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Ksivusya [100]
2 years ago
10

What is the volume of 1.60 grams of O2 gas at STP? (5 points)

Chemistry
2 answers:
tiny-mole [99]2 years ago
5 0

Answer:

  • <u><em>1.12 liters</em></u>

Explanation:

<u>Calculating number of moles</u>

  • Molar mass of O₂ = 32 g
  • n = Given weight / Molar mass
  • n = 1.6/32
  • n = 0.05 moles

<u>At STP</u>

  • One mole of O₂ occupies 22.4 L
  • Therefore, 0.05 moles will occupy :
  • 22.4 L x 0.05 = <u><em>1.12 L</em></u>
VladimirAG [237]2 years ago
4 0

1mol of any gas at STP contains 22.4L

moles of O_2

  • Given mass/molar mass
  • 1.6/32
  • 0.05mol

Now

Volume of O_2:-

  • 0.05(22.4)
  • 1.120L

Option D

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2 years ago
a 125 g chunk of aluminum at 182 degrees Celsius was added to a bucket filled with 365 g of water at 22.0 degrees Celsius. Ignor
Diano4ka-milaya [45]
<h3>Answer:</h3>

32.98°C

<h3>Explanation:</h3>

We are given the following;

Mass of Aluminium as 125 g

Initial temperature of Aluminium as 182°C

Mass of water as 265 g

Initial temperature of water as 22°C

We are required to calculate the final temperature of the two compounds;

First, we need to know the specific heat capacity of each;

Specific heat capacity of Aluminium is 0.9 J/g°C

Specific heat capacity of water is 4.184 J/g°C

<h3>Step 1: Calculate the Quantity of heat gained by water.</h3>

Assuming the final temperature is X°C

we know, Q = mcΔT

Change in temperature, ΔT = (X-22)°C

therefore;

Q = 365 g × 4.184 J/g°C × (X-22)°C

    = (1527.16X-33,597.52) Joules

<h3>Step 2: Calculate the quantity of heat released by Aluminium </h3>

Using the final temperature, X°C

Change in temperature, ΔT = -(X°- 182°)C (negative because heat was lost)

Therefore;

Q = 125 g × 0.90 J/g°C × (182°-X°)C

  = (20,475- 112.5X) Joules

<h3>Step 3: Calculating the final temperature</h3>

We need to know that the heat released by aluminium is equal to heat absorbed by water.

Therefore;

(20,475- 112.5X) Joules = (1527.16X-33,597.52) Joules

Combining the like terms;

1639.66X = 54072.52

             X = 32.978°C

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Therefore, the final temperature of the two compounds will be 32.98°C

7 0
3 years ago
A student is instructed to make 1 L of a 2.0 M solution of CaCl2 using dry salt. How should he do this?
Pachacha [2.7K]
<span>The student should follow following steps to make 1 L of </span>2.0 M CaCl₂.<span>
<span>
1. First he should calculate the number of moles of 2.0 M CaCl</span></span>₂ in 1 L solution.<span>

</span>Molarity of the solution = 2.0 M<span>
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Molarity = number of moles / volume of the solution

Hence, number of moles in 1 L = 2 mol

2. Find out the mass of dry CaCl</span>₂ in 2 moles.<span>

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3. Weigh the mass accurately 

4. Then take a cleaned and dry1 L volumetric flask and place a funnel top of it. Then carefully add the salt into the volumetric flask and finally wash the funnel and watch glass with de-ionized water. That water also should be added into the volumetric flask.

5. Then add some de-ionized water into the volumetric flask and swirl well until all salt are dissolved.

<span>6. Then top up to mark of the volumetric flask carefully. 
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</span>
7. As the final step prepared solution should be labelled.
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