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shepuryov [24]
3 years ago
8

Radio waves carry information from

Chemistry
1 answer:
Vlad [161]3 years ago
6 0
The correct answer is B
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Question 8 4 pts What would be the resulting molarity of a solution made by dissolving 31.3 grams of Ca(OH)2 in enough water to
Gennadij [26K]
Answer is: <span>molarity of a solution is 0,401 M.
</span>m(Ca(OH)₂) = 31,3 g.
n(Ca(OH)₂) = m(Ca(OH)₂) ÷ M(Ca(OH)₂).
n(Ca(OH)₂) = 31,3 g ÷ 74 g/mol.
n(Ca(OH)₂) = 0,422 mol.
V(solution) = 1050 mL · 0,001 L/mL = 1,050 L.
c(Ca(OH)₂) = n(Ca(OH)₂) ÷ V(solution).
c(Ca(OH)₂) = 0,422 mol ÷ 1,050 L.
n(Ca(OH)₂) = 0,401 mol/L = 0,401 M.
3 0
4 years ago
Read 2 more answers
ICMP scanning involves in checking for the live systems, which can be done by sending the following ping scan request to a host.
abruzzese [7]

Answer:

ICMP Echo Request

Explanation:

ICMP Echo Request is a form of probe or message sent by a user to a destination system.

5 0
4 years ago
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
What is the charge of a Gamma particle?
Igoryamba

Answer:

+2

Explanation:

I just looked it up, but it says the Alpha Particle charge is +2 so I hope that helps.

3 0
3 years ago
Why are paperclips attracted to magnets?
choli [55]
Iron. Well actually steel, but iron is the closest.
3 0
3 years ago
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