Which expression correctly describes energy using SI units? A. 1 J=1kg•m^2/s^2 B. 1 J= 1kg•m/s^2 C. 1 J= 1kg• m/s D. 1 J= 1kg•m^2/s

Answers

Answer 1

Answer:

A. 1 J=1kg•m^2/s^2

Explanation:

Energy refers to the capacity to do work. According to the International System of units (SI units), energy is measured in Joules.

Energy is represented by the force applied over a distance. Force is measured in Newton (N) and distance in metres (m). Hence, energy is Newton × metre (N.m)

Newton is derived from the SI units of mass (Kilograms), and acceleration (metres per seconds^2) i.e Kg.m/s^2, since Force = mass × acceleration.

Since; Energy = Newton × metres

If Newton = Kg.m/s^2 and metres = m

Energy (J) will therefore be; Kg.m/s^2 × m

1J = Kg.m^2/s^2

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1. Potassium (K) has an atomic mass of 39.0983 amu and only two naturally-occurring isotopes. The K-41 isotope (40.9618 amu) has a natural abundance of 6.7302%. What is the mass (in amu) of the other isotope

At 25.0 ⁰C the henry's law constant for hydrogen sulfide(H2S) gas in water is 0.087 M/atm. Caculate the mass in grams of H2S gas that can be dissolved in 400.0 ml of water at 25.00 C and a H2S partial pressure of 2.42atm.

Answers

Answer: The mass of hydrogen sulfide that can be dissolved is 2.86 grams.

Explanation:

Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the gas.

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_(H_2S)=K_H* p_(liquid)$

where,

$K_H$ = Henry's constant = $0.087M/atm$

$p_(H_2S)$ = partial pressure of hydrogen sulfide gas = 2.42 atm

Putting values in above equation, we get:

$C_(H_2S)=0.087M/atm* 2.42atm\n\nC_(H_2S)=0.2105M$

To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}* 1000}{\text{Molar mass of solute}* \text{Volume of solution (in mL)}}$

We are given:

Molarity of solution = 0.2105 M

Molar mass of hydrogen sulfide = 34 g/mol

Volume of solution = 400.0 mL

Putting values in above equation, we get:

$0.2105M=\frac{\text{Mass of hydrogen sulfide}* 1000}{34g/mol* 400.0mL}\n\n\text{Mass of }H_2S=(0.2105* 34* 400)/(1000)=2.86g$

Hence, the mass of hydrogen sulfide that can be dissolved is 2.86 grams.

Dinitrogen monoxide has a structural formula of NNO and requires resonance structures in order to draw the Lewis structures of the molecule. Based on formal charge distributions, themostsignificant (stable) resonance structure for this molecule exhibits the order of formal charges for the 1st N, the central N, and the O atoms, respectively, as:A. 0,+1,-1
B. -1,+1,0
C. -2,+3,-1
D. 0,0,0

Answers

Three resonance structures contribute to the structure of dinitrogen monoxide.

The resonance structure is invoked when a single structure can not sufficiently explain all the bonding properties of a compound. All the various contributing structures contribute to the final structure of the compound but not all to the same degree.

There are three resonance structures of dinitrogen monoxide. The most stable structure is always the structure that has the formal charges as -1, +1 and zero as shown.

Learn more: brainly.com/question/14283892

Answer:

A. 0, +1, -1

Explanation:

You can draw the lewis structure for NNO 3 ways: With two double bonds N=N=O, with a triple bond between the N and O and single bond between the two N's, or a triple bond between the two N's and a single bond between the N and O.

The goal is to have formal charges that are as small as possible, to have no identical formal charges on adjacent atoms, and to have the most negative formal charge on the most electronegative atom. The most stable structure is the one with the triple bond between the two N's because it gives the formal charges 0, 1, and -1 respectively. Unlike the other two structures, the negative formal charge is correctly placed on O, the most electronegative atom.

