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3 years ago

15

ELECTRICITY IS ALLOWED TO PASS THROUGH ACIDIFIED WATER USING THE FOLLOWING EXPERIMENTAL SETUP.

1 answer:

zhuklara [117]3 years ago

6 0

Answer: Electrolysis is passing an electric current through a liquid containing ions so that decomposition reaction takes place.

Electrolysis is passing an electric current through a liquid containing ions so that decomposition reaction takes place.When an electric current is passed through acidified water, it decomposes to give hydrogen and oxygen gas. The hydrogen gas is obtained at the cathode and the oxygen gas is obtained at the anode.

Electrolysis is passing an electric current through a liquid containing ions so that decomposition reaction takes place.When an electric current is passed through acidified water, it decomposes to give hydrogen and oxygen gas. The hydrogen gas is obtained at the cathode and the oxygen gas is obtained at the anode.The reaction for decomposition of water into oxygen and hydrogen is as follows:

Electrolysis is passing an electric current through a liquid containing ions so that decomposition reaction takes place.When an electric current is passed through acidified water, it decomposes to give hydrogen and oxygen gas. The hydrogen gas is obtained at the cathode and the oxygen gas is obtained at the anode.The reaction for decomposition of water into oxygen and hydrogen is as follows:2H2O(l) -------> 2H2(g) + O2(g)

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How to solve 2 and 3

GarryVolchara [31]

Use the Heat formula for both problems.

q=m*c*∆t

Where
q= heat in Joules
m= mass in grams
c= specific heat which is a constant 4.18
∆t= change in temperature

3 0

3 years ago

The solubility of N2 in blood at 37°C and a partial pressure of 0.80 atm is 5.6 ✕ 10−4 mol·L−1. A deep-sea diver breathes compre

marysya [2.9K]

Answer:

0.0126 moles are released

Explanation:

Using Henry's law, where the amount of dissolved gas in a liquid is proportional to its partial pressure above the liquid:

S = k×P

<em>Where S is solubility (5.6x10⁻⁴molL⁻¹), k is Henry's constant and P is partial pressure (0.80atm)</em>

Replacing:

<em>5.6x10⁻⁴molL⁻¹ / 0.80atm = 7x10⁻⁴molL⁻¹atm⁻¹</em>

Thus, with Henry's constant, solubility of N₂ when partial pressure is 3.8atm is:

S = 7x10⁻⁴molL⁻¹atm⁻¹ × 3.8atm

S = 2.66x10⁻³molL⁻¹

Thus, when the deepd-sea diver has a pressure of 3.8amt, moles dissolved are:

6.0L × 2.66x10⁻³molL⁻¹ = <em>0.01596 moles of N₂</em>

At the surface, pressure is 0.80atm and solubility is 5.6x10⁻⁴molL⁻¹, moles dissolved are:

6.0L × 5.6x10⁻⁴molL⁻¹ = <em>3.36x10⁻³mol</em>

Thus, released moles are:

0.01596mol - 3.36x10⁻³mol = <em>0.0126 moles are released</em>

8 0

3 years ago

What common item can detect the presence of radiation? A. Bleach B. A roll of photographic film C. LED watch D. Baking soda

Alenkinab [10]

Answer: B- a roll of photographic film

Explanation:

Founders education answer

6 0

3 years ago

Read 2 more answers

The diagrams are showing a cell undergoing a mitotic division at different points of Anaphase. Use the ABC labels on the drawing

Advocard [28]

Answer:

where is the drawing? I can't help if I can't see it sorry

7 0

3 years ago

Read 2 more answers

Determine the boiling point of a 3.70 m solution of phenol in benzene. Benzene has a boiling point of 80.1°C and a boiling point

xeze [42]

Answer: The boiling point of a 3.70 m solution of phenol in benzene is $89.5^0C$

Explanation:

Elevation in boiling point:

$\Delta T_b=i\times k_b\times m$

where,

$\Delta T_b$ = change in boiling point

i= vant hoff factor = 1 (for benzene which is a non electrolyte )

$k_b$ = boiling point constant = $2.53^0C/kgmol$

m = molality = 3.70

$T_{solution}-T_{solvent}=i\times k_b\times m$

$T_{solution}-80.1^0C=1\times 2.53\times 3.70$

$T_{solution}=89.5^0C$

Thus the boiling point of a 3.70 m solution of phenol in benzene is $89.5^0C$

8 0

3 years ago