Answer:
Star's
Explanation:
Stars have their own brightness
5 g of potassium oxalate react to produce 0.03 moles of calcium oxalate.
Calcium oxalate (CaC₂O₄) is obtained by the reaction of 5 g of potassium oxalate (K₂C₂O₄).
We can calculate the moles of CaC₂O₄ obtained considering the following relationships.
- The molar mass of K₂C₂O₄ is 184.24 g/mol.
- The mole ratio of K₂C₂O₄ to CaC₂O₄ is 2:1.

5 g of potassium oxalate react to produce 0.03 moles of calcium oxalate.
Learn more: brainly.com/question/15288923
The mass of oxygen collected from the thermal decomposition of potassium chlorate at a temperature of 297 K and 762 mmHg is 0.16 g
<h3>How to determine the mole of oxygen produced </h3>
We'll begin by obtaining the number of mole of oxygen gas produced from the reaction. This can be obtained by using the ideal gas equation as illustrated below:
- Volume (V) = 0.128 L
- Temperature (T) = 297 K
- Pressure (P) = 762 – 22.4 = 739.6 mmHg
- Gas constant (R) = 62.363 mmHg.L/Kmol
- Number of mole (n) =?
PV = nRT
739.6 × 0.128 = n × 62.363 × 297
Divide both sides by 62.363 × 297
n = (739.6 × 0.128) / (62.363 × 297)
n = 0.0051 mole
Thus, the number of mole of oxygen gas produced is 0.0051 mole
<h3>How to determine the mass of oxygen collected</h3>
Haven obtain the number of mole of oxygen gas produced, we can determine the mass of the oxygen produced as follow:'
- Mole = 0.0051 mole
- Molar mass of oxygen gas = 32 g/mole
- Mass of oxygen =?
Mole = mass / molar mass
0.0051 = mass of oxygen / 32
Cross multiply
Mass of oxygen = 0.0051 × 32
Mass of oxygen = 0.16 g
Thus, we can conclude that the mass of oxygen gas collected is 0.16 g
Learn more about ideal gas equation:
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Answer:
To live for God.
Explanation:
We are Gods creation. We are here to live for the Lord, to do love& to help save each other so when time comes we will be forever blessed.
Answer:
0.107 mole of SO2.
Explanation:
1 mole of a gas occupy 22.4 L at standard temperature and pressure (STP).
With the above information, we can simply calculate the number of mole of SO2 that will occupy 2.4 L at STP.
This can be obtained as follow:
22.4 L contains 1 mole of SO2.
Therefore, 2.4 L will contain = 2.4/22.4 = 0.107 mole of SO2.
Therefore, 0.107 mole of SO2 is present in 2.4 L at STP.