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3 years ago

In what way are chemicals a part of our everyday lives

Chemistry

2 answers:

dsp733 years ago

8 0

Answer:

Everywhere

Explanation:

Chemicals are a significant part of our lives.

We use it in our cloths (as the dyeing), for cooking (as salt and the fire we use), to clean our houses (as detergents, disinfectant), to our hygiene (as deodorants, toothpaste, soap).

But chemicals move our bodies, as the breathing process and all the other biological process we need to live.

vazorg [7]3 years ago

3 0

There are chemicals in lots of things we use every day such as mouthwash, toothpaste, and soap.

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A flexible container at an initial volume of 6.13 L contains 2.51 mol of gas. More gas is then added to the container until it r

aleksley [76]

Answer:

2.12 moles of gas were added.

Explanation:

We can solve this problem by using<em> Avogadro's law</em>, which states that at constant temperature and pressure:

V₁n₂=V₂n₁

Where in this case:

V₁ = 6.13 L

n₂ = ?

V₂ = 11.3 L

n₁ = 2.51 mol

We <u>input the data</u>:

6.13 L * n₂ = 11.3 L * 2.51 mol

n₂ = 4.63

As <em>4.63 moles is the final number of moles</em>, the number of moles added is:

4.63 - 2.51 = 2.12 moles

5 0

2 years ago

Sonia was experimenting with electric charges. She tied two inflated balloons together, held them next to each other, and rubbed

Sladkaya [172]

Sonia observed that the two balloons repelled each other. This is because both balloons acquired the same charge when she rubbed them with the piece of wool, and like charges repel each other.

6 0

3 years ago

Read 2 more answers

Alex dragged a log across the yard in 30 the log weighted 400n she did 900 j of work how much power did I she have

In-s [12.5K]

The power used by Alex to drag the log across the yard is determined as 2,656 W.

The mass of the log is calculated as follows;

W = mg

m = W/g

m = (400)/9.8

m = 40.82 kg

<h3>Velocity of the log</h3>

K.E = ¹/₂mv²

v² = 2K.E/m

v² = (2 x 900)/(40.82)

v² = 44.096

v = 6.64 m/s

<h3>Power used by Alex</h3>

P = Fv

P = 400 x 6.64

P = 2,656 W

Learn more about power here: brainly.com/question/13881533

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5 0

2 years ago

How can you distinguish between crystalline allotropic modifications of Sulphur from those of amorphous allotrops?

Burka [1]

The crystalline allotropes of sulfur are very strong and have a high melting and boiling point while the amorphous allotropes of sulfur are brittle and breaks easily.

<h3>What is a crystalline substance?</h3>

A crystalline substance is one that has a definite arrangement of the atoms in the substance. An amorphous substance lacks this definite arrangement. We can see this arrangement when we conduct an X-ray crystallography of the sulfur.

Also, the crystalline allotropes of sulfur are very strong and have a high melting and boiling point while the amorphous allotropes of sulfur are brittle and breaks easily.

Learn more about sulfur:brainly.com/question/13469437

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4 0

2 years ago

If E=mc^2,solve for both m and c. Also, if m=80 and c=0.40 what is the value of E?

Pie

<u><em>Answer:</em></u>

$m = \frac{E}{c^2}$

$c = \sqrt{\frac{E}{m} }$

$E = 12.8 J$

<u><em>Explanation:</em></u>

<u>Part 1: Solving for m</u>

<u>We are given that:</u>

E = mc²

To solve for m, we will need to isolate the m on one side of the equation

This means that we will simply divide both sides by c²

$m = \frac{E}{c^2}$

<u>Part 2: Solving for c</u>

<u>We are given that:</u>

E = mc²

To solve for c, we will need to isolate the m on one side of the equation

This means that first we will divide both sides by m and then take square root for both sides to get the value of c

$c^2 = \frac{E}{m}\\ \\ c=\sqrt{\frac{E}{m}}$

<u>Part 3: Solving for E</u>

<u>We are given that:</u>

m = 80 and c = 0.4

<u>To get the value of E, we will simply substitute in the given equation: </u>

E = mc²

E = (80) × (0.4)²

E = 12.8 J

Hope this helps :)

4 0

3 years ago