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Feliz [49]
3 years ago
5

The picture represents the process that produces most of the energy used by living organisms on Earth. Which process is represen

ted in the picture?

Chemistry
2 answers:
forsale [732]3 years ago
6 0
Wheres the pic i cant answer without the pic
Andreyy893 years ago
6 0

Answer:

here is the image

Explanation:

You might be interested in
Enter your answer in the provided box. Consider the reaction: N2(g) + 3H2(g) → 2NH3(g) Suppose that a particular moment during t
Mashcka [7]

Answer:

0.0253 M/s

Explanation:

From the reaction

N₂ + 3H₂ → 2NH₃

The rate of reaction can be written as

Rate = - \frac{d[N_2]}{dt} = - \frac{1}{3} \frac{d[H_2]}{dt} = + \frac{1}{2} \frac{d[NH_3]}{dt}

From the above rate equation we can conclude that the rate of reaction of N₂ is equal to one third of the rate of reaction of H₂,

So,

Rate of reaction of molecular nitrogen = \frac{1}{3} \times 0.0759

Upon calculation, we get rate of reaction of molecular nitrogen = 0.0253 M/s

4 0
3 years ago
What is the pH at the equivalence point in the titration of a 25.7 mL sample of a 0.370 M aqueous nitrous acid solution with a 0
expeople1 [14]

Answer:

pH = 8.24

Explanation:

Nitrous acid (HNO₂) reacts with KOH, thus:

HNO₂ + KOH → KNO₂ + H₂O

Moles of HNO₂ are:

0.0257mL ₓ (0.370mol / L) = 0.00951moles.

In equivalence point, the complete moles of nitrous acid reacts with KOH producing potassium nitrite. There are needed:

0.00951mol ₓ (1L / 0.491mol) = 0.01937L ≡ 19.4mL of 0.491M KOH to reach equivalence point.

Total volume in equivalence point is: 19.4mL + 25.7mL = <em>45.1mL</em>

Potassium nitrite is in equilibrium with water, thus:

NO₂⁻ + H₂O ⇄ HNO₂ + OH⁻

Where equilibrium constant, Kb, is defined as:

Kb = 1.41x10⁻¹¹ = \frac{[OH^-][HNO_2]}{[NO_2]}

In equilibrium, molarity of each compound are:

[NO₂⁻]: 0.00951mol/0.00451L - X = 0.211M - X

[HNO₂]: X

[OH⁻]: X

<em>Where X is reaction coordinate</em>

Replacing in Kb:

1.41x10⁻¹¹ = \frac{[X][X]}{[0.211 -X]}

0 = X² + 1.41x10⁻¹¹X - 2.97x10⁻¹²

Solving for X:

X = -1.72x10⁻⁶ <em>FALSE ANSWER. There is no negative concentrations.</em>

X = 1.72x10⁻⁶. <em>Right answer.</em>

That means:

[OH⁻]: 1.72x10⁻⁶M

As pOH is -log [OH⁻] and pH = 14-pOH:

pOH = 5.76; <em>pH = 8.24</em>

3 0
3 years ago
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
mr_godi [17]

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>
4 0
3 years ago
7. In which environment would liverworts and hornworts grow?
Lelechka [254]
Didjtntbbfjchfjfnt thriving tkfkdkalemrnnfnfjcjfjfjrj
5 0
3 years ago
if the Celsius temperature of a gas at constant pressure is increased from 10 Celsius to 20 Celsius the volume is
svet-max [94.6K]

Answer:

               The volume is increased.

Explanation:

                     According to <em>Charles' Law</em>, " <em>at constant pressure the volume and temperature of the gas are directly proportional to each other</em>". Mathematically this law is presented as;

                                                    V₁ / T₁ = V₂ / T₂   -----(1)

In statement the data given is,

T₁  =  10 °C  =  283.15 K                    ∴  K  =  273.15 + °C

T₂  =  20 °C  =  293.15 K

So, it is clear that the temperature is being increased hence, we will find an increase in volume. Let us assume that the starting volume is 100 L, so,

V₁  =  100 L

V₂  =  Unknown

Now, we will arrange equation 1 for V₂ as,

                                                    V₂  =  V₁ × T₂ / T₁

Putting values,

                                     V₂  = 100 L × 293.15 K / 283.15 K

                                     V₂  = 103.52 L

Hence, it is proved that by increasing temperature from 10 °C to 20 °C resulted in the increase of Volume from 100 L to 103.52 L.

3 0
3 years ago
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