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lesantik [10]
3 years ago
10

How long will a current of 0.995 A need to be passed through water (containing H2SO4) for 5.00 L of O2 to be produced at STP

Chemistry
1 answer:
DIA [1.3K]3 years ago
3 0

Answer:

24 hours

Explanation:

The computation is shown below:

The needed mole of O_2 is

= 5 ÷22.4 = n

Also 1 mole of O_2 required four electric charge

Now the charge needed is

= n × 4 × 96,500 C

= 4 × 96,500 × 5 c ÷ 22.4

= 86160.714 C

Now

q = i t

t = q ÷ i

= 86160.714 C ÷ 0.995

= 86593.7 seconds

= 24 hours

Hence, the correct option is A.

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40.0 L of oxygen were collected at 10°C and<br> 758 mm of Hg. Calculate its volume at STP.
marta [7]

Answer:

189.5

Explanation:

its very very very very easy

3 0
3 years ago
Will ethane ionize in a solution
Jet001 [13]

Answer:

Ethane can be ionized

Explanation:

Ionization or ionisation is the process by which an atom or a molecule acquires a negative or positive charge by gaining or losing electrons, often in conjunction with other chemical changes. The resulting electrically charged atom or molecule is called an ion.

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5 0
3 years ago
The atomic masses of 151eu and 153eu are 150.919860 and 152.921243 amu, respectively. The average atomic mass of europium is 151
vitfil [10]

Answer:-  The natural abundance of ^1^5^1_E_u is 0.478 or 47.8% and ^1^5^3_E_u is 0.522 or 52.2% .

Solution:- Average atomic mass of an element is calculated from the atomic masses of it's isotopes and their abundances using the formula:

Average atomic mass = mass of first isotope(abundance) + mass of second isotope(abundance)

We have been given with atomic masses for ^1^5^1_E_u and ^1^5^3_E_u as 150.919860 and 152.921243 amu, respectively.  Average atomic mass of Eu is 151.964 amu.

Sum of natural abundances of isotopes of an element is always 1. If we assume the abundance of ^1^5^1_E_u as n then the abundance of ^1^5^3_E_u would be 1-n .

Let's plug in the values in the formula:

151.964=150.919860(n)+152.921243(1-n)

151.964=150.919860n+152.921243-152.921243n

on keeping similar terms on same side:

151.964-152.921243=150.919860n-152.921243n

-0.957243=-2.001383n

negative sign is on both sides so it is canceled:

0.957243=2.001383n

n=\frac{0.957243}{2.001383}

n=0.478

The abundance of ^1^5^1_E_u is 0.478 which is 47.8%.  

The abundance of ^1^5^3_E_u is = 1-0.478

= 0.522 which is 52.2%

Hence, the natural abundance of ^1^5^1_E_u is 0.478 or 47.8% and ^1^5^3_E_u is 0.522 or 52.2% .


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Which organelle this illustration represent?
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What 6 elements in the periodic table are named after real people?
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4 0
3 years ago
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