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victus00 [196]
2 years ago
6

If the volume of an object increases but the mass stays the same, which will happen to the density?

Chemistry
2 answers:
Reil [10]2 years ago
8 0

Answer: density will decrease.

Explanation:

Density is defined as the mass contained per unit volume.

\rho=\frac{m}{V}

where ,

\rho= Density of the object

m= Mass of object

v= volume of the object

As density is inversely proportional to volume, if volume of an object increases, the density of the object decreases when mass remains the same.

Thus the density of the object decreases if the volume of an object increases but the mass stays the same.

nikklg [1K]2 years ago
5 0
We all know, Density = Mass/ volume. When volume increases and the mass remains the same , the density will decrease considerably.
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We have the masses of two reactants, so this is a <em>limiting reactant</em> problem.

We know that we will need a <em>balanced equation</em> with masses, moles, and molar masses of the compounds involved.

<em>Step 1</em>. <em>Gather all the informatio</em>n in one place with molar masses above the formulas and everything else below them.

MM: ___28.01  2.016 ___32.04

_______CO + 2H_2 → CH_3OH

Mass/g: 7.0 __2.5

<em>Step 2</em>. Calculate the <em>moles of each reactant</em>

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Moles of H_2 =2.5 g H_2 × (1 mol H_2/2.016 g H_2) = 1.24 mol H_2

<em>Step 3. </em>Identify the<em> limiting reactan</em>t

Calculate the <em>moles of CH_3OH</em> we can obtain from each reactant.

<em>From CO</em>: Moles of CH_3OH = 0.250 mol CO  × (1 mol CH_3OH /1 mol CO)

= 0.250 mol CH_3OH

<em>From H_2</em>: Moles of CH_3OH = 1.24 mol H_2 × (1 mol CH_3OH /2 mol H_2)

= 0.620 mol CH_3OH

<em>CO is the limiting reactant</em> because it gives the smaller amount of CH_3OH.

<em>Step 4</em>. Calculate the <em>mass of CH_3OH</em>

Mass of CH_3OH = 0.250 mol CH_3OH × (32.04 g CH_3OH /1 mol CH_3OH) = 8.0 g CH_3OH

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