boiling point
[tex]k_b[/tex] = boiling point constant
m = molality
For the given options:
Option 1: 0.050 m [tex]Mg(NO_3)_2[/tex]Value of i = 3
So, molal concentration will be = [tex](0.05)\times 3=0.15m[/tex]
Option 2: 0.100 m ethanolValue of i = 1 (for non-electrolytes)
So, molal concentration will be = [tex](0.100)\times 1=0.100m[/tex]
Option 3: 0.090 m NaClValue of i = 2
So, molal concentration will be = [tex](0.09)\times 2=0.18m[/tex]
As, the molal concentration of NaCl is the highest, so its boiling point will be the highest.
Thus, the order of increasing boiling points follows:
[tex]\text{Ethanol }<Mg(NO_3)_2<\text{ NaCl}[/tex]
Calculate the pH of: (a) 0.1M HCl; (b) 0.1M NaOH; (c) 3 X 10% M HNO3; (d) 5 X 10-10 M HCIO.; and (e) 2 x 10-8 M KOH.
Answer:
(a) [tex]pH = -Log (0.1M) = 1[/tex]
(b) [tex]pH = -Log (10^{-13}M) = 13[/tex]
(c) [tex]pH = -Log (3x10^{-3}M) = 2.5[/tex]
(d) [tex]pH = -Log (4.93x10^{-10}M) = 9.3[/tex]
(e) [tex]pH = -Log (5^{-7}M) = 6.3[/tex]
Explanation:
To calculate de pH of an acid solution the formula is:
[tex]pH = -Log ([H^{+}]) = 1[/tex]
were [H^{+}] is the concentration of protons of the solution. Therefore it is necessary to know the concentration of the protons for every solution in order to solve the problem.
(a) and (c) are strong acids so they dissociate completely in aqueous solution. Thus, the concentration of the acid is the same as the protons.
(b) and (e) are strong bases so they dissociate completely in aqueous solution too. Thus, the concentration of the base is the same as the oxydriles. But in this case it is necessary to consider the water autoionization to calculate the protons concentration:
[tex]K_{w} =[H^{+} ][OH^{-}]=10^{-14}[/tex]
clearing the [tex][H^{+} ][/tex]
[tex][H^{+} ]=\frac{10^{-14}}{[OH^{-}]}[/tex]
(d) is a weak base so it is necessary to solve the equilibrium first, knowing [tex]Ka=3.24x10^{-8}[/tex]
The reaction is [tex]HClO[/tex] → [tex]H^{+} + CO^{-}[/tex] so the equilibrium is
[tex]Ka=3.24x10^{-8}=\frac{x^{2}}{5x10^{-8}-x}[/tex]
clearing the x
[tex]{x^{2}={1.62x10^{-17}-3.24x10^{-8}x}[/tex]
[tex]x=[H^{+}]=4.93x10^{-10}[/tex]
How many significant figures are in the following? a) 0.1111 b) 2000 c) 35.6 d) 180,701
Answer:
a) 0.1111 : 4 significant figures
b) 2000 : 1 significant figure
c) 35.6 : 3 significant figures
d) 180,701 : 6 significant digits
Explanation:
Significant figures or digits of a number are the digits that carry meaning and contribute to the precision of the number.
a) 0.1111 : 4 significant figures, leading zeros are not significant digits.
b) 2000 : 1 significant figure, trailing zeros are not significant.
c) 35.6 : 3 significant figures, all digits are significant.
d) 180,701 : 6 significant digits, zeros between mom-zero digits are significant.
The number of significant figures in the given examples are:
a) 4 significant figures for 0.1111
b) 1 significant figure for 2000 (assuming no decimal)
c) 3 significant figures for 35.6
d) 6 significant figures for 180,701
To determine the number of significant figures in a number, certain rules are followed:
Non-zero digits are always significant.Any zeros between significant digits are significant.Leading zeros (zeros before non-zero digits) are not significant.Trailing zeros in a decimal number are significant.In a number without a decimal point, trailing zeros may or may not be significant, depending on whether a decimal point is specified or assumed.Applying these rules, we have:
0.1111 - All digits are significant because they are non-zero, giving 4 significant figures.2000 - Without additional context or a decimal point, we cannot be certain if the trailing zeros are significant, so this number has 1 significant figure.35.6 - All digits including the zero are significant, giving 3 significant figures.180,701 - All digits are significant, giving 6 significant figures.Henry law is obeyed by a gas when gas has high
a) Pressure
b) Temperature
c) Solubility
d) Non of the above
Answer:
Option b
Explanation:
Henry law describes solubility of gases in liquids.
