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polet [3.4K]
3 years ago
5

What is the amplitude of the above wave?

Chemistry
1 answer:
Alinara [238K]3 years ago
8 0

Answer:Amplitude in a longitudinal wave

Explanation:Wave amplitude of a longitudinal wave is the distance between particles of the medium where it is compressed by the wave. Wave amplitude is determined by the energy of the disturbance that causes the wave. A wave caused by a disturbance with more energy has greater amplitude.

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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
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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
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