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boyakko [2]
3 years ago
8

Gastric acid pH can range from 1 to 4, and most of the acid is HCl.

Chemistry
1 answer:
erma4kov [3.2K]3 years ago
5 0

Given the molarity of the solution is3.03*10^{-2}.

The volume of the solution is 10.9/1000 L

Number of moles of HCl = Molarity * volume of the solution in L

Number of moles of HCl =3.03*10^{-2}*\frac{1000}{10.9}

Number of moles of HCl = 2.7 moles.

Therefore number of moles of HCl in 10.9 ml of the stomach acid is 2.7 moles


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Which of the following best helps to account for the fact that the F- ion is smaller than the O2- ion?
katrin [286]

Answer: Option (B) is the correct answer.

Explanation:

Both oxygen and fluorine are period 2 elements and when we move across a period then there occurs a decrease in atomic size of the atoms. Hence, the atomic radius of a neutral fluorine atom is smaller than a neutral oxygen atom.

Moreover, atomic number of fluorine is 9 and it has higher nuclear charge due to which it will cause more attraction of electrons. As a result, size of a fluorine ion will be smaller.  

On the other hand, size of oxygen atom is larger and has small nuclear charge due to which attraction of electrons by its nucleus will not be strong enough. Hence, the size of O^{2-} will be larger.

Thus, we can conclude that the statement F^{-} has a larger nuclear charge than O^{2-} has, is correct for the fact that the F^{-} ion is smaller than the O^{2-} ion.

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The rate of a reaction is measured by how fast a reactant is used up or how fast a product is formed.
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Read 2 more answers
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
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