According to Henry's law, amount of gas dissolved in a liquid depends upon its partial pressure above the liquids.
Mathematically, Henry's law is represented as:
C = K × P
Where,
C = Solubility of gas or concentration of gas in liquids
K = Henry's constant
P = Partial pressure of the gas over the liquid
For, Henry's law to be valid, pressure should be not too high and temperature should not be too low. Henry's law is also valid in case of low dissolved gas concentrations.
So, among the given options, option b, temperature is correct.
The half-life of a pesticide determines its persistence in the environment. A common pesticide degrades in a first-order process with a half-life of 2 days. What fraction (in decimal notation) of the pesticide remains in the environment after 18 days? Enter to 4 decimal places.
Answer:
0.0020 fraction of the pesticide remains in the environment after 18 days
Explanation:
For a first order reaction, [tex]\frac{N}{N_{0}}=(\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}[/tex]
where N is remaining mass after "t" time , [tex]N_{0}[/tex] is initial mass, [tex]\frac{N}{N_{0}}[/tex] represents fraction of mass remains after "t" time and [tex]t_{\frac{1}{2}}[/tex] is half-life
Here t is 18 days and [tex]t_{\frac{1}{2}}[/tex] is 2 days
So, [tex]\frac{N}{N_{0}}=(\frac{1}{2})^{\frac{18}{2}}=(\frac{1}{2})^{9}=0.0020[/tex]
Hence 0.0020 fraction of the pesticide remains in the environment after 18 days
The fraction (in decimal notation) of the pesticide that remains in the environment after 18 days is 0.0020
We'll begin by calculating the number of half-lives that has elapsed. This can be obtained as follow:
Half-life (t½) = 2 days
Time (t) = 18 days
Number of half-lives (n) =?n = t / t½
n = 18 / 2
n = 9Thus, 9 half-lives has elapsed.
Finally, we shall determine the fraction of the pesticide that remains in the environment. This can be obtained as follow:Original amount (N₀) = 1
Number of half-lives (n) = 9
Fraction remaining (N / N₀) =?N = 1/2ⁿ × N₀
Divide both side by N₀
N / N₀ = 1/2ⁿ
N / N₀ = 1/2⁹
N / N₀ = 0.0020Thus, the fraction of the pesticide that remains in the environment is 0.0020
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In an evaporator 25 ton / h of a solution of 10% NaOH, 10% NaCl, and 80% water by weight. During evaporation, the water evaporates and the salt precipitates like crystals They are allowed to settle and are removed. The outgoing concentrated solution of the Evaporator contains 50% NaOH, 2% NaCl and 48% water. Based on This information is requested:
1. Draw the process flow diagram, indicating each of its streams and compositions (known and unknown).
2. Calculate the kilograms of precipitated salt and the kilograms of solution concentrated for every hour of work.
Answer:
1- Flow Diagram (file attached)
2- Kilograms of precipitated salt: 2400kg/h
Kilograms of solution concentrated per hour: 5000kg/h
Explanation:
We have an Evaporator with 25ton/h of a solution with: 10% NaOH, 10% NaCl, and 80% water. That means that we have an input which is a stream with the following composition fractions:
Stream 1:
NaOH= 0.1
NaCl= 0.1
Water= 0.8
Then we have 3 outputs, 3 streams that leave the evaporator, which are:
Stream 2:
Only contains water so its composition fraction of water is 1.
Stream 3:
Only contains NaCl so its composition fraction of NaCL is 1.
