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2 years ago

A student makes a sound by blowing into the mouth of an empty soda bottle. How could the student make the sound louder?

Chemistry

1 answer:

natima [27]2 years ago

6 0

Answer:

A. Increase the amplitude by blowing with more force.

Explanation:

I just know this because i love doing that with glass soda bottles

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How could you put the understanding of the size of a drop of water to good use in Chemistry?

Alinara [238K]

Raindrops start to form in a roughly spherical structure due to the surface tension of water. This surface tension is the "skin" of a body of water that makes the molecules stick together. The cause is the weak hydrogen bonds that occur between water molecules

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3 years ago

3 complete paragraphs on Static Electricity

Digiron [165]

Answer: Employments of friction based electricity incorporate contamination control, Xerox machines, and painting. They utilize the property that inverse electrical charges pull in. There are different utilizations including the properties of aversion and the making of electricity produced via friction flashes

friction based electricity is an awkwardness of electric charges inside or on the outside of a material. The charge stays until it can move away through an electric flow or electrical release.

In view of similar kinds of examinations like the one you performed, researchers had the option to set up three laws of electrical charges: Inverse charges pull in one another. Like charges repulse one another. Charged items pull in nonpartisan articles.

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3 years ago

What are OBSERVABLE PROPERTIES

svp [43]

Answer:Observable properties are characteristics or things about materials or objects that we can describe using our five senses. Color, texture, hardness, and flexibility are all things we can determine with our senses.

Explanation:

5 0

3 years ago

compound consists of carbon, hydrogen and fluorine. In one experiment, combustion of 2.50 g of the compound produced 3.926 g of

arlik [135]

Answer: The empirical and molecular formula of the compound is $CH_2F$ and $C_{14}H_{28}F_{14}$ respectively

Explanation:

We are given:

Mass of $CO_2=3.926g$

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 3.926 g of carbon dioxide, $\frac{12}{44}\times 3.926=1.071g$ of carbon will be contained.

To calculate the percentage composition of element in sample, we use the equation:

$\%\text{ composition of element}=\frac{\text{Mass of element}}{\text{Mass of sample}}\times 100$ ......(1)

For Carbon:

Mass of carbon = 1.071 g

Mass of sample = 2.50 g

Putting values in equation 1, we get:

$\%\text{ composition of carbon}=\frac{1.071g}{2.50g}\times 100=42.84\%$

For Fluorine:

Mass of fluorine = 2.54 g

Mass of sample = 5.00 g

Putting values in equation 1, we get:

$\%\text{ composition of fluorine}=\frac{2.54g}{5.00g}\times 100=50.8\%$

Percent composition of hydrogen = [100 - 42.84 - 50.8] % = 6.36 %

We are given:

Percentage of C = 42.84 %

Percentage of F = 50.8 %

Percentage of H = 6.36 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 42.84 g

Mass of F = 50.8 g

Mass of H = 6.36 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{42.84g}{12g/mole}=3.57moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{6.36g}{1g/mole}=6.36moles$

Moles of Fluorine = $\frac{\text{Given mass of Fluorine}}{\text{Molar mass of Fluorine}}=\frac{50.8g}{19g/mole}=2.67moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 2.67 moles.

For Carbon = $\frac{0.072}{2.67}=1.34\approx 1$

For Hydrogen = $\frac{6.36}{2.67}=2.38\approx 2$

For Fluorine = $\frac{2.67}{2.67}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : F = 1 : 2 : 1

The empirical formula for the given compound is $CH_2F$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 448.4 g/mol

Mass of empirical formula = $12+(2\times 1)+19]=33g/mol$

Putting values in above equation, we get:

$n=\frac{448.4g/mol}{33g/mol}=13.6\approx 14$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(14\times 1)}H_{(14\times 2)}F_{(14\times 1)}=C_{14}H_{28}F_{14}$

Hence, the empirical and molecular formula of the compound is $CH_2F$ and $C_{14}H_{28}F_{14}$ respectively

3 0

3 years ago

Lastly, Snape thinks we should try one more calculation.

EleoNora [17]

Answer:

E) 0.90

Explanation:

In TLC (Thin-Layer-Chromatography) retention factor RF is defined as the ratio between distance of the sample and the solvent front. RF is very important in chemistry to know the composition of any sample by comparison.

In the problem, as the sample has a distance of 0,20cm from the solvent front, the distance of the sample is:

2,0cm - 0,20 cm = 1,8 cm. Thus, RF is:

RF = 1,8cm / 2,0cm = 0,90

E) 0,90

I hope it helps!

3 0

3 years ago