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3 years ago

12

Which of the following is an endothermic reaction?. . A) an explosion. B) condensation of steam. C) freezing water. D) thawing i

ce.

2 answers:

Feliz [49]3 years ago

8 0

thawing ice because endothermic reactions makes things colder

Bezzdna [24]3 years ago

7 0

Endothermic reaction : A reaction in which the system absorb energy, usually in the form of heat

the one that is an endothermic reaction is : D. Thawing ice

hope this helps

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Which of the following statements is incorrect? Group of answer choices The product of wavelength and frequency of electromagnet

morpeh [17]

Answer: Option (b) is the correct.

Explanation:

It is known that energy is directly proportional to frequency.

Mathematically, $E \propto \nu$

or, E = $h \times \nu$ ............ (1)

where, E = energy

h = Plank's constant

$\nu$ = frequency

This relation shows that higher is the energy then more will be the frequency of electrmagnetic radiation.

Also, $\nu = \frac{c}{\lambda}$ ......... (2)

where, c = speed of light

$\lambda$ = wavelength

Therefore, substituting value of $\nu$ from equation (2) into equation (1) as follows.

E = $h \times \nu$

E = $h \times \frac{c}{\lambda}$

This relation shows that higher is the energy then lower will be the wavelength.

Thus, we can conclude that as the energy of a photon increases, its frequency decreases, is incorrect statement.

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3 years ago

Give an example of a physical model of the circulatory system

GaryK [48]

Arteries carry blood away from the heart and veins carry blood back to the heart. Hope that helps (:

6 0

3 years ago

Which half-reaction is most easily oxidized?

netineya [11]

From the ones that you are showing me <span>the more positive the potential the more likely: </span>

<span>Fe+3 + e- ---> Fe+2
I hope this is something very useful</span>

7 0

4 years ago

Read 2 more answers

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

What is the mass of bromine gas if it has a pressure of 720 mmHg, volume of 25mL and a temperature of 18.3 degrees Celsius

Annette [7]

Answer:

Mass = 0.158 g

Explanation:

Formula used,

P V = n R T

Or,

n = P V / R T

Putting values,

n = 0.948 atm . 0.025 L / 0.0821 L.atm.K⁻¹.mol⁻¹ . 291.45

n = 0.00099 mol

Note: we have changed pressure from mmHg to atm, volume from mL to L and temperature from C to K)

Also,

Mass = n . Molecular Mass

Mass = 0.00099 mol × 159.808 g/mol

Mass = 0.158 g

7 0

2 years ago