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irga5000 [103]
2 years ago
14

How were noble gases discovered?

Chemistry
1 answer:
AURORKA [14]2 years ago
8 0

Answer:

Lord Rayleigh's observation that the density of nitrogen extracted from the air was always greater than nitrogen released from various chemical compounds.

Explanation:

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Calculate the missing variables in each experiment below using Avogadro’s law.
blagie [28]

Answer:

The answer to your question is: letter c

Explanation:

Data

V1 = 612 ml    n1 = 9.11 mol

V2 = 123 ml    n2 = ?

Formula

                               \frac{V1}{n1}  =  \frac{V2}{n2}

                                         n2 = \frac{n1V2}{V1}

                                         n2 = \frac{(9.11)((123)}{(612)}

                                                n2 = 1.83 mol                                                

5 0
2 years ago
How many moles of carbon are there in 5 g of carbon?
Umnica [9.8K]

Answer:

0.416666667

Explanation:

number of moles= mass of sample ÷ molar mass

=5÷12

=0.41666667

8 0
2 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
I need help plz ill give u Brainliest plzz i need help...Two students are discussing gravity. One student says that there is a g
DIA [1.3K]

Answer:

the first.

Explanation:

Newton's law of universal gravitation, in part, states that every particle attracts every other particle in the universe.

4 0
2 years ago
What is the effect of substituting soluble metal salts such as NaCl and K2CO3 for HCl or (NH4)2CO3 in the qualitative analysis s
larisa [96]

Answer:

D. You can no longer tell if your original sample contained Na+ or K+.

Explanation:

Group reagents are the reagents which are used to separate a whole group of cations from the other groups of cations.

For example, for group 1 , the group reagent is HCl and for group 5, the group reagent is ammonium carbonate.

<u>When sodium chloride and potassium carbonate is used, it means we are contaminating our sample with foreign cations which may interfere in the process of detection. Also, we will not be able to decipher whether the sample has these ions in the initial or not as we have added them.</u>

8 0
3 years ago
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