Stream 4:
NaOH= 0.5
NaCl= 0.02
Water= 0.48
We know that stream 1 is 25 ton/h and enter the evaporator but we do not know the flow rate of stream 4 that is the concentrated solution leaving the evaporator, so we will make a particular mass balance of the component NaOH that is present in both streams:
fraction of NaOH in stream 1 ×flow rate of stream 1= fraction of NaOH in stream 4× flow rate of stream 4
0.1×25 ton/h = 0.5× flow rate of stream 4
flow rate of stream 4= (0.1×25 ton/h)/0.5= 5ton/h= 5000kg/h
Now to know the kilograms of precipitated salt, which is the flow rate of stream 3 we make a particular mass balance of the component NaCl:
(fraction of NaCl in stream 1 ×flow rate of stream 1)- (fraction of NaCl in stream 3×flow rate of stream 3)=flow rate of stream 4× fraction of NaCl in stream 4
(0.1×25 tn/h)- (1×flow rate of stream 3)= 5tn/h × 0.02
flow rate of stream 3= 2.4 tn/h =2400 kg/h
Consider an atom with 6 protons, 6 neutrons, and 6 electrons. State the number of protons, neutrons, and electrons that would occur in a different isotope of the same element, and explain what changed by relating the change to the definition of an isotope.
Explanation:
An isotope is defined as the specie which contains same number of protons but different number of neutrons.
For example, [tex]^{12}_{6}C[/tex] and [tex]^{13}_{6}C[/tex] are isotopes.
In a neutral carbon atom, there are 6 protons and 6 neutrons.
As it is known that in a neutral atom the number of protons equal to the number of electrons.
This means that in a neutral carbon atom there are also 6 electrons.
Whereas [tex]^{13}_{6}C[/tex] is an isotope of carbon atom whose atomic number is 6 and atomic mass is 13.
Hence, calculate the number of neutrons in [tex]^{13}_{6}C[/tex] as follows.
Atomic mass = no. of protons + no. of neutrons
13 = 6 + no. of neutrons
no. of neutrons = 13 - 6
= 7
Hence, in a [tex]^{13}_{6}C[/tex] isotope of carbon atom there are 6 protons, 6 electrons and 7 neutrons.
This shows that change in number of neutrons take place according to the definition of an isotope.
Forces between similar molecules are said to be
---Select---cohesiveadhesiveconcaveconvex while those between
different types ofmolecule are said to be
---Select---cohesiveadhesiveconcaveconvex. Water 'beads' due to its
strong---Select---cohesiveadhesiveconcaveconvex forces. The
meniscus of water in a glasstube is
---Select---cohesiveadhesiveconcaveconvex because the
---Select---cohesiveadhesiveconcaveconvex forces are strong.
Answer:
Forces between similar molecules are said to be cohesive while those between different types of molecules are said to be adhesive.
Water 'beads' due to its strong cohesive forces. The meniscus of water in a glass tube is concave because the adhesive forces are strong.
Explanation:
The water in a tube has stronger adhesive forces between the water and glass molecules, so the cohesive forces between water molecules are weaker. That makes the water 'ascend' through the tube, giving a concave form of the meniscus. Another example is mercury, which is the opposite. In this case, the cohesive forces are stronger than the adhesive ones, thus the meniscus is convex.
How many ml of concentrated hydrochloric acid would be required to make 1 L of a 0.2 M solution? Assume that concentrated hydrochloric acid is 12.1 M
Answer: The volume of concentrated hydrochloric acid required is 16.53 mL
Explanation:
To calculate the volume of concentrated solution, we use the equation:
[tex]M_1V_1=M_2V_2[/tex]
where,
[tex]M_1\text{ and }V_1[/tex] are the molarity and volume of the concentrated solution
[tex]M_2\text{ and }V_2[/tex] are the molarity and volume of diluted solution
We are given:
Conversion factor: 1 L = 1000 mL
[tex]M_1=12.1M\\V_1=?mL\\M_2=0.2M\\V_2=1L=1000mL[/tex]
Putting values in above equation, we get:
[tex]12.1\times V_1=0.2\times 1000\\\\V_1=16.53mL[/tex]
Hence, the volume of concentrated hydrochloric acid required is 16.53 mL
Describe where the "Reference Carbon" (the one that determines if the structure is D or L) is located in a carbohydrate with more than one asymmetric carbon
Answer:
Farthest from the carbonyl carbon.
