Whether or not the process is observed in nature, which of the following could account for the transformation of phosphorus-28 to silicon-28?
a. beta decay.
b. eletron capture
c. positron emission
d. alpha decay
Answer:
positron emission
Explanation:
Positron emission,involves the decrease in the nuclear positive charge thus the atomic number of the daughter nucleus decreases by one unit compared to that of the parent nucleus while the mass number of the daughter nucleus remains the same as that of the parent nucleus.
Thus, when phosphorus-28 is transformed into to silicon-28, a positron emission has taken place because the mass number of the daughter nucleus remained the same while the atomic number of the daughter nucleus was decreased by one unit.
Lab 2: paper chromatography of organic dyes
Picture of questions below.
Answer:
The three primary colors used when mixing dyes or paints are red, yellow, and blue. Other colors are often a mixture of these three colors. Try running a chromatography test again with non-primary-color markers, like purple, brown, and orange.
Explanation:
Mixtures that are suitable for separation by chromatography include inks, dyes and colouring agents in food. ... As the solvent soaks up the paper, it carries the mixtures with it. Different components of the mixture will move at different rates. This separates the mixture out.
2.67 Determine the density (g/mL) for each of the following:
a. A 20.0-mL sample of a salt solution has a mass of 24.0 g.
The density (g/mL) for a 20.0-mL sample of a salt solution has a mass of 24.0 g is 1.2 g/ml.
What is density?Density is the mass per unit volume. Density is a scalar quantity. It is denoted by d and the symbol for density is given as rho, a Greek symbol. Density is calculated as mass divided by volume.
Mass is the quantity of matter in a physical body. The product of the molar mass of the compound and the moles of the substance are defined as mass.
Volume is the space occupied by a three-dimensional object. Volume is calculated by dividing mass by density.
Given, the 20.0-mL sample of a salt solution, which is the volume.
The mass of the solution is 24.0 g
To calculate the density
mass/volume
24.0 / 20.0 = 1.2 g/ml
Thus, the density of the given salt solution is 1.2 g/ml.
To learn more about density, refer to the link:
https://brainly.com/question/17851361
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Hydrogen and chlorine gases react to form HCl. You and a friend are on opposite sides of a long hallway, you with H2 and your friend with Cl2. You both want to form HCl in the middle of the room. Which of the following is true?
a) You should release the H2 first.
b) Your friend should release the Cl2 first.
c) You both should release the gases at the same time.
d) You need to know the length of the room to answer this question.
e) You need to know the temperature to answer this question.
Answer: It is true that Your friend should release the [tex]Cl_{2}[/tex] first.
Explanation:
The velocity of particles of a gas is inversely proportional to the mass of gas. This means that more is the mass of gas less will be its velocity.
Or, more will be the mass of gas more slowly it will move from one place to another.
The molar mass of chlorine gas is more than the molar mass of hydrogen gas. Therefore, chlorine gas will move slowly.
So, your friend should release the [tex]Cl_{2}[/tex] gas first and then according to the length of room you should release the [tex]H_{2}[/tex] gas.
Thus, we can conclude that it is true that Your friend should release the [tex]Cl_{2}[/tex] first.
what is the maximum number of grams of sodium chloride that you could dissolve in 500 grams of water at 20 C
Answer:
178.5g of NaCl is the maximum amount that can be dissolved
Explanation:
Solubility is defined as the maximum amount of solute that can be dissolved in aAt 20°C, the solubility of NaCl is 35.7g per 100g of water at 20°C. That means the maximum amount that can be dissolved of NaCl is:
500g water * (35.7g NaCl / 100g water) =
178.5g of NaCl is the maximum amount that can be dissolvedConvert the following:
1) 367.5 mg = _______ g
2) 367 mL = _______ L
3) 28.59 in =______ cm
4) 8 0z =_______lb
5) 0.671 mm =_____m
To prepare a sample in a capillary tube for a melting point determination, gently tap the tube into the sample with the ___________ end of the tube down. Continue tapping until the sample. _____________ Then, with the __________ end of the tube down, tap the sample down slowly or __________ to move the sample down faster. Finally, make sure that you can see ___________ in the magnifier when placed in the melting point apparatus before turning on the heat.
Answer: Hello I was able to find the missing options related to your question online but I couldn't attach them hence I just used them to provide the relevant solution
answer:
Open Is a couple of millimeters highclosed Drop tube into a longer tube sample clearlyExplanation:
OPEN end of the tube down. Continue tapping until the sample. Is a couple of mm high Then, with the Closed end of the tube down, tap the sample down slowly or drop tube into a longer/larger tube to move the sample down faster. Finally, make sure that you can see sample clearly in the magnifier when placed in the melting point apparatus before turning on the heat.
0.159 mol 2.25 M = A L of HCI
Answer:
hmmm?what?
