Answer:
Δ S° > 0
Explanation:
Entropy -
In a system, the randomness is measured by the term entropy .
Randomness basically refers as a form of energy that can not be used for any work.
The change in entropy is given by the change in heat per change in temperature.
When solid is converted to gas entropy increases,As the molecules in solid state are tightly packed and has more force of attraction between the molecules, but as it is converted to gas, the force of attraction between the molecule decreases and hence entropy increases.
So,
The particles of the substance , if are tightly held by strong force of attraction will decrease the entropy ,
And
If the particles are loosely held , the entropy will increase .
From the reaction given in the question ,
Cu₂O (s) + C (s) → 2 Cu (s) + CO (g)
Considering the states of the reactant and the product ..
The product formed i.e. CO is in gaseous species , hence , the entropy increases .
An infant ibuprofen suspension contains 100 mg/5.0mL suspension. The recommended dose is 10 mg/kg body weight.How many milliliters of this suspension should be given to an infant weighing 15 lb ? (Assume two significant figures.)
Answer : 3.4 mL of this suspension should be given to an infant weighing 15 lb.
Explanation :
First we have to convert weight of infant from 'lb' to 'kg'.
Conversion used :
As, 1 lb = 0.454 kg
So, 15 lb = [tex]\frac{15lb}{1lb}\times 0.454kg=6.81kg[/tex]
Now we have to calculate the recommended dose.
As, 1 kg weight recommended dose 10 mg
So, 6.81 kg weight recommended dose [tex]\frac{6.81kg}{1kg}\times 10mg=68.1mg[/tex]
Now we have to calculate the milliliter of ibuprofen suspension.
As, 100 mg dose present in 5.0 mL
So, 68.1 mg dose present in [tex]\frac{68.1mg}{100mg}\times 5.0mL=3.4mL[/tex]
Therefore, 3.4 mL of this suspension should be given to an infant weighing 15 lb.
The type(s) of intermolecular forces exhibited by hydrogen bromide molecules, HBr is/are. Fill in the blank
Answer:
London dispersion forces and dipole-dipole interactions
Explanation:
Intermolecular forces are the interactive forces that are present between the molecules. These intermolecular forces can be attractive or repulsive.
Hydrogen bromide is a diatomic covalent polar molecule, having chemical formula: HBr. Since, hydrogen bromide is a polar molecule, it has a permanent dipole.
Therefore, the intermolecular forces present between the hydrogen bromide molecules are dipole-dipole interactions and the london dispersion forces.
A water treatment plant has 4 settling tanks that operate in parallel (the flow gets split into 4 equal flow streams), and each tank has a volume of 600 m3 . If the total flow to the plant is 12 MGD, what is the retention time in each of the settling tanks? b) If, instead, the tanks operated in series (the entire flow goes first through one tank, then the second, and so on), what would be the retention time in each tank?
Answer:
a) When the 4 tanks operate in parallel the retention time is 1.26 hours.
b) If the tanks are in series, the retention time would be 0.31 hours
Explanation:
The plant has 4 tanks, each tank has a volume V = 600 [tex]m_3[/tex]. The total flow to the plant is Ft = 12 MGD (Millions of gallons per day)
When we use the tanks in parallel, it means that the total flow will be divided in the total number of tanks. F1 will be the flow of each tank.
Firstly, we should convert the MGD to [tex]m_3 /day[/tex]. In that sense, we can calculate the retention time using the tank volume in [tex]m_3[/tex].
[tex]Ft =12 \frac{MG}{day} (\frac{1*10^6 gall}{1MG} ) (\frac{3.78 L}{1 gall} ) (\frac{1 dm^3}{1L} ) (\frac{1 m^3}{10^3 dm^3} ) = 45360 \frac{m^3}{day}[/tex]
After that, we should divide the total flow by four, because we have four tanks.
[tex]F1 = Ft/4 =(45360 \frac{m^3}{d} )/4 = 11 340 \frac{m^3}{d}[/tex]
To calculate the retention time we divide the total volume V by the flow of each tank F1.
[tex]t1 =\frac{V1}{F1} = \frac{600 m^3}{11340 \frac{m^3}{d} } = 0.0529 day\\t1 = 0.0529 d (\frac{24 h}{1d} ) = 1.26 h[/tex]
After converting t1 to hours we found that the retention time when the four reactors are in parallel is 1.26 hours.
b)
If the four reactors were working in series, the entire flow goes first through one tank, then the second and so on. It means the total flow will be the flow of each tank.
In that order of ideas, the flow for reactors in series will be F2, and will have the same value of F0.
F2 = F0
To calculate the retention time t2 we divide the total volume V by the flow of each tank F2.
[tex]t2 =\frac{V1}{F2} = \frac{600 m^3}{45360 \frac{m^3}{d} } = 0.01 day\\t2 = 0.01 d (\frac{24 h}{1d} ) = 0.31 h[/tex]
After converting t2 to hours we found that the retention time when the four reactors are in series is 0.31 hours.
a) Retention time in each settling tank for parallel operation is approximately 317.0 hours.
b) Retention time in each settling tank for series operation is also approximately 317.0 hours.
