After moving down a group in the periodic table, the number of valence electrons

The calculation of quantities in chemical equations are called...

Anni [7]

The calculation of quantities in chemical equations are called Stoichiometry. Stoichiometry is a branch of chemistry which deals with relative quantities of reactants and products in chemical reactions. The correct answer is 'Stoichoimetry'. I hope this helps you.

5 0

2 years ago

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 :

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

7 0

3 years ago

A study of the decomposition reaction 3RS2  3R + 6S yields the following initial rate dat

shutvik [7]

Missing question: What is the rate constant for the reaction?
[RS2](mol L-1) Rate (mol/(L·s))
0.150 0.0394
0.250 0.109
0.350 0.214
0.500 0.438
Chemical reaction: 3RS₂ → 3R + 6S.
Compare second and fourth experiment, when concentration is doubled, rate of concentration is increaced by four. So rate is:
rate = k·[RS₂]².
k = 0,438 ÷ (0,500)².
k = 1,75 L/mol·s.

3 0

2 years ago

Identify the limiting reactant when 1.22 g of O2 reacts with 1.05 g H2 to produce water.

kupik [55]

The reaction between oxygen, O2, and hydrogen, H2, to produce water can be expressed as,

2H2 + O2 --> 2H2O

The masses of each of the reactants are calculated below.

2H2 = 4(1.01 g) = 4.04 g
O2 = 2(16 g) = 32 g

Given 1.22 grams of oxygen, we determine the mass of hydrogen needed.
(1.22 g O2)(4.04 g H2 / 32 g O2) = 0.154 g of O2

Since there are 1.05 grams of O2 then, the limiting reactant is 1.22 grams of oxygen.

Answer: 1.22 g of oxygen

4 0

3 years ago

A student places three ice cubes in a beaker and allows them to partially melt. if she measures the temperature of the water in

atroni [7]

She will most likely observe that the temperature does not change during melting because the heat absorbed is used to overcome intermolecular forces rather than to increase the kinetic energy of the particles if she measures the temperature of the water in the beaker.

8 0

3 years ago

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