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Marianna [84]
3 years ago
7

6. Which of the following could be a base in a water solution?

Chemistry
1 answer:
Elza [17]3 years ago
5 0

Answer:

h20

Explanation:

H2O is an acid and a base, and H20 is water.

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Nicole has a 5 lb bag of carrots and a 2 oz bag<br>of carrots. How much does Nicole have in total?​
svp [43]

Answer:

5 pounds 2 ounces.

Explanation:

1 pound = 16 ounces

So just add the 2 ounces to the pounds.

4 0
3 years ago
What is meant by Critical temperatue?
yuradex [85]

Answer:

The critical temperature of a substance is the temperature at and above which vapour of the substance cannot be liquefied, no matter how much pressure is applied.

4 0
3 years ago
1. For each of the molecules below, determine the electron geometry, molecule geometry, and bond
Alexxx [7]

Answer:

CCl4- tetrahedral bond angle 109°

PF3 - trigonal pyramidal bond angles less than 109°

OF2- Bent with bond angle much less than 109°

I3 - linear with bond angles = 180°

A molecule with two double bonds and no lone pairs - linear molecule with bond angle =180°

Explanation:

Valence shell electron-pair repulsion theory (VSEPR theory) helps us to predict the molecular shape, including bond angles around a central atom, of a molecule by examination of the number of bonds and lone electron pairs in its Lewis structure. The VSEPR model assumes that electron pairs in the valence shell of a central atom will adopt an arrangement which tends to minimize repulsions between these electron pairs by maximizing the distance between them. The electrons in the valence shell of a central atom are either bonding pairs of electrons, located primarily between bonded atoms, or lone pairs. The electrostatic repulsion of these electrons is reduced when the various regions of high electron density assume positions as far apart from each other as possible.

Lone pairs and multiple bonds are known to cause more repulsion than single bonds and bond pairs. Hence the presence of lone pairs or multiple bonds tend to distort the molecular geometry geometry away from that predicted on the basis of VSEPR theory. For instance CCl4 is tetrahedral with no lone pair and four regions of electron density around the central atom. This is the expected geometry. However OF2 also has four regions of electron density but has a bent structure. The molecule has four regions of electron density but two of them are lone pairs causing more repulsion. Hence the observed bond angle is less than 109°.

8 0
3 years ago
What will be the pressure of 2.00 mol of an ideal gas at a temperature of 20.5 degrees Celsius and a volume of 62.3L?
erastova [34]

.774atm

First, look at what you have and look at the equations you can use to solve this problem. The best equation would be PV=nRT.

P being pressure, V being volume, n being moles, R being the gas constant, and T being temperature.

Before you start doing any of the math, make sure of two things. Since you're looking for pressure, you'll need a gas constant. When I did the problem, I used the gas constant of atm or atmospheres which is .0821.

Also! Remember to always convert celsius into kelvin, to do this, add 273 to the given celsius degree. After this is all set and done, your equation should look like this:

P = \frac{2 x .0821 x 293.5}{62.3}

The reason that the equation is divided by the volume is due to the fact that you need to isolate the variable or pressure.

Multiply everything on the top and divide by the bottom and you should receive the final answer of .774atm.

Hope this helps!

7 0
3 years ago
You have a 25.2 L sample of gas at 1.25 atm and 25.0 degrees Celsius. How many moles are present in this gas. For your answer, p
Elenna [48]

Answer:

  • <u>1.29 mol</u>

Explanation:

This is a direct application of the equation for ideal gases.

  • PV=nRT

Where:

  • P = pressure = 1.25 atm
  • V = volume = 25.2 liter
  • R = Universal constant of gases = 0.08206 atm-liter/K-mol
  • T = absolute temperature = 25.0ºC = 25 + 273.15 K = 298.15 K
  • n = number of moles

Solving for n:

  • n=\frac{PV}{RT}

Substituting:

n=\frac{1.25atm\times 25.2liter}{0.08206atm-liter/K-mol\times298.15K }\\\\n=1.29mol

8 0
3 years ago
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