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2 years ago

6

A simple pendulum of length 1.00 m has a mass of 100 g attached. It is drawn back 30.0° and then released. What is the maximum s

peed of the mass?

1 answer:

Softa [21]2 years ago

4 0

Answer:

4.16ms-1

Explanation:

Vmax=√2gl(1-cos30)

=√2x 10x 1(1-cos30)

=√20x (1-cos 30)

√17.3205

=4.16ms-1

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What is the concentration of a 1:10 dilution of a 2.5 M solution of NaCl? How would you prepare exactly 100 ml of such a solutio

Marina CMI [18]

<u>Answer:</u> The concentration of the solution is 0.25 M

<u>Explanation:</u>

Let the volume of solution of 2.5 M NaCl be 10 mL

We are given:

Dilution ratio = 1 : 10

So, the solution prepared will have a volume of = $\frac{10}{1}\times 1000=100mL$

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated NaCl solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted NaCl solution

We are given:

$M_1=2.5M\\V_1=10mL\\M_2=?M\\V_2=100mL$

Putting values in above equation, we get:

$2.5\times 10=M_2\times 100\\\\M_2=\frac{2.5\times 10}{100}=0.25M$

Hence, the concentration of the solution is 0.25 M

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3 years ago

The Density of pure carbon in Diamond form is 3.52 g/cm^3. How many cubic inches would 23.7 moles of pure diamond occupy?

Vedmedyk [2.9K]

Answer : The volume of pure diamond is $0.493inch^3$

Explanation : Given,

Density of pure carbon in diamond = $3.52g/cm^3$

Moles of pure diamond = 23.7 moles

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First we have to calculate the mass of carbon or pure diamond.

$\text{ Mass of carbon}=\text{ Moles of carbon}\times \text{ Molar mass of carbon}$

Molar mass of carbon = 12 g/mol

$\text{ Mass of carbon}=(23.7moles)\times (12g/mole)=284.4g$

Now we have to calculate the volume of carbon or pure diamond.

Formula used:

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Now putting all the given values in this formula, we get:

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As we know that:

$1cm^3=0.061inch^3$

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Volume = $0.061\times 80.8inch^3$

Volume = $0.493inch^3$

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2 years ago

At 350 k, kc = 0.142 for the reaction 2 brcl(g) ⇀↽ br2(g) + cl2(g) an equilibrium mixture at this temperature contains equal con

djyliett [7]

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Kc = ([Br2][Cl2])/[BrCl]^2
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[BrCl]^2 * Kc = ([Br2][Cl2])
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[BrCl] = sqrt(([Br2][Cl2])/Kc)
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