Calculate the mass percent of oxygen in KMnO4.

Answers

Answer: 40.496%

Hope this helps! (:

Identify the true statements about introns.a- they code for polypeptide proteinsb- they have a branch site located 20 to 50 nucleotides upstream of the 3' splice sitec- they end with the nucleotides AG at the 3' endd- they begin with the nucleotides GU at the 5' ende- they tend to be common in bacterial genes

Answers

Answer:

The answer is "Option b, c, and d".

Explanation:

In such a gene, Autosomes are also the sequence for code and transposable elements, not the series of encoding. Through the expression of genes, such fragments of its introns are split through protein complexes throughout the translation process. There has been no kenaf fiber in the genomes of prokaryotic cells.

Develop expressions for the mole fraction of reacting species functions of the reaction coordinate for: A system initially containing 2 mol NH3 and 5 mol O2 and undergoing the reaction: 4NH3 (g) + 5O2 (g) ® 4NO (g) + 6 H20 (g) A system initially containing 3 mol NO2, 4 mol NH3, and 1 mol N2 and undergoing the reaction: 6NO2 (g) + 8NH3 (g) ® 7N2 (g) +12H2O (g)

Answers

Answer:

Individual mole fractions of all the species of the all reaction is as follows.

(a)

$y_{NH_(3)}=(2+(-4)\epsilon)/(7+\epsilon)$

$y_{O_(2)}=(5+(-5)\epsilon)/(7+\epsilon)$

$y_(NO)=(0+(4)\epsilon)/(7+\epsilon)=(4\epsilon)/(7+\epsilon)$

$y_{H_(2)O}=(0+6\epsilon)/(7+\epsilon)$

(b)

$y_{H_(2)S}=(3+(-2)\epsilon)/(8- \epsilon)$

$y_{O_(2)}=(5+(-3)\epsilon)/(8- \epsilon)$

$y_{H_(2)O}=(2\epsilon)/(8- \epsilon)$

$y_{SO_(2)}=(2\epsilon)/(8- \epsilon)$

(c)

$y_{NO_(2)}=(3+(-6)\epsilon)/(8+5\epsilon)$

$y_{NH_(3)}=(4+(-8)\epsilon)/(8+5\epsilon)$

$y_{N-{2}}=(1+7\epsilon)/(8+5 \epsilon)$

$y_{H_(2)O}=(12\epsilon)/(8+5\epsilon)$

Explanation:

(a)

Initial number of moles of $NH_(3)$ and $O_(2)$ are 2 mol and 5 mol respectively.

The given chemical reaction is as follows.

$4NH_(3)(g)+5O_(2)(g)\rightarrow 4NO(g)+6H_(2)O$

The stoichiometric numbers are as follows.

$v_{NH_(3)}=-4$

$v_{O_(2)}=-5$

$v_(NO)=4$

$v_{H_(2)O}=6$

The total number of moles initially present -7

$\Sigma v_(i)\epsilon = (-4-5+4+6)= 1\epsilon$

The expression for the mole fraction of species"i" is as follows.

$y_(i)=\frac{(n_{i_(o)})+(v_(i)\epsilon) }{n_(o)+v\epsilon}$

The individual mole fractions of all the species are as follows.

$y_{NH_(3)}=(2+(-4)\epsilon)/(7+\epsilon)$

$y_{O_(2)}=(5+(-5)\epsilon)/(7+\epsilon)$

$y_(NO)=(0+(4)\epsilon)/(7+\epsilon)=(4\epsilon)/(7+\epsilon)$

$y_{H_(2)O}=(0+6\epsilon)/(7+\epsilon)$

(b)

Initial number of moles of $H_(2)S$ and $O_(2)$ are 3 mol and 5 mol respectively.

The given chemical reaction is as follows.

$2H_(2)S(g)+3O_(2)(g)\rightarrow 2H_(2)O(g)+2SO_(2)$

The stoichiometric numbers are as follows.