To find the retention time in each settling tank, we'll use the formula:
[tex]\[ \text{Retention Time} = \frac{\text{Volume of Tank}}{\text{Flow Rate}} \][/tex]
Where:
- Volume of Tank is the volume of each settling tank.
- Flow Rate is the total flow rate.
We'll first convert the flow rate to cubic meters per day (m\(^3\)/day) to match the units of the tank volume:
[tex]\[ 1 \, \text{MGD} = 1 \, \text{million gallons} \times \frac{3.785 \times 10^3 \, \text{liters}}{1 \, \text{m}^3} \times \frac{1 \, \text{day}}{24 \, \text{hours}} \]\[ = 1 \, \text{MGD} \times \frac{3.785 \times 10^3 \, \text{liters}}{1 \, \text{m}^3} \times \frac{1 \, \text{day}}{24 \, \text{hours}} \times \frac{10^{-6} \, \text{m}^3}{1 \, \text{liter}} \]\[ = \frac{3.785 \times 10^3}{24 \times 10^6} \, \text{m}^3/\text{hour} \]\[ = 0.1577 \, \text{m}^3/\text{hour} \][/tex]
Now we can calculate the retention time for both scenarios:
a) For parallel operation:
[tex]\[ \text{Retention Time (Parallel)} = \frac{V_{\text{tank}}}{Q} = \frac{600 \, \text{m}^3}{1.8924 \, \text{m}^3/\text{hour}} \]\[ \text{Retention Time (Parallel)} \approx 317.0 \, \text{hours} \][/tex]
b) For series operation:
In series, the total flow rate remains the same, but the volume is effectively multiplied by the number of tanks in series. So, the retention time for each tank remains the same as in the parallel case:
[tex]\[ \text{Retention Time (Series)} = \frac{V_{\text{tank}}}{Q} = \frac{600 \, \text{m}^3}{1.8924 \, \text{m}^3/\text{hour}} \]\[ \text{Retention Time (Series)} \approx 317.0 \, \text{hours} \][/tex]
Therefore, the retention time in each settling tank is approximately 317.0 hours for both parallel and series operation.
Which one of the following statements concerning the Boltzmann distribution is TRUE?
Select one
a)The Boltzmann equation expresses a statistical population distribution corresponding to maximum entropy
b)The Boltzmann equation applies to all systems, whether or not they are in thermal equilibrium
c)The Boltzmann distribution is not valid at very low temperatures.
d)Upper energy levels are always less populated than lower energy levels, regardless of degeneracy.
e)At room temperature, a Boltzmann distribution predicts few rotational levels are populated but many electronic levels.
Answer:
c)The Boltzmann distribution is not valid at very low temperatures.
Explanation:
Option cis correct because the Boltzman distribution can be applied when the temperature is high enough or the particle density is low enough to render quantum effects negligible.
The Boltzman distribution in statistics describe the distribution of particles over different energy states when there is thermal equilibrium (that is why option b is INCORRECT).
if tempreture dry=25 C , PH=45% , what it actual temperature? by using Psychrometric Chart
1) 19 C
2) 29 C
3) 39 C
4) 49 C
Answer:
19°C
Explanation:
When you use a psychrometric chart the thing that you need to know is the value of a least two variables of your system, then you need to intercept these variables and then you can find the other variables seeing the others axis. In your case, you know the dry temperature and the percentage of humidity, so as you see in the attach image your temperature is between 15°C and 20°C. The only choice in that range is 19°C.
A second order reaction of the type A+B>P was carried out in a solution that was initially .075 mol dm-3 in A and .03 mol dm-3 in B. After 1 hour, the concentration of A had fallen to .02 mol dm-3. a. Calculate the rate constant. b. What is the half life of the reactant? Answers: a. 16.2 dm3mol-hr-, 4.5E-3 dm3mol-s- b. 5.1E3s, 2.1E3 s
Answer:
The rate constant of the reaction is [tex]1.0185\times 10^{-2} dm^3/ mol s[/tex].
[tex]1.31\times 10^3 s[/tex] is the half life of the reactant.
Explanation:
A+B → P
Integrated rate law for second order kinetics is given by:
[tex]\frac{1}{A}=kt+\frac{1}{A_0}[/tex]
A = concentration left after time t = [tex]0.02 mol /dm^3[/tex]
[tex]A_o[/tex] = Initial concentration = [tex]0.075 mol /dm^3[/tex]
t = 1 hour = 3600 seconds
[tex]\frac{1}{0.02 mol /dm^3}=k\times 3600 s+\frac{1}{0.075 mol /dm^3}[/tex]
[tex]k=\frac{1}{3600 s}\times (\frac{1}{0.02 mol /dm^3}-\frac{1}{0.075 mol /dm^3)}[/tex]
[tex]k = 1.0185\times 10^{-2} dm^3/ mol s[/tex]
Half life for second order kinetics is given by:
[tex]t_{\frac{1}{2}}=\frac{1}{k\times a_0}[/tex]
[tex]t_{\frac{1}{2}}=\frac{1}{1.0185\times 10^{-2} dm^3/ mol s\times 0.075 mol /dm^3}=1.31\times 10^3 s[/tex]
The rate constant of the reaction is [tex]1.0185\times 10^{-2} dm^3/ mol s[/tex].
