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GarryVolchara [31]
3 years ago
12

What is the state of matter of bromine ?

Chemistry
1 answer:
SSSSS [86.1K]3 years ago
6 0
Bromine is in gaseous state
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What is the pressure of 4 moles of helium in a 50 L tank at 308k using PV=nRT
Ilya [14]

Hey there!:

p = ??

n = 4 moles

V = 50 L

T = 308 K

R =  0.082 atm

p*V = n * R * T

p * 50 = 4 * 0.082 * 308

p*50 = 101.024

p = 101.024 / 50

p = 2.020 atm


Hope that helps!


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why is nitrous oxide sedation especially useful for control of soft tissue discomfort during dental hygiene procedure
Stolb23 [73]

Answer:

Because it has a quickly sedative effect and it has antimicrobial effect.

Explanation:

This gas is useful because is used to trait pain, reduce anxiety and promote relaxation, slow down the body reaction, so the dentist can use it to calm down the patients.

If the patient present some injuries, this gas can help in wound healing.

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How many protons does the neutral atom pictured have?<br> A) 8<br> B) 18<br> C) 2<br> D) 20
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b 18

Explanation:

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Theory is often used as a synonym for idea in everyday life.
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How much thermal energy is added to 10.0 g of ice at −20.0°C to convert it to water vapor at 120.0°C?
Sonbull [250]

Answer:

7479 cal.

31262.2 joules

Explanation:

This is a calorimetry problem where water in its three states changes from ice to vapor.

We must use, the calorimetry formula and the formula for latent heat.

Q = m . C . ΔT

Q = Clat . m

First of all, let's determine the heat for ice, before it melts.

10 g . 0.5 cal/g°C ( 0° - (-20°C) = 100 cal

Now, the ice has melted.

Q = Clat heat of fusion . 10 g

Q = 79.7 cal/g . 10 g → 797 cal

We have water  at 0°, so this water has to receive heat until it becomes vapor. Let's determine that heat.

Q = m . C . ΔT

Q = 10 g . 1 cal/g°C (100°C - 0°C) → 1000 cal

Water is ready now, to become vapor so let's determine the heat.

Q = Clat heat of vaporization . m

Q = 539.4 cal/g . 10 g → 5394 cal

Finally we have vapor water, so let's determine the heat gained when this vapor changes the T° from 100°C to 120°

Q = m . C . ΔT

Q = 10 g . 0.470 cal/g°C . (120°C - 100°C) → 94 cal

Now, we have to sum all the heat that was added in all the process.

100 cal + 797 cal + 1000 cal + 5394 cal + 94 cal =7479 cal.

We can convert this unit to joules, which is more acceptable for energy terms.

1 cal is 4.18 Joules.

Then, 7479 cal are (7479 . 4.18) = 31262.2 joules

6 0
3 years ago
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