What is the molecular geometry if you have a double bond, a single bond and 1 lone pair around the central atom?

Answers

Answer 1
Answer:

Answer:

4 pairs are needed for the bonds, leaving 1 lone pair. Each double bond uses 2 bond pairs and can be thought of as a single unit. There are 2 double bond units and 1 lone pair, which will try to get as far apart as possible - taking up a trigonal planar arrangement.


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What are the two formulas for the ions that's make up AgC2H3O2 bromide?

Answers

Answer:

KCI+AgC2H3O2

Explanation:

It is the outermost layer of the solid portion of earth

Answers

Answer:

Crust.

Explanation:

Its is made up hard rocks

Answer:lithosphere

The lithosphere is the solid, outer part of the Earth. The lithosphere includes the brittle upper portion of the mantle and the crust, the outermost layers of Earth's structure.

Explanation:

A 3.0 L flask containing helium at 145 mmHg is connected by a closed valve to a 2.0 L flask containing argon at 355 mmHg. When the valve is opened and the gases are allowed to mix equally in the two flasks, what is the total pressure (in mmHg) in the two connected flasks after mixing ?

Answers

Answer:

Assuming that both helium and argon act like ideal gases, the total pressure after mixing would be approximately 229\; \rm mmHg.  

Explanation:

By the ideal gas equation, P\cdot V = n \cdot R \cdot T, where

  • P is the pressure of the sample.
  • V is the volume of the container.
  • n is the number of moles of gas particles in the sample.
  • R is the ideal gas constant.
  • T is the temperature of the sample.

Rewrite to obtain:

  • \displaystyle n = (P \cdot V)/(R\cdot T), and
  • \displaystyle P = (n \cdot R \cdot T)/(V).

Assume that the two samples have the same temperature, T. Also, assume that mixing the two gases did not affect the temperature.

Apply the equation \displaystyle n = (P \cdot V)/(R\cdot T) to find the number of moles of gas particles in each container:

  • In the helium container, V = 3.0\; \rm L and P = \rm 145\; mmHg. Hence, \displaystyle n_1 = (P\cdot V)/(R \cdot T) = \frac{(3.0\; \text{L}) \cdot (145\; \text{mmHg})}{R\cdot T}.
  • In the argon container, V = 2.0\; \rm L and P = 355\; \rm mmHg. Hence, \displaystyle n_2 = (P\cdot V)/(R \cdot T) = \frac{(2.0\; \text{L}) \cdot (355\; \text{mmHg})}{R\cdot T}.

After mixing, V = 2.0 + 3.0 = 5.0\; \rm L. Assuming that temperature T stays the same.

\displaystyle n_1 + n_2 = \frac{(3.0\; \text{L}) \cdot (145\; \text{mmHg})}{R\cdot T} + \frac{(2.0\; \text{L}) \cdot (355\; \text{mmHg})}{R\cdot T}.

Apply the equation \displaystyle P = (n \cdot R \cdot T)/(V) to find the pressure after mixing.

\begin{aligned}P &= \displaystyle \frac{\displaystyle \displaystyle \left(\frac{(3.0\; \text{L}) \cdot (145\; \text{mmHg})}{R\cdot T} + \frac{(2.0\; \text{L}) \cdot (355\; \text{mmHg})}{R\cdot T}\right) \cdot R \cdot T}{5.0\; \rm L} \n &= (3.0\; \rm L * 145\; \rm mmHg + 2.0\; \rm L * 355\; \rm mmHg)/(5.0\; \rm L) \n &\approx 229\; \rm mmHg\end{aligned}.

Answer:

The total pressure is 229 atm

Explanation:

Step 1: Data given

Volume of helium flask = 3.0 L

Pressure helium flask = 145 mm Hg

Volume of argon flask = 2.0 L

Pressure argon flask = 355 mmHg

total volume = 5.0 L

Step 2: Partial pressure helium

pHe = 145 *(3/5) = 87.0 atm

Step 3: Calculate pressure argon

pAr = 355*(2/5) = 142.0 atm

Step 4: Calculate total pressure

Total pressure = 87.0 + 142.0 atm

Total pressure = 229 atm

The total pressure is 229 atm

22. What is the mass in grams of each of the following?a. 3.011 x 1023 atoms F
b. 1.50 x 1023 atoms Mg
c. 4.50 x 1012 atoms Cl
d. 8.42 x 1018 atoms Br
e. 25 atoms W
f. 1 atom Au

Answers

The mass in grams of 3.011 x 10²³ atoms of F is 9.5 g.

The mass in grams of  1.50 x 10²³ atoms of Mg is 5.98 g.

The mass in grams of  4.50 x 10¹² atoms of Cl is 2.65 x 10⁻¹⁰ g.

The mass in grams of  8.42 x 10¹⁸ atoms of Br is 1.12 x 10⁻³ g.

The mass in grams of  25 atoms of W is 3.1 x 10⁻²¹ g.

The mass in grams of  1 atom of Au is 3.27 x 10⁻²² g.

What is the mass in grams of 3.011 x 10²³ atoms F?

The mass in grams of 3.011 x 10²³ atoms of F is calculated as follows;

6.023 x 10²³ atoms = 19 g of F

3.011 x 10²³ atoms F  = ?

