What's mole in 34g of H2O<br><br><br>

Convert the composition of the following alloy from atom percent to weight percent (a) 44.9 at% of silver, (b) 46.3 at% of gold,

vesna_86 [32]

Answer:

Mass percentage of gold : 33.35%

Mass percentage of silver: 62.80%

Mass percentage of copper : 3.85%

Explanation:

$N=n\times N_A$

Where : N = Number of atoms

n = moles

$N_A=6.022\times 10^{23}$ = Avogadro number

Let the total atoms present in the alloy be 100.

Atom percentage of silver = 44.9%

Atoms of of silver = 44.9% of 100 atoms = 44.9

Atomic weight of silver = 107.87 g/mol

Mass of 44.9 silver atoms = 107.87 g/mol × 44.9 = 4,843.363 g/mol

Atom percentage of gold= 46.3%

Atoms of of gold = 46.3% of 100 atoms = 46.3

Atomic weight of gold = 196.97 g/mol

Mass of 44.9 gold atoms = 196.97 g/mol × 46.3 = 9,119.711 g/mol

Atom percentage of copper = 8.8 %

Atoms of of copper = 8.8% of 100 atoms = 8.8

Atomic weight of copper = 63.55 g/mol

Mass of 44.9 copper atoms = 63.55 g/mol × 8.8 = 559.24 g/mol

Total mass of an alloy :4,843.363 g/mol + 9,119.711 g/mol + 559.24 g/mol

Mass percentage of gold :

$\frac{4,843.363 g/mol}{4,843.363 g/mol + 9,119.711 g/mol + 559.24 g/mol}\times 100=33.35 \%$

Mass percentage of silver:

$\frac{9,119.711 g/mol}{4,843.363 g/mol + 9,119.711 g/mol + 559.24 g/mol}\times 100=62.80\%$

Mass percentage of copper :

$\frac{559.24 g/mol}{4,843.363 g/mol + 9,119.711 g/mol + 559.24 g/mol}\times 100=3.85\%$

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Consider the following equilibrium reaction having gaseous reactants and products. 4HCl + O2 ⇌ 2H2O + Cl2 Which of the following

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<h3><u>Answer</u>;</h3>

B.The rate of forward reaction increases.

<h3><u>Explanation;</u></h3>

Le Chatelier's principle states that changing a factor such as concentration, temperature, or pressure of a reaction at equilibrium will cause the reaction to shift in the direction that counteracts the effect of that change.
<em><u>Therefore, when reactants are added to a reaction at equilibrium shift when more reactants are added then the reaction shifts to the right to make more products.</u></em>

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The compound adrenaline contains 56.79% c, 6.56% h, 28.37% o, and 8.28% n by mass. what is the empirical formula for adrenaline?

Phoenix [80]

To determine the empirical formula of the compound given, we need to determine the ratio of each element in the compound. To do that we assume to have 100 grams sample of the compound with the given composition. Then, we calculate for the number of moles of each element. We do as follows:<span>
mass    moles
C 56.79 4.73
H 6.56 6.50
O 28.37 1.77
N 8.28 0.59

Dividing the number of moles of each element with the smallest value, we will have the empirical formula:

</span> moles ratio
C 4.73 / 0.59 8
H 6.50   / 0.59 11
O   1.77   / 0.59 3
N    0.59   / 0.59 1<span>
</span><span>
The empirical formula would be C8H11O3N.</span>

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Why weren’t scientists initially concerned that the ocean absorbs excess carbon dioxide?

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They only discovered

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What's mole in 34g of H2O​

What's mole in 34g of H2O