[tex]1.31\times 10^3 s[/tex] is the half life of the reactant.
Report the answer with the correct number of significant figures: a. 12.3 mm x 15.853 mm = b. 72.98 · 15.830 ml = c. 12.3 mm + 15.853 mm = d. 172.3 cm - 15.853 cm - 12.5 pts
Explanation:
The rule apply for the multiplication and division is :The least number of significant figures in any number of the problem determines the number of significant figures in the answer.
The rule apply for the addition and subtraction is :The least precise number present after the decimal point determines the number of significant figures in the answer.
a) 12.3 mm × 15.853 mm :
[tex] 12.3 mm\times 15.853 mm :194.9919 mm^2\approx 195 mm^2[/tex]
Three significant figures.
b) 72.98 · 15.830 ml
[tex]72.98\times 15.830 ml=1,155.2734 ml\approx 1,155 mL[/tex]
Three significant figures.
c) 12.3 mm + 15.853 mm
[tex]12.3 mm + 15.853 mm = 28.153 mm\approx 28.1 mm[/tex]
Three significant figures.
d) 172.3 cm - 15.853 cm
[tex] 172.3 cm - 15.853 cm= 156.447 cm\approx 156.4 cm[/tex]
Four significant figures.
Final answer:
The student's Mathematics question involves calculating with significant figures. When multiplying or dividing values, the result is rounded to the number of significant figures in the least precise value, while addition and subtraction are rounded to match the least number of decimal places.
Explanation:
The subject of this question is Mathematics, specifically related to significant figures in calculations, which is a concept typically covered in High School mathematics or physics courses.
Report the answer with the correct number of significant figures:
(a) To multiply 12.3 mm by 15.853 mm, we should round our answer to the same number of significant figures as the number with the fewest significant figures. In this case, 12.3 has three significant figures. Therefore, the result should also be reported with three significant figures.
(b) When multiplying 72.98 by 15.830 ml, the number with the fewest significant figures is 72.98, which has four significant figures. The answer should be reported with four significant figures.
(c) When adding 12.3 mm to 15.853 mm, we look for the least precise value in terms of decimal places, which is 12.3 mm (it is precise to the tenths place). Therefore, our result should also be rounded to the tenths place.
(d) In the subtraction 172.3 cm - 15.853 cm - 12.5 cm, since 12.5 cm has the least number of decimal places (one decimal place), our final result should also be rounded to one decimal place.
When it comes to significant figures, the rule for multiplication and division is to match the result to the number with the fewest significant figures. For addition and subtraction, we round the result to the least number of decimal places in the values used.
You are given a protein solution with a concentration of 0.15 mg/ml.
v. Suppose that we want to prepare 100 microliters of 0.1 micrograms/microliters solution. How much of H2O and protein stock do we need to add to obtain the target concentration and volume?
Explanation:
The given data is as follows.
Initial concentration = 0.15 mg/ml,
Final concentration = 0.1 mg/ml, (as [tex]\frac{0.1 microgram}{1 microliter} \times \frac{10^{-3}}{1 microgram} \times \frac{1 microliter}{10^{-3}ml}[/tex])
Final volume = [tex]100 microliter \times \frac{10^{-3} ml}{1 microliter}[/tex] = 0.1 ml
According to the dilution formula we get the following.
[tex]C_{i} \times V_{i} = C_{f} \times V_{f}[/tex]
or, [tex]V_{i}[/tex] = [tex]\frac{C_{f} \times V_{f}}{C_{i}}[/tex]
Putting the given values into the above formula we get the following.
[tex]V_{i}[/tex] = [tex]\frac{C_{f} \times V_{f}}{C_{i}}[/tex]
= [tex]\frac{0.1 mg/ml \times 0.1 ml}{0.15 mg/ml}[/tex]
= 0.0667 ml
= [tex]6.67 \times 10^{-2} ml \times \frac{1 microliter}{10^{-3} ml}[/tex]
= 66.7 microliter
This means that volume of protein stock which is required is 66.7 ml. Hence, calculate the volume of water required as follows.
Volume of water required = Total volume - volume of protein stock
= (100 - 66.7) microliter
= 33.3 microliter
Thus, we can conclude that we need 33.3 microliter of water and 66.7 microliter of protein.
Calculate the pH of a solution prepared by mixing: (Show your work for these calculations) pk of acetic acid is 4.75 a. two moles of sodium acetate with two moles of acetic acid. b. two moles of sodium acetate with one mole of acetic acid.
Answer:
Explanation:
To calculate pH you need to use Henderson-Hasselbalch formula:
pH = pka + log₁₀ [tex]\frac{[A^-]}{[HA]}[/tex]
Where HA is the acid concentration and A⁻ is the conjugate base concentration.
