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Taya2010 [7]
3 years ago
7

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

Chemistry
1 answer:
olga2289 [7]3 years ago
6 0

Answer:

  • <u>609g </u>

<u></u>

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>

<em></em>

<h2>Solution</h2>

<em></em>

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

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There are ________ unpaired electrons in the Lewis symbol for a nitride ion.
Anna71 [15]

Answer:

There are no unpaired electrons.

Explanation:

There are  no unpaired electrons in the Lewis symbol for a nitride ion({ N }^{ 3- }).The nitride ion has a charge of -3. The negative charge on the Nitride ion indicates a gain in electrons . Nitrogen has 5 valence electrons that is the number of electrons that are in its outer shell .The total number of electrons that the nitride ion has is equal to 5+3 = 8 electrons . Electrons usually appear in pairs and obey the octet rule therefore the nitride ion has four electron pairs no unpaired electrons.

5 0
3 years ago
50.0 mL of an HNO^3 solution were titrated with 36.90 mL of a 0.100 M LiOH solution to reach the equivalence point. What is the
NISA [10]

Answer:

0.0738 M

Explanation:

HNO3 +LiOH = LiNO3 + H2O

Number of moles HNO3 = number of moles LiOH

M(HNO3)*V(HNO3) = M(LiOH)*M(LiOH)

M(HNO3)*50.0mL = 0.100M*36.90 mL

M(HNO3) = 0.100*36.90/50.0 M = 0.0738 M

6 0
3 years ago
how many kilograms of a 35% m/m sodium chlorate solution is needed to react completely with 0.29 l of a 22% m/v aluminum nitrate
Stolb23 [73]

Answer:- 0.273 kg

Solution:- A double replacement reaction takes place. The balanced equation is:

3NaClO_3+Al(NO_3)_3\rightarrow 3NaNO_3+Al(ClO_3)_3

We have 0.29 L of 22% m/v aluminum nitrate solution. m/s stands for mass by volume. 22% m/v aluminium nitrate solution means 22 g of it are present in 100 mL solution. With this information, we can calculate the grams of aluminum nitrate present in 0.29 L.

0.29L(\frac{1000mL}{1L})(\frac{22g}{100mL})

= 63.8 g aluminum nitrate

From balanced equation, there is 1:3 mol ratio between aluminum nitrate and sodium chlorate. We will convert grams of aluminum nitrate to moles and then on multiplying it by mol ratio we get the moles of sodium chlorate that could further be converted to grams.

We need molar masses for the calculations, Molar mass of sodium chlorate is 106.44 gram per mole and molar mass of aluminum nitrate is 212.99 gram per mole.

63.8gAl(NO_3)_3(\frac{1mol}{212.99g})(\frac{3molNaClO_3}{1molAl(NO_3)_3})(\frac{106.44g}{1mol})

= 95.7gNaClO_3

sodium chlorate solution is 35% m/m. This means 35 g of sodium chlorate are present in 100 g solution. From here, we can calculate the mass of the solution that will contain 95.7 g of sodium chlorate  and then the grams are converted to kg.

95.7gNaClO_3(\frac{100gSolution}{35gNaClO_3})(\frac{1kg}{1000g})

= 0.273 kg

So, 0.273 kg of 35% m/m sodium chlorate solution are required.

7 0
3 years ago
Sodium carbonate (Na2CO3) is used to neutralize the sulfuric acid spill. How many kilograms of sodium carbonate must be added to
valina [46]

5.451 X 10³ kg of sodium carbonate must be added to neutralize 5.04×103 kg of sulfuric acid solution.

<u>Explanation</u>:

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The atomic mass is determined by the number of protons and neutrons in a atoms for example oxygen has eight protons and neutrons which gives oxygen and atomic mass of 16
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