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Temka [501]
3 years ago
6

14 point. Please help need this ASAP.

Chemistry
1 answer:
mezya [45]3 years ago
7 0
What are you asking
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Warm air and water both tend to rise while cooler air and water sink. When different parts of the oceans are heated unevenly, th
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Answer:

a

Explanation:

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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
A solution of hydrocloric acid has a molarity of 2.25 M HCl. If a reaction requires 5.80 g of HCI, what
Natali [406]

Answer:

0.071L

Explanation:

From the question given, we obtained the following data:

Molarity of HCl = 2.25 M

Mass of HCl = 5.80g

Molar Mass of HCl = 36.45g/mol

Number of mole of HCl =?

Number of mole = Mass /Molar Mass

Number of mole of HCl = 5.8/36.45 = 0.159mole

Now, we can obtain the volume required as follows:

Molarity = mole /Volume

Volume = mole /Molarity

Volume = 0.159mole/ 2.25

Volume = 0.071L

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