Answer:
Use a line graph
Explanation:
Which factor decide the reactivity of alkyl halide?
Answer:
The reactivity order reflects both the strength of the C-X bond and the stability of X(-) as a leaving group and leads to the general conclusion that alkyl iodides are the most reactive members of this functional class
Detects radio waves from objects in space:
a) Radio Telescopes
b) Reflection Telescope
c) Compound telescope
d) Refraction telescope
Answer:
The correct answer is A
Explanation:
A radio telescope is used to detect, collect and focus radio waves from distant objects in the sky or space. While a compound telescope, by design, can both refract and reflect waves. A reflection telescope uses curved mirrors to reflect light to form an image. A refraction telescope forms an image using a lens as its objective.
From the above definitions, it can be deduced that option A is the correct answer.
How does an indicator show the difference between an acid and an alkali?
Answer: There are various indicators like litmus paper, phenolphthalein, universal indicator etc. that changes its colour to show the difference between acid and alkali.
Which alkane is the isomer of butane called 2-methylpropane? A skeletal model of a 4-carbon chain has a straight line extending from the second and third carbons in the chain. A three-carbon chain has a straight line extending from the center carbon. A 5-carbon chain. A three-carbon chain has 2 lines extending from the same side of the center carbon.
Answer:
A three-carbon chain has a straight line extending from the center carbon.
Explanation:
Isomers are compounds having the same molecular formula but different structural formulas.
Butane and 2-methylpropane are constitutional isomers. Constitutional isomers differ in the way that the constituent atoms are connected to each other.
Butane is a straight chain compound while the compound 2-methylpropane consists of a three-carbon chain which has a straight line extending from the center carbon.
Answer:
it is b on edge 2021
Explanation:
Calculate the total amount of energy required in calories to convert 50.0 g of ice at 0.00 degrees Celsius to steam at 100. degrees Celsius.
Specific heat capacity of water is 1.00 cal/g OC
Hfusion = 80 cal/g OC and Hvap = 540 cal/g OC
Write the complete equation you will use.
Substitute the values in the equation in step 1.
Report the math answer with 3 sig figs and the correct unit.
Answer: The amount of heat absorbed is [tex]36.0\times 10^3Cal[/tex]
Explanation:
Few processes involved are:
(1): [tex]H_2O (s) (0^oC, 273K) \rightleftharpoons H_2O(l) (0^oC,273K)[/tex]
(2): [tex]H_2O (l) (0^oC, 273K) \rightleftharpoons H_2O(l) (100^oC,373K)[/tex]
(3): [tex]H_2O (l) (100^oC, 373K) \rightleftharpoons H_2O(g) (100^oC,373K)[/tex]
Calculating the heat absorbed for the process having same temperature:
[tex]q=m\times \Delta H_{(f , v)}[/tex] ......(i)
where,
q is the amount of heat absorbed, m is the mass of sample and is the enthalpy of fusion or vaporization
Calculating the heat released for the process having different temperature:
[tex]q=m\times C_{s,l}\times (T_2-T_1)[/tex] ......(ii)
where,
[tex]C_{s,l}[/tex] = specific heat of solid or liquid
[tex]T_2\text{ and }T_1[/tex] are final and initial temperatures respectively
For process 1:We are given:
[tex]m=50.0g\\\Delta H_{fusion}=80Cal/g[/tex]
Putting values in equation (i), we get:
[tex]q_1=50.0g\times 80Cal/g\\\\q_1=4000Cal[/tex]
For process 2:We are given:
[tex]m=50.0g\\C=1.00Cal/g^oC\\T_2=100^oC\\T_1=0^oC[/tex]
Putting values in equation (i), we get:
[tex]q_2=50g\times 1Cal/g^oC\times (100-0)\\\\q_2=5000Cal[/tex]
For process 3:We are given:
[tex]m=50.0g\\\Delta H_{vap}=540Cal/g[/tex]
Putting values in equation (i), we get:
[tex]q_3=50.0g\times 540J/g\\\\q_3=27000Cal[/tex]
Calculating the total amount of heat released:
[tex]Q=q_1+q_2+q_3[/tex]
[tex]Q=[(4000)+(5000)+(27000)]Cal=36000Cal=36.0\times 10^3Cal[/tex]
Hence, the amount of heat absorbed is [tex]36.0\times 10^3Cal[/tex]
Which diagram shows how the covalent bonds most likely form in a phosphorous trichloride molecule?
