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3 years ago

8

Afirmar que el h20 es la formula que corresponde al agua, esto es una teoría, idea, ley, hipotesis o experimentación

1 answer:

frozen [14]3 years ago

6 0

Answer:

Ley.

Explanation:

En la teoría de la ciencia, la regularidad de los procesos en la naturaleza se denomina ley de la naturaleza. Las leyes naturales se diferencian de otras leyes en que los seres humanos no pueden ponerlas en vigor ni anularlas a su discreción. En tal sentido, la composición química del agua es indudablemente una ley natural, en tanto el hombre no puede modificarla sin modificar las características inherentes del agua como tal.

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43 milliliters of water weighs 43 g. what is the density of the water?

Anastaziya [24]

Answer:

$\rho =1g/mL$

Explanation:

Hello,

In this case, since the density is defined as the ratio between the mass and the volume as shown below:

$\rho =\frac{m}{V}$

We can compute the density of water for the given 43 g that occupy the volume of 43 mL:

$\rho =\frac{43g}{43mL}=1g/mL$

Regards.

4 0

3 years ago

An element has three stable isotopes with masses of 27.977 amu, 28.976 amu, and 29.973 amu. the heavier two isotopes have an abu

Lera25 [3.4K]

Check attached file for solution.

4 0

4 years ago

For the reaction 2Co3+(aq)+2Cl−(aq)→2Co2+(aq)+Cl2(g). E∘=0.483 V what is the cell potential at 25 ∘C if the concentrations are [

sveta [45]

Explanation:

The given data is as follows.

$E^{o}$ = 0.483, $[Co^{3+}]$ = 0.173 M,

$[Co^{2+}]$ = 0.433 M, $[Cl^{-}]$ = 0.306 M,

$P_{Cl_{2}}$ = 9.0 atm

According to the ideal gas equation, PV = nRT

or, P = $\frac{n}{V}RT$

Also, we know that

Density = $\frac{mass}{volume}$

So, P = MRT

and, M = $\frac{P}{RT}$

= $\frac{9.0 atm}{0.0820 L atm/mol K \times 298 K}$

= $\frac{9.0}{24.436}$

= 0.368 mol/L

Now, we will calculate the cell potential as follows.

E = $E^{o} - \frac{0.0591}{n} log \frac{[Co^{2+}]^{2}[Cl_{2}]}{[Co^{3+}][Cl^{-}]^{2}}$

= $0.483 - \frac{0.0591}{2} log \frac{(0.433)^{2}(0.368)}{(0.173)(0.306)^{2}}$

= $0.483 - 0.02955 log \frac{0.0689}{0.0162}$

= $0.483 - 0.02955 \times 0.628$

= 0.483 - 0.0185

= 0.4645 V

Thus, we can conclude that the cell potential of given cell at $25^{o}C$ is 0.4645 V.

4 0

4 years ago

List the six types of energy and provide an example of each.

givi [52]

Answer:

This is very confusing

Explanation:

lol

6 0

3 years ago

Read 2 more answers

Giúp em với mị người

alukav5142 [94]

Answer:that is so hard

Explanation:

6 0

3 years ago