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kupik [55]
3 years ago
9

Un mol de amoniaco Tiene una masa molar de 17 g y ocupa un volumen de 22.4 l qué volumen ocupa el 50 g amoniaco en condiciones n

ormales de temperatura y presión
Chemistry
1 answer:
mr_godi [17]3 years ago
4 0

Answer:

V  = 65.81 L

Explanation:

En este caso, debemos usar la expresión para los gases ideales, la cual es la siguiente:

PV = nRT  (1)

Donde:

P: Presion (atm)

V: Volumen (L)

n: moles

R: constante de gases (0.082 L atm / mol K)

T: Temperatura (K)

De ahí, despejando el volumen tenemos:

V = nRT / P   (2)

Sin embargo como estamos hablando de condiciones normales de temperatura y presión, significa que estamos trabajando a 0° C (o 273 K) y 1 atm de presión. Lo que debemos hacer primero, es calcular los moles que hay en 50 g de amoníaco, usando su masa molar de 17 g/mol:

n = 50 / 17 = 2.94 moles

Con estos moles, reemplazamos en la expresión (2) y calculamos el volumen:

V = 2.94 * 0.082 * 273 / 1

<h2>V = 65.81 L</h2>
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1. A 2-kg bowling ball sits on top of a building that is 40 meters tall. (5 pts) Circle one: KE / GPE / both ( can you show work
masha68 [24]
Potential energy is energy due to an object's height above the ground.
Potential energy = mass x gravity x height
Kinetic energy is energy due to the motion of the object.
Kinetic energy = 1/2 x mass x velocity²

1.
The ball is not moving and is at a height above the ground so it has only potential energy.
P.E = 2 x 9.81 x 40
P.E = 784.8 J

2.
The ball is moving and has a height above the Earth's surface so it has both kinetic and potential energy.
P.E = same as part 1 = 784.8 J
K.E = 1/2 x 2 x 5²
K.E = 25 J

3.
The ball has no height above the Earth's surface and is moving so it has only kinetic energy.
K.E = 1/2 x 2 x 10²
K.E = 100 J

4.
50000 = 1/2 x 1000 x v²
v = 10 m/s

5.
39200 = 200 x 9.81 x h
h = 20.0 m

6.
12.5 = 1/2 x 1 x v²
v = 5 m/s
98 = 1 x 9.81 x h
h = 10.0 m
3 0
3 years ago
What is the minimum volume of 5.50 M HCl necessary to neutralize completely the hydroxide in 718.0 mL of 0.183 M NaOH
Deffense [45]

The volume of HCl required is 23.89 mL

Calculation of volume:

The reaction:

HCl + NaOH \rightarrow NaCl + H_2O

As HCl and NaOH react in 1 : 1 ratio.

Volume of NaOH= 718 mL

Concentration= 0.183M

Volume of HCl= ?

Concentration= 5.50M

Using the dilution formula:

M_1\times V_1(NaOH)= M_2\times V_2(HCl)

0.183\times 718= 5.50 \times V_2\\V_2=\frac{131.394}{5.50} \\V_2 = 23.89 \,mL

Therefore,

Volume of HCl required will be 23.89 mL.

Learn more about neutralization reaction here,

brainly.com/question/1822651

#SPJ4

6 0
2 years ago
.
Natasha_Volkova [10]

Answer:

(C) through the atmosphere

Explanation:

6 0
3 years ago
Read 2 more answers
Calculate the molality of isoborneol in the product if, a) the melting point of pure camphor is 179°C and the melting point take
tresset_1 [31]

Answer:

The molality of isoborneol in camphor is 0.53 mol/kg.

Explanation:

Melting point of pure camphor= T =179°C

Melting point of sample = T_f = 165°C

Depression in freezing point = \Delta T_f=?

\Delta T_f=T- T_f=179^oC-165^oC=14^oC

Depression in freezing point  is also given by formula:

\Delta T_f=i\times K_f\times m

K_f = The freezing point depression constant

m = molality of the sample

i = van't Hoff factor

We have: K_f = 40°C kg/mol

i = 1 (  organic compounds)

\Delta T_f=14^oC

14^oC=1\times 40^oC kg/mol\times m

m=\frac{14^oC}{1\times 40^oC kg/mol}=0.35 mol/kg

The molality of isoborneol in camphor is 0.53 mol/kg.

8 0
3 years ago
A chemical reaction produces 3.81g potassium chloride (KCl). How many formula units of potassium chloride are there? *Find Molar
Aleks04 [339]

Answer:

The number of formula units in 3.81 g of potassium chloride (KCl) is approximately 3.08 × 10²²

Explanation:

The given parameters is as follows;

The mass of potassium chloride produced in the chemical reaction (KCl) = 3.81 g

The required information = The number of formula units of potassium chloride (KCl)

The Molar Mass of KCl = 74.5513 g/mol

The \ number \ of \ moles \ of \ a \ substance, n = \dfrac{The  \  mass  \  of  \  the substance}{The   \  Molar  \   Mass   \  of  \   the   \  substance}

Therefore, we have;

The \ number \ of \ moles \ of \ KCl= \dfrac{3.81 \ g}{74.5513 \ g/mol} \approx 0.051106 \ moles

1 mole of a substance, contains Avogadro's number (6.022 × 10²³) of  formula units

Therefore;

0.051106 moles of KCl contains 0.051106 × 6.022 × 10²³  ≈ 3.077588 × 10²² formula units

From which we have, the number of formula units in 3.81 g of potassium chloride (KCl) ≈ 3.08 × 10²² formula units.

8 0
3 years ago
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