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s2008m [1.1K]
3 years ago
14

Sodium carbonate (na2co3) is used to neutralize the sulfuric acid spill. how many kilograms of sodium carbonate must be added to

neutralize 6.05×103 kg of sulfuric acid solution?
Chemistry
1 answer:
gayaneshka [121]3 years ago
8 0
THe balanced chemical reaction would be:
Na2CO3 + H2SO4 = Na2SO4 + H2O + CO2

We first convert the mass of sulfuric acid solution to moles by the molar mass. Then, we relate H2SO4 with Na2CO3 from the reaction. We do as follows:

<span>6.05×10^3 kg H2SO4 ( 1 kmol / 98.08 kg) ( 1 kmol Na2CO3 / 1 kmol H2SO4 ) ( 105.99 kg / kmol ) = 6537.92 kg Na2CO3 needed</span>
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3 0
3 years ago
Read 2 more answers
3. What is the mass % of a solution that contains 36g KCl in 475g of water?
stira [4]

Answer:

%KCl = 7.05%

%Water = 92.95%

Explanation:

Step 1: Given data

  • Mass of KCl (solute): 36 g
  • Mass of water (solvent): 475 g

Step 2: Calculate the mass of the solution

The mass of the solution is equal to the sum of the masses of the solute and the solvent.

m = 36 g + 475 g = 511 g

Step 3: Calculate the mass percentage of the solution

We will use the following expression.

%Component = mComponent/mSolution × 100%

%KCl = 36 g/511 g × 100% = 7.05%

%Water = 475 g/511 g × 100% = 92.95%

7 0
3 years ago
How many milligrams of AgNO3 is required to completely react with 81.5 mg LiOH?
BartSMP [9]

Answer:

m_{AgNO_3}=577.6mg

Explanation:

Hello there!

In this case, according to the given chemical reaction, it is first necessary to compute the moles of reacting LiOH given its molar mass:

n_{LiOH}=81.5mg*\frac{1g}{1000mg}*\frac{1mol}{23.95g}  =0.0034molLiOH

Thus, since there is a 1:1 mole ratio between lithium hydroxide and silver nitrate (169.87 g/mol) the resulting milligrams turn out to be:

m_{ AgNO_3}=0.0034molLiOH*\frac{1molAgNO_3}{1molLiOH} *\frac{169.87gAgNO_3}{1molAgNO_3} *\frac{1000gAgNO_3}{1gAgNO_3} \\\\m_{AgNO_3}=577.6mg

Best regards!

7 0
3 years ago
HELP: MODELING MOLECULES!!!!!!!!!!!!!!!!
garik1379 [7]
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(see the figure)

and 
</span><span>So we made 6 bonds, that means we used up 12 electrons, so if you're counting (AND YOU SHOULD BE!) you have 36 electrons or simply 18 electron pairs left to place. Now let's give chlorine a neutral charge.</span>

8 0
3 years ago
Question 3 of 6
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Answer:

c,d,e

Explanation:

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