The equilibrium of acetic acid is:
CH₃COOH ⇄ CH₃COO⁻ + H⁺ pka: 4,75
Where CH₃COOH is the acid and CH₃COO⁻ is the conjugate base.
Thus, Henderson-Hasselbalch formula for acetic acid equilibrium is:
pH = 4,75 + log₁₀ [tex]\frac{[CH_{3}COO^-]}{[CH_{3}COOH]}[/tex]
a) The pH is:
pH = 4,75 + log₁₀ [tex]\frac{[2 mol]}{[2 mol]}[/tex]
pH = 4,75
b) The pH is:
pH = 4,75 + log₁₀ [tex]\frac{[2 mol]}{[1mol]}[/tex]
pH = 5,05
I hope it helps!
Name the simplest "aldose" and "ketone"
Explanation:
Aldose is monosaccharide sugar in which the carbon backbone chain has carbonyl group on endmost carbon atom which corresponds to an aldehyde, and the hydroxyl groups are connected to all other carbon atoms.
Example - Glucose
Ketone is functional group with structure RC(=O)R', where the groups, R and R' can be variety of the carbon-containing substituents.
Example, Ethylmethylketone
Final answer:
The simplest aldose is glyceraldehyde, and the simplest ketone is acetone. These classifications are based on the presence of a carbonyl group, with aldoses having the group at the end of the chain and ketones within the chain. This naming follows IUPAC nomenclature rules with the suffixes -al for aldehydes and -one for ketones.
Explanation:
The simplest aldose is glyceraldehyde, which is a three-carbon aldehyde with the molecular formula C₃H₆O₃. It contains a carbonyl group (CHO) at the end of the carbon chain, making it an aldehyde. The simplest ketone is acetone (propanone), which has the molecular formula C₃H₆O. It contains a carbonyl group (CO) within the carbon chain, which classifies it as a ketone.
Both aldehydes and ketones contain a carbonyl group, a functional group with a carbon-oxygen double bond. The nomenclature rules for naming these compounds involve the use of the suffixes -al for aldehydes and -one for ketones. For instance, the name glyceraldehyde is derived from glycerol (the alcohol form), with the ending changed to -al to indicate it's an aldehyde, while the name acetone is derived from the root word for two carbon groups (acet-) with the ending -one to signify it's a ketone.
How many moles of disulfur decafluoride are present in 2.45 grams of this compound? moles
Answer:
Take account the molar mass of this compound (first of all u should know the formula, S2F10). As the molar mass is 254,1 g/mol you will know that in 1 mol, you have 254,1 g so make a rule of three to solve it. If we find 254,1 g of S2F10 in 1 mol, 2,45 g of it are in 9,64 *10^-3 moles. That is the right number.
Explanation:
Which of the following is true about the following chemical reaction at equilibrium? PICK ALL THAT APPLY.
C6H12O6 + 6O2 6CO2 + 6H2O + Energy
A. The concentration of H2O equals the concentration of O2.
B. The rate of the forward reaction, reactants to products, equals the rate of the reverse reaction, products to reactants.
C. The C-C bonds hold more energy in C6H12O6 than the C-O bonds in CO2. Hint: think teter-toter.
D. The concentration of O2 is changing at equilibrium.
Answer:
Option B and C are correct
Explanation:
Chemical reaction at equillibrium means: the rate of forward reaction equals the rate of reverse reaction. This means no product will be consumed, neither formed.
A) The concetration of H2O equals the concentration of O2. This is false.
At the equillibrium the concetnrations are constant. But not necessarily equal.
B) The rate of the forward reaction, reactants to products, equals the rate of the reverse reaction, products to reactants. This is true. IT's the definition of a the equillibrium.
C) The C-C bonds hold more energy in C6H12O6 than the C-O bonds in CO2. Hint: think teter-toter.
This is to, C-C bonds hold more energy than C-O, that's why C6H12O6 is burned, energy is released.
D) The concentration of O2 is changing at equilibrium. This is false. The concentration of O2 is constant.
how many protons are contained in the nucleus of the atom?
A mixture of ethyl acetate vapour and air has a relative saturation of 50% at 303 K and a total pressure of 100 kPa. If the vapour pressure of ethyl acetate at 303 K is 16 kPa, the molal saturation is (a) 0.080 (b) 0.087 (c) 0.264 (d) 0.029
Answer : The correct option is, (b) 0.087
Explanation :
The formula used for relative saturation is:
[tex]\text{Relative saturation}=\frac{P_A}{P_A^o}[/tex]
where,
[tex]P_A[/tex] = partial pressure of ethyl acetate
[tex]P_A^o[/tex] = vapor pressure of ethyl acetate
Given:
Relative saturation = 50 % = 0.5
Vapor pressure of ethyl acetate = 16 kPa
Now put all the given values in the above formula, we get:
[tex]0.5=\frac{P_A}{16kPa}[/tex]
[tex]P_A=8kPa[/tex]
Now we have to calculate the molar saturation.
