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3 years ago

Penny has a pencil that is 19 cm long. How long the pencil In inches

Chemistry

2 answers:

JulijaS [17]3 years ago

8 0

You divide 19 centimeters by 2.54 making 19 centimeters equivalent to 7.48 inches

marysya [2.9K]3 years ago

3 0

Answer : The length of pencil in inches are 7.48 inches.

Explanation :

The conversion used form centimeter to inch is :

$1in=2.54cm$

As we are given that the length of the pencil is 19 cm. Now we have to determine the length of the pencil in inches.

As, 2.54 centimeter length of pencil = 1 inch

So, 19 centimeter length of pencil = $\frac{19cm}{2.54cm}\times 1inch=7.48inches$

Therefore, the length of pencil in inches are 7.48 inches.

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2 years ago

The half-life of nitrogen-13 is 10.0 minutes. if you begin with 53.3 mg of this isotope, what mass remains after 25.9 minutes ha

zimovet [89]

Hello!

The half-life is the time of half-disintegration, it is the time in which half of the atoms of an isotope disintegrate.

We have the following data:

mo (initial mass) = 53.3 mg

m (final mass after time T) = ? (in mg)

x (number of periods elapsed) = ?

P (Half-life) = 10.0 minutes

T (Elapsed time for sample reduction) = 25.9 minutes

Let's find the number of periods elapsed (x), let us see:

$T = x*P$

$25.9 = x*10.0$

$25.9 = 10.0\:x$

$10.0\:x = 25.9$

$x = \dfrac{25.9}{10.0}$

$\boxed{x = 2.59}$

Now, let's find the final mass (m) of this isotope after the elapsed time, let's see:

$m = \dfrac{m_o}{2^x}$

$m = \dfrac{53.3}{2^{2.59}}$

$m \approx \dfrac{53.3}{6.021}$

$\boxed{\boxed{m \approx 8.85\:mg}}\end{array}}\qquad\checkmark$

I Hope this helps, greetings ... DexteR! =)

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3 years ago

Denver, Colorado (altitude above sea level is 5,280 ft) has an atmospheric pressure of 633. mm Hg. Calculate the boiling point o

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Answer:

T2 = 94.6 C

Explanation:

Use Clausius-Clayperyon equation.

ln P1/P2 = ∆Hvap/R (1/T2 - 1/T1) where R = 8.314 J/mol-K and T is in degrees K

P1 = 760 mmHg

P2 = 630 mmHg

T1 = 373 K

T2 = ?

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R = 0.008314 kJ/mole-K (NOTE: change R to units of kJ)

Plug in and solve for T2

ln 760 mmHg/630 mmHg = 40.7 kJ/mole (1/T2 - 1/373K)

T2 = 367.74 K = 94.6 C

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