Explanation:
Reference carbon that determined the absolute D and L configuration is located farthest from the carbonyl carbon.
In other words, reference carbon is that assymentric carbon which is located farthest from the carbolyl carbon and has configuration similar to D- or L-glyceraldehyde isomers.
D and L configuration is decided by the direction of -OH group attached to the reference carbon.
In L-isomer, -OH group is attached to the left side of the reference carbon and in D-isomer, -OH group is attached to the right side of the reference carbon.
Menthol (molar mass = 156.3 g/mol), the strong-smelling substance in many cough drops, is a compound of carbon, hydrogen and oxygen. When 0.1595g of menthol was burned in a combustion apparatus, 0.449g of CO2 and 0.184g of H2O formed. What is menthol’s molecular formula?
Answer: The molecular formula for the menthol is [tex]C_{10}H_{20}O[/tex]
Explanation:
The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:
[tex]C_xH_yO_z+O_2\rightarrow CO_2+H_2O[/tex]
where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.
We are given:
Mass of [tex]CO_2=0.449g[/tex]
Mass of [tex]H_2O=0.184g[/tex]
We know that:
Molar mass of carbon dioxide = 44 g/mol
Molar mass of water = 18 g/mol
For calculating the mass of carbon:In 44g of carbon dioxide, 12 g of carbon is contained.
So, in 0.449 g of carbon dioxide, [tex]\frac{12}{44}\times 0.449=0.122g[/tex] of carbon will be contained.
For calculating the mass of hydrogen:In 18g of water, 2 g of hydrogen is contained.
So, in 0.184 g of water, [tex]\frac{2}{18}\times 0.184=0.0204g[/tex] of hydrogen will be contained.
Mass of oxygen in the compound = (0.1595) - (0.122 + 0.0204) = 0.0171 gTo formulate the empirical formula, we need to follow some steps:
Step 1: Converting the given masses into moles.Moles of Carbon = [tex]\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.122g}{12g/mole}=0.0102moles[/tex]
Moles of Hydrogen = [tex]\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.0204g}{1g/mole}=0.0204moles[/tex]
Moles of Oxygen = [tex]\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.0171g}{16g/mole}=0.00107moles[/tex]
Step 2: Calculating the mole ratio of the given elements.For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.00107 moles.
For Carbon = [tex]\frac{0.0102}{0.00107}=9.53\approx 10[/tex]
For Hydrogen = [tex]\frac{0.0204}{0.00107}=19.54\approx 20[/tex]
For Oxygen = [tex]\frac{0.00107}{0.00107}=1[/tex]
Step 3: Taking the mole ratio as their subscripts.The ratio of C : H : O = 10 : 20 : 1
Hence, the empirical formula for the given compound is [tex]C_{10}H_{20}O_1=C_{10}H_{20}O[/tex]
For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.
The equation used to calculate the valency is :
[tex]n=\frac{\text{molecular mass}}{\text{empirical mass}}[/tex]
We are given:
Mass of molecular formula = 156.3 g/mol
Mass of empirical formula = 156 g/mol
Putting values in above equation, we get:
[tex]n=\frac{156.3g/mol}{156g/mol}=1[/tex]
Multiplying this valency by the subscript of every element of empirical formula, we get:
[tex]C_{(1\times 10)}H_{(1\times 20)}O_{(1\times 1)}=C_{10}H_{20}O[/tex]
Thus, the molecular formula for the menthol is [tex]C_{10}H_{20}O[/tex]
Consider the formation of nitryl fluoride: 2NO2(g)+F2(g)⇌2NO2F(g) The reaction is first order in F2 and second order overall. What is the rate law? View Available Hint(s) Consider the formation of nitryl fluoride: The reaction is first order in and second order overall. What is the rate law? rate=k[NO2]2[F2]2 rate=k[NO2][F2]2 rate=k[F2] rate=k[NO2] rate=k[NO2][F2] rate=k[NO2]2[F2]
Answer:
Rate = [tex]k[NO_{2}][F_{2}][/tex]
Explanation:
Two reactants are present in this reaction which are [tex]NO_{2}and F_{2}[/tex]We know overall order of a reaction is summation of individual order with respect to reactants present in rate law equation.Here, overall order of reaction is 2 including first order with respect to [tex]F_{2}[/tex]So, rate of reaction should also be first order with respect to another reactant i.e. first order with respect to [tex]NO_{2}[/tex]So, rate law: rate = [tex]k[NO_{2}][F_{2}][/tex]If ine mole of pennies were divided amung 250 million peoplein
the US, how many dollars would each person recieve?
