Cuantos gramos de nitrato de potasio (KNO3) son necesarias para preparar 800ml de solucion al 12 % m/v

Chemistry

1 answer:

maxonik [38]2 years ago

3 0

Answer:

96 g

Explanation:

Tomando en cuenta que estamos hablando de una solución de Nitrato de potasio cuya concentración es 12% m/V, debemos entonces usar la expresión de %m/V:

%m/V = msto / Vsol * 100 (1)

De esta expresión, debemos despejar la masa de soluto (msto) que es precisamente el valor que estamos buscando. Al hacer el despeje tenemos:

msto = %m/V * Vsol / 100 (2)

Ahora solo nos queda sustituir los valores dados y resolver:

msto = 12 * 800 / 100

Espero te sirva.

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Match each substance with the correct designation for the equation HSO3- + CH3NH2 <=> SO32- + CH3NH3+ HSO3- CH3NH2 SO32- C

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Answer:

$HSO_3^-$ : conjugate acid of $SO_3^{2-}$

$CH_3NH_2$ : conjugate base of $CH_3NH_3^+$

$SO_3^{2-}$ : conjugate base of $HSO_3^-$

$CH_3NH_3^+$ : conjugate acid of $CH_3NH_2$

Explanation:

According to the Bronsted-Lowry conjugate acid-base theory, an acid is defined as a substance which looses donates protons and thus forming conjugate base and a base is defined as a substance which accepts protons and thus forming conjugate acid.

$HSO_3^-+CH_3NH_2\rightleftharpoons SO_3^{2-}+CH_3NH_3^+$

Here in forward reaction $CH_3NH_2$ is accepting a proton, thus it is considered as a base and after accepting a proton, it forms $CH_3NH_3^+$ which is a conjugate acid.

And $HSO_3^-$ is losing a proton, thus it is considered as an acid and after loosing a proton, it forms $SO_3^{2-}$ which is a conjugate base.

Similarly in the backward reaction, $CH_3NH_3^+$ is loosing a proton, thus it is considered as a acid and after loosing a proton, it forms $CH_3NH_2$ which is a conjugate base.

And $SO_3^{2-}$ is accepting a proton, thus it is considered as a base and after accepting a proton, it forms $HSO_3^{-}$ which is a conjugate acid.