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3 years ago

11

For the following Bronsted acid-base reaction, complete the reaction, identify the acid, base, conjugate

1 answer:

madam [21]3 years ago

7 0

Answer:

algun moderador puede acabar a este otro este me borró respuesta que no debió borrar

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The calculated yield for water was 176 g but the measured yield what is 140 g what percent is the yield

IrinaK [193]

Divide measured yield by calculated yield
140/176 = 79.55%

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4 years ago

If an iron atom loses two electrons, the name of the iron ion is

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3 years ago

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When Rutherford performed his experiment, only 1 in 20,000 alpha particles bounced straight back or were

Anna007 [38]

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when rutherford performed his experiment, only 1 in 20,000 alpha particles bounced straight back or were deflected greatly. the rest went straight through the gold foil. e) based on this evidence, what is in atom's center? positively charged particles.

Explanation:

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3 years ago

Identify the oxidizing agent and the reducing agent for 4Li(s) + O_2 (g) to 2Li_2O(s).

Doss [256]

The reaction is:

<span>4Li(s) + O2 (g) = 2Li+ + 2O-2(s).

The oxidizing agent is the one that is being reduced which is oxygen where the charge changed from neutral to -2 while the reducing agent is the on being oxidized which is lithium where the charge change from neutral to +1.</span>

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3 years ago

What should be the mole fraction of O2 in the gas mixture the diver breathes in order to have the same partial pressure of oxyge

Nutka1998 [239]

The question is incomplete, here is the complete question:

A scuba diver is at a depth of 355 m, where the pressure is 36.5 atm.

What should be the mole fraction of $O_2$ in the gas mixture the diver breathes in order to have the same partial pressure of oxygen in his lungs as he would at sea level? Note that the mole fraction of oxygen at sea level is 0.209.

<u>Answer:</u> The mole fraction of oxygen in the gas mixture is 0.00573

<u>Explanation:</u>

To calculate the partial pressure, we use the equation given by Raoult's law, which is:

$p_{A}=p_T\times \chi_{A}$ ........(1)

where,

$p_A$ = partial pressure of oxygen at sea level = ?

$p_T$ = total pressure at sea level = 1.00 atm

$\chi_A$ = mole fraction of oxygen at sea level = 0.209

Putting values in equation 1, we get:

$p_{O_2}=1.00atm\times 0.209\\\\p_{O_2}=0.209atm$

As, partial pressure of the oxygen in the diver's lungs is equal to the partial pressure of oxygen at sea level

We are given:

$p_T=36.5atm\\p_{O_2}=0.209atm$

Putting values in equation 1, we get:

$0.209atm=36.5atm\times \chi_{O_2}\\\\\chi_{O_2}=\frac{0.209}{36.5}=0.00573$

Hence, the mole fraction of oxygen in the gas mixture is 0.00573

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3 years ago