Answer:
[tex]2.4088\times 10^{13}[/tex] dollars each person will receive.
Explanation:
Number of people in which 1 mole of pennies is distributed = 250 million =
[tex]1 million = 10^6 [/tex]
250 million = [tex]2.5\times 10^8 [/tex] persons
Number of pennies in 1 mole = [tex]6.022\times 10^{23}[/tex]
Pennies per person:
[tex]\frac{6.022\times 10^{23} pennies}{2.5\times 10^8 persons}=2.4088\times 10^{15} pennies/person[/tex]
1 penny = 0.01 $
[tex]2.4088\times 10^{15} pennies/person=2.4088\times 10^{15}\times 0.01 \$/person=2.4088\times 10^{13} \$/person[/tex]
[tex]2.4088\times 10^{13}[/tex] dollars each person will receive.
A 1577 kg car is traveling down the road at 96.6 km/h. While traveling at this rate of speed, what is the kinetic energy of this vehicle in kilojoules? E = kJ contect prvacy policy e of e h about us careen MacBook Pro Q Search or enter website name & 7 4 5 6 E P Y U R K G H C B V ontion Command V M 00 N T
Answer:
The kinetic energy of the car is 566 kJ
Explanation:
The kinetic energy is calculated using this equation:
E = 1/2 m*v²
where:
E = kinetic energy
m = mass of the object
v = speed of the object
First, let´s convert the units of the speed:
96.6 km/h *(1 h / 3600s) * (1000 m / 1 km) =26.8 m/s
The kinetic energy will be:
E = 1/2 1577 kg * (26.8 m/s)² = 5.66 x 10⁵ kg m²/s² = 5.66 x 10⁵ J
E = 5.66 x 10⁵ J * (1 kJ /1000 J ) = 566 kJ
The kinetic energy of this vehicle : 567.600 kJ
Further explanationEnergy is the ability to do work
Energy because its motion is expressed as Kinetic energy (KE) which can be formulated as:
[tex] \displaystyle KE = \frac {1} {2} mv ^ 2 [/tex]
So for two objects that have the same speed, the greater the mass of the object, the greater the kinetic energy
Whereas for a stationary object it has kinetic energy = 0
While the effort required by an object to change the velocity from v1 to v2 is the change in the kinetic energy of the object
W = Ek2 - Ek1
A 1577 kg car is traveling down the road at 96.6 km / h
then known
m = 1577 kg
v = 96.6 km / h
[tex]=96.6\times \dfrac{1000}{3600} = 26.83 \frac{m}{s}[/tex]
Kinetik energy :
[tex]\displaystyle KE=\frac{1}{2}mv^2\\\\KE=\frac{1}{2}.1577.26.83^2\\\\KE=567.600~kJ[/tex]
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Dietary Supplement Fact Sheets is hosted by a. Food & Drug Administration d. National Institute of Health b. Private agency c. Center for Disease Control
Answer:
The answer is d. National Institute of Health.
Explanation:
Knowing the information about supplements that you include in your daily diet is one of the most important functions of the institute of health due to it helps to control the safety information about all products you consume.
Other reason is that by law since 1994 it is required to define what is a dietary supplement which is basically a product that contains vitamins, minerals or other substances needed in a diet.