The formula used for molar saturation is:
[tex]\text{Molar saturation}=\frac{P_{vapor}}{P_{\text{vapor free}}}[/tex]
and,
P(vapor free) = Total pressure - Vapor pressure
P(vapor) = [tex]P_A[/tex] = 8 kPa
So,
P(vapor free) = 100 kPa - 8 kPa = 92 kPa
The molar saturation will be:
[tex]\text{Molar saturation}=\frac{P_{vapor}}{P_{\text{vapor free}}}[/tex]
[tex]\text{Molar saturation}=\frac{8kPa}{92kPa}=0.087[/tex]
Therefore, the molar saturation is 0.087
Write 0.70894 in Engineering Notation with 3 significant figures.
Final answer:
The number 0.70894 is written as 709 × 10⁻³ in Engineering Notation with 3 significant figures.
Explanation:
To write the number 0.70894 in Engineering Notation with 3 significant figures, we first identify the significant figures in the number. Since leading zeros are not significant, the significant figures are 708. We then adjust the number to have a multiple of three as its exponent for the base unit, which is standard in Engineering Notation. The number can be written as 709 (rounded up from 708.94 to maintain 3 significant figures) multiplied by 10 to the power of -3. Therefore, the number in Engineering Notation is 709 × 10⁻³.
Convert 1.248×1010 g to each of the following units.
a) kg Express your answer using four significant figures.
b) Mg Express your answer using four significant figures.
c) mg Express your answer using four significant figures.
d) metric tons (1 metric ton = 1000 kg) . Express your answer using four significant figures.
Final answer:
To convert 1.248×1010 g to different units, one must use specific conversion factors for each unit, resulting in 1.248×107 kg, 1.248×104 Mg or metric tons, and 1.248×1013 mg, all while maintaining four significant figures.
Explanation:
To convert 1.248×1010 g to various units, we'll use the relationships between grams, kilograms, milligrams, and metric tons.
Kilograms (kg): Since 1 kg = 103 g, divide by 103 to convert. 1.248×1010 g = 1.248×107 kg.Megagrams (Mg) or metric tons: Since 1 Mg = 106 g, divide by 106 to convert. 1.248×1010 g = 1.248×104 Mg or metric tons.Milligrams (mg): Since 1 mg = 10-3 g, multiply by 103 to convert. 1.248×1010 g = 1.248×1013 mg.Metric tons: Utilize the same conversion as for Megagrams, since 1 metric ton = 1 Mg = 1.248×104 metric tons.It's important to keep significant figures in mind, maintaining four significant figures through each conversion to ensure precision.
What is the difference between natural capital and natural resources?
Answer: A resource which we get naturally from the enviornment such as, air water, soil etc. is known as natural resources. Natural resources can be categories on the basis of renewability and it is of two types; renewable and non-renewable.
A natural capital is the stock of natural resources; which includes soil, air and all the living organisms present on this planet
Final answer:
Natural resources are raw materials found in nature used for economic gain, while natural capital includes these resources plus the ecosystem services that sustain life. Whereas natural resources can be renewable or nonrenewable, natural capital focuses on the sustainable management of these resources and ecosystem services.
Explanation:
The difference between natural capital and natural resources is found in their definitions and applications. Natural resources are materials or substances that occur in nature and can be used for economic gain; they are the raw inputs we find in our environment that have not been altered by human hands. Examples of natural resources include water, minerals, timber, and fossil fuels. They are essential for producing goods and services, but they vary in their abundance and renewability. Renewable resources, such as solar energy and timber, can be replenished over time, while nonrenewable resources, like fossil fuels and minerals, are finite and diminish with use.
On the other hand, natural capital encompasses not just the natural resources themselves, but also the ecosystem services that sustain and fulfill human life. Natural capital refers to the world's stocks of natural assets which include geology, soil, air, water and all living organisms. There is a growing recognition of the need to manage natural capital sustainably to avoid depleting these vital assets. The concept of natural capital broadens the scope of what is considered valuable in the natural environment beyond just the extractive resources.
It's important to note that while all natural capital includes natural resources, not all natural resources serve as natural capital. This distinction lies in the broader ecological value natural capital provides, benefiting economic processes as well as the environment in ways that maintain or enhance our stock of natural assets.
Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used.
The number above the element’s symbol is the element’s atomic number. Match each element with the number of valence electrons it has.
Answer:
4= Si
5=P
3= Al
6= S
2= Mg
Explanation:
4 → Si
Silicon is part of the carbon group and has 14 electrons, spread over 3 shells:
⇒ on the first shell (K) 2 electrons
⇒on the second shell (L) 8 electrons
⇒ on the third shell (M) the remaining electrons : 14-8-2 = 4
5 → P
phosphorus is part of the nitrogen group and has 15 electrons, spread over 3 shells:
⇒ on the first shell (K) 2 electrons
⇒on the second shell (L) 8 electrons
⇒ on the third shell (M) the remaining electrons : 15-8-2 = 5
3 → Al
Aluminium is part of the Boron group and has 13 electrons, spread over 3 shells:
⇒ on the first shell (K) 2 electrons
⇒on the second shell (L) 8 electrons
⇒ on the third shell (M) the remaining electrons : 13-8-2 = 3
6 → S
Sulfur is part of the oxygen group and has 16 electrons, spread over 3 shells:
⇒ on the first shell (K) 2 electrons
⇒on the second shell (L) 8 electrons
⇒ on the third shell (M) the remaining electrons : 16-8-2 = 6
2 → Mg
Magnesium is part of the Beryllium or alkaline earth metals group and has 12 electrons, spread over 3 shells:
⇒ on the first shell (K) 2 electrons
⇒on the second shell (L) 8 electrons
⇒ on the third shell (M) the remaining electrons : 12-8-2 = 2
The completed pairs of element with the number of valence electrons are:
aluminum ⇒ 3sulfur ⇒ 6sodium ⇒ 1silicon ⇒ 4magnesium ⇒ 2phosphorus ⇒ 5What are these valence electrons?Valence electrons are the electrons in the outermost shell of an atom that can participate in chemical bonding. The number of valence electrons an element has determines its chemical properties.
