All categories

2 years ago

Correct forms of the equation of Charles’s law is (are)

1 answer:

8 0

According to Charles' Law the volume of an ideal gas is directly proportional to its absolute temperature in Kelvin keeping the pressure constant.

V∝ T, P is constant

where V, T and P are volume, temperature and pressure

$\frac{V1}{T1 }$ = $\frac{V2}{T2}$

where V₁, T₁, V₂ and T₂ are initial volume, initial temperature, final volume and final temperature.

You might be interested in

What type of heat transfer occurs when you touch a metal

Galina-37 [17]

Answer:

heat conduction

Explanation:

7 0

2 years ago

Read 2 more answers

Which sub-layer thins out into space, where there is no air

KengaRu [80]

The "exosphere" is the most distant and tenuous "layer" of our atmosphere.

3 0

2 years ago

Read 2 more answers

I will give brainliest!! How long is the pencil? Make sure to use the appropriate amount of significant figures.

8090 [49]

Answer: Option (c) is the correct answer.

Explanation:

The bottom of pencil is placed at the starting point of scale. Whereas the tip of pencil depicts the end point of its length.

The bottom of pencil is at 0 mm and tip of pencil is at 18.73 mm. The appropriate amount of significant figures is 18.73 mm.

Therefore, we can conclude that out of the given options, pencil is 18.73 mm long.

5 0

3 years ago

Regarding the Lyman series of hydrogen, calculate the frequency of the n = 5 line.

zvonat [6]

The Lyman series can be expressed in the formula 1/λ=RH(1−1/n2) where RH=1.0968×107m−1=13.6eVhc

Where n is a natural number greater than or equal to 2 (i.e. n = 2,3,4,...). Therefore, the lines seen in the image above are the wavelengths corresponding to n=2 on the right, to n=∞on the left (there are infinitely many spectral lines, but they become very dense as they approach to n=∞ (Lyman limit), so only some of the first lines and the last one appear).
The wavelengths (nm) in the Lyman series are all ultraviolet
:2 3 4 5 6 7 8 9 10 11
Wavelength (nm) 121.6 102.6 97.3 95 93.8 93.1 92.6 92.3 92.1 91.9 91.18 (Lyman limit)
In your case for the n=5 line you have to replace "n" in the above formula for 5 and you should get a value of 95 x 10^-9 m for the wavelength. then you have to use the other equation that convert wavelength to frequency.

4 0

2 years ago

Aluminum reacts with sulfur gas to produce aluminum sulfide. a) What is the limiting reactant? What is the excess reagent? b) Ho

Sophie [7]

Answer:

a) Limiting: sulfur. Excess: aluminium.

b) 1.56g Al₂S₃.

c) 0.72g Al

Explanation:

Hello,

In this case, the initial mass of both aluminium and sulfur are missing, therefore, one could assume they are 1.00 g for each one. Thus, by considering the undergoing chemical reaction turns out:

$2Al(s)+3S_2(g)\rightarrow 2Al_2S_3(s)\\$

a) Thus, considering the assumed mass (which could be changed based on the one you are given), the limiting reagent is identified as shown below:

$n_S^{available}=1.00gS_2*\frac{1molS_2}{64gS_2} =0.0156molS_2\\n_S^{consumed\ by \ Al}=1.00gAl*\frac{1molAl}{27gAl}*\frac{3molS_2}{2molAl}=0.0556molS_2$

Thereby, since there 1.00g of aluminium will consume 0.0554 mol of sulfur but there are just 0.0156 mol available, the limiting reagent is sulfur and the excess reagent is aluminium.

b) By stoichiometry, the produced grams of aluminium sulfide are:

$m_{Al_2S_3}=0.0156molS_2*\frac{2molAl_2S_3}{3molS_2} *\frac{150gAl_2S_3}{1molAl_2S_3} =1.56gAl_2S_3$

c) The leftover is computed as follows:

$m_{Al}^{excess}=(0.0556-0.0156)molS_2*\frac{2molAl}{3molS_2}*\frac{27gAl}{1molAl} =0.72 gAl\\$

NOTE: Remember I assumed the quantities, they could change based on those you are given, so the results might be different, but the procedure is quite the same.

Best regards.

7 0

2 years ago