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vodomira
17 days ago
12

what is the general formula for straight-chain hydrocarbons with n carbon atoms and m (multiple) double bonds?

Chemistry
1 answer:
VMariaS [2.6K]17 days ago
8 0
The formula is expressed as where

indicates the number of carbon atoms and x represents the number of double bonds. For instance, if we take a hydrocarbon with 4 carbon atoms, varying x yields: x = 1 results in C_{4} H_{2*4+2-2*1}, while x = 2 leads to C_{4} H_{2*4+2-2*2} and x = 3 gives C_{4} H_{2*4+2-2*3}.

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At the boiling point, the density of the liquid is 809 g/l and that of the gas is 4.566 g/l. how many liters of liquid nitrogen
KiRa [2711]

Result: 1.68 L of liquid nitrogen is generated during the gas liquefaction process.

Clarification:

This process involves transforming gaseous nitrogen into its liquid form.

The two states possess distinct densities, thus occupying varying volumes; however, the mass remains constant.

Step 1: Calculate the mass of nitrogen gas

Let’s determine the mass of nitrogen gas associated with 297 L.

The density formula is:

Density = \frac{Mass}{Volume}

With a density of nitrogen gas at 4.566 g/L and a volume of 297 L, we can compute the mass of nitrogen gas as follows:

Using these values yields:

Mass = Density \times Volume

Mass = \frac{4.566g}{L} \times 297L

Mass = 1356g

The mass of nitrogen gas calculates to be 1356 g.

Step 2: Derive the volume of liquid nitrogen from the mass obtained

The mass for liquid nitrogen remains the same.

With the density of liquid nitrogen at 809 g/L, we can substitute this into our formula to find the volume of liquid.

Volume = \frac{Mass}{Density}

Volume = \frac{1356g}{809g/L}

Therefore, the volume of liquid nitrogen is 1.68 L.


6 0
1 month ago
Researchers stationed at different areas on a mountain and in a tunnel midway through the mountain boiled water at the same time
VMariaS [2690]

Moving on to the second issue

Let's tackle the second question first. Once you grasp that, the first question will be simpler. By the way, this is an excellent question to clarify. The concepts of less than and more than can be quite tricky in the sciences. Every question you encounter that utilizes less or more should be approached with caution.

As altitude increases, air pressure decreases (essential term: less highlight this sentence in color. Take a moment to reflect on it.)

As the pressure declines, less energy (again, key term) is required for water molecules to escape the surface. Thus, the boiling temperature is lower than it would be at sea level.

Answer to problem two: Lower

Problem One

Water reaches its boiling point when the greatest number of molecules can leave the water's surface. Equal to is the right answer.  Although pinpointing the exact answer can be challenging, equal to is indeed the correct response.

6 0
17 days ago
You want to prepare a solution with a concentration of 200.0μM from a stock solution with a concentration of 500.0mM. At your di
lorasvet [2515]

Answer:

1) This dilution plan will yield a 200μM solution.

2) This dilution plan will not yield a 200μM solution.

3) This dilution plan will not yield a 200μM solution.

4) This dilution plan will yield a 200μM solution.

5) This dilution plan will yield a 200μM solution.

Explanation:

Convert the initial molarity into molar form as shown.

500mM = 500mM \times (\frac{1M}{1000M})= 0.5M

Let's examine the following serial dilution processes.

1)

Dilute 5.00 mL of the stock solution to 500 mL. Then take 10.00 mL of this new solution and dilute it further to 250 mL.

Concentration of 500 mL solution:

M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(5.00mL)}{500 mL}= 5 \times 10^{-3}M

10 mL of this solution is further diluted to 250 mL

M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(5 \times 10^{-3}M)(10.0mL)}{250 mL}= 2 \times 10^{-4}M

Convert μM:

2 \times 10^{-4}M = (2 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 200 \mu M

Thus, this dilution scheme will yield a 200μM solution.

