Answer is: 2,42·10²² atoms of gold.
<span>1 ounce (oz) is equal to 28,34952 grams (g).
</span>0,375 oz = 10,631 g.
75% gold = 10,631 g · 0,75 = 7,973 g.
n(Au) = m(Au) ÷ M(Au).
n(Au) = 7,973 g ÷ 197 g/mol.
n(Au) = 0,0404 mol.
N(Au) = n(Au) · Na.
N(Au) = 0,0404 mol · 6,023·10²³ 1/mol.
N(Au) = 2,42·10²².
n - amount of substance.
Na - Avogadro number.

8 0

3 years ago

A sample of substance X that has a mass of 326.0 g releases 4325.8 call when it freezes at its freezing point. If substance X ha

o-na [289]

Answer : The molar heat of fusion for substance X is, 775.6 cal/mol

Explanation :

First we have to calculate the moles of substance X.

$\text{Moles of substance X}=\frac{\text{Mass of substance X}}{\text{Molar mass of substance X}}$

Given:

Molar mass of substance X = 58.45 g/mole

Mass of substance X = 326.0 g

Now put all the given values in the above expression, we get:

$\text{Moles of substance X}=\frac{326.0g}{58.45g/mol}=5.577mol$

Now we have to calculate the molar heat of fusion for substance X.

$\Delta H=\frac{Q}{n}$

where,

Q = heat releases = 4325.8 cal

$\Delta H$ = molar heat of fusion for substance X = ?

n = moles = 5.577 mol

Now put all the given values in the above expression, we get:

$\Delta H=\frac{4325.8cal}{5.577mol}=775.6cal/mol$

Thus, the molar heat of fusion for substance X is, 775.6 cal/mol

6 0

3 years ago

Predict whether the reaction given is product favored or reactant favored by calculating delta G from the entropy and enthalpy c

GenaCL600 [577]

ΔG= ΔH - TΔS

YOU need to change the values to the same unit.. 42.08 J to Kj which is 42.08 x 10-3.

ΔG= (41.17) - (298K)(42.08x10-3)= 28.6 Kj/mol

because delta G is positive, it means this reaction is reactant favored.

4 0

3 years ago

True or False some synthetics are safer than their natural counterparts

worty [1.4K]

Answer:

The answer is False

hope this helpes

6 0

3 years ago