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prisoha [69]
3 years ago
11

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

Chemistry
1 answer:
maxonik [38]3 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

<h2>msto = 96 g de KNO₃</h2>

Espero te sirva.

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2 years ago
The hydrogen gas produced is collected over water at 25.0 oC. The volume of the gas is 7.80 L, and the total pressure in the gas
Eva8 [605]

Answer:

20.4 grams Zn

Explanation:

To find the mass, you first need to find the moles. This can be found using the Ideal Gas Law equation:

PV = nRT

In this equation,

-----> P = pressure (atm)

-----> V = volume (L)

-----> n = moles

-----> R = Ideal Gas Constant (0.08206 atm*L/mol*K)

-----> T = temperature (K)

Before you can plug the values into the equation, you need to convert Celsius to Kelvin.

P = 0.980 atm                   R = 0.08206 atm*L/mol*K

V = 7.80 L                          T = 25.0 °C + 273.15 = 298.15 K

n = ? moles

PV = nRT

(0.980 atm)(7.80 L) = n(0.08206 atm*L/mol*K)(298.15 K)

7.644 = n(24.466)

0.312 moles = n

Now that you have the number of moles, you can convert it to grams using the atomic mass of zinc. The final answer should have 3 sig figs to match the sig figs in the given values.

Atomic Mass (Zn): 65.380 g/mol

0.312 moles Zn           65.380 grams
-------------------------  x  -------------------------  =  20.4 grams Zn
                                         1 mole

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1 year ago
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