The first diagram shows the covalent bonds most likely form in a phosphorous trichloride molecule. Hence, option A is correct.
What is a covalent bond?Covalent bonds can be polar or nonpolar depending on the electronegativity difference between the atoms involved. In a nonpolar covalent bond, the electrons are shared equally between the atoms. In a polar covalent bond, the electrons are shared unequally.
Phosphorus trichloride (PCl₃) is a covalent compound that consists of one phosphorus atom and three chlorine atoms. The three chlorine atoms are located in a plane that is perpendicular to the phosphorus atom, forming an equilateral triangle with bond angles of 120°.
The molecule has a trigonal pyramidal shape, with the phosphorus atom at the apex and the three chlorine atoms at the base of the pyramid. The fourth hybridized orbital of phosphorus contains a lone pair of electrons that is located above the trigonal plane, giving the molecule a distorted tetrahedral geometry.
Thus, option A is correct.
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How many grams are in 562 dg?
56.2 g
56,200 g
5.620 g
5.62 g
Answer:
56.2
Explanation:
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tell me the name please
Answer:
molecules
Explanation:
they are molecule
NaBr
ions in
an aqueous solution.
A
forms
B
does not form
Answer:
A
Explanation:
i had to answer this question a few weeks ago
how to determine the aspects for core and valence electrons in a periodic table. How comes in some elements like cobalt the 3d is on the valence side and arsenic the 3d is on the core side.
Answer:
See explanation
Explanation:
The electron configuration of an atom in an element determines the property of the atom. The core electrons are found inside the atom while the valence electrons are found on the outermost shell of the atom.
For cobalt, the outermost shell electron configuration is; [Ar] 3d7 4s2. The 3d7 and 4s2 are found in the valence shell of cobalt.
For arsenic, the electronic configuration is [Ar] 3d¹⁰ 4s² 4p³. The valence electrons are 4s2, 4p3. The 3d electrons are found inside the arsenic atom.
The values used in the scale of pH and pOH are derived from a system designed by ______.
Curie
Gordonsen
Dalton
Sorenson
What determines the strength of a dipole dipole force?
Answer:
The more polar the molecules, the stronger the force. The more hydrogen in the molecule, the stronger the force.
Explanation:
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Calculate the heat energy, in kJ, required to melt 1.0kg of ice at 0 degrees celcius.
Show all working out
Answer:
here is the answer. Feel free to ask for more chem help <3
The absorption spectrum of sodium has a line at 589 nm. What is the energy of this line? (The speed of light in a vacuum is 3.00 x 10^8 m/s, and Planck's constant is 6.626 x 10^-34 Jos.)
A. 8.89 x 10^26 J
B. 5.09 x 10^14 J
C. 3.37 x 10^-19 J
D. 2.96 x 10^18 J
Answer:
C. 3.37 x 10^-19 J
Explanation:
Which of the following most likely happens when the volume of a gas increases?
-The number of collisions of gas particles remains same.
-The number of collisions of gas particles increases.
-The pressure of the gas remains same.
-The pressure of the gas decreases.
the pressure of the gas increases
Answer:
The number of collisions of gas particles increases.
Explanation:
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please i need help ……
Answer:
H2
Explanation:
(l) means liquid
(s) means solid
(g) means gaseous
I hope this helped [:
What is a molecule containing only carbon and hydrogen called?
O A. An inorganic molecule
O B. A hydrocarbon
O c. A metallic molecule
D. An ionic molecule
(answer is provided) :)
Answer:
A
Explanation:
did that on my test
The density of aluminum is 2.7 grams per cm'.
How many grams does 12 cm of aluminum
weigh?
Answer:
32.4grams
Explanation:
Density of a substance can be calculated as follows:
Density = mass ÷ volume
According to this question, the density of aluminum is 2.7g/cm³ while its volume is 12 cm³. The mass of aluminum is therefore,
2.7 = m/12
m = 12 × 2.7
m = 32.4grams.
The mass of aluminum is 32.4grams
Summarize your findings in a short report of 150 words. Include your hypothesis, observations, data, calculations, and conclusion. The answers to the questions below should be included in your conclusion.
Why didn't the box slide off the table before you put enough coins in the bag?
What was the coefficient for the smooth surface? For the rough surface?
What was the difference between the two coefficients? What caused this difference?
