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andriy [413]
2 years ago
5

Use the conversion factor 1amu=6.66054 x 10^-24 to answer the following questions.. . a) 1.674 x 10^-24 g of neutrons is how man

y amu?. . b)what is the mass in grams of one lithium atom(lithium has an atomic weight of 6.94 amu). . c)6.492 x 10^-23 g of potassium is how many amu?
Chemistry
2 answers:
Wewaii [24]2 years ago
6 0
A.) The conversion factor is 1 amu = 1.66054 ^{-24}g

To know how many amu in 1.674x10^{-24}g grams of neutrons:
1.674x10^{-24}g ( \frac{x (amu) }{ 1.66054x10^{-24} g)}

=1.00811 amu

b.) The mass in grams of one lithium ion which has an atomic weight of 6.94 amu.

6.94 amu (\frac{6.66054 ^{-24} g}{1 amu}) = 4.62241x10 ^{23} g

c.) How many amu in 6.492x10^-23g potassium?

6.492x10^{-23}g ( \frac{x (amu) }{ 1.66054x10^{-24}g }) = 39.0957 amu


saw5 [17]2 years ago
3 0

Answer :

(a) The number of amu is 1.008 amu.

(b) The mass of one lithium atom is 11.5\times 10^{-24}g

(c) The number of amu is 39.09 amu.

Explanation :

As we are given that conversion factor:

1amu=1.66054\times 10^{-24}g

(a) To calculate the number of amu.

As, 1.66054\times 10^{-24}g of neutrons = 1 amu

So, 1.674\times 10^{-24}g of neutrons = \frac{1.674\times 10^{-24}g}{1.66054\times 10^{-24}g}\times 1amu=1.008amu

Thus, the number of amu is 1.008 amu.

(b) To calculate the mass of one lithium atom.

As, 1 amu of lithium atom = 1.66054\times 10^{-24}g

So, 6.94 amu of lithium atom = 6.94\times 1.66054\times 10^{-24}g=11.5\times 10^{-24}g

Thus, the mass of one lithium atom is 11.5\times 10^{-24}g

(c) To calculate the number of amu.

As, 1.66054\times 10^{-24}g of neutrons = 1 amu

So, 6.492\times 10^{-23}g of neutrons = \frac{6.492\times 10^{-23}g}{1.66054\times 10^{-24}g}\times 1amu=39.09amu

Thus, the number of amu is 39.09 amu.

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Explanation;

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Answer:

0.3023 M

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Let Picric acid = H_{picric}

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The ICE table can be given as:

                          H_{picric}     +       H_2}O          ⇄      H_3}O^+     +     Picric^-

Initial:                0.52                                               0                  0

Change:             - x                                                 + x                 + x

Equilibrium:      0.52 - x                                        + x                 + x

Given that;

acid dissociation constant  (K_a) = 0.42

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0.42 = \frac{[x][x]}{0.52-x}}

0.42 = \frac{[x]^2}{0.52-x}}

0.42(0.52-x) = x²

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x²  + 0.42x - 0.2184 = 0                   -------------------- (quadratic equation)

Using the quadratic formula;

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= \frac{0.6045}{2}                 OR    -\frac{1.4445}{2}

= 0.30225          OR     - 0.72225

So, we go by the +ve integer that says:

x =  0.30225

x = [ H_3}O^+ ] = [   Picric^- ] =  0.3023  M

∴  the value of  [H3O+] for an 0.52 M solution of picric acid  = 0.3023 M     (to 4 decimal places).

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