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elena-14-01-66 [18.8K]
2 years ago
15

What is the largest area of a circle that can be made from a piece of thread 25.2cm long assume no knot

Chemistry
1 answer:
Zarrin [17]2 years ago
7 0

Answer:

50.2 cm^2

Explanation:

Here we want to find the largest area of a circle that can be made from a piece of thread 25.2 cm long, assume no knot .

This means that the circumference of the circle must equal to the length of the thread, so we can write:

L = 25.2 cm

And the circumference can be written as:

2\pi r = L = 25.2

where

r is the radius of the circle

Solving for r, we find the radius:

r=\frac{L}{2\pi}=\frac{25.2}{2\pi}=4.0 cm

The area of a circle is given by the formula

A=\pi r^2

So in this case, we find:

A=\pi (4.0)^2=50.2 cm^2

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DNA is the answer for the question
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Boris is interested in conducting his first "real" scientific research. However, he is a bit overwhelmed with all of the possibl
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Questionnaire

Explanation:

This is an instrument for research usually in the form of questions with multiple choice answers with sole purpose of gathering information from respondents for research.

5 0
2 years ago
5. A Dumas bulb is filled with chlorine gas at the ambient pressure and is found to contain 7.1 g of chlorine when the temperatu
kati45 [8]

Answer:

a. The original temperature of the gas is 2743K.

b. 20atm.

Explanation:

a. As a result of the gas laws, you can know that the temperature is inversely proportional to moles of a gas when pressure and volume remains constant. The equation could be:

T₁n₁ = T₂n₂

<em>Where T is absolute temperature and n amount of gas at 1, initial state and 2, final states.</em>

<em />

<em>Replacing with values of the problem:</em>

T₁n₁ = T₂n₂

X*7.1g = (X+300)*6.4g

7.1X = 6.4X + 1920

0.7X = 1920

X = 2743K

<h3>The original temperature of the gas is 2743K</h3><h3 />

b. Using general gas law:

PV = nRT

<em>Where P is pressure (Our unknown)</em>

<em>V is volume = 2.24L</em>

<em>n are moles of gas (7.1g / 35.45g/mol = 0.20 moles)</em>

R is gas constant = 0.082atmL/molK

And T is absolute temperature (2743K)

P*2.24L = 0.20mol*0.082atmL/molK*2743K

<h3>P = 20atm</h3>

<em />

7 0
2 years ago
A 7.591-9 gaseous mixture contains methane (CH4) and butane
mestny [16]

Answer:

65.71%

Explanation:

First, we can write the mass of the mixture, thus:

7.519g = X + Y <em>(1)</em>

<em>Where X is the mass of methane and Y the mass of butane</em>

<em />

Also, the reactions of combustion are:

CH₄ + 2O₂ → CO₂ + 2H₂O

<em>2 moles of oxygen react per mole of methane</em>

C₄H₁₀ + 13/2 O₂ → 4CO₂ + 5H₂O

<em>13/2 moles of oxygen react per mole of methane</em>

<em />

That means, in therms of moles of oxygen we can write:

0.9050 moles = 2X/16.04 + 13/2Y/ 58.12

0.9050 = 0.12469X + 0.11184Y <em>(2)</em>

<em>Where 16.04 and 58.12 are molar masses of methane and butane</em>

That is because if X is the mass of methane:

X g Methane * (1mol / 16.04g) = Moles methane

Moles methane * (2 moles Oxygen / mole methane) = Moles oxygen

Replacing (1) in (2):

0.9050 = 0.12469X + 0.11184 (7.519 - X)

0.9050 = 0.12469X + 0.841 - 0.11184X

0.0641 = 0.01285X

X = 4.988g = Mass of methane.

And mass percent of methane is:

4.988g / 7.591g * 100

<h3>65.71%</h3>

7 0
2 years ago
The accepted value is 1.43. Which correctly describes this student’s experimental data?
Natalija [7]

Answer:

Neither accurate nor precise

Explanation:

The values were not near or even the same as the accepted value thus making it neither accurate nor precise.

4 0
2 years ago
Read 2 more answers
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