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timurjin [86]
3 years ago
6

This part insulates the reaction chamber from the transfer of heat to or from the surrounding environment. a.b.c.

Chemistry
2 answers:
gizmo_the_mogwai [7]3 years ago
7 0

The correct answers are:

1. This part insulates the reaction chamber from the transfer of heat to or from the surrounding environment. C

2. This part contains a known amount of water, which is placed there before the experiment starts. B

3. This part is where the metal is placed at the beginning of the experiment. B

4. This part is used for measuring the initial and final temperatures of the water. A

Correct on Edg.enuity/E2020

|Huntrw6|

Fiesta28 [93]3 years ago
4 0

Answer:

C

B

B

A

Explanation:

  1. This part insulates the reaction chamber from the transfer of heat to or from the surrounding environment. (C)
  2. This part contains a known amount of water, which is placed there before the experiment starts. (B)
  3. This part is where the metal is placed at the beginning of the experiment. (B)
  4. This part is used for measuring the initial and final temperatures of the water. (A)

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<u>Answer:</u> The longest wavelength of light is 656.5 nm

<u>Explanation:</u>

For the longest wavelength, the transition should be from n to n+1, where: n = lower energy level

To calculate the wavelength of light, we use Rydberg's Equation:

\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )

Where,

\lambda = Wavelength of radiation

R_H = Rydberg's Constant  = 1.096776\times 10^7m^{-1}

n_f = Higher energy level = n_i+1=(2+1)=3

n_i= Lower energy level = 2    (Balmer series)

Putting the values in above equation, we get:

\frac{1}{\lambda }=1.096776\times 10^7m^{-1}\left(\frac{1}{2^2}-\frac{1}{3^2} \right )\\\\\lambda =\frac{1}{1.5233\times 10^6m^{-1}}=6.565\times 10^{-7}m

Converting this into nanometers, we use the conversion factor:

1m=10^9nm

So, 6.565\times 10^{-7}m\times (\frac{10^9nm}{1m})=656.5nm

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I'm stuck on this assignment, +20 PTS and BRAINLIEST for step by step:
dolphi86 [110]

Answer:

2. 2.74 L

3. 488 K

4. 7.47 L

5. 38.6 L

6. 2.85 mol

7. 319 K

8. 3.43 kPa

Explanation:

Generally speaking, you want to convert units to SI units, but in this case, we are working with ratios.  This makes up for using the units that wouldn't appropriate elsewhere.

2.  Use the equation P₁V₁ = P₂V₂.  Solve for V₂.  

(3.05 L)(870 kPa) = (969 kPa)(V₂)  

V₂ = 2.74 L.  

3.  Use the equation V₁/T₁ = V₂/T₂. Solve for T₂.  

(3.32 L)/(360 K) = (4.50 L)/(T₂)  

T₂ = 488 K.

4.  Do the same as above, but for V₂.  

(5.10 L)/(-56°C) = V₂/(-82°C)  

V₂ = 7.47 L

5.  Use the equation V₁/n₁ = V₂/n₂.  Solve for V₂.  

(37.2 L)/(0.750 mol) = (V₂)/(0.778 mol)  

V₂ = 38.6 L

6.  Do the same as above, but for n₂.

(86.0 L)/(2.65 mol) = (92.5 L)(n₂)  

n₂ = 2.85 mol

7.  Use the equation P₁/T₁ = P₂/T₂.  Solve for T₂.  

(3.00 atm)/(390 K) = (2.45 atm)/(T₂)  

T₂ = 319 K

8.  Do the same as above, but for P₂.  

In this specific case, however you will need to convert units.  Since both temperatures don't have the same sign, the ratio won't come out right.  Convert to Kelvin.  Add 273.15 to the temperature in Celsius to convert to Kelvin -12.3°C = 260.85 K  25°C = 298.15 K.

(3.00 kPa)/(260.85 K) = P₂/(298.15 K)

P₂ = 3.43 kPa

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