For example, sodium (Na) has 1 valence electron, which it can donate to other atoms to form bonds. This is why sodium is a reactive metal.
Sulfur (S) has 6 valence electrons, which it can share with other atoms to form bonds. This is why sulfur can form a variety of compounds with other elements.
The number of valence electrons an element has can also be determined by looking at its periodic table group. Elements in the same group have the same number of valence electrons.
Find out more on element here: https://brainly.com/question/18096867
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What is the main difference between laboratory scale distillation and industrial distillation?
Answer: Distillation is a process in which we use the boiling point, condensation of the substances to separate them from each other, it is a physical separation not the chemical because physical property of substance is used in it.
Now, the distinction in between a laboratory distillation and an industrial distillation is that, the distillation process in laboratory distillation works in batches, while in the industrial distillation it occurs in continuous manner, industrial distillation is the large scale distillation.
Final answer:
The main difference between laboratory scale distillation and industrial distillation lies in the scale and complexity of the process.
Explanation:
The main difference between laboratory scale distillation and industrial distillation lies in the scale and complexity of the process. Laboratory scale distillation is typically used for smaller quantities and is carried out using simple apparatus. It is commonly used for research, quality control, and small-scale production.
On the other hand, industrial distillation is conducted on a much larger scale and involves more complex equipment. It is used in industries such as oil refining and chemical manufacturing, where large quantities of materials need to be processed. Industrial distillation may include fractionating columns and other specialized equipment to separate components from mixtures.
The combustion of 44.45 grams of a hydrocarbon gas mixture releases 2,649 kJ of heat. How much energy, in kWh, will the combustion of exactly one pound of this hydrocarbon gas produce? There are 453.592 g in 1 lb. Please express your answer to the appropriate number of significant figures. Your Answer:
Explanation:
It is given that 44.45 g of hydrocarbon gas produces 2649 kJ or [tex]2649 \times 10^{3} J[/tex] of heat.
Also here, 1 lb = 453.592 g.
Therefore, amount of energy released by 453.592 g of hydrocarbon gas will be calculated as follows.
[tex]\frac{2649 \times 10^{3} J \times 453.592 g}{44.45 g}[/tex]
= [tex]27.03 \times 10^{6} J[/tex]
It is known that 1 J = [tex]2.778 \times 10^{-7} Kwh[/tex].
Hence, [tex]27.03 \times 10^{6} J[/tex] = [tex]27.03 \times 10^{6} J \times 2.778 \times 10^{-7} Kwh/J[/tex]
= 7.508 Kwh
Thus, we can conclude that the combustion of exactly one pound of this hydrocarbon gas produce 7.508 Kwh energy.
35S has a half-life of 88 days. You start with 128 micrograms of 3Ss. How much 35s is left after 264 days? O a. 32 micrograms b.8 micrograms O c. 64 micrograms O d. 16 micrograms O e. 4 micrograms
Answer:
Option d, Amount left = 16 μg
Explanation:
Half life = 88 days
Initial concentration = 128 μg
[tex]No.\ of\ half\ life=\frac{Total\ days}{t_{1/2}}[/tex]
[tex]No.\ of\ half\ life=\frac{264}{88} = 3[/tex]
Amount of material left after n half life = [tex]\frac{Amount\ of\ starting\ material}{2^n}[/tex]
Where, n = No. of half life
Amount of material left = [tex]\frac{128}{2^3}[/tex]
= [tex]\frac{128}{8} = 16[/tex]μg
SO, option d is correct
Solve for x, where M is molar and s is seconds.
x=(2.4×103M−2s−1)(0.38M)3
Answer:
x = 130 Ms⁻¹
Explanation:
The equation is:
x = (2.4 x 10³ M⁻²s⁻¹)(0.38 M)³
The expression raised to the power of 3 is simplified:
x = (2.4 x 10³ M⁻²s⁻¹)(0.054872 M³)
The two values are multiplied together and units canceled:
x = 130 Ms⁻¹
Answer:
x = 131.69Ms⁻¹
Explanation:
x = (2.4×10³M⁻²s⁻¹) (0.38M)³
Upon expanding, we have;
x = (2.4 × 10³M⁻²s⁻¹) (0.38M) * (0.38M) * (0.38M)
x = (2.4 × 10³M⁻²s⁻¹) (0.054872 M³)
x = 2.4 * 0.054872 * 10³ * M⁻²s⁻¹ * M³
x = 0.13169 * 10³ * M⁻²s⁻¹ * M³
x = 131.69 * M⁽⁻²⁺³ ⁾* s⁻¹
x = 131.69 * Ms⁻¹
x = 131.69Ms⁻¹
Gallium oxide, Ga.Oy, forms when gallium is combined with oxygen. A 1.25 g of Ga is allowed to react with excess oxygen and 1.68 g of the gallium oxide formed. What is the chemical formula of the gallium oxide if the molar mass is 187.44 g/mol?