Identify the molecules with a dipole moment: (a) SF (b) CF (c) CCCB (d) CHCI (e) H.CO
Answer: An atom has either tendency of accepting or losing electron, on the basis of this virtue it is named as
Electronegative: An atom has tendency to attract the shared pair of electron towards itself is called electronegative.Electropositive: An atom that has tendency to give the shared pair of electron towards an electronegative atom is called electropositive.For the existence of dipole within a molecule there must be a difference in electronegativity of the atoms participating in it. All the molecules here have dipole moment because the atoms participating in them have a difference in electronegative.
You have a sample of water that contains the organic compound
C7H12ON2 at a concentration of 50
mg/L. The compound can be oxidized by bacteria to form carbon
dioxide, water, and ammonia. How many mg/L of oxygen is needed to
biodegrade the compound? Note: Determine only carbonaceous
demand.
Answer:
131.4 mg/L of oxygen is needed to biodegrade the organic compound.
Explanation:
The chemical reaction will be written as:
[tex]2C_7H_{12}ON_2+23O_2\rightarrow 14CO_2+12H_2O+4NO_2[/tex]
Concentration of the organic compound = 50 mg/L
This means that 50 milligrams of organic compound in present in 1 L of the solution.
50 mg = 0.050 g
1 mg = 0.001 g
Moles of organic compound = [tex]\frac{0.050 g}{140 g/mol}=0.0003571 mol[/tex]
According to reaction, 2 moles of organic compound reacts with 23 moles of oxygen gas.
Then 0.0003571 moles of an organic compound will react with:
[tex]\frac{23}{2}\times 0.0003571 mol=0.004107 mol[/tex] oxygen gas.
Mass of 0.004107 moles of oxygen gas:
0.004107 mol × 32 g/mol = 0.1314 g = 131.4 mg
131.4 mg/L of oxygen is needed to biodegrade the organic compound.
A 0.1153-gram sample of a pure hydrocarbon was burned in a C-H combustion train to produce 0.3986 gram of CO2and 0.0578 gram of H2O. Determine the masses of C and H in the sample and the percentages of these elements in this hydrocarbon.
Answer: The mass of carbon and hydrogen in the sample is 0.1087 g and 0.0066 g respectively and the percentage composition of carbon and hydrogen in the sample is 94.27 % and 5.72 % respectively.
Explanation:
The chemical equation for the combustion of hydrocarbon having carbon and hydrogen follows:
[tex]C_xH_y+O_2\rightarrow CO_2+H_2O[/tex]
where, 'x' and 'y' are the subscripts of carbon and hydrogen respectively.
We are given:
Mass of [tex]CO_2=0.3986g[/tex]
Mass of [tex]H_2O=0.0578g[/tex]
Mass of sample = 0.1153 g
We know that:
Molar mass of carbon dioxide = 44 g/mol
Molar mass of water = 18 g/mol
For calculating the mass of carbon:In 44g of carbon dioxide, 12 g of carbon is contained.
So, in 0.3986 g of carbon dioxide, [tex]\frac{12}{44}\times 0.3986=0.1087g[/tex] of carbon will be contained.
For calculating the mass of hydrogen:In 18g of water, 2 g of hydrogen is contained.
So, in 0.0578 g of water, [tex]\frac{2}{18}\times 0.0578=0.0066g[/tex] of hydrogen will be contained.
To calculate the percentage composition of a substance in sample, we use the equation:
[tex]\%\text{ composition of substance}=\frac{\text{Mass of substance}}{\text{Mass of sample}}\times 100[/tex] ......(1)
For Carbon:Mass of sample = 0.1153 g
Mass of carbon = 0.1087 g
Putting values in equation 1, we get:
[tex]\%\text{ composition of carbon}=\frac{0.1087g}{0.1153g}\times 100=94.27\%[/tex]
For Hydrogen:Mass of sample = 0.1153 g
Mass of hydrogen = 0.0066 g
Putting values in equation 1, we get:
[tex]\%\text{ composition of hydrogen}=\frac{0.0066g}{0.1153g}\times 100=5.72\%[/tex]
Hence, the mass of carbon and hydrogen in the sample is 0.1087 g and 0.0066 g respectively and the percentage composition of carbon and hydrogen in the sample is 94.27 % and 5.72 % respectively.
