Carbon and Oxygen react to form Carbon Dioxide. What are the reactants?

Can someone help me answer this

Ostrovityanka [42]

Answer:sorry

Explanation:

3 0

3 years ago

How do you find the amount of moles is .032 grams of water and whats the answer

masha68 [24]

Answer:

$\boxed {\boxed {\sf 0.0018 \ mol \ H_2 O }}$

Explanation:

First, we need to find the molecular mass of water (H₂O).

H₂O has:

2 Hydrogen atoms (subscript of 2)
1 Oxygen atom (implied subscript of 1)

Use the Periodic Table to find the mass of hydrogen and oxygen. Then, multiply by the number of atoms of the element.

Hydrogen: 1.0079 g/mol
Oxygen: 15.9994 g/mol

There are 2 hydrogen atoms, so multiply the mass by 2.

2 Hydrogen: (1.0079 g/mol)(2)= 2.0158 g/mol

Now, find the mass of H₂O. Add the mass of 2 hydrogen atoms and 1 oxygen atom.

2.0158 g/mol + 15.9994 g/mol = 18.0152 g/mol

Next, find the amount of moles using the molecular mass we just calculated. Set up a ratio.

$0.032 \ g \ H_2 O* \frac{ 1 \ mol \ H_2 O}{18.0152 \ g \ H_2 O}$

Multiply. The grams of H₂O will cancel out.

$0.032 * \frac{1 \ mol \ H_2 O}{18.0152 }$

$\frac{0.032 *1 \ mol \ H_2 O}{18.0152 }$

$0.00177627781 \ mol \ H_2 O$

The original measurement given had two significant figures (3,2). We must round to have 2 significant figures. All the zeroes before the 1 are not significant. So, round to the ten thousandth.

The 7 in the hundred thousandth place tells us to round up.

$0.0018 \ mol \ H_2 O$

There are about <u>0.0018 moles in 0.032 grams.</u>

6 0

3 years ago

At a certain temperature the vapor pressure of pure benzene is measured to be . Suppose a solution is prepared by mixing of benz

Marianna [84]

Answer:

P(C₆H₆) = 0.2961 atm

Explanation:

I found an exercise pretty similar to this, so i'm gonna use the data of this exercise to show you how to do it, and then, replace your data in the procedure so you can have an accurate result:

<em>"At a certain temperature the vapor pressure of pure benzene (C6H6) is measured to be 0.63 atm. Suppose a solution is prepared by mixing 79.2 g of benzene and 115. g of heptane (C7H16) Calculate the partial pressure of benzene vapor above this solution. Round your answer to 2 significant digits. Note for advanced students: you may assume the solution is ideal".</em>

<em />

Now, according to the data, we want partial pressure of benzene, so we need to use Raoul's law which is:

P = Xₐ * P° (1)

Where:

P: Partial pressure

Xₐ: molar fraction

P°: Vapour pressure

We only have the vapour pressure of benzene in the mixture. We need to determine the molar fraction first. To do this, we need the moles of each compound in the mixture.

To get the moles: n = m / MM

To get the molar mass of benzene (C₆H₆) and heptane (C₇H₁₆), we need the atomic weights of Carbon and hydrogen, which are 12 g/mol and 1 g/mol:

MM(C₆H₆) = (12*6) + (6*1) = 78 g/mol

MM(C₇H₁₆) = (7*12) + (16*1) = 100 g/mol

Let's determine the moles of each compound:

moles (C₆H₆) = 79.2 / 78 = 1.02 moles

moles (C₇H₁₆) = 115 / 100 = 1.15 moles

moles in solution = 1.02 + 1.15 = 2.17 moles

To get the molar fractions, we use the following expression:

Xₐ = moles(C₆H₆) / moles in solution

Xₐ = 1.02 / 2.17 = 0.47

Finally, the partial pressure is:

P(C₆H₆) = 0.47 * 0.63

Hope this helps

7 0

3 years ago

Why does mass of popcorn stay the same after popping it

Varvara68 [4.7K]

Answer:

Explanation:

it stays the same because the seeds or whatever in the bag was still the pop corn just not fully devloped

3 0

3 years ago

Hexane and heptane are miscible with each other in all proportions. In making a solution of them, is the entropy of the system i

Damm [24]

Answer: Option (a) is the correct answer.

Explanation:

A miscible solution is defined as the one in which two or more than two components are soluble with each other.

So, when hexane is added to heptane then both of them will mix with each other and hence they are miscible. This mixing will lead to formation of more number of ions into the solution as a result, more will be the disorder present in the solution.

As entropy is the degree of randomness present.

Hence, we can conclude that a solution of hexane and heptane will lead to an increase in entropy.

8 0

4 years ago