What factors caused a margin for human error in this investigation?
Would the coefficient vary if you tied the string to a similar object of different masses, but on the same surface? Explain.
What did you learn from this investigation? Be thoughtful in your answer.
Answer:
Follow these steps.
1. Fill the matchbox with pebbles. Weigh the matchbox with the pebbles inside. Record that weight.
2. Tie the string to the box. Allow the string to hang over the edge of the table.
3. Tie the other end of the string to a corner of the plastic bag, leaving an opening to put in coins.
4. Add coins one by one until the box is pulled off the table.
5. Count and record the number of coins and the weight of the bag with the coins in it.
6. Lay the round sticks on the table about 1 inch apart and about 2 inches from the edge of the table.
7. Put the matchbox on the rollers farthest from the edge of the table.
8. Now add coins one by one to the bag until the box is pulled off the table.
9. Count and record the number of coins and the weight of the bag with the coins in it.
10. Repeat the experiment. Determine your margin of error if your results vary. For accuracy, repeat the experiment if desired.
11. Using the equation for the coefficient of friction in the text above, determine the coefficient of friction for the matchbox in each experiment. Include this data in your summary.
Explanation:
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Answer:Follow these steps.
1. Fill the matchbox with pebbles. Weigh the matchbox with the pebbles inside. Record that weight.
2. Tie the string to the box. Allow the string to hang over the edge of the table.
3. Tie the other end of the string to a corner of the plastic bag, leaving an opening to put in coins.
4. Add coins one by one until the box is pulled off the table.
5. Count and record the number of coins and the weight of the bag with the coins in it.
6. Lay the round sticks on the table about 1 inch apart and about 2 inches from the edge of the table.
7. Put the matchbox on the rollers farthest from the edge of the table.
8. Now add coins one by one to the bag until the box is pulled off the table.
9. Count and record the number of coins and the weight of the bag with the coins in it.
10. Repeat the experiment. Determine your margin of error if your results vary. For accuracy, repeat the experiment if desired.
11. Using the equation for the coefficient of friction in the text above, determine the coefficient of friction for the matchbox in each experiment. Include this data in your summary.
Explanation:
the rate of diffusion of gas having molecules mass 4 times that ouch4 is 50 cc.the rate of diffusion of mithane is??
Answer: The rate of diffusion of methane is 100 cc.
Explanation:
Graham's law states that the rate of diffusion of a gas is inversely proportional to the square root of the molar mass of the gas. The equation for this follows:
[tex]\text{Rate}=\frac{1}{\sqrt{M}}[/tex] .....(1)
where M is the molar mass of the gas
We are given:
[tex]Rate_{X}=50cc[/tex]
[tex]M_{X}=4\times M_{CH_4}[/tex]
Since the molar mass of methane = 16 g/mol
Using equation 1:
[tex]\frac{Rate_{CH_4}}{Rate_{X}}=\sqrt{\frac{M_X}{M_{CH_4}}}[/tex]
Putting values in above equation, we get:
[tex]\frac{Rate_{CH_4}}{50cc}=\sqrt{\frac{(4\times 16)}{16}}\\\\Rate_{CH_4}=50\times \sqrt{4}\\\\Rate_{CH_4}=50times 2=100cc[/tex]
Hence, the rate of diffusion of methane is 100 cc.
(1) The solubility of Salt AB2(S) IS 5mol/dm^3.
a. Obtain an expression for the solubility product of AB2(S),in terms of s.
b. Calculate the Ksp of AB2,given that solubility is 2.4x10^3mol/dm^3
Answer:
a. Ksp = 4s³
b. 5.53 × 10⁴ mol³/dm⁹
Explanation:
a. Obtain an expression for the solubility product of AB2(S),in terms of s.