The masses of C and H in the sample are 0.1083 grams and 0.00324 grams, respectively. The percentages of C and H in the hydrocarbon are approximately 93.94% and 2.80%, respectively.
All the carbon in the hydrocarbon is converted to [tex]CO_2[/tex], we can use the mass of [tex]CO_2[/tex] to find the mass of carbon:
[tex]\[ \text{Mass of carbon} = \text{Mass of CO2} \times \frac{\text{Molar mass of carbon}}{\text{Molar mass of CO2}} \][/tex]
[tex]\[ \text{Mass of carbon} = 0.3986 \text{ g} \times \frac{12.01 \text{ g/mol}}{44.01 \text{ g/mol}} \][/tex]
[tex]\[ \text{Mass of carbon} = 0.3986 \text{ g} \times \frac{12}{44} \][/tex]
[tex]\[ \text{Mass of carbon} = 0.3986 \text{ g} \times 0.2727 \][/tex]
[tex]\[ \text{Mass of carbon} = 0.1083 \text{ g} \][/tex]
All the hydrogen in the hydrocarbon is converted to [tex]H_2O[/tex], we can use the mass of [tex]H_2O[/tex] to find the mass of hydrogen:
[tex]\[ \text{Mass of hydrogen} = \text{Mass of H2O} \times \frac{\text{Molar mass of hydrogen}}{\text{Molar mass of H2O}} \][/tex]
[tex]\[ \text{Mass of hydrogen} = 0.0578 \text{ g} \times \frac{1.008 \text{ g/mol}}{18.015 \text{ g/mol}} \][/tex]
[tex]\[ \text{Mass of hydrogen} = 0.0578 \text{ g} \times \frac{1}{18} \][/tex]
[tex]\[ \text{Mass of hydrogen} = 0.0578 \text{ g} \times 0.05597 \][/tex]
[tex]\[ \text{Mass of hydrogen} = 0.00324 \text{ g} \][/tex]
Now that we have the masses of carbon and hydrogen, we can calculate the percentages of these elements in the hydrocarbon:
[tex]\[ \text{Percentage of carbon} = \frac{\text{Mass of carbon}}{\text{Mass of hydrocarbon}} \times 100\% \][/tex]
[tex]\[ \text{Percentage of carbon} = \frac{0.1083 \text{ g}}{0.1153 \text{ g}} \times 100\% \][/tex]
[tex]\[ \text{Percentage of carbon} \approx 93.94\% \][/tex]
[tex]\[ \text{Percentage of hydrogen} = \frac{\text{Mass of hydrogen}}{\text{Mass of hydrocarbon}} \times 100\% \][/tex]
[tex]\[ \text{Percentage of hydrogen} = \frac{0.00324 \text{ g}}{0.1153 \text{ g}} \times 100\% \][/tex]
[tex]\[ \text{Percentage of hydrogen} \approx 2.80\% \][/tex]
Balance the equation in the box. Click in the answer box to activate the palette. N2(g) + H2(g) → NH3(g)
Answer:
N2(g) + 3H2(g) → 2 NH3(g)
Explanation:
N2(g) + H2(g) → NH3(g)
We start equaling the number of N atoms in both sides multiplying by 2 the NH3.
N2(g) + H2(g) → 2 NH3(g)
So we equals the H atoms (there are six in products sites)
N2(g) + 3 H2(g) → 2 NH3(g)
You are given a protein solution with a concentration of 0.15 mg/ml.
v. Suppose that we want to prepare 100 microliters of 10 micrograms/microliters solution. How much of H2O and protein stock do we need to add to obtain the target concentration and volume? (THE CONCENTRATION IS NOT GREATER IN THE QUESTION, YOU NEED TO CONVERT IT TO MICROGRAMS/MICROLITERS.... THEN IT SHOULD MAKE SENSE)
Answer:
6,666.66 micro liter of water and protein stock we will need to add to obtain the target concentration and volume.
