Humans have the ability to change the characteristics, or traits, of organisms over time. What is this process called?

How many moles are in 4.233X 10^21

krek1111 [17]

0.00702906176 moles
hope this helps :)

6 0

3 years ago

Illustrate: The Avogadro constant is so large it is normally written in scientific notation. To

Licemer1 [7]

Answer:

602200000000000000000000

Explanation:

move the decimal place over 23 places to the right

5 0

3 years ago

Read 2 more answers

If the energy difference between two electronic states is 214.68 kJ / mol , calculate the frequency of light emitted when an ele

atroni [7]

${\qquad\qquad\huge\underline{{\sf Answer}}}$

Here we go ~

Energy difference btween the two electronic states can be expressed as :

${ \qquad \sf \dashrightarrow \: \Delta E = h\nu}$

[ h = planks constant, ${\: \nu }$ = frequency ]

$\qquad \sf \dashrightarrow \:214.68 = 39.79 \times 10 {}^{ - 14} \times \nu$

$\qquad \sf \dashrightarrow \: \nu = \cfrac{214.68}{39.79 \times 10 {}^{ - 4} }$

$\qquad \sf \dashrightarrow \: \nu = \cfrac{214.68}{39.79 } \times 10 {}^{14}$

$\qquad \sf \dashrightarrow \: \nu \approx 5.395 \times10 {}^{14} \:\:hertz$

5 0

2 years ago

Propane can be turned into hydrogen by the two-step reforming process. In the first step, propane and water react to form carbon

JulsSmile [24]

Answer:

C₃H₈(g) + 6 H₂O(g) ⇒ + 10 H₂(g) + 3 CO₂(g)

Explanation:

Propane can be turned into hydrogen by the two-step reforming process.

In the first step, propane and water react to form carbon monoxide and hydrogen. The balanced chemical equation is:

C₃H₈(g) + 3 H₂O(g) ⇒ 3 CO(g) + 7 H₂(g)

In the second step, carbon monoxide and water react to form hydrogen and carbon dioxide. The balanced chemical equation is:

CO(g) + H₂O(g) ⇒ H₂(g) + CO₂(g)

In order to get the net chemical equation for the overall process, we have to multiply the second step by 3 and add it to the first step. Then, we cancel what is repeated.

C₃H₈(g) + 3 H₂O(g) + 3 CO(g) + 3 H₂O(g) ⇒ 3 CO(g) + 7 H₂(g) + 3 H₂(g) + 3 CO₂(g)

C₃H₈(g) + 6 H₂O(g) ⇒ + 10 H₂(g) + 3 CO₂(g)

4 0

3 years ago

At room temperature (20°C} and pressure, the density of air is 1.189 g/L. An object will float in air if its density is less tha

Alekssandra [29.7K]

Explanation:

$Density =\frac{Mass}[Volume}$

Density of the air ,d= 1.189 g/L

(a) Density of the evacuated ball

Mass of the ball ,m = 0.12 g

Volume of the ball = $V=560 cm^3=560 ml=0.560 L$

$D =\frac{0.12 g}{0.560 L}=0.214 g/L$

D<d, teh evacuated ball will flaot in air.

(b) Density of the evacuated ball D = 0.214 g/L

Density of carbon dioxide gas = $d_1=1.830 g/L$

Mass of the carbon dioxide gas :

$1.830 g/L\times 0.560 L=1.0248 g$

Total density of filled ball with carbon dioxide gas:

$\frac{0.12 g+1.0248 g}{0.560 L}==2.044 g/L$

The ball filled with carbon dioxide will not float in the air because total density of filled ball is greater than the density of an air.

(c) Density of the evacuated ball D = 0.214 g/L

Density of hydrogen gas = $d_2=0.0899 g/L$

Mass of the hydrogen gas :

$1.830 g/L\times 0.560 L=0.050344 g$

Total density of filled ball with hydrogen gas:

$\frac{0.12 g+0.050344 g}{0.560 L}==0.3041 g/L$

The ball filled with hydrogen will float in the air because total density of filled ball is lessor than the density of an air.

(d) Density of the evacuated ball D = 0.214 g/L

Density of oxygen gas = $d_3=1.330 g/L$

Mass of the oxygen gas :

$1.330 g/L\times 0.560 L=1.7448 g$

Total density of filled ball with oxygen gas:

$\frac{0.12 g+1.7448 g}{0.560 L}=1.5442 g/L$

The ball filled with oxygen will not float in the air because total density of filled ball is greater than the density of an air.

(e) Density of the evacuated ball D = 0.214 g/L

Density of nitrogen gas = $d_4=1.165 g/L$

Mass of the nitrogen gas :

$1.165 g/L\times 0.560 L=0.6524 g$

Total density of filled ball with nitrogen gas:

$\frac{0.12 g+0.6524 g}{0.560 L}==1.3792 g/L$

The ball filled with nitrogen will not float in the air because total density of filled ball is greater than the density of an air.

f) Mass must be added to sink the ball = m

Density of ball > Density of the air ; to sink the ball.

$\frac{0.12g +m}{0.560L}>1.189 g/L$

m > 0.54584 g

For any case weight added to ball to make it sink in an air should be grater than the value of 0.54584 grams.

5 0

3 years ago