All categories

3 years ago

Help please please thanks

Chemistry

1 answer:

9966 [12]3 years ago

4 0

Answer:

the agricultural revolution is the time where people went from mostly being Hunters and Gatherers to mostly farmers, a very long time ago

You might be interested in

4. Juan tiene un cilindro de 40 libras herméticamente cerrado con gas propano y lo conecta mediante una válvula a otro cilindro

ch4aika [34]

Answer:

Ver explicacion

Explanation:

Cuando el cilindro de 40 libras está conectado al cilindro de 100 libras, generamos una presión que continúa hasta que los dos cilindros alcanzan la misma presión de gas en equilibrio. Recuerde que el cilindro de 100 libras estaba inicialmente vacío. Esto significa que su presión inicial es 0. El cilindro de 40 libras ya estaba lleno, por lo que dividimos esta cantidad en dos para tener en cuenta su distribución entre los dos cilindros.

Ahora tenemos 20 libras de gas propano presentes en cada cilindro. La implicación de esto es que, en el cilindro de 100 libras, necesitamos 80 libras adicionales para completar las 100 libras.

5 0

3 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! =)

3 0

3 years ago

How many moles of aluminum oxide (Al2O3) can be produced from 12.8 moles of oxygen gas (02)

zhannawk [14.2K]

Answer:

Theoretical Yield

Percent yield

Example stoichiometry problem

How much oxygen can be prepared from 12.25 g KClO3 . (Use molar mass KClO3 = 122.5 g.)

Most stoichiometry problems can be solved using the following steps.

Step 1.

Write and balance the equation for the decomposition of KClO3 with heat (∆). 2KClO3 + ∆ → 2KCl + 3O2

Step 2.

Convert what you have (in this case g KClO3) to moles.

# moles = grams/molar mass = 12.25 g /122.5 = 0.100 mole KClO3.

Step 3.

Using the coefficients in the balanced equation, convert moles of what you have (moles KClO3) to moles of what you want (in this case moles oxygen).

0.100 mol KClO3 x (3 moles O2/2 moles KClO3) = 0.100 x (3/2) = 0.150 mole O2.

Step 4.

Convert moles from step 3 to grams.

moles x molar mass = grams

0.150 mole O2 x (32.0 g O2/mole O2) = 4.80 g O2 produced from 12.25 g KClO3. This is the theoretical yield. If the ACTUAL yield is 4.20 grams, calculate percent yield. Percent yield = (actual yield/theoretical yield) x 100 = (4.20/4.80) x 100 = 87.5% yield

NOTE: In step 1, moles can be obtained other ways; in step 4 moles can be converted to other units.

a. For solutions, M x L = moles (or mL x M = millimoles).

b. For gases, L/22.4 = moles

4 0

3 years ago

Ight i'm soo bored just wanna talk hit me up

ElenaW [278]

Answer:

hii

Explanation:

8 0

3 years ago

HELP ASAP

arlik [135]

Answer:

He's flying and was looking

Explanation:

6 0

3 years ago