AB₂ dissociates to give
AB₂ ⇄ A²⁺ + 2B⁻
Since 1 mole of AB₂ gives 1 mole of A and 2 moles of B, we have the mole ratio as
AB₂ ⇄ A²⁺ + 2B⁻
1 : 1 : 2
Since the solubility of AB₂ is s, then the solubility of A is s and that of B is 2s
So, we have
AB₂ ⇄ A²⁺ + 2B⁻
[s] [s] [2s]
So, the solubility product Ksp = [A²⁺][B⁻]²
= (s)(2s)²
= s(4s²)
= 4s³
b. Calculate the Ksp of AB₂, given that solubility is 2.4 × 10³ mol/dm³
Given that the solubility of AB is 2.4 × 10³ mol/dm³ and the solubility product Ksp = [A²⁺][B⁻]² = 4s³ where s = solubility of AB = 2.4 × 10³ mol/dm³
Substituting the value of s into the equation, we have
Ksp = 4s³
= 4(2.4 × 10³ mol/dm³)³
= 4(13.824 × 10³ mol³/dm⁹)
= 55.296 × 10³ mol³/dm⁹
= 5.5296 × 10⁴ mol³/dm⁹
≅ 5.53 × 10⁴ mol³/dm⁹
Ksp = 5.53 × 10⁴ mol³/dm⁹
A brass metal bar is made up of copper and zinc. (Between 50% to 70% is copper.) A brass metal bar weighs 20.15 ounces. 35% of the bar is zinc. How many ounces of copper are in the bar?
Answer:
The correct answer would be - 13.0975 ounces.
Explanation:
We know that brass is made up of zinc and copper. The copper part of the brass is approximately around 50 to 70% and the remaining is the zinc part.
Given:
Zinc percentage in a brass = 35%
copper percentage would be - 65% (100 - 35 = 65)
Weight of the brass metal bar = 20.15 ounce
ounces of copper are in the bar = ?
Solution:
The weight of the copper in the brass metal in ounces-
= weight of brass* percentage of copper/100
= 20.15*65/100
= 13.0975 ounces of copper.
2nd method =
The weight of the zinc in the brass metal in ounces-
= weight of brass* percentage of zinc/100
= 20.15*35/100
= 7.0525
The weight of the copper = total weight - weight of zinc
= 13.0975 ounces of copper.
True or false one example of a chemical reaction is when H20 melts and changes state ice to liquid
Answer:
faulse
Explanation:
this is an example of a change in proporties
At 40km up in the atmosphere the pressure is 3.8 kPa. I release a balloon with a volume of 30.0
liters from sea level. The balloon ascends up into to atmosphere 40 km. What will be the new
volume of the balloon, assuming no change in temperature?
Answer in liters
What is the y-intercept of the function f(x) = -x+ ?
2 름
The points on the y-intercepts are (0,y) whereas the point on the x-intercept are (x,0). Here the y-intercept of the function f(x) = -x + 2 is 2.
What is y-intercept?When the line touches the y-axis, we can say that these are y-intercepts. The y-intercept can be found out when x = 0. This represents the point at which the graph crosses through the y-axis. The y-intercept can also be referred as y-value.
In order to find the intercepts of any function, we must find the coordinates of the point where the function intersects the x and y axis. This can also be done by equating the independent or dependent variable value as 0 and then finding out the other value.
Here the given function is:
f(x) = -x + 2
When we put x = 0, then
f(x) = -0 + 2 = 2
Thus the y-intercept is 2.
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Which of the following molecules is an
alcohol? There is only one correct answer.
CH3COOH
CH3OH
CH2NH2
C2H5CI
CH3OH IS MOLECULES OF ALCOHOL
số oxi hóa của các nguyên tố: Fe(OH)2; Fe(OH)3; FexOy; Zn(OH)2 là gì
Answer:
esfsef
Explanation:
esfsefseF
Pu-238 (Plutonium, 238) decays by α emission to form an atom, which atom is this?
Answer:
Helium atom
Explanation:
For the RTGs flown by the United States, the radioisotope has been some chemical form of plutonium-238 (238Pu or Pu-238). Plutonium-238 decays primarily by the emission of an alpha particle (a helium atom without its electrons).
304 cm x m and 67kg to gram please helpppp!!!!!!!!
Answer:
[tex]304cm = 3.04m \\ 67kg = 67000g[/tex]
Explanation:
[tex]100cm \: = 1m \\ 304cm = x \\ use \: \: cross \: multipication \\ 100x = 304 \\ \frac{100x}{100} = \frac{304}{100} \\ x = 3.04m[/tex]
-------------------------------------------------
[tex]1kg = 1000g \\ 67kg = x \\ \\ use \: cross \: \: multipication \\ 1x = 67 \times 1000 \\ x = 67000g[/tex]
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Which substance is a basic oxide?
The oxide that gives a base in water is known as a basic oxide. An amphoteric solution is a substance that can chemically react as either acid or base. However, it is also possible for an oxide to be neither acidic nor basic, but is a neutral oxide.