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3 years ago

12

How many grams of solid sodium cyanide should be added to 1.50 L of a 0.175 M hydrocyanic acid solution to prepare a buffer with

a pH of 10.210?

1 answer:

marshall27 [118]3 years ago

7 0

I have -12 points thx a lot
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3 years ago

Can any one help on this one

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3 years ago

Which energy profile best shows that the enthalpy of formation of CS2 is 89.4 KJ/mol?

Anna11 [10]

Answer:

Option C. Energy Profile D

Explanation:

Data obtained from the question include:

Enthalpy change ΔH = 89.4 KJ/mol.

Enthalpy change (ΔH) is simply defined as the difference between the heat of product (Hp) and the heat of reactant (Hr). Mathematically, it is expressed as:

Enthalpy change (ΔH) = Heat of product (Hp) – Heat of reactant (Hr)

ΔH = Hp – Hr

Note: If the enthalpy change (ΔH) is positive, it means that the product has a higher heat content than the reactant.

If the enthalpy change (ΔH) is negative, it means that the reactant has a higher heat content than the product.

Now, considering the question given, the enthalpy change (ΔH) is 89.4 KJ/mol and it is a positive number indicating that the heat content of the product is higher than the heat content of the reactant.

Therefore, Energy Profile D satisfy the enthalpy change (ΔH) for the formation of CS2 as it indicates that the heat content of product is higher than the heat content of the reactant.

7 0

3 years ago

Read 2 more answers

Coca cola is similar to tomato juice because

olya-2409 [2.1K]

B.) Because they both are Acids

Hope this helps!

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3 years ago

Read 2 more answers

Water (with density of 1000 kg/m3) with the mass flowrate of 10 kg/sec is flowing into an empty tank. The outlet volumetric flow

Montano1993 [528]

Explanation:

Apply the mass of balance as follows.

Rate of accumulation of water within the tank = rate of mass of water entering the tank - rate of mass of water releasing from the tank

$\frac{d}{dt}(\rho V) = 10 - \rho \times (0.01 h)$

$\rho A_{c} \frac{dh}{dt} = 10 - (0.01) \rho h$

$\frac{dh}{dt} + \frac{0.01 \rho h}{\rho A_{c}} = \frac{10}{\rho A_{c}}$

[/tex]\frac{dh}{dt} + \frac{0.01}{0.01}h[/tex] = $\frac{10}{\rho A_{c}}$

$A_{c} = 0.01 m^{2}$

$\frac{dh}{dt}$ + h = 1

$\frac{dh}{dt}$ = 1 - h

$\frac{dh}{1 - h}$ = dt

$\frac{ln(1 - h)}{-1}$ = t + C

Given at t = 0 and V = 0

$A \times h$ = 0

or, h = 0

-ln(1 - h) = t + C

Initial condition is -ln(1) = 0 + C

C = 0

So, -ln(1 - h) = t

or, t = $ln (\frac{1}{1 - h})$ ........... (1)

(a) Using equation (1) calculate time to fill the tank up to 0.6 meter from the bottom as follows.

t = $ln (\frac{1}{1 - h})$

t = $ln (\frac{1}{1 - 0.6})$

= $ln (\frac{1}{0.4})$

= 0.916 seconds

(b) As maximum height of water level in the tank is achieved at steady state that is, t = $\infty$ .

1 - h = exp (-t)

1 - h = 0

h = 1

Hence, we can conclude that the tank cannot be filled up to 2 meters as maximum height achieved is 1 meter.

8 0

3 years ago