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3 years ago

9

A vessel of 356 cm3 capacity contains oxygen at a pressure of 760 mmHg and temperature of 15 oC. Assuming that the volume remain

s constant, calculate: (10 points)
(a) the pressure inside the vessel when it was warmed to 100 oC

(b). the temperature to which it must be raised to give a pressure of 2 atm.

1 answer:

liq [111]3 years ago

8 0

(a) the pressure inside the vessel when it was warmed to 100 oC

(

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What can you say about the activity?

maks197457 [2]

Answer:

I would put, A substance has two different parts of it, a pure substance and a compound and mixtures have two as well called, homogenous mixtures and hetrogeneous mixtures.

Explanation:

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3 years ago

What type of bond will form<br><br> Nitrogen and Tellurium

sineoko [7]

Answer: Vander waals

Explanation:

Tellurium is a chemical element with atomic number 52. It's symbol is Te.

Nitrogen is a chemical element with atomic number 14. Its symbol is N

VANDER WAALS bond is found between Nitrogen and Tellurium, just as in Tellurium(II) Nitrogen.

This bond is relatively weak compared to some other kinds of bonds.

That is why tellurium and most of its compounds are usually brittle and easily grounded.

So, Vander Waals bonds is the answer

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3 years ago

Consider the reaction

SOVA2 [1]

Answer :

(a) The average rate will be:

$\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s$

(b) The average rate will be:

$\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s$

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

$5Br^-(aq)+BrO_3^-(aq)+6H^+(aq)\rightarrow 3Br_2(aq)+3H_2O(l)$

The expression for rate of reaction :

$\text{Rate of disappearance of }Br^-=-\frac{1}{5}\frac{d[Br^-]}{dt}$

$\text{Rate of disappearance of }BrO_3^-=-\frac{d[BrO_3^-]}{dt}$

$\text{Rate of disappearance of }H^+=-\frac{1}{6}\frac{d[H^+]}{dt}$

$\text{Rate of formation of }Br_2=+\frac{1}{3}\frac{d[Br_2]}{dt}$

$\text{Rate of formation of }H_2O=+\frac{1}{3}\frac{d[H_2O]}{dt}$

Thus, the rate of reaction will be:

$\text{Rate of reaction}=-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{d[BrO_3^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}=+\frac{1}{3}\frac{d[H_2O]}{dt}$

<u>Part (a) :</u>

<u>Given:</u>

$\frac{1}{5}\frac{d[Br^-]}{dt}=1.56\times 10^{-4}M/s$

As,

$-\frac{1}{5}\frac{d[Br^-]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}$

and,

$\frac{d[Br_2]}{dt}=\frac{3}{5}\frac{d[Br^-]}{dt}$

$\frac{d[Br_2]}{dt}=\frac{3}{5}\times 1.56\times 10^{-4}M/s$

$\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s$

<u>Part (b) :</u>

<u>Given:</u>

$\frac{1}{5}\frac{d[Br^-]}{dt}=1.56\times 10^{-4}M/s$

As,

$-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}$

and,

$-\frac{1}{6}\frac{d[H^+]}{dt}=\frac{3}{5}\frac{d[Br^-]}{dt}$

$\frac{d[H^+]}{dt}=\frac{6}{5}\times 1.56\times 10^{-4}M/s$

$\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s$

5 0

3 years ago

By how much will the water temperature increases if 1046 J of heat energy are added. The specific heat of water is 4.184 J/g • °

Brilliant_brown [7]

Answer:

Option A

250 degrees Celcius

Explanation:

If 1046J of heat energy is added to water, the water will experience a rise in temperature, at a rate that is directly proportional to its specific heat capacity.

Mathematically, this can be seen as $Q=C\Delta T$

Where C = specific heat of water = 4.184 J/g • °C.

Q = heat energy = 1046 J

$\Delta T =1046/4.148=250 degrees Celcius$

Therefore, the increase in temperature that will be experienced, is for 250 degrees Celcius

8 0

2 years ago

After an investigation, what do scientists often do?

den301095 [7]

Record observations, pose a question, create a test for an individual variable, test, come to a conclusion

7 0

3 years ago