Answer:
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Explanation:
The translated question is:
<em>What maximum amount of grams of potassium nitrate (V) can be dissolved in 300g of water at 90 °C</em>
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<h2>Solution</h2>
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To answer the question you need to consultate the solubiity information for potassium nitrate (V), KNO₃.
The attached table contains the solutibility table for KNO₃ at different temperatures.
At 90ºC it is 203g / 100g water.
Then, to calculate the <em>maximum amount of grams of potassium nitrate (V) that can be dissolved in 300g of water at 90 °C</em>, just multiply by the amount of water:
- 203g / 100g water × 300 g water = 609g ← answer
The answer to this question is Released
This is the law of energy conservation that being done in order to reduce the total consumption of energy by using less energy.
In most cases, the conservation requires either by using the intial energy more efficiently or by reducing the total result of the energy formed.
Answer:
the second question should be along the staircase
Answer: 8.59 L of oxygen gas are needed to produce 100 kJ of energy at STP
Explanation:
According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number 
of particles.
Standard condition of temperature (STP) is 273 K and atmospheric pressure is 1 atmosphere respectively.
1 mole of every gas occupy volume at STP = 22.4 L
The balanced chemical reaction is:
3909.9 kJ of of energy is produced by = 
100 kJ of oxygen gas are needed to produce = 
Answer:
9.91 mL
Explanation:
Using the combined gas law equation as follows;
P1V1/T1 = P2V2/T2
Where;
P1 = initial pressure (torr)
P2 = final pressure (torr)
V1 = initial volume (mL)
V2 = final volume (mL)
T1 = initial temperature (K)
T2 = final temperature (K)
According to the information provided in this question;
V1 = 15.0mL
V2 = ?
P1 = 760 torr
P2 = 1252 torr
T1 = 10°C = 10 + 273 = 283K
T2 = 35°C = 35 + 273 = 308K
Using P1V1/T1 = P2V2/T2
760 × 15/283 = 1252 × V2/308
11400/283 = 1252V2/308
Cross multiply
11400 × 308 = 283 × 1252V2
3511200 = 354316V2
V2 = 3511200 ÷ 354316
V2 = 9.91 mL