Answer : The chemical formula of a compound is, [tex]Ga_2O_3[/tex]
Solution : Given,
Mass of gallium = 1.25 g
Mass of gallium oxide = 1.68 g
Mass of oxygen = Mass of gallium oxide - Mass of gallium
Mass of oxygen = 1.68 - 1.25 = 0.43 g
Molar mass of Ga = 69.72 g/mole
Molar mass of O = 16 g/mole
Step 1 : convert given masses into moles.
Moles of Ga = [tex]\frac{\text{ given mass of Ga}}{\text{ molar mass of Ga}}= \frac{1.25g}{69.72g/mole}=0.0179moles[/tex]
Moles of O = [tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{0.43g}{16g/mole}=0.027moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For Ga = [tex]\frac{0.0179}{0.0179}=1[/tex]
For O = [tex]\frac{0.027}{0.0179}=1.5[/tex]
The ratio of Ga : O = 1 : 1.5
To make in whole number we multiple ratio by 2, we get:
The ratio of Ga : O = 2 : 3
The mole ratio of the element is represented by subscripts in empirical formula.
The Empirical formula = [tex]Ga_2O_3[/tex]
The empirical formula weight = 2(69.72) + 3(16) = 187.44 gram/eq
Now we have to calculate the molecular formula of the compound.
Formula used :
[tex]n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}[/tex]
[tex]n=\frac{187.44}{187.44}=1[/tex]
Molecular formula = [tex](Ga_2O_3)_n=(Ga_2O_3)_1=Ga_2O_3[/tex]
Therefore, the chemical of the compound is, [tex]Ga_2O_3[/tex]
The empirical formula of gallium oxide, formed by the reaction of gallium with excess oxygen, is Ga₂O₃.
To determine the chemical formula of gallium oxide formed, follow these steps:
The mass of Ga used is 1.25 g and the mass of Ga oxide produced is 1.68 g. First, find the mass of oxygen in the oxide:Mass of O = 1.68 g (mass of Ga oxide) - 1.25 g (mass of Ga) = 0.43 g
Next, convert these masses to moles:Moles of Ga = 1.25 g / 69.72 g/mol (molar mass of Ga) = 0.0179 moles
Moles of O = 0.43 g / 16.00 g/mol (molar mass of O) = 0.0269 moles
Determine the simplest whole number ratio of Ga to O:Ratio of Ga to O = 0.0179 : 0.0269 ≈ 2:3
The given molar mass of the gallium oxide (187.44 g/mol) confirms the empirical formula Ga₂O₃.
What are the Lewis definitions of an acid and a base? In what way are Bronsted definitions? they more general than the In terms of orbitals and electron arrangements, what must be present for a molecule or an ion to act as a Lewis acid (use HT and BF3 as examples)? What must be present for a molecule or ion to act as a Lewis base (use OH and NH3 as examples)?
Explanation:
Lewis definition of Acids and Bases
Chemical species which are capable of accepting electron pairs or donating protons are called Lewis acid.
Chemical species which are capable of donating electron pairs or accepting protons are called Lewis base.
Bronsted definition of acids and bases
Chemical species which are capable of donating H+ are called Bronsted acids.
Chemical species which are capable of accepting H+ are called Bronsted bases.
So all Bronsted acids are Lewis acids but all Lewis acids are not Bronsted acids.
For a chemical species to behave as Lewis acid, they must have:
Incomplete octetDouble bondVacant d-orbitalsFor example, in BF3, octet of boron is incomplete, so it can accept a pair of electron and behaves as Lewis acid.
For a chemical species to behave as Lewis base, they must have:
lone pair of electronsFor example, NH3 and OH, both N and O have lone pairs of electrons, hence behave as Lewis base.
In Lewis theory, an acid is an electron pair acceptor and a base is an electron pair donor. Bronsted definitions are a subset of this concept, and are therefore less general. Lewis acids require an empty orbital, while Lewis bases need a lone pair of electrons.
Explanation:The Lewis definition of an acid is a species that can accept an electron pair, whereas a Lewis base is a species that has an electron pair available for donation to a Lewis acid. A clear example of a Lewis acid is BF3 since it can accept an electron pair due to its empty orbital. On the other hand, a Lewis base, such as OH-, has a lone pair of electrons that it can donate.
In regards to the Bronsted definitions, they define an acid as a proton donor and a base as a proton acceptor but they represent a subcategory within Lewis theory and are therefore less general.
