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3 years ago

How many milliliters of 0.25M H2SO4 can be prepared from 57 mL of a 3.0M solution of H2SO4?

Chemistry

1 answer:

DochEvi [55]3 years ago

5 0

Answer:

Why ? Because 1 molecule of H2SO4 gives 2 H+ ions per molecule while only one H+ ion is required to neutralize 1 molecule of KOH. So, 1 molecule of H2SO4 can neutralize 2 molecules of KOH. Hence, we would require 525 ml of 0.03 M H2SO4 to neutralize 525 ml of 0.06 M KOH. How will we prepare 525 ml of 0.03 M H2SO4 ?

Explanation:

Now, we have 0.025 M H2SO4 and we do not know how much volume we have.

We will use the standard N1 X V1 = N2 X V2 for this calculation.

N1=0.025 M; V1=unknown; N2=0.03 M and V2=525 ml.

So V1= (0.03 X 525)/(0.025) = 630 ml.

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The actual density of iron is 7.874 g/mL. In a laboratory investigation, Jason finds the density of a piece of iron to be 7.921

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The percent error associated with Jason’s measurement is 0.596%.

HOW TO CALCULATE PERCENTAGE ERROR:

The percentage error of a measurement can be calculated by following the following process:

Find the difference between the true value and the measured value of a quantity.
Then, divide by the true value and then multiplied by 100

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Difference = 0.047 g/mL

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1 year ago

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3 years ago

(a) Calculate the energy released by the alpha decay of 222 86Rn.

Alinara [238K]

Answer:

a). The energy released by the alpha decay of 222 Rn is to 218 Po :

= 5.596 MeV

b).The energy of the alpha particle is 5.5 MeV

c) The recoil energy of Po = 0.096 MeV

Explanation:

The equation for the alpha decay of Rn to Polonium is:

$_{86}^{222}\textrm{Rn}\rightarrow$ $_{2}^{4}\textrm{He}+_{84}^{218}\textrm{Po}$

The energy of the decay process can be calculated using:

$\Delta E=\Delta mc^{2}$

$\Delta m$ = Change in the mass

Mass of Po = 218.008965 u

Mass of He = 4.002603 u

Mass of Rn = 222.017576 u

$\Delta m$ = mass of Po + mass of He - mass of Rn

= 4.002603 + 218.008965-222.017576

= - 0.006008

$\Delta E=m(931.5MeV)$

$\Delta E=-0.006008\times 931.5MeV$

= -5.596 MeV

The negative sign means energy is released during the process.

b) The energy of the alpha particle is :

$\frac{Po\ mass }{Rn\ mass}\times E$

$\frac{218.0089}{222.0175}\times \Delta E$

$\frac{218.0089}{222.0175}\times 5.596$

= 5.494 MeV

= 5.5 MeV

c).

Recoil energy: When the parent nucleus is at rest before the decay then , there must be the some recoil of daughter nucleus to conserve the momentum.This is termed as recoil energy.

The energy of the recoil polonium atom :

The formula for recoil energy is :

The total energy - the kinetic energy

= 5.596 - 5.5

= 0.096 MeV

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3 years ago