Explanation:
Concentration of given solution[tex]C_1 = 0.15 mg/mL[/tex]
1 mL = 1000 μL , 1 mg = 1000 μg
[tex]C_1=0.15 mg/mL=\frac{0.15\times 1000 \mu g}{1\times 1000 \mu L}=0.15 \mu g/\mu L[/tex]
The volume of the given solution =[tex]V_1= V[/tex]
Concentration of required solution = [tex]C_2=10 \mu g/\mu L[/tex]
Volume of required solution = [tex]V_2=100 \mu L[/tex]
[tex]C_1V_1=C_2V_2[/tex]
[tex]V=\frac{C_2V_2}{C_1}=\frac{10 \mu g/\mu L\times 100 \mu L}{0.15 \mu g/\mu L}=6,666.66 \mu L[/tex]
6,666.66 micro liter of water and protein stock we will need to add to obtain the target concentration and volume.
Nault 25000L 250 our = 25000L 15040 8. How many grams of CaCl2 are needed to make 150.0 mL of a 0.500 M CF solution? (Note: CaCl2 is a soluble salt. The molar mass of CaCl2 is 110.98 g/mol.)
Answer:
You need 8,324 g of CaCl₂ yo make this solution
Explanation:
Molarity is a way to express concentration in a solution, in units of moles of solute per liter of solution.
To know the grams of CaCl₂ it is necessary to know, first, the moles of this substance with the desired volume and concentration , thus:
0,1500 L × [tex]\frac{0,500 mol}{L}[/tex] = 0,075 CaCl₂ moles
Now, with the molar mass of CaCl₂ you will obtain the necessary grams, thus:
0,075 CaCl₂ moles × [tex]\frac{110,98 g}{mol}[/tex] = 8,324 g of CaCl₂
So, you need 8,324 g of CaCl₂ to make 150,0 mL of a 0,500M solution
I hope it helps!
A 0.4657 g sample of a pure soluble bromide compound is dissolved in water, and all of the bromide ion is precipitated as AgBr by the addition of an excess of silver nitrate. The mass of the resulting AgBr is found to be 0.8878 g. What is the mass percentage of bromine in the original compound?
Answer:
The mass percentage of bromine in the original compound is 81,12%
Explanation:
Step 1: Calculate moles AgBr
moles AgBr = mass AgBr / molar mass AgBr
= 0.8878 g / 187.77 g/mol
= 0.00472812 moles AgBr
⇒
Since 1 mol AgBr contains 1 mol Br-
Then the amount of moles Br- in the original sample must also have been 0.00472812 moles
Step 2: Calculating mass Br-
mass Br- = molar mass Br x moles Br-
= 79.904 g/mol x 0.00472812 mol
= 0.377796 g Br-
⇒
There were 0.377796 g Br- in the original sample
Step 3: Calculating mass percentage Br-
⇒mass percentage = actual mass Br- / total mass x 100%
% mass Br = 0.377796 g / 0.4657 g x 100 %
= 81.12%
In the winter, a heated home in the Northeast might be maintained at a temperature of 78°F. What is this temperature on the Celsius and Kelvin scales? T.r 1.8T.c+32 °C Submit Answer &Next
Answer: [tex]25.6^0C[/tex], [tex]298.6K[/tex]
Explanation:
Temperature of the gas is defined as the degree of hotness or coldness of a body. It is expressed in units like [tex]^0C[/tex] and [tex]K[/tex]
These units of temperature are inter convertible.
We are given:
Temperature of the gas = [tex]78^0F[/tex]
Converting this unit of temperature into [tex]^0C[/tex] by using conversion factor:
[tex]^oC=\frac{5}{9}\times (^oF-32)[/tex]
[tex]^0C=\frac{5}{9}\times (78^oF-32)[/tex]
[tex]25.6^0C[/tex]
Converting this unit of temperature into [tex]K[/tex] by using conversion factor:
[tex]K=t^0C+273[/tex]
[tex]K=25.6+273[/tex]
[tex]298.6K[/tex]
Thus the temperature on the Celsius and Kelvin scales are [tex]25.6^0C[/tex] and [tex]298.6K[/tex] respectively.