As for the electron arrangements, a molecule or ion can act as a Lewis acid if it has an empty orbital to accept an electron pair. On the contrary, for a molecule or ion to behave as a Lewis base, the presence of a lone pair of electrons is required, as seen in NH3.
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Calculate AS° for the reaction below: N2(g)+202(g) 2NO2(g) where ASo for N2(g), O2(g), & NO2(g), respectively, is 191.5, 205.0, & 240.5 J/mol-K -156.0 J/K 156.0 J/K 120.5 J/K -120.5 J/K O OOO
Answer:
ΔS Rx = -120, 5 J/K
Explanation:
The ΔS in a reaction is defined thus:
ΔS Rx = ∑ n S°products - ∑ m S°reactants
For the reaction:
N₂(g) + 2 O₂(g) → 2NO₂(g)
ΔS Rx = 2 mol × 240,5 [tex]\frac{J}{mol.K}[/tex] - [ 1 mol × 191,5 [tex]\frac{J}{mol.K}[/tex] + 2 mol × 205,0 [tex]\frac{J}{mol.K}[/tex]]=
ΔS Rx = -120, 5 J/K
A negative value in ΔS means a negative entropy of the process. Doing this process entropycally unfavorable.
I hope it helps!
Name the source of lactose and name the type of enzyme that is required to digest it
Answer:
Lactose is found in animal milk is lactase is the enzyme that is required to digest it.
Explanation:
Lactose is a disaccharide made of glucose and galactose units. This sugar is naturally present in animal milk and it is digested (broken in its units) by the lactase enzyme.
Some people are lactose intolerant because their bodies is not able to produce the enzyme. If they ingest dairy products it may cause health issues. Nowadays it is possible to purchase the lactase enzyme and ingest it with a dairy food to avoid any health effects.
In a demonstration of strong electrolytes, weak electrolytes, and non-electrolytes, Professor Popsnorkle used a lightbulb apparatus that showed solution conduction of electricity by the brightness of the lightbulb. He tested H2SO4(aq), NH3 and a 1M sugar solution. What was the result?
Answer:
The result was that H₂SO₄ is a strong electrolyte, NH₃ is a weak electrolyte and sugar is a non-electrolyte.
Explanation:
The experiment will show that the H₂SO₄ solution will light the lightbulb brightly, the NH₃ solution will light it weakly and the sugar solution will not light it at all.
This happens because to light the lightbulb we need free charges moving in the solution. Therefore, the substance that produce more ions will be the best electrolyte.
H₂SO₄ → 2H⁺ + SO₄²⁻
NH₃ + H₂O → NH₄⁺ + OH⁻
Sugar will not generate ions
Upon decomposition, one sample of magnesium fluoride produced 2.15 kg of magnesium and 3.36 kg of fluorine. A second sample produced 1.00 kg of magnesium. How much fluorine (in grams) did the second sample produce?
Answer:
The second sample will produce 1563 grams of fluorine (F2)
Explanation:
The reaction will be MgF2 → Mg + F2
The stoichiometry ratio of MgF2 and F2 is 1 : 1.
That means for 1 mole of MgF2 consumed there is 1 mole of F2 produced.
The first sample produces 2.15 kg of magnesium and 3.36 kg of fluorine
The second sample produced 1 kg of magnesium and x kg of fluorine
This we can show in the following equation =
2.15kg / 3.36 kg = 1kg / x
2.15/3.36 = 0.63988
0.63988 = 1/ x
x= 1/0.63988 = 1.563 kg
1.563kg = 1563 grams
The second sample will produce 1563 grams of fluorine (F2)
One liter of ocean water contains 35.06 g of salt. How much salt is present in 3.79 L of ocean water? Express your answer in grams using the correct number of significant figures. Do not enter your answer using scientific notation.
Answer:
133g
Explanation:
If one litre of ocean water contains 35.06g of salt, all we have to to to get the mass of salt in 3.79 L of ocean water is multiply 35.06g/L by 3.79 L.
35.06g/L × 3.79 L = 133g
The amount of salt in 3.79 L of ocean water can be found by multiplying the volume of the water by the constant ratio of 35.06 g of salt per liter. This gives 132.8774 grams, which when we round to 4 significant figures is equivalent to 132.9 g.
Explanation:To find the amount of salt in 3.79 L of ocean water, we use the given ratio of salt to water: 1 liter of ocean water contains 35.06 grams of salt. So to find the amount of salt in any given volume of ocean water, we just multiply the volume (in liters) by this constant ratio of 35.06 grams of salt per liter.
Applying this to the volume of 3.79 L we get: 3.79 L * 35.06 g/L = 132.8774 grams. Since we should express our answer using the correct number of significant figures (4 in this case, because there are 4 significant figures in the 3.79 L volume measurement), the answer is rounded to 132.9 g.
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The chemical formula for sucrose (also known as table sugar) is C12H22011. How many carbon atoms are in 1 molecule of sucrose? O a. 22 O b. 13 Oc. 12 O d. 11 Oe. 72