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steposvetlana [31]
2 years ago
10

Explain why baking soda, nahco3, is commonly used as an emergency safety remedy for lab spills that are acids or bases. support

your answer by writing an equation for baking soda cast onto: a spill of naoh(aq) a spill of h2so4(aq)
Chemistry
2 answers:
Bond [772]2 years ago
8 0

Answer:

The question is partially wrong. For spills of strong bases, such as sodium hydroxide, it must be neutralized with sulfuric or hydrochloric acid. Otherwise, for acid spills, neutralize them with sodium bicarbonate. The reactions are as follows:

Base spills: H2SO4 + 2NaOH → Na2SO4 + 2H2O

Sulfuric acid reacts with sodium hydroxide forming sodium sulfate and water.

Acid spills: H2SO4 + 2NaHCO3 → Na2SO4 + 2CO2 + 2H2O

Sulfuric acid reacts with sodium bicarbonate forming sodium sulfate, water and carbon dioxide.

Explanation:

For the destruction of strong inorganic bases such as sodium hydroxide (NaOH) they must be neutralized with a dilute solution of sulfuric acid or a dilute solution of hydrochloric acid.

For the destruction of strong inorganic acids such as sulfuric acid, they must be neutralized with an inorganic base such as sodium bicarbonate (NaHCO3)

miskamm [114]2 years ago
7 0
NAHCO3 <span>doesn't help basic spills. It's slightly basic itself. It's good for acid spills though. It's not a very strong base so any leftover NaHCO3 won't be a hazard.
I DONT KNOW about h2so4</span>
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decomposition of hydrogen peroxide produce water and oxygen . calculate the volume of O2 formed from the decomposition of 150 mL
Ugo [173]

Explanation:

2H2O2 => 2H2O + O2

Moles of hydrogen peroxide = 0.150dm³ * (0.02mol/dm³) = 0.003mol .

Moles of oxygen = 0.0015mol.

Volume of oxygen = 0.0015mol * (22.4dm³/mol) = 0.0336dm³.

5 0
2 years ago
Using your understanding about the patterns between vaporization and condensation, explain why the boiling point and the condens
eimsori [14]

Answer:

Vaporization and Condensation When a liquid vaporizes in a closed container, gas molecules cannot escape. As these gas phase molecules move randomly about, they will occasionally collide with the surface of the condensed phase, and in some cases, these collisions will result in the molecules re-entering the condensed phase.

7 0
2 years ago
On a class trip to the coast, you and your lab partner find a rock in the spot marked in red on the diagram. You show your teach
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8 0
3 years ago
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What is the pH of a solution made by mixing 200 mL of 0.025M HCl and 150 mL of 0.050 M HCl?
seraphim [82]

Answer:

The answer to your question is pH = 1.45  

Explanation:

Data

pH = ?

Volume 1 = 200 ml

[HCl] 1 = 0.025 M

Volume 2 = 150 ml

[HCl] 2 = 0.050 M

Process

1.- Calculate the number of moles of each solution

Solution 1

                Molarity = moles / volume

-Solve for moles

                moles = 0.025 x 0.2

result

               moles = 0.005

Solution 2

               moles = 0.050 x 0.15

-result

                moles = 0.0075

2.- Sum up the number of moles

Total moles = 0.005 + 0.0075

                   = 0.0125

3.- Sum up the volume

total volume = 200 + 150

                     350 ml or 0.35 l

4.- Calculate the final concentration

Molarity = 0.0125 / 0.35

              = 0.0357

5.- Calculate the pH

pH = -log [H⁺]

-Substitution

pH = -log[0.0357]

-Result

pH = 1.45    

8 0
3 years ago
Determine the mass of CaCO3 required to produce 40.0 mL CO2 at STP. Hint use molar volume of an ideal gas (22.4 L)
cupoosta [38]

Answer:

m_{CaCO_3}=0.179gCaCO_3

Explanation:

Hello,

In this case, since the undergoing chemical reaction is:

CaCO_3(s)\rightarrow CaO(s)+CO_2(g)

The corresponding moles of carbon dioxide occupying 40.0 mL (0.0400 L) are computed by using the ideal gas equation at 273.15 K and 1.00 atm (STP) as follows:

PV=nRT\\\\n=\frac{PV}{RT}=\frac{1.00 atm*0.0400L}{0.082\frac{atm*L}{mol*K}*273.15 K})=1.79x10^{-3} mol CO_2

Then, since the mole ratio between carbon dioxide and calcium carbonate is 1:1 and the molar mass of the reactant is 100 g/mol, the mass that yields such volume turns out:

m_{CaCO_3}=1.79x10^{-3}molCO_2*\frac{1molCaCO_3}{1molCO_2} *\frac{100g CaCO_3}{1molCaCO_3}\\ \\m_{CaCO_3}=0.179gCaCO_3

Regards.

3 0
3 years ago
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