Explanation:
Objectives and goals serve as a foundation for successful organizations since they establish the business's objective and aid in the identification of required measures. Goals are broad assertions of what you want to achieve, whereas goals are the precise stages or steps you need to take to get there.
The values used in the scale of pH and pOH are derived from a system designed by ______.
Curie
Gordonsen
Dalton
Sorenson
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]
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.
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:
The element copper has 29 protons. What is the atomic number of copper?
Answer:
29
Explanation:
since protons are a constant and unique characteristic of an element it's considered as he atomic number of that element.
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
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]
Hope this helps
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Cl2 + 2NO --> 2NOCI
What mass of NOCI will be produced if 2.58 X 10^22 molecules of CI2 react?
Answer:
Explanation:
Find the answer below
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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Why phenol is more acidic than alcohol?
Answer:
Phenol is more acidic than cyclohexanol and acyclic alcohols because the phenoxide ion is more stable than the alkoxide ion. ... But in a phenoxide ion, the negative charge is delocalized over the benzene ring. Phenoxide ion is therefore resonance stabilized.
Answer:
Phenol is more acidic than cyclohexanol and acyclic alcohols because the phenoxide ion is more stable than the alkoxide ion. ... But in a phenoxide ion, the negative charge is delocalized over the benzene ring. Phenoxide ion is therefore resonance stabilized.
Explanation:
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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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What is the colour of Copper sulphate solution
Answer:
blue
Explanation:
What is the percent by mass of chlorine CI in NaCI to 3 significat digits please help
Answer:
60.7%
Explanation:
Molar mass of NaCl = 23 +35.5
= 58.5 g/mol
Molar mass of Cl = 35.5 g/mol
Percentage by mass of Cl =?
The percentage by mass of Cl in NaCl can be obtained as shown below'
Percentage by mass = mass of element / mass of the compound × 100
Percentage by mass of Cl
= 35.5 / 58.5 × 100
= 3550 / 58.5
= 60.7%
Therefore, the percentage by mass of Cl in NaCl is 60.7%
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 type of molecule is 3-hexyne?
The type of molecule in which 3-hexyne is grouped under is referred to an alkyne.
What is Alkyne?This is an organic compound which has a triple bond. It falls under the group of hydrocarbons present on earth.
They are usually represented with the suffix -yne which is why 3-hexyne falls under such category.
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what is a chemical name for the compound with the formula Na2s? (sodium (na) has one valance electron and sulfur (s) has 7 valence electrons.)
a: sulfur sodate
b: sulfur bisulfate
c: sodium sulfate
d: sodium sulfide
Answer: sodium sulfide
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
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:
I hope this helps and please ask if you need further help!
tell me the name please
Answer:
molecules
Explanation:
they are molecule
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).
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.
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
số oxi hóa của các nguyên tố: Fe(OH)2; Fe(OH)3; FexOy; Zn(OH)2 là gì
Answer:
esfsef
Explanation:
esfsefseF
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.
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
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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A sample of gas occupies a volume of 55.3 L, has a temperature of 23.3 °C and a pressure of .658 atm. Calculate the number of moles of gas which are present in the sample. R= .0821 atm L/mol K
Answer: The number of moles of gas which are present in the sample are 1.49 mol.
Explanation:
Given: Volume = 55.3 L
Temperature = [tex]23.3^{o}C[/tex] = (23.3 + 273) K = 296.3 K
Pressure = 0.658 atm
Formula used is as follows.
PV = nRT
where,
P = pressure
V = volume
n = no. of moles
R = gas constant = 0.0821 L atm/mol K
T = temperature
Substitute the values into above formula as follows.
[tex]PV = nRT\\0.658 atm \times 55.3 L = n \times 0.0821 L atm/mol K \times 296.3 K\\n = \frac{0.658 atm \times 55.3 L}{0.0821 L atm/mol K \times 296.3 K}\\= \frac{36.3874}{24.32623}\\= 1.49 mol[/tex]
Thus, we can conclude that the number of moles of gas which are present in the sample are 1.49 mol.
Can someone please solve this for me and explain it
83.3% yield
Explanation:
First, we need to convert 240 g of [tex]Fe_{2}O_{3}[/tex] into moles:
[tex]240 \:g \:Fe_{2}O_{3} \:\times(\frac{1\:\text{mol}\:Fe_{2}O_{3}}{159.69\: \text {g}\:Fe_{2}O_{3}})[/tex]
[tex]=1.50 \:\text{mol}\:Fe_{2}O_{3}[/tex]
Next, find the theoretical Fe yield using molar ratios.
[tex]1.50 \: \text {mol} \: Fe_{2}O_{3}\: \times (\frac{2\: \text{mol} \: Fe}{1 \:\text{mol} \: Fe_{2}O_{3}})[/tex]
[tex] = 3.00 \: \text{mol} \: Fe[/tex]
Then convert this back into grams:
[tex]3.00 \: \text{mol} \:Fe \times (\frac{55.845 \: \text{g} \: Fe}{1 \: \text{mol} \: Fe}) = 168 \: \text{g} \: Fe[/tex]
Note that actual yield is only 140 g Fe so percentage yield is
[tex]\dfrac{140\:\text{g}\:Fe}{168\:\text{g}\:Fe} \times 100[/tex]%= 83.3%
83.3%
The answer is 83.3%
BRAINILIEST PLEASESummarize 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: