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2 years ago

In the reaction 2H2O (l) + 2Cl– (aq) → H2 (g) + Cl2 (g) + 2OH– (aq), which substance is oxidized?

Chemistry

1 answer:

Vlad1618 [11]2 years ago

4 0

Answer:

Explanation:

because i asked alexa and it told me hehr

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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

Which of the following actions will lead to an increase in pressure?

MatroZZZ [7]

By decreasing n we can increase presure because decrease in n will shift equilibrium to either forward or reverse direction

3 0

2 years ago

Which of these is NOT a property of water?

Elis [28]

Answer:

-low specific heat

3 0

3 years ago

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Determine the number of ions produced in the dissociation of the compound listed. BaS

11Alexandr11 [23.1K]

BaS → Ba²⁺ + S²⁻
two types of ions

4 0

3 years ago

Read 2 more answers

Use the following balanced reaction to solve:

Naily [24]

Answer: 60.7 g of $PH_3$ will be formed.

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given volume}}{\text{Molar volume}}$

$\text{Moles of} H_2=\frac{60L}{22.4L}=2.68moles$

The balanced chemical reaction is

$P_4(s)+6H_2(g)\rightarrow 4PH_3(g)$

$H_2$ is the limiting reagent as it limits the formation of product and $P_4$ is the excess reagent.

According to stoichiometry :

6 moles of $H_2$ produce = 4 moles of $PH_3$

Thus 2.68 moles of $H_2$ will produce= $\frac{4}{6}\times 2.68=1.79moles$ of $PH_3$

Mass of $PH_3=moles\times {\text {Molar mass}}=1.79moles\times 33.9g/mol=60.7g$

Thus 60.7 g of $PH_3$ will be formed by reactiong 60 L of hydrogen gas with an excess of $P_4$

3 0

2 years ago