$v_{H_(2)S}=-2$

$v_{O_(2)}=-3$

$v_{H_(2)O}=2$

$v_{SO_(2)=2$

The total number of moles initially present -8

$\Sigma v_(i)\epsilon = (-2-3+2+2)= - \epsilon$

The expression for the mole fraction of species"i" is as follows.

$y_(i)=\frac{(n_{i_(o)})+(v_(i)\epsilon) }{n_(o)+v\epsilon}$

The individual mole fractions of all the species are as follows.

$y_{H_(2)S}=(3+(-2)\epsilon)/(8- \epsilon)$

$y_{O_(2)}=(5+(-3)\epsilon)/(8- \epsilon)$

$y_{H_(2)O}=(2\epsilon)/(8- \epsilon)$

$y_{SO_(2)}=(2\epsilon)/(8- \epsilon)$

(c)

Initial number of moles of $NO_(2)$ , $NH_(3)$ and $N_(2)$ are 3 mol,4 mol and 1 mol respectively.

The given chemical reaction is as follows.

$6NO_(2)(g)+8NH_(3)(g)\rightarrow 7N_(2)(g)+12H_(2)O$

The stoichiometric numbers are as follows.

$v_{NO_(2)}=-6$

$v_{NH_(3)}=-8$

$v_{N_(2)}=7$

$v_{H_(2)O}=12$

The total number of moles initially present -8

$\Sigma v_(i)\epsilon = (-6-8+7+12)= 5 \epsilon$

The expression for the mole fraction of species"i" is as follows.

$y_(i)=\frac{(n_{i_(o)})+(v_(i)\epsilon) }{n_(o)+v\epsilon}$

The individual mole fractions of all the species are as follows.

$y_{NO_(2)}=(3+(-6)\epsilon)/(8+5\epsilon)$

$y_{NH_(3)}=(4+(-8)\epsilon)/(8+5\epsilon)$

$y_{N-{2}}=(1+7\epsilon)/(8+5 \epsilon)$

$y_{H_(2)O}=(12\epsilon)/(8+5\epsilon)$

Final answer:

Expressions for the mole fractions of reacting species are determined using stoichiometry and the initial molar amounts, taking into account the stoichiometric coefficients of the chemical reactions.

Explanation:

To develop expressions for the mole fraction of reacting species as functions of the reaction coordinate for the given systems, we will examine each reaction individually. For the reaction 4NH3 (g) + 5O2 (g) ® 4NO (g) + 6 H2O (g), we can use stoichiometry to correlate the molar amounts of each species with reaction progress. Given the initial amounts, we will track how the molar amount changes for each mole of NH3 reacted.

Starting with 2 mol NH3 and 5 mol O2, the mole ratio from NH3 to NO and H2O is 1:1 and 1:1.5, respectively. The mole ratio from NH3 to O2 is 4:5. If x moles of NH3 react, the mole fractions for each species at any point in the reaction can be expressed as follows:

Mole fraction of NH3: (2 - x)/(Total moles)
Mole fraction of O2: (5 - 5x/4)/(Total moles)
Mole fraction of NO: (4x)/(Total moles)
Mole fraction of H2O: (6x)/(Total moles)

Note that 'Total moles' is the sum of the ongoing moles of all species. The mole fractions must always add up to 1 at any point during the reaction.

For the second reaction 6NO2 (g) + 8NH3 (g) ® 7N2 (g) +12H2O (g), with initial amounts of 3 mol NO2, 4 mol NH3, and 1 mol N2, similar steps are taken. For every mole of NH3 reacted, the corresponding changes in molar amounts can be calculated from the stoichiometry of the balanced equation.

Learn more about mole fractions here:

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To create an image, the lens of an eye ________focuses light on the cornea
reflects light away from the retina
bends the cornea to correct vision
focuses light on the retina

Answers

It focuses light on the retina.

Answer:

focuses light on the retina

Explanation:

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