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2 years ago

How many liters of water would it require to 36 grams of the substance?

Chemistry

2 answers:

WARRIOR [948]2 years ago

6 0

It would required 3 liters of substance

Bas_tet [7]2 years ago

3 0

3 liters

________________

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The work function of an element is the energy required to remove an electron from the surface of the solid. The work function fo

uranmaximum [27]

Answer:

λ = 2.38 × 10^(-7) m

Explanation:

We are given the work function for palladium as 503.7 kJ/mol.

Now let's convert this to KJ/electron.

We know from avogadro's number that;

1 mole of electron = 6.022 × 10^(23) electrons

Thus,

503.7 kJ/mol = 503.7 × 1/(6.022 × 10^(23)) = 8.364 × 10^(-22) KJ/electron = 8.364 × 10^(-19) J/electron

Formula for energy of a photon is;

E = hv

Where;

h is Planck's constant = 6.626 × 10^(-34) J.s

v is velocity

Now, v = c/λ

Where;

c is speed of light = 3 × 10^(8) m/s

λ is wavelength of light.

Thus;

E = hc/λ

Making λ the subject, we have;

λ = hc/E

λ = (6.626 × 10^(-34) × 3 × 10^(8))/(8.364 × 10^(-19))

λ = 2.38 × 10^(-7) m

4 0

2 years ago

13.50 grams of Pb(NO3)4 are dissolved in enough water to make 250 mL of solution. What is the molar its of the resulting solutio

likoan [24]

Data:
M (molarity) = ? (M or Mol/L)
m (mass) = 13.50 g
V (volume) = 250 mL → 0.25 L
MM (Molar Mass) of Lead(IV) Nitrate $Pb(NO_3)_4$
Pb = 1*207 = 207 amu
N = (1*14)*4 = 14*4 = 56 amu
O = (3*16)*4 = 48*4 = 192 amu
------------------------------------
MM of $Pb(NO_3)_4$ = 207+56+192 = 455 g/mol

Formula:
$M = \frac{m}{MM*V}$

Solving:
$M = \frac{m}{MM*V}$
$M = \frac{13.50}{455*0.25}$
$M = \frac{13.50}{113.75}$
$M = 0.118681318...\:\:\to\:\:\boxed{\boxed{M \approx 0.119\:Mol/L}}\end{array}}\qquad\quad\checkmark$

Answer:
<span>B. 0.119 M</span>

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