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Bad White [126]
3 years ago
15

What are the units of density

Chemistry
1 answer:
Nataliya [291]3 years ago
6 0

Answer:

Kg/ ml

Explanation:

Hope I help you

thank you

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Forming a hypothesis is accomplished through___ reasoning
mixer [17]
Scientific would be the word to fill in the blank
5 0
3 years ago
21 mL It required 42.35 mL of H2SO4 to neutralize 21.17 mL of 0.5000 M NaOH. Calculate the concentration of H2SO4
bija089 [108]
The balanced equation for the above reaction is 
2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O
stoichiometry of NaOH to H₂SO₄ is 2:1
number of NaOH moles required-0.5000 M / 1000 mL/L x 21.17 mL = 0.010585 mol
According to stoichiometry, acid moles required are 1/2 of the base moles reacted
Therefore number of H₂SO₄ moles reacted - 0.010585 /2  mol
Number of moles in 42.35 mL of H₂SO₄ - 0.010585 /2 mol
Therefore in 1 L solution - (0.010585) /2 / 42.35 mL x 1000 mL/L = 0.125 M
Molarity of H₂SO₄ - 0.125 M
5 0
3 years ago
Balance chemical reaction of oxalic acid with conc. sulphuric acid​
seropon [69]

Answer:

Explanation: since oxalic acid is a weak acid it wont provide a strong acidic medium. So in order to provide a strong acidic medium dilute sulphuric acid is added.

3 0
2 years ago
Which option draws the correct conclusion from the following case study?
trasher [3.6K]

Answer:

I believe the answer The case study was influenced by bias, and led to incorrect conclusions being drawn. plz correct me if I am wrong

Explanation:

4 0
3 years ago
Read 2 more answers
For the reaction
s2008m [1.1K]

Answer:

0.558mole of SO₃

Explanation:

Given parameters:

Molar mass of SO₃ = 80.0632g/mol

Mass of S = 17.9g

Molar mass of S = 32.065g/mol

Number of moles of O₂ = 0.157mole

Molar mass of O₂ = 31.9988g/mol

Unknown:

Maximum amount of SO₃

Solution

  We need to write the proper reaction equation.

           2S + 3O₂ → 2SO₃

We should bear in mind that the extent of this reaction relies on the reactant that is in short supply i.e limiting reagent. Here the limiting reagent is the Sulfur, S. The oxygen gas would be in excess since it is readily availbale.

So we simply compare the molar relationship between sulfur and product formed to solve the problem:

First, find the number of moles of Sulfur, S:

   Number of moles of S = \frac{mass }{molar mass}

   Number of moles of S =  \frac{17.9 }{32.065} = 0.558mole

Now to find the maximum amount of SO₃ formed, compare the moles of reactant to the product:

       2 mole of Sulfur produced 2 mole of SO₃

   Therefore; 0.558mole of sulfur will produce 0.558mole of SO₃

5 0
3 years ago
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