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faust18 [17]
3 years ago
12

What pressure is required to compress 196.0 liters of air at 1.00 atmosphere into a cylinder whose volume is 26.0 liters?

Chemistry
1 answer:
Leto [7]3 years ago
3 0
7.54 atm kf im correct
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For the reaction shown, calculate how many grams of oxygen form when each quantity of reactant completely reacts.
Nikolay [14]

Explanation:

Molar mass of KClO_{3} = 39.1 + 35.5 + 3(16.0) = 122.6 g

Molar mass of KCl = 39.1 + 35.5 = 74.6 g

Molar mass of O_{2} = 32.0 g

According to the equation, 2 moles of KClO_{3} reacts to give 3 moles of oxygen.

Therefore, 2 (122.6) = 245.2 g of KClO_{3} will give 3 (32.0) = 96.0 g of oxygen. Thus, 245.2 g of KClO_{3} gives 96.0 g of oxygen.

(a) Calculate the amount of oxygen given by 2.72 g of KClO_{3} as follows.

       \frac {96g}{245.2g} \times 2.72 g = 1.06 g of O_{2}


(b) Calculate the amount of oxygen given by 0.361 g of KClO_{3} as follows.

    \frac {96g}{245.2g} \times 0.361 g = 0.141 g of O_{2}


c) Calculate the amount of oxygen given by 83.6 kg KClO_{3} as follows.  

    \frac {96g}{245.2g} \times 83.6 g = 32.731 kg of O_{2}

Convert kg into grams as follows.

    \frac{32.731 kg}{1 kg} \times 1000 g = 32731 g of O_{2}


(d) Calculate the amount of oxygen given by 22.5 mg of KClO_{3} as follows.  

    \frac {96g}{245.2g} \times 22.5 mg = 8.79 mg

Convert mg into grams as follows.

  \frac{8.79 mg}{1 mg}\times 10^{-3} g = 8.79 \times 10^{-3} g of O_{2}

8 0
3 years ago
The _________ indicates that 70% of the Sun's radiation is contained on Earth and 30% is reflected back into space.
vichka [17]

i think it would be  Earth's albedo hope this helps

4 0
4 years ago
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Upon balancing the equation what is the smallest possible integers that goes in front O₂? HBr + O₂ ➞ H₂O + Br₂ *
Leokris [45]

Answer:

1

Explanation:

4 HBr + O2 → 2H 20 + 2Br 2

...............

7 0
3 years ago
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How do you solve this ??
kotegsom [21]
Answer : Option A) 2.00 eV

Explanation : The conversion of J to eV is done with the following formula;

E_{eV} = E_{J} X (6.241 X 10^{18})

Here, we have the value of particle in terms of Joules which is 3.2 X 10^{-19}

So, on substituting we get,
E_{eV} = 3.2 X 10^{-19}  X  (6.241 X 10^{18} )


E_{eV} = 1.99 eV so, it can be rounded off to 2.00 eV.
3 0
3 years ago
Using the periodic table, how would you find elements with chemical properties similar to helium?
rosijanka [135]
<span>The state of the helium in its natural form is gaseous and is a chemical element of colorless aspect and belongs to the group of noble gases. The atomic number of helium is 2. The chemical symbol of helium is He. For the following we focus on those elements and relate it with similar chemical properties. Then we find that; Neon, Hydrogen, Boron and Carbon are related to helium, either by proximity in their atomic number or period or by their group.</span>
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4 years ago
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