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3 years ago

If the volume of a solid is 100cm^3 and its mass is 400 g, what is it’s density

Chemistry

1 answer:

Aleksandr-060686 [28]3 years ago

4 0

Answer:

The answer to your question is: density = 4 g/cm³

Explanation:

Data

Volume = 100 cm³

Mass = 400 g

Density = ?

Formula

density = mass/volume

substitution

density = 400/100 = 4 g/cm³

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Brodation<br>number of oxygen in F2O is.

larisa86 [58]

Answer:

Oxidation number of F2O = 0−(−1×2)

State of oxygen will be=+2

4 0

3 years ago

BARSIC [14]

The number of C atoms in 0.524 moles of C is 3.15 atoms.

The number of $SO_2$ molecules in 9.87 moles $SO_2$ is 59.43 molecules.

The moles of Fe in 1.40 x $10^{22}$ atoms of Fe is 0.23 x $10^{-1}$

The moles of $C_2H_6O$ in 2.30x $10^{24}$ molecules of $C_2H_6O$ is 3.81.

<h3>What are moles?</h3>

A mole is defined as 6.02214076 × $10^{23}$ of some chemical unit, be it atoms, molecules, ions, or others. The mole is a convenient unit to use because of the great number of atoms, molecules, or others in any substance.

A. The number of C atoms in 0.524 mole of C:

6.02214076 × $10^{23}$ x 0.524 mole

3.155601758 atoms =3.155 atoms

B. The number of $SO_2$ molecules in 9.87 moles of $SO_2$ :

6.02214076 × $10^{23}$ x 9.87

59.4385293 molecules= 59.43 molecules

C. The moles of Fe in 1.40 x $10^{22}$ atoms of Fe:

1.40 x $10^{22}$ ÷ 6.02214076 × $10^{23}$

0.2324754694 x $10^{-1}$ moles.

0.23 x $10^{-1}$ moles.

D. The moles of $C_2H_6O$ in 2.30x $10^{24}$ molecules of $C_2H_6O$ :

2.30x $10^{24}$ ÷ 6.02214076 × $10^{23}$

3.819239854 moles=3.81 moles

Learn more about moles here:

brainly.com/question/8455949

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6 0

2 years ago

2) A common "rule of thumb" -- for many reactions around room temperature is that the

babunello [35]

The question is incomplete. The complete question is :

A common "rule of thumb" for many reactions around room temperature is that the rate will double for each ten degree increase in temperature. Does the reaction you have studied seem to obey this rule? (Hint: Use your activation energy to calculate the ratio of rate constants at 300 and 310 Kelvin.)

Solutions :

If we consider the activation energy to be constant for the increase in 10 K temperature. (i.e. 300 K → 310 K), then the rate of the reaction will increase. This happens because of the change in the rate constant that leads to the change in overall rate of reaction.

Let's take :

$T_1=300 \ K$