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slavikrds [6]
2 years ago
8

How many moles of sodium bicarbonate is needed to neutralize 0.8 ml of sulphuric acid?

Chemistry
1 answer:
svet-max [94.6K]2 years ago
6 0

Answer:

n NaHCO3 = 9.6 E-3 mol

Explanation:

balanced reaction:

  • 2 NaHCO3(s) + H2SO4(ac) ↔ Na2SO4(ac) + 2 CO2(g) + 2 H2O(l)
  • assuming a concentration of H2SO4 6M....normally worked in the lab

⇒ n H2SO4 = 8 E-4 L * 6 mol/L = 4.8 E-3 mol H2SO4

according to balanced reaction, we have that for every mol of H2SO4 there are two mol of NaHCO3 ( sodium bicarbonate)

⇒ mol NaHCO3 = 4.8 E-3 mol H2SO4 * ( 2 mol NaHCO3 / mol H2SO4 )

⇒ ,mol NaHCO3 = 9.6 E-3 mol

So 9.6 E-3 mol NaHCO3,  are the minimun moles necessary to neutralize the acid.

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An object has a mass of 5.52 g and a volume of 12 mL. What is the density of the object?
Rudik [331]

Answer:

<h2>Density = 0.46 g/mL</h2>

Explanation:

Density of a substance can be found by using the formula

<h3>Density =  \frac{mass}{volume}</h3>

From the question

mass = 5.52 g

volume = 12 mL

Substitute the values into the above formula and solve for the Density

That's

<h3>Density =  \frac{5.52}{12}</h3>

We have the final answer as

<h3>Density = 0.46 g/mL</h3>

Hope this helps you

4 0
3 years ago
If a sample of radioactive isotopes takes 60 minutes to decay from 200 grams to 50 grams, what is the half-life of the isotope
qaws [65]

Answer:

<u><em>I HOPE IT WILL BE YOUR ANSWER:</em></u>

Explanation:

As given

60 min = 50 gm        (1)

then we know half-life mean half amount decay time

so we can write as the half of 200 is 100 gm hence

T 1/2 = 100               (2)

solving these two equation by cross multiplication we will get

T 1/2 = 120 min

<em><u>THANKS FOR ASKING QUESTION</u></em>

6 0
3 years ago
PLEASE SHOW WORK and inlcude units on each number
Sedbober [7]

 The  limiting  reagent  is   <u>H₂SO₄</u>

   <u><em>calculation</em></u>

<u><em> </em></u>Step 1 :write the equation for reaction

2 NaOH + H₂SO₄  → Na₂SO₄   + 2 H₂O

Step  2: use the mole ratio to determine the  moles of product  produced from each reactant

that is   from  equation above,

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= 10.0 moles x 1/2 =  5.0 moles

H₂SO₄ :Na₂SO₄  is 1:1 therefore the moles  of Na₂SO₄  is also = 3.50 moles


H₂SO₄  is the limiting reagent since  it produces  less amount of Na₂SO₄



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