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2 years ago

Explain how to calculate the density of a regular shaped object such as a rectangular box.

Chemistry

1 answer:

belka [17]2 years ago

4 0

The density of a regular shaped object such as a rectangular box is calculated as thus:

The amount of matter is its mass, and the space it takes up is its volume.

So therefore, to calculate the density of an object such as the a regular shaped one like a rectangular box;

Its density = mass / volume.

<h3>Derived quantities</h3>

These are physical quantities which are obtained from fundamental quantities. I'm order words, derived quantities are quantities which are obtained from the combination of of fundamental quantities either by addition, division or multiplication

Some few examples of derived quantities are as follows:

Density

Velocity
Upthrust
Volume
Acceleration

So therefore, the density of a regular shaped object such as a rectangular box is calculated as thus:

The amount of matter is its mass, and the space it takes up is its volume.

So therefore, to calculate the density of an object such as the a regular shaped one like a rectangular box;

Its density = mass / volume.

Learn more about derived quantities:

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Select the correct formula for the polyatomic ion.

Helga [31]

The poly atomic ion formula for ammonium would be NH4+

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4 years ago

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Dovator [93]

Answer : The number of grams of calcium perchlorate is, 0.00253 grams.

Explanation :

Molar mass of calcium perchlorate = 238.9 g/mol

As, $6.022\times 10^{23}$ formula units present in 238.9 g of calcium perchlorate

So, $6.37\times 10^{18}$ formula units present in $\frac{6.37\times 10^{18}}{6.022\times 10^{23}}\times 238.9=0.00253g$ of calcium perchlorate

Therefore, the number of grams of calcium perchlorate is, 0.00253 grams.

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3 years ago

Which is the smallest unit of a compound that still has the properties of that compound? A. an element B. a molecule C. an atom

olasank [31]

Answer:

c. an atom

Explanation:

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3 years ago

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[50 Points] In the experiment below, Beaker 1 contains 12.0 M HCl. Each subsequent beaker is diluted by 50%. What is the [OH-] i

Sidana [21]

Answer:

2.67x10⁻¹⁴M = [OH-]

Explanation:

To solve this question we need first to find the HCl concentration of the beaker 6. Then, using:

Kw = [OH-][H+]

<em>Where Kw = 1x10⁻¹⁴</em>

<em>[OH-] is our incognite</em>

<em />

Beaker 2 = 12.0M / 2 = 6.0M

Beaker 3 = 6.0M / 2 = 3.0M

Beaker 4 = 3.0M / 2 = 1.5M

Beaker 5 = 1.5M / 2 = 0.75M

Beaker 6 = 0.75M / 2 = 0.375M

HCl = 0.375M = [H+]

1x10⁻¹⁴ / [H+] = [OH-]

1x10⁻¹⁴ / 0.375M = 2.67x10⁻¹⁴M

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3 years ago

2. Suppose 13.7 g of C2H2 reacts with 18.5 g O2 according to the reaction below. C2H2(g) + O2(g) → CO2(g) + H2O(ℓ) a. What is th

max2010maxim [7]

Answer:

Mass of CO₂ produced = 20.328 g

Oxygen is limiting reagent.

Explanation:

Given data:

Mass of C₂H₂ = 13.7 g

Mass of O₂ = 18.5 g

Mass of CO₂ produced = ?

What is limiting reagent = ?

Solution:

Chemical equation:

2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O

Number of moles of C₂H₂:

Number of moles = mass /molar mass

Number of moles = 13.7 g/ 26.04 g/mol

Number of moles = 0.526 mol

Number of moles of O₂:

Number of moles = mass /molar mass

Number of moles = 18.5 g/ 32 g/mol

Number of moles = 0.578 mol

Now we will compare the moles of CO₂ with C₂H₂ and O₂

C₂H₂ : CO₂

2 : 4

0.526 : 4/2×0.526 = 1.052

O₂ : CO₂

5 : 4

0.578 : 4/5×0.578 = 0.462

The number of moles of CO₂ produced by O₂ are less thus oxygen will be limiting reactant.

Mass of CO₂ produced:

Mass = number of moles × molar mass

Mass = 0.462 mol × 44 g/mol

Mass = 20.328 g

7 0

3 years ago