Explanation:
ayusin mo flece:>
According to hydrogen’s chemical properties, it belongs in the same group as:
1. alkali metals
2. alkaline earth metals
3 .halogens
4. both a and c
---
Select the following group of elements that contains elements with similar chemical properties
A.
N, As, Bi
B.
Rb, Sn, In
C.
Li, Mg, Ga
D.
W, Tc, Fe
---
What would you use to measure
to measure an amount of solid sodium chloride(NaCI)
crystals to add to a 30 mL solution?
A. Triple Beam Balance that measures to the nearest 0.1 grams.
) B. 10 ml graduated cylinder.
Scale that can weigh up to 250 lbs.
D. Spectrophotometer that measures between 100 and 900 nanometers.
once water reach its boiling point what does you notice about its temperature??
You HAVE 1L of 2.0 mol/L HCI. You want to dilute the solution to a concentration of 0.1 mol/L. What volume of the new solution (V2)?
0.05 L
O 20L
0.2 L
M₁ = 2.0 mol/L
V₁ = 1 L
M₂ = 0.1 mol/L
Required:V₂
Solution:M₁V₁ = M₂V₂
V₂ = M₁V₁ / M₂
V₂ = (2.0 mol/L)(1 L) / (0.1 L)
V₂ = 20 L
Therefore, the volume of the new solution will be 20 L.
#ILoveChemistry
#ILoveYouShaina
M₁V₁ = M₂V₂
V₂ = M₁V₁ / M₂
V₂ = (2.0 mol/L)(1 L) / (0.1 L)
V₂ = 20 L
THE BEST ANSWER IS
O 20L
What are the lengths of the diagonals of the kite?
The answer ( 13 and 8 )
x²=5²+12²
x²=25+144
x²=169
x=13
x²=5²+6²
x²=25+36
x²=61
x=7.8
x=8
12.0: A
Mention three body fluids that are alkaline in nature
Draw the predominant product(s) of the following reactions including stereochemistry when it is appropriate. CH3CH2 C C CH3 H2O/H2SO4/HgSO4
Answer:
Draw the predominant product(s) of the following reactions including stereochemistry when it is appropriate.
CH3CH2 C C CH3 H2O/H2SO4/HgSO4
Explanation:
The given compound is: pent-2-yne.
When it reacts with water, in presence of sulphuric acid and mercuric sulphate then a ketone is formed as shown below:
This reaction is an example of nucleophilic attack of water on carbon carbon triple bond.
The general mechanism of the reaction is hsown below:
Pent-2-yne reacts with water and form 3-pentanone.
The reaction is shown below:
张
ci Metal displaces 50cm of water
when completely immersed in water if the mos
of the metal is 35.0q, calculate its
density
Answer:
0.70 g/cm³
Explanation:
Step 1: Given data
Volume of water displaced by the metal (V): 50 cm³ (this is equal to the volume of the metal)Mass of the metal (m): 35.0 gStep 2: Calculate the density (ρ) of the metal
Density is an intrinsic property of matter. It is equal to the mass of the metal divided by its volume.
ρ = m/V
ρ = 35.0 g/50 cm³ = 0.70 g/cm³
2.
A solution contains 50 g of KCl per 100 g of water at
90°C. Is the solution unsaturated, saturated or
supersaturated?
unsaturated, saturated or supersaturated?
What is the volume of 0.410 moles of co2 at STP
Answer: 9.18 Litres
Standard Temperature and Pressure (STP). Think of this as the perfect environment where the Temp. is 0°C or 273 Kelvin and Pressure is always 1 atm. This is only true in STP.
This question uses the Ideal Gas Equation:
PV=nRT
P= 1 atm
V = ??
T = 273 K (always convert to Kelvin unless told otherwise)
n = 0.410 mol
R = 0.0821 L.atm/mol.K
What R constant to use depends on the units of the other values. (look at the attachments) The units cancel out and only Litres is left. You simply multiply the values.
Answer:
9.18 liters .............
heya
Identify the indicated protons in the following molecules as unrelated, homotopic, enantiotopic, or diastereotopic. a) Methyls a & b: _________ b) Ha & Hc: _________
Answer:
Identify the indicated protons in the following molecules as unrelated, homotopic, enantiotopic, or diastereotopic. a) Methyls a & b: _________ b) Ha & Hc: ________
Explanation:
Homotopic hydrogens:
Consider two hydrogens in the given molecule and replace one by one with a different atom say for example deuterium, then if the two molecules formed by replacing hydrogens are the same then the two hydrogens are called homotopic hydrogens.
After replacing the two hydrogens with a different atom then, enantiomers are formed then, the two hydrogens are called enantiotopic hydrogens.
After replacing the two hydrogens with a different atom then, diastereomers are formed then, the two hydrogens are called diastereotopic hydrogens.
In the methyl group, select two hydrogens and replace one hydrogen atom with a D-atom name the compound.
Again replace another hydrogen atom with D-atom.
Name it.
If both are the same then, the hydrogens are homotopic and they are shown below:
Hence, they are homotopic protons.
