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3 years ago

1. Is O2 a compound? Explain.

1 answer:

4 0

1. O2 is not a compound because it only contains one or more type of the same element atom.
2. O2 is a molecule because a molecule is one or more of the same element atom.
3. The law of conversion is that the mass of the system will stay the same when transfer takes place. Like if you had an equation O+H2—> H2O the mass will remain the same.

4. It will be equal to 10 because of law of conservation of matter.

5. One observation can be that the compound, reaction you’re observing, has change states.

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What is the name of The compound?

Liono4ka [1.6K]

Answer:

But-2-ene is your answer i guess

5 0

3 years ago

What type of chemical bond is hydrogen?

oksian1 [2.3K]

Hydrogen is a covalent bond. (A bond where one or more pairs of electrons are shared by two atoms)

7 0

3 years ago

How much 2.0 M phenylmagnesium chloride solution is needed if adding 2.9 mmol of Grignard reagent to a reaction? Your answer sho

Vladimir [108]

Answer:

1,45 mL

Explanation:

The Grignard reaction is a very important organometallic chemical reaction where the Grignard reagent ( alkyl, vinyl, or aryl-magnesium halides) acts as nucleophile in order to the formation of Carbon-Carbon bonds.

In the problem, the phenylmagnesium chloride is the grignard reagent. The volume of 2.0 M phenylmagnesium chloride solution you need to add 2.9 mmol is:

2,9 mmol × $\frac{1mL}{2,0mM}$ = <em>1,45 mL</em>

I hope it helps!

3 0

3 years ago

SHOW WORK

horsena [70]

10.0gNaCl/2.0Lsolution= 5.0g/L

4 0

3 years ago

Will give lots of points if answered correctly. Determine the kb for chloroform when 0.793 moles of solute in 0.758 kg changes t

Liono4ka [1.6K]

Answer: The value of $K_{b}$ for chloroform is $3.62^{o}C/m$ when 0.793 moles of solute in 0.758 kg changes the boiling point by 3.80 °C.

Explanation:

Given: Moles of solute = 0.793 mol

Mass of solvent = 0.758

$\Delta T_{b} = 3.80^{o}C$

As molality is the number of moles of solute present in kg of solvent. Hence, molality of given solution is calculated as follows.

$Molality = \frac{no. of moles}{mass of solvent (in kg)}\\= \frac{0.793 mol}{0.758 kg}\\= 1.05 m$

Now, the values of $K_b$ is calculated as follows.

$\Delta T_{b} = i\times K_{b} \times m$

where,

i = Van't Hoff factor = 1 (for chloroform)

m = molality

$K_{b}$ = molal boiling point elevation constant

Substitute the values into above formula as follows.

$\Delta T_{b} = i\times K_{b} \times m\\3.80^{o}C = 1 \times K_{b} \times 1.05 m\\K_{b} = 3.62^{o}C/m$

Thus, we can conclude that the value of $K_{b}$ for chloroform is $3.62^{o}C/m$ when 0.793 moles of solute in 0.758 kg changes the boiling point by 3.80 °C.

7 0

3 years ago