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3 years ago

Choose all the answers that apply.

1 answer:

max2010maxim [7]3 years ago

6 0

Answer: has properties similar to other elements in group 18, does not react readily with other elements, is part of the noble gas group

Explanation: I’ve done on edg before

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The water in a pressure cooker boils at a temperature greater than 100°C because it is under pressure. At this higher temperatur

vazorg [7]

Answer:

the activation energy Ea = 179.176 kJ/mol

it will take 7.0245 mins for the same food to cook in an open pot of boiling water at an altitude of 10000 feet.

Explanation:

From the given information

$T_1 = 100^0 C = 100+273 = 373 \ K \\ \\ T_2 = 113^0 C = 113 + 273 = 386 \ K$

$R_1 = \dfrac{1}{7}$

$R_2 = \dfrac{1}{49}$

Thus; $\dfrac{R_2}{R_1} = 7$

Because at 113.0°C; the rate is 7 time higher than at 100°C

Hence:

$In (7) = \dfrac{Ea}{8.314}( \dfrac{1}{373}- \dfrac{1}{386})$

1.9459 = $\dfrac{Ea}{8.314}* 9.0292 *10^{-5}$

$1.9459*8.314 = Ea * 9.0292*10^{-5}$

$16.1782126= Ea * 9.0292*10^{-5}$

$Ea = \dfrac{16.1782126}{ 9.0292*10^{-5}}$

Ea = 179.176 kJ/mol

Thus; the activation energy Ea = 179.176 kJ/mol

here;

$T_2 = 386 \ K \\ \\T_1 = (89.8 + 273)K = 362.8 \ K$

$In(\dfrac{R_2}{R_1})= \dfrac{Ea}{R}(\dfrac{1}{T_1}- \dfrac{1}{T_2})$

$In(\dfrac{R_2}{R_1})= \dfrac{179.176}{8.314}(\dfrac{1}{362.8}- \dfrac{1}{386})$

$In (\dfrac{R_2}{R_1}) = 0.00357$

$\dfrac{R_2}{R_1}= e^{0.00357}$

$\dfrac{R_2}{R_1}= 1.0035$

where ;

$R_2 = \dfrac{1}7{}$

$R_1 = \dfrac{1}{t}$

Now;

$\dfrac{t}{7}= 1.0035$

t = 7.0245 mins

Therefore; it will take 7.0245 mins for the same food to cook in an open pot of boiling water at an altitude of 10000 feet.

4 0

3 years ago

Which structure could a scientist look for in a plant that would identify it as a club moss rather than a liverwort?

Marianna [84]

<h3>Answer;</h3>

Phloem

<h3>Explanation;</h3>

Club moss plant belongs to the the family Lycopodiaceae, Lycophyte includes any spore-bearing vascular plant.
Liverworts on the other hand are bryophytes which belongs to the division bryophyta. Bryophytes are small, non-vascular plants which includes mosses, hornworts and liverworts.
Vascular plants contain vascular tissues which play an important role of transportation in plants. The major vascular tissues are phloem and xylem. Non-vascular plants on the other hand lacks the vascular tissues for transportation of substances.

7 0

3 years ago

Read 2 more answers

At a given temperature, the elementary reaction A --->B in the forward direction is first order in A with a rate constant of

Pani-rosa [81]

Answer:

The value of the equilibrium constant for the reaction A ⇒ B is Kc = 1.72 × 10³.

The value of the equilibrium constant for the reaction B ⇒ A is K'c = 5.81 × 10⁻⁴.

Explanation:

For the reaction A ⇒ B, the equilibrium constant (Kc) is equal to the forward rate constant (kf) divided by the reverse rate constant (ki).

$Kc=\frac{kf}{ki} =\frac{1.60 \times 10^{2} s^{-1} }{ 9.30 \times 10^{-2} s^{-1}} =1.72 \times 10^{3}$

If we consider the inverse reaction B ⇒ A, its equilibrium constant (K'c) is the inverse of the forward reaction equilibrium constant.

$K'c=\frac{1}{Kc} =\frac{1}{1.72 \times 10^{3} } =5.81 \times 10^{-4}$

4 0

4 years ago

What amount of energy is required to change a spherical drop of water with a diameter of 1.80 mm to three smaller spherical drop

Gekata [30.6K]

This is a straightforward question related to the surface energy of the droplet.

You know the surface area of a sphere is 4π r² and its volume is (4/3) π r³. 

With a diameter of 1.4 mm you have an original droplet with a radius of 0.7 mm so the surface area is roughly 6.16 mm² (0.00000616 m²) and the volume is roughly 1.438 mm³. 

The total surface energy of the original droplet is 0.00000616 * 72 ~ 0.00044 mJ 

The five smaller droplets need to have the same volume as the original. Therefore 

5 V = 1.438 mm³ so the volume of one of the smaller spheres is 1.438/5 = 0.287 mm³. 

Since this smaller volume still has the volume (4/3) π r³ then r = cube_root(0.287/(4/3) π) = cube_root(4.39) = 0.4 mm. 

Each of the smaller droplets has a surface area of 4π r² = 2 mm² or 0.0000002 m². 

The surface energy of the 5 smaller droplets is then 5 * 0.000002 * 72.0 = 0.00072 mJ 
From this radius the surface energy of all smaller droplets is 0.00072 and the difference in energy is 0.00072- 0.00044 mJ = 0.00028 mJ. 

Therefore you need roughly 0.00028 mJ or 0.28 µJ of energy to change a spherical droplet of water of diameter 1.4 mm into 5 identical smaller droplets.

7 0

4 years ago

how would you write equation to show how buffers H2CO3 and NaHCO3 behave when (a) HCl is added and (b) NaOH is added.

Vlad1618 [11]

H2CO3 <---> H+ + HCO3-
NaHCO3 <---> Na+ + HCO3-

When acid is added in the buffer, the excess H+ of that acid reacts with HCO3- to form H2CO3, and due to this NaHCO3 dissociates into HCO3- to attain the equilibrium. and hence there is no net effect of H+ due to pH remain almost constant. when a base is added to the buffer, the OH- ion of base react eith H+ ion present in buffer, then to attain equilibrium of H+ ion, the H2CO3 dissociates to produce H+ ion, but now there is the excess of HCO3- due to which Na+ ion react with them to attain equilibrium of HCO3-. hence there is again no net change in H+ ion due to which pH remain constant.....

8 0

4 years ago