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Artist 52 [7]
3 years ago
11

Is the osmotic pressure of a 0.10 M solution of NaCl greater than, less than, or equal to that of a 0.10 M solution of KBr?a. eq

ual tob. greater thanc. less thand. insufficient information
Chemistry
1 answer:
jonny [76]3 years ago
4 0

Answer:

a. equal to

Explanation:

The <em>osmotic pressure</em> is calculated by the formula:

π = <em>i</em> * M * R * T

Where π is the osmotic pressure, M is the concentration, R is a constant, T is temperature and <em>i</em> is the van't Hoff's factor (the number of ions a compound forms when dissolved in water,<u> for both NaCl and KBr is 2</u>).

Because R is always the same, and <u>Temperature and Concentration are equal between the two solutions</u>, the osmotic pressure of both solutions are also equal.

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As the temperature of a liquid increases, the solubility of a gas in the liquid
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How are elements similar to those in the same group on the periodic table ​
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Answer:

All the members of a group of elements have the same number of valence electrons and similar chemical properties.

Explanation:

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3 years ago
The diagram below show an enlarged view of the beams of a triple-beam balance
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Answer:

1) 455.2 g. 2) 545.2 g.

Explanation:

i think this is right

6 0
3 years ago
Naturally occurring silicon has an atomic mass of 28.086 and consists of three isotopes. The major isotope is 28Si, natural abun
Elden [556K]

Answer:

29Si has a natural abundance of 4.68%.

30Si has a relative atomic mass of 29.99288 and a natural abundance of 3.09%.

Explanation:

The atomic mass of silicon is given by:

Si=Si²⁸×A₁+Si²⁹×A₂+Si³⁰×A₃

Where:

Si: atomic mass of silicon (28.086)

Si²⁸: relative atomic mass of 28Si (27.97693)

A₁: natural abundance of 28Si (92.23%)

Si²⁹: relative atomic mass of 29Si (28.97649)

A₂: natural abundance of 29Si

Si³⁰: relative atomic mass of 30Si

A₃: natural abundance of 30Si

We also know that 30Si natural abundance is in the ratio of 0.6592 to that of 29Si.

We have to set up a system of three equations in three unknowns:

Si=Si²⁸×A₁+Si²⁹×A₂+Si³⁰×A₃

A₃=0.6592×A₂

A₁+A₂+A₃=1

First, we find substitute the value of A₃ in the third equation and solv teh value of A₂:

A₁+A₂+0.6592×A₂=1

A₁+1.6592×A₂=1

1.6592×A₂=1-A₁

A₂=\frac{1-A₁}{1.6592}=\frac{1-0.9223}{1.6592}=0.0468

Then, we find the value of A₃:

A₃=0.6592×A₂

A₃=0.6592×0.0468=0.0309

Finally, we find the value of Si³⁰ in the first equation:

Si=Si²⁸×A₁+Si²⁹×A₂+Si³⁰×A₃

28.086=27.97693×0.9223+28.97649×0.0468+Si³⁰×0.0309

28.086=27.15922+Si³⁰×0.0309

28.086-27.15922=Si³⁰×0.0309

\frac{0.92678}{0.0309}=Si³⁰

Si³⁰=29.99288

8 0
3 years ago
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