Which tool would give the most precise measurement of 15 milliliters of water

A solution is prepared by dissolving 0.26 mol of hydrazoic acid and 0.26 mol of sodium azide in water sufficient to yield 1.00 L

Pavel [41]

Answer:

The pH does not decrease drastically because the HCl reacts with the <u>sodium azide (NaN₃)</u> present in the buffer solution.

Explanation:

The buffer solution is formed by 0.26 moles of the weak acid, hydrazoic acid (HN₃), and by 0.26 moles of sodium azide (NaN₃). The equilibrium reaction of this buffer solution is the following:

HN₃(aq) + H₂O(l) ⇄ N₃⁻(aq) + H₃O⁺(aq)

The pH of this solution is:

$pH = pka + log(\frac{[N_{3}^{-}]}{[HN_{3}]}) = -log(2.5 \cdot 10^{-5}) + log(\frac{0.26 mol/1 L}{0.26 mol/1 L}) = 4.60$

When 0.05 moles of HCl is added to the buffer solution, the following reaction takes place:

H₃O⁺(aq) + N₃⁻(aq) ⇄ HN₃(aq) + H₂O(l)

The number of moles of NaN₃ after the reaction with HCl is:

$\eta_{NaN_{3}} = \eta_{i} - \eta_{HCl} = 0.26 moles - 0.05 moles = 0.21 moles$

Now, the number of moles of HN₃ is:

$\eta_{HN_{3}} = \eta_{i} + \eta_{HCl} = 0.26 moles + 0.05 moles = 0.31 moles$

Then, the pH of the buffer solution after the addition of HCl is:

$pH = pka + log(\frac{[N_{3}^{-}]}{[HN_{3}]}) = -log(2.5 \cdot 10^{-5}) + log(\frac{0.21 mol/V_{T}}{0.31 mol/V_{T}}) = 4.43$

The pH of the buffer solution does not decrease drastically, it is 4.60 before the addition of HCl and 4.43 after the addition of HCl.

Therefore, the pH does not decrease drastically because the HCl reacts with the sodium azide (NaN₃) present in the buffer solution.

I hope it helps you!

6 0

3 years ago

Uranus orbits the sun at more than 60% of the speed of Saturn? True or false

Anestetic [448]

Answer:

the answer is FALSE it is less than 60% not more

3 0

3 years ago

Read 2 more answers

First: Calculate the number of moles of Kool-Aid powder needed to make 100mL of a 0.1M solution.

morpeh [17]

For this question, we can use the molarity formula, which is as follows:

$Molarity= \frac{moles}{Liters}$

We know that we want a 0.1M solution in 100 mL, so we can plug those into the equation to find the number of moles required:

$0.1M= \frac{moles}{0.1L}$
$moles=0.01$

So now we know that the number of moles of the Kool-Aid powder is 0.01. We also know that the chemical formula for this powder is $C_{12} H_{22} O_{11}$ .

We can find the molar weight of this powder by taking the atomic weight from the periodic table and multiply it by the number of atoms in the compound. Let's find the molar weight of the powder:

Find the atomic weight of carbon:

$C=12.01 g$

There are 12 C's, so we multiply by 12:

$12.01g*12=144.12g$

Then we have 22 H's, each having an atomic weight of 1.008g:

$1.008g*22=22.18g$

And finally there are 11 O's, each with an atomic weight of 15.99g:

$15.99g*11=175.89g$

Then we add up all of the weights:

$144.12g+22.18g+175.89g=342.19g$ -> So we know that the molar weight of the Kool-Aid powder is 342.19g.

Now we can find the amount of grams that we need for the desired solution as follows:

$\frac{342.19g}{1mol}* \frac{0.01mol}{1}=3.42g$

So now we know that we need 0.01 moles of the powder, and we need 3.42g of the powder to make the desired solution.

7 0

3 years ago

While investigating greenhouse gases, I measured 0.1875 grams of carbon dioxide in a 500-gram sample of the atmosphere. What is

Harlamova29_29 [7]

Answer:

[CO₂] = 375 ppm

Explanation:

We can determine the concentration in value of ppm

ppm = mass of solute (mg) / mass of solution (kg)

Solution: atmosphere

Solute: CO₂

We convert the mass of CO₂ from g to mg → 0.1875 g . 1000 mg / 1g = 187.5 mg

We convert the mass of the atmosphere from g to kg: 500 g . 1 kg/1000g = 0.500 kg

ppm = 187.5 mg / 0.500kg = 375

6 0

3 years ago

The table provided shows the properties of four elements.

Zielflug [23.3K]

By my guess, It would either be A) 1 because by research and what i’ve learned, Metal is a good conductor for Electricity and Heat. It’s also shiny and bendable, so that would be my best guess.
Hope this helped. :)

4 0

3 years ago