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lawyer [7]
3 years ago
10

What happens when you swallow acid

Chemistry
2 answers:
aleksklad [387]3 years ago
4 0

Some of the symptoms from swallowing hydrochloric acid:

<span><span>
</span></span><span><span>-Abdominal pain 
</span>-Breathing difficulty due to swelling of throat
-Chest pain
-Drooling
-Fever
-Mouth pain
-Rapid drop in blood pressure
-Throat pain <span>
-Vomiting blood</span></span>
Dmitrij [34]3 years ago
3 0
It would depend on the type of acid.. If you were to swallow <span>Hydrochloric </span>Acid you would likely have very bad pain in your abdominal, difficulty breathing, bad chest pain, fever, bad pain in your mouth, pain in your throat, and it's also possible to be vomiting blood.
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We can preserve natural resources by
DedPeter [7]
The answer is c, relying on renewable energy sources
3 0
3 years ago
Determine the number of molecules in a 100. gram sample of CCl4
enyata [817]
100. g CCl4* (1 mol CCl4/ 153.8 g CCl4)* (6.02*10^23 CCl4 molecules/ 1 mol CCl4)= 3.91*10^23 CCl4 molecules.
(Note that the units cancel out so you get the answer)

Hope this helps~
8 0
3 years ago
A sample of argon gas has a volume of 73 mL at a pressure of 1.20 atm and a temperature of 112 degrees Celsius. What is the fina
NeX [460]

Answer:

V₂ → 106.6 mL

Explanation:

We apply the Ideal Gases Law to solve the problem. For the two situations:

P . V = n . R . T

Moles are still the same so → P. V / R. T = n

As R is a constant, the formula to solve this is: P . V / T

P₁ . V₁ / T₁ = P₂ .V₂ / T₂   Let's replace data:

(1.20 atm . 73mL) / 112°C = (0.55 atm . V₂) / 75°C

((87.6 mL.atm) / 112°C) . 75°C = 0.55 atm . V₂

58.66 mL.atm = 0.55 atm . V₂

58.66 mL.atm / 0.55 atm = V₂ → 106.6 mL

3 0
3 years ago
1.33 dm3 of water at 70°C are saturated by 2.25
astraxan [27]

Given that 4.50 dm³ of Pb(NO₃)₂ is cooled from 70 °C to 18 °C, the

amount amount of solute that will be deposited is 1,927.413 grams.

<h3>How can the amount of solute deposited be found?</h3>

The volume of water 1.33 dm³ of water 70 °C.

The number of moles of Pb(NO₃)₂ that saturates 1.33 dm³ of water at 70 °C  = 2.25 moles

At 18 °C, the number of moles of Pb(NO₃)₂ that saturates 1.33 dm³ of water = 0.53 moles

Therefore;

Number of moles of Pb(NO₃)₂ in 4.50 dm³ at 70 °C is therefore;

1.33 dm³ contains 2.25 moles.

Number \ of \ moles \ in \ 4.50 \ dm^3 = \dfrac{2.25}{1.33} \times 4.50 \approx \mathbf{7.613 \, moles}

Number of moles of Pb(NO₃)₂ in 4.50 dm³ at 70 °C ≈ 7.613 moles

Number of moles of Pb(NO₃)₂ in 4.50 dm³ at 18 °C is therefore;

1.33 dm³ contains 0.53 moles

Number \ of \ moles \ in \ 4.50 \ dm^3 = \dfrac{0.53}{1.33} \times 4.50 \approx \mathbf{1.79 \, moles}

Number of moles of Pb(NO₃)₂ in 4.50 dm³ at 18 °C ≈ 1.79 moles

The number of moles that precipitate out = The amount of solute deposited

Which gives;

Amount of solute deposited = 7.613 moles - 1.79 moles = 5.823 moles

The molar mass of Pb(NO₃)₂ = 207 g + 2 × (14 g + 3 × 16 g) = 331 g

The molar mass of Pb(NO₃)₂ = 331 g/mol

The amount of solute deposited = Number of moles × Molar mass

Which gives;

The amount of solute deposited = 5.823 moles × 331 g/mol =<u> 1,927.413 g </u>

Learn more about saturated solutions here:

brainly.com/question/2624685

5 0
2 years ago
Which state of matter undergoes changes in volume most easily
AlladinOne [14]
Solid should be the anwser
3 0
3 years ago
Read 2 more answers
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