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Sedaia [141]
3 years ago
6

I’ll give brainliest!!!

Chemistry
1 answer:
vodka [1.7K]3 years ago
4 0
Is this prier to a lab you've done?
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How long does it take for a 12.62g sample of ammonia to heat from 209K to 367K if heated at a constant rate of 6.0kj/min? The me
Georgia [21]
First, consider the steps to heat the sample from 209 K to 367K.

1) Heating in liquid state from 209 K to 239.82 K

2) Vaporaizing at 239.82 K

3) Heating in gaseous state from 239.82 K to 367 K.


Second, calculate the amount of heat required for each step.

1) Liquid heating

Ammonia = NH3 => molar mass = 14.0 g/mol + 3*1g/mol = 17g/mol

=> number of moles = 12.62 g / 17 g/mol = 0.742 mol

Heat1 = #moles * heat capacity * ΔT

Heat1 = 0.742 mol * 80.8 J/mol*K * (239.82K - 209K) = 1,847.77 J

2) Vaporization

Heat2 = # moles * H vap

Heat2 = 0.742 mol * 23.33 kJ/mol = 17.31 kJ = 17310 J

3) Vapor heating

Heat3 = #moles * heat capacity * ΔT

Heat3 = 0.742 mol * 35.06 J / (mol*K) * (367K - 239.82K) = 3,308.53 J

Third, add up the heats for every steps:

Total heat = 1,847.77 J + 17,310 J + 3,308.53 J = 22,466.3 J

Fourth, divide the total heat by the heat rate:

Time = 22,466.3 J / (6000.0 J/min) = 3.7 min

Answer: 3.7 min


3 0
3 years ago
The Brahman cattle have good resistance to high temperature, but its meat is poor and tough to eat. The English shorthorn cattle
N76 [4]

Answer:

d

Explanation:

4 0
3 years ago
Read 2 more answers
A Y particle is a what ​
elena-s [515]

Answer:

The answer is <em>G</em><em>a</em><em>m</em><em>m</em><em>a</em><em> </em><em>r</em><em>a</em><em>y</em><em> </em><em>.</em>

(Correct me if I am wrong)

3 0
3 years ago
If the take off velocity of an airplane on a runway is 300 km /hr with an acceleration of 1 m/s2. What is the take off time of t
Rama09 [41]
<h3>Answer:</h3>

83.33 seconds.

<h3>Explanation:</h3>

<u>We are given;</u>

  • Take off velocity as 300 km/hr
  • Acceleration as 1 m/s²

We are required to calculate the take off time of the airplane.

<h3>Step 1: Convert velocity from km/hr to m/s </h3>

We are going to use the conversion factor.

The conversion factor is 3.6 km/hr per m/s

Therefore;

Velocity = 300 km/hr ÷ 3.6 km/hr per m/s

             = 83.33 m/s

<h3>Step 2: Calculate the take off time</h3>

We know that;

v = u + at

where, u is the initial velocity, v the final velocity, a the acceleration and t is time.

But, initial velocity is Zero

Therefore;

83.33 m/s = 1 m/s² × t

Thus;

time = 83.33 m/s ÷ 1 m/s²

       = 83.33 seconds

Therefore, the take off time is 83.33 seconds.

5 0
2 years ago
A geochemist in the field takes a 36.0 mL sample of water from a rock pool lined with crystals of a certain mineral compound X.
NeTakaya

Answer:

solubility of X in water at 17.0 ^{0}\textrm{C} is 0.11 g/mL.

Explanation:

Yes, the solubility of X in water at 17.0 ^{0}\textrm{C} can be calculated using the information given.

Let's assume solubility of X in water at 17.0 ^{0}\textrm{C} is y g/mL

The geochemist ultimately got 3.96 g of crystals of X after evaporating the diluted solution made by diluting the 36.0 mL of stock solution.

So, solubility of X in 1 mL of water = y g

Hence, solubility of X in 36.0 mL of water = 36y g

So, 36y = 3.96

   or, y = \frac{3.96}{36} = 0.11

Hence solubility of X in water at 17.0 ^{0}\textrm{C} is 0.11 g/mL.

3 0
3 years ago
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