All categories

3 years ago

#12 please! worth 40 points

Chemistry

1 answer:

Serga [27]3 years ago

3 0

Okay so because of the difference in density a simple method for telling the difference between the two is to put a sample in a container with oil, because water has a higher density than the oil it would sink to the bottom but alcohol on the other hand is lighter than oil and would float on top of the oil.

However with this question I think that what you would do is use the ice to find out what the substance is, it would float on top of the liquid if it were water because the water is denser than ice but the ice would sink if it was alcohol because the alcohol is less dense than ice.

I hope this helps you, good luck : )

You might be interested in

the spectral lines observed for hydrogen arise from transitions from excited states back to the n=2 principle quantum level. Cal

Sunny_sXe [5.5K]

Rydberg formula is given by:

$\frac{1}{\lambda } = R_{H}\times (\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} )$

where, $R_{H}$ = Rydberg constant = $1.0973731568508 \times 10^{7} per metre$

$\lambda$ = wavelength

$n_{1}$ and $n_{2}$ are the level of transitions.

Now, for $n_{1}$ = 2 and $n_{2}$ = 6

$\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{6^{2}} )$

= $1.0973731568508 \times 10^{7} \times (\frac{1}{4}-\frac{1}{36} )$

= $1.0973731568508 \times 10^{7} \times (0.25-0.0278 )$

= $1.0973731568508 \times 10^{7} \times 0.23$

= $0.2523958\times 10^{7}$

$\lambda = \frac{1}{0.2523958\times 10^{7}}$

= $3.9620\times 10^{-7} m$

= $396.20\times 10^{-9} m$

= $396.20 nm$

Now, for $n_{1}$ = 2 and $n_{2}$ = 5

$\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{5^{2}} )$

= $1.0973731568508 \times 10^{7} \times (0.25-0.04 )$

= $1.0973731568508 \times 10^{7} \times (0.21 )$

= $0.230 \times 10^{7}$

$\lambda= \frac{1}{0.230 \times 10^{7}}$

= $4.3478 \times 10^{-7} m$

= $434.78\times 10^{-9} m$

= $434.78 nm$

Now, for $n_{1}$ = 2 and $n_{2}$ = 4

$\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{4^{2}} )$

= $1.0973731568508 \times 10^{7} \times (0.25-0.0625 )$

= $1.0973731568508 \times 10^{7} \times (0.1875 )$

= $0.20575 \times 10^{7}$

$\lambda= \frac{1}{0.20575 \times 10^{7}}$

= $4.8602 \times 10^{-7} m$

= $486.02 \times 10^{-9} m$

= $486.02 nm$

Now, for $n_{1}$ = 2 and $n_{2}$ = 3

$\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{3^{2}} )$

= $1.0973731568508 \times 10^{7} \times (0.25-0.12 )$

= $1.0973731568508 \times 10^{7} \times (0.13 )$

= $0.1426585\times 10^{7}$

$\lambda= \frac{1}{0.1426585\times 10^{7}}$

= $7.0097 \times 10^{-7} m$

= $700.97 \times 10^{-9} m$

= $700.97 nm$

5 0

2 years ago

Read 2 more answers

What is the volume in (a) liters and (b) cubic yards of a room that is 10. meters wide by 15 meters long and 8.0 ft high?

deff fn [24]

Answer:

V = 364500 L, 476.748 yard³

Explanation:

Given that,

The dimensions of a room are 10 meters wide by 15 meters long and 8.0 ft high.

l = 10 m, b = 15 m, h = 8 ft = 2.43 m

The volume of the room is :

V = lbh

So,

V = 10×15×2.43

V = 364.5 m³

As 1 m³ = 1000 L

364.5 m³ = 364500 L

Also, 1 m³ = 1.30795 yard³

364.5 m³ = 476.748 yard³

Hence, this is the required solution.

7 0

2 years ago

Which description best characterization the motion of particles in a solid

marta [7]

Are there any choices? Because from what the question is it seems like we need choices to help

4 0

3 years ago

Read 2 more answers

Which of the following statements is true of an aqueous solution of sodium chloride?

Ilya [14]

Answer:

What are the statements please

Explanation:

4 0

3 years ago

I was going to ask if you can my help friend so can you answer his question, Biodiversity is important because living things dep

USPshnik [31]

True, biodiversity creates different species animals depend on each other for food and other reasons so the answer is true.

4 0

2 years ago

Read 2 more answers