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klio [65]
3 years ago
8

The convex lens of a magnifying glass is used to focus sunlight, as shown. Which of the lettered parts of the lens diagram repre

sents the focal point?
Part A - the imaginary vertical line that divides the lens in half
Part B - the point at which the light converges on the other side of the lens
Part C - the distance between the optical center and where light converges
Part D - the point in the center of the lens

Chemistry
1 answer:
vesna_86 [32]3 years ago
5 0

This is just a guess, but I do believe it is B

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What pressure is exerted by 932.3 g of CH4 in a 0.560 L steel container at 136.2 K?
Roman55 [17]
Molar mass CH4 = 16.0 g/mol

* number of moles:

932.3 / 16 => 58.26875 moles

T = 136.2 K

V = 0.560 L

P = ?

R = 0.082

Use the clapeyron equation:

P x V = n x R x T

P x 0.560 =  58.26875 x 0.082 x 136.2

P x 0.560 = 650.76

P = 650.76 / 0.560

P = 1162.07 atm
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4 years ago
When an electron of a hydrogen atom travels from n = 4 to n = 2, a photon of energy 4.165 × 10-19 joules is released. Calculate
gulaghasi [49]

Answer:

Explanation:

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4 0
3 years ago
Read 2 more answers
Caed for this question.
OLga [1]

Answer:

0.962 atm.

97.4 kPa.

731 torr.

14.1 psi.

97,434.6 Pa.

Explanation:

Hello.

In this case, given the available factors equaling 1 atm of pressure, each required pressure turns out:

- Atmospheres: 1 atm = 760 mmHg:

p=731mmHg*\frac{1atm}{760mmHg} =0.962atm

- Kilopascals:: 101.3 kPa = 760 mmHg:

p=731mmHg*\frac{101.3kPa}{760mmHg} =97.4kPa

- Torrs: 760 torr = 760 mmHg:

p=731mmHg*\frac{760 torr}{760mmHg} =731 torr

- Pounds per square inch: 14.69 psi = 760 mmHg:

p=731mmHg*\frac{14.69}{760mmHg} =14.1psi

- Pascals: 101300 Pa = 760 mmHg:

p=731mmHg*\frac{101300Pa}{760mmHg} \\\\p=97,434.6Pa

Best regards.

5 0
3 years ago
Assuming that sea water is a 3.5 wt % solution of NaCl in water, calculate its osmotic pressure at 20°C. The density of a 3.5% N
olga nikolaevna [1]

Answer:

π = 14.824 atm

Explanation:

wt % = ( w NaCL / w sea water ) * 100 = 3.5 %

assuming w sea water = 100 g = 0.1 Kg

⇒ w NaCl = 3.5 g

osmotic pressure ( π ):

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∴ T = 20 °C  + 273 = 293 K

∴ C ≡ mol/L

∴ density sea water = 1.03 Kg/L....from literature

⇒ volume sea water = 0.1 Kg * ( L / 1.03 Kg ) = 0.097 L sln

⇒ mol NaCl = 3.5 g NaCL * ( mol NaCL / 58.44 g ) = 0.06 mol

⇒ C NaCl = 0.06 mol / 0.097 L = 0.617 M

⇒ π = 0.617 mol/L * 0.082 atm L / K mol * 293 K

⇒ π = 14.824 atm

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4 years ago
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