Round 5578 L to three significant figures.
a. 557 L
b. 558 L
c. 5.58 x 10(3) L
d. 5.58 x 10(-3) L
Answer:
b
Explanation:
The figures of a number that are noteworthy in high accuracy or precision are known as significant figures. They are as follows:
Any digit that isn't zeroAs in 20012 or 64.60007, there are zeros between non-zero numbers.Only use leading zeros when a decimal point is present, such as in 5640.0 or 532.330.If the figure is less than 5, remove it and leave the rest of the number unaffected but if more than 5 add, round it up to 1, and add it to the next figure in the line.From the given information:
To (3) sig. fig: 5578 L = 558 L
A flexible vessel contains 65.8 L of gas at a pressure of 2.07 atm. Under the conditions of constant temperature and constant number of moles of gas, what is the pressure of the gas (in atm) when the volume of the vessel increased by a factor of 16.00
Answer: Pressure of the gas is 0.129375 atm when the volume of the vessel increased by a factor of 16.00.
Explanation:
The formula for ideal gas equation is as follows.
[tex]PV = Nk_{b}T[/tex]
where,
[tex]k_{b}[/tex] = Boltzmann constant
N = number of moles
That can also be written as:
[tex]\frac{PV}{T} = constant[/tex]
As pressure and volume are inversely proportional to each other. So, if one of the state variable is increased then the other one will decrease or vice-versa.
So, if volume of the vessel increased by a factor of 16.00 then it means pressure is decreased by a factor of 16.00
Therefore, final volume is as follows.
[tex]65.8 L \times 16.00\\= 1052.8 L[/tex]
Now, final pressure is as follows.
[tex]\frac{2.07}{16.00}\\= 0.129375 atm[/tex]
Initially the product of pressure and volume is as follows.
[tex]PV = 2.07 \times 65.8\\= 136.206[/tex]
Hence, if volume of the vessel increased by a factor of 16.00 and pressure is decreased by a factor of 16.00 then its product is as follows.
[tex]PV = 0.129375 \times 1052.8\\= 136.206[/tex]
Here, product of pressure and volume remains the same.
Thus, we can conclude that pressure of the gas is 0.129375 atm when the volume of the vessel increased by a factor of 16.00.
A solution of KMnO4 has an absorbance of 0.539 when measured in the colorimeter. Determine the concentration of the KMnO4 given the following data for a calibration plot.
Concentration of KMNO4 (M) Absorbance
0.0150 0.081
0.0300 0.159
0.0450 0.260
0.0600 0.334
Answer:
Concentration of unknown solution is 0.0416 M
Explanation:
As we know
Absorbance is equal to the product of molar absorptivity of KMnO4 m, path length and concentration
From the given set of graphical data, it is clear that the absorbance vs concentration is a straight line.
From the graph, we can obtain-
Y = 5.73 X – 0.0065
Absorbance = 0.232
0.232 = 5.73 X – 0.0065
X = 0.0416
Concentration of unknown solution is 0.0416 M
Hey what is your guyses fear, I am terrified of wasp, especially those huge red ones I call red goobers.
When a liquid is heated the average what? Of energy of its particles will increase
what is a colloidal with liquid as both dispersed phase and dispersion medium called?
Answer:
A colloidal system in which liquid is a dispersed phase and solid is dispersion medium is known as Gel.
Explanation:
What is the specific heat value of brass?
Answer:
Nhiệt dung riêng của đồng là
380
J
/
k
g
.
K
Explanation:
Which state of matter is characterized by having an indefinite shape, but a definite volume?
solid
gas
liquid
Answer:
liquid is the right answer k
Answer:
liquid
Explanation:
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Which of the following is matter?
O A. Light
O
B. Sound
O C. Atoms
O D. Radio waves
One source of aluminum metal is alumina, Al2O3. a. Determine the percentage composition of Al in alumina. b. How many pounds of aluminum can be extracted from 2.0 tons of alumina.?
Answer:
sorry i cant give the answer but you can gi end check in answer sheet of this becoz i had same question in exam in chemistry so i revised then i checked the answer if you want to check answer go to www.coachscotchemistry.com there you xan find the answer
The oxygen atom of a ketone (such as cyclohexanone) contains 2 lone pairs of electrons. These pairs of electrons most likely reside in what type of orbital
Answer: The given pairs of electrons most likely reside in [tex]sp^{2}[/tex] type of orbital.
Explanation:
As it is given that two lone pair of electrons are present on the oxygen atom of ketone (such as cyclohexanone).
Also, there will be one bond pair between carbon and oxygen atom.
Hence, total electrons present in the domain are as follows.
2 lone pairs + 1 bon pair of electron = 3 electron domains
This means that there will be [tex]sp^{2}[/tex] type of orbital present.
Thus, we can conclude that given pairs of electrons most likely reside in [tex]sp^{2}[/tex] type of orbital.