= (3.011 x 10²³ x 19 g)/(6.023 x 10²³)

= 9.5 g

The mass in grams of  1.50 x 10²³ atoms of Mg is calculated as follows;

6.023 x 10²³ atoms = 24g of Mg

1.5 x 10²³ atoms F  = ?

= (1.5 x 10²³ x 24 g)/(6.023 x 10²³)

= 5.98 g

The mass in grams of  4.50 x 10¹² atoms of Cl is calculated as follows;

6.023 x 10²³ atoms = 35.5 g of Cl

4.5 x 10²³ atoms Cl  = ?

= (4.5 x 10¹² x 35.5 g)/(6.023 x 10²³)

= 2.65 x 10⁻¹⁰ g

The mass in grams of  8.42 x 10¹⁸ atoms of Br is calculated as follows;

6.023 x 10²³ atoms = 80 g of Br

8.42 x 10¹⁸ atoms Br = ?

= (8.42 x 10¹⁸  x 80 g)/(6.023 x 10²³)

= 1.12 x 10⁻³ g

The mass in grams of  25 atoms of W is calculated as follows;

6.023 x 10²³ atoms = 74 g of W

25 atoms W = ?

= (25  x 74 g)/(6.023 x 10²³)

= 3.1 x 10⁻²¹ g

The mass in grams of  1 atom of Au is calculated as follows;

6.023 x 10²³ atoms = 197 g of Au

1 atom of Au = ?

= (1  x 197 g)/(6.023 x 10²³)

= 3.27 x 10⁻²² g

Learn more about atomic mass here: brainly.com/question/338808

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Final answer:

This solution provides the calculations necessary to convert the number of atoms of various elements (smallest particle of an element) to grams. It does so by using the molar mass of each element and Avogadro's number.

Explanation:

The mass of atoms can be determined by using Avogadro's number (6.022 x 1023 atoms/mol) and the molar mass of the specific element (g/mol). We use these to create a conversion factor and multiply by the number of atoms given.

  1. For F (fluorine), which has a molar mass of about 18.9984 g/mol, 3.011 x 1023 atoms F is 9.00 g F.
  2. For Mg (magnesium), with molar mass of about 24.3050 g/mol, 1.5 x 1023 atoms Mg is 6.07 g Mg.
  3. For Cl (chlorine), with molar mass of about 35.453 g/mol, 4.50 x 1012 atoms Cl is 2.67 x 10-10 g Cl.
  4. For Br (bromine), with molar mass about 79.904 g/mol, 8.42 x 1018 atoms Br is 0.12 g Br.
  5. For W (tungsten), with molar mass about 183.84 g/mol, 25 atoms W is 7.65 x 10-22 g W.
  6. For Au (gold), with molar mass about 197.0 g/mol, 1 atom Au is 3.28 x 10-22 g Au.

Learn more about Atoms to Mass Conversion here:

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The vapor pressure of liquid octane, C8H18, is 100 mm Hg at 339 K. A sample of C8H18 is placed in a closed, evacuated 537 mL container at a temperature of 339 K. It is found that all of the C8H18 is in the vapor phase and that the pressure is 68.0 mm Hg. If the volume of the container is reduced to 338 mL at constant temperature, which of the following statements are correct?a. No condensation will occur.
b. Some of the vapor initially present will condense.
c. The pressure in the container will be 100. mm Hg.
d. Only octane vapor will be present.
e. Liquid octane will be present.

Answers

Answer:

the final pressure (108.03 mmHg ) inside the container at 339 K is more than the vapor pressure of liquid octane (100 mmHg) at 339 K.

Hence,

b. Some of the vapor initially present will condense.

e. Liquid octane will be present.

Explanation:

Given that;

The vapor pressure of liquid octane, C8H18, is 100 mm Hg at 339 K

Initial volume of the container, V1 = 537 mL

Initial vapor pressure, P1 = 68.0 mmHg

Final volume of the container, V2 = 338 mL

Let us say that the final vapor pressure = P2  

From Boyle's law,

P2V2 = P1V1

P2 * 338 = 68.0  * 537

338P2 = 36516

P2 = 36516 / 338

P2 = 108.03 mmHg

 

Thus, the final pressure (108.03 mmHg ) inside the container at 339 K is more than the vapor pressure of liquid octane (100 mmHg) at 339 K.

Hence,

b. Some of the vapor initially present will condense.

e. Liquid octane will be present.

What is the electron configuration for26 / 12 Mg^+2

1s2 2s2 2p6 3s2 3p2


1s2 2s2 2p6 3s2 3p6 4s2 3d4


1s2 2s2 2p6


1s2 2s2 2p6 3s2 3p6 4s2 3d6


1s2 2s2 2p6 3s2

Answers

Answer:

The number of electrons for the Mg atom are 12 electrons. The electron configuration of magnesium is,

Mg (Z= 12) = 1s2 2s2 2p6 3s2

Explanation:

The first two electrons is placed in the 1s orbital. The 1s orbital can accommodate two electrons.

The next 2 electrons for magnesium go in the 2s orbital.

The next six electrons will go in the 2p orbital. The p orbital can hold up to six electrons.

We’ll put six in the 2p orbital and then put the remaining two electrons in the 3s.

Therefore, the Magnesium electron configuration will be 1s22s22p63s2.

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