2)

Dilute 5.00 mL of the stock solution to 100 mL. Then take 10.00 mL of this new solution and dilute to 1000 mL.

Concentration of 100 mL solution:

M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(5.00mL)}{100 mL}= 2.5 \times 10^{-2}M

10 mL of this solution is further diluted to 1000 mL

M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(2.5 \times 10^{-2}M)(10.0mL)}{1000 mL}= 2.5 \times 10^{-4}M

Convert μM:

2.5 \times 10^{-4}M = (2.5 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 250 \mu M

Thus, this dilution scheme will not yield a 200μM solution.

3)

Dilute 10.00 mL of the stock solution to 100 mL, followed by diluting 5 mL of that new solution to 100 mL.

Concentration of 100 mL solution:

M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(10mL)}{100 mL}= 0.05M

5 mL of this solution is diluted to 1000 mL

M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(0.05M)(5mL)}{1000 mL}= 0.25 \times 10^{-4}M

Convert μM:

0.25 \times 10^{-4}M = (0.25 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 25 \mu M

Thus, this dilution scheme will not yield a 200μM solution.

4)

Dilute 5 mL of the stock solution to 250 mL. Then take 10 mL of this new solution and further dilute it to 500 mL.

Concentration of 250 mL solution:

M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(5mL)}{250 mL}= 0.01M

10 mL of this solution is further diluted to 500 mL

M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(0.01M)(10mL)}{500 mL}= 2 \times 10^{-4}M

Convert μM:

2 \times 10^{-4}M = (2 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 200 \mu M

Thus, this dilution scheme will yield a 200μM solution.

5)

Dilute 10 mL of the stock solution to 250 mL. Then take another 10 mL of this new solution and dilute it to 1000 mL.

Concentration of 250 mL solution:

M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(10mL)}{250 mL}= 0.02M

10 mL of this solution is further diluted to 1000 mL

M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(0.02M)(10mL)}{1000 mL}= 2 \times 10^{-4}M

Convert μM:

2 \times 10^{-4}M = (2 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 200 \mu M

Thus, this dilution scheme will yield a 200μM solution.

7 0
1 month ago
(a) calculate the %ic of the interatomic bond for the intermetallic compound tial3. (b) on the basis of this result, what type o
Tems11 [2400]

Answer :

The percentage ionic character (%IC) equals 10%, indicating the bond is mostly covalent with slight polarity.

Percent Ionic Character:

This reflects the fraction of ionic nature within a polar covalent bond. The formula for %IC (% ionic character) is:

Percent Ionic character = 1 - e^-^0^.^2^5 ^*^(^X^a^-^X^b^) * 100

Here, Xa is the electronegativity of atom A and Xb is that of atom B.

Given: The compound is TiAl₃.

Electronegativity of Ti = 2.0

Electronegativity of Al = 1.6 (as shown in the provided image)

Substitute these values into the formula:

Percent Ionic character = 1 - e^-^0^.^2^5 ^*^(^2^.^0^-^1^.^6^) * 100

Percent Ionic character = 1 - e^(^-^0^.^2^5 ^*^0^.^4^) * 100

Percent Ionic character = 1 - e^(^-^0^.^1^) * 100

The value of e⁻¹ equals 0.90.

Therefore, percent ionic character = (1 - 0.90) × 100

Percent Ionic Character = 10%

Because the % IC is only 10%, which is relatively low, the bond is classified as covalent with minimal polarity.

8 0
1 month ago
1. Bailey wants to find out which frozen solid melts the fastest: soda, gatorade, or orange juice. She pours each of the three l
VMariaS [2690]

First scenario:

IV: soda, gatorade, orange juice, and water

DV: state of the liquids listed above

Control: freezer and ice tray

Second scenario:

IV: laundry detergent, water

DV: cleanliness of the squares post-wash

Control: chocolate, cloth type, cloth squares

Third scenario:

IV: type of water used, pea plant

DV: growth of the pea plant

Control: pots and daily water amount for the plant

4 0
1 month ago
Read 2 more answers
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