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Answer:

esfsef

Explanation:

esfsefseF

Answer:

A

Explanation:

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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

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]

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]

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]

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

Answer:

blue

Explanation:

Answer:

29

Explanation:

since protons are a constant and unique characteristic of an element it's considered as he atomic number of that element.

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:

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.

Answer:

H2

Explanation:

(l) means liquid

(s) means solid

(g) means gaseous

I hope this helped [:

The first diagram shows the covalent bonds most likely form in a phosphorous trichloride molecule. Hence, option A is correct.

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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Answer:

C. 3.37 x 10^-19 J

Explanation:

Answer:

molecules

Explanation:

they are molecule

Answer: sodium sulfide

Answer:

faulse

Explanation:

this is an example of a change in proporties

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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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

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).

Answer:

56.2

Explanation:

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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%

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%

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.

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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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.

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.

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.

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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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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Answer:

Explanation:

Find the answer below

The type of molecule in which 3-hexyne is grouped under is referred to an 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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