In the conclusion to the snake mole lab, you have to convert 3.45 moles of (NH4)2S04 to

Air is made up of different gases, such as oxygen, nitrogen, and car

tia_tia [17]

The correct option would be that oxygen, nitrogen, and carbon dioxide cannot react with one another.

<h3>Why air components cannot react</h3>

The components of atmospheric air, nitrogen, oxygen, carbon dioxide, etc., cannot react with one another because there is not enough energy in the atmosphere to set the reaction rolling.

For a reaction to take place, there must be enough energy to break the bonds in each air component. This is why the air components will not spontaneously react with one another, except during special events such as lightning and thunder.

More on air components can be found here: brainly.com/question/17288850

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7 0

2 years ago

Why is DNA important to every living thing?

attashe74 [19]

Answer: sorry for the late answer, I just took the test today.

It provides instructions for processes of the cells

Explanation:

7 0

4 years ago

Read 2 more answers

What is a covalent bond?

Archy [21]

A covalent bond, also called a molecular bond, is a chemical bond that involves the sharing of electron pairs between atoms.

4 0

4 years ago

Read 2 more answers

Calculate the final temperature of the system: A 50.0 gram sample of water initially at 100 °C and a 100 gram sample initially a

katrin2010 [14]

Answer:

The final temperature of the system is 42.46°C.

Explanation:

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c\times (T_f-T_1)=-(m_2\times c\times (T_f-T_2))$

where,

c = specific heat of water= $4.18J/g^oC$

$m_1$ = mass of water sample with 100 °C= 50.0 g

$m_2$ = mass of water sample with 13.7 °C= 100.0 g

$T_f$ = final temperature of system

$T_1$ = initial temperature of 50 g of water sample= $100^oC$

$T_2$ = initial temperature of 100 g of water = $13.7^oC$

Now put all the given values in the given formula, we get

$50.0 g\times 4.184 J/g^oC\times (T_f-100^oC)=-(100 g\times 4.184 J/g^oC\times (T_f-13.7^oC))$

$T_f=42.46^oC$

The final temperature of the system is 42.46°C.

5 0

3 years ago

What amount of a nonelectrolytic solute should be dissolved in 28.9 moles of benzene (C₆H₆) at 25°C to change the vapor pressure

RideAnS [48]

Answer:

The amount of a nonelectrolytic solute that has to be dissolved in 28.9 moles to change the vapor pressure of benzene by 17.9 is 6.31 moles.

Explanation:

Let's see how to solve this, step by step applying Raoult's Law (Colligative property).

Pressure solution = χsolvent . Pressure solvent

You have to consider that Pressure in solution is, Pressure Pure - Pressure in Solution (Solute+Solvent). The statement says, that pressure has fallen 17.9%, so let's find out that

100% ....... 94.2 Torr

17.9 % ...... x

16.86 Torr (after the Rule of 3) - This is the value of decline.

So, 94.2Torr - 16.86 Torr = 77.34 Torr

Now the formula

77.34 Torr = 94.2 Torr . X (where x is molar fraction)

x = moles of benzene / moles of benzene + moles of nonelectrolytic solute

77.34 Torr / 94.2 Torr = X

0.821 = moles of benzene / moles of benzene + moles of nonelectrolytic solute

0.821 = 28.9 moles / 28.9 moles + nonelectrolytic

0.821 (28.9 moles + nonelectrolytic) = 28.9 moles

23.72 moles + 0.821 nonelectrolytic = 28.9 moles

0.821 nonelectrolytic = 28.9 moles - 23.72 moles

0.821 nonelectrolytic = 5.18 moles

nonelectrolytic moles = 5.18 / 0.821

nonelectrolytic moles = 6.31

8 0

3 years ago