All categories

2 years ago

Jaką maksymalną ilość gramów azotanu (V) potasu można rozpuscic w 300g wody w temperaturze 90 C

Chemistry

1 answer:

olga2289 [7]2 years ago

6 0

Answer:

609g

Explanation:

The translated question is:

What maximum amount of grams of potassium nitrate (V) can be dissolved in 300g of water at 90 °C

<h2>Solution</h2>

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 maximum amount of grams of potassium nitrate (V) that can be dissolved in 300g of water at 90 °C, just multiply by the amount of water:

203g / 100g water × 300 g water = 609g ← answer

You might be interested in

Describe how to find the area of a triangle (do not just give the formula)

NemiM [27]

When you have the bottom measurement of the triangle (base) and the height of the triangle in the center, you multiply them together and then divide by 2.

~Do you need more explaining~?

8 0

2 years ago

In an experiment, 4.14 g of phosphorus combined with chlorine to produce 27.8 g of a white solid compound. what is the empirical

Umnica [9.8K]

Grams of Phosphorus = 4.14 grams
Grams of white compound = 27.8 grams
Grams of Chlorine would be = 27.8 - 4.14 = 23.66 grams
Calculating moles which would be grams / molar mass
Molar mass of P = 30.97 grams / moles; Molar mass of Cl = 35.45 grams / moles
Moles of Phosphorus = 4.14 grams / 30.97 grams / moles = 0.1337 moles
Moles of Chlorine = 23.66 grams / 35.45 grams / moles = 0.6674 moles
Calculating the ratios by dividing with the small entity
P = 0.1337 moles / 0.1337 moles = 1
Cl = 0.6674 moles / 0.1337 moles = 5
So the empirical formula would be PCl5

3 0

2 years ago

Hiii please help me to balance the chemical equation:

Fed [463]

Answer:

2Na + 2H2O = 2NaOH + H2

Explanation:

hope this helps

4 0

2 years ago

Which molecule has the same shape and hybridization as methane

Korvikt [17]

Hi, the answer is CF2Cl2 :)

6 0

3 years ago

Find the number of molecules in 27 grams of water

Aleks04 [339]

In this question we have given the gram of water and we know that 1 mole of water = 18 gram of water and 27 g of water contain 1.5 g of water 27 / 18 = 1.5 g

As we know that avogandro'S no is equal 6.022*1023

1.5g * 6.022*1023 = 9.0 * 1023 molecules present in each 27 g of water.

I hope you will understand better now if you like then comment below and tell me. Best of luck.

5 0

2 years ago