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Tomtit [17]
4 years ago
11

Identify the atomic number and the chemical symbol for lead, silver, and gold. drag each item to the appropriate bin.

Chemistry
2 answers:
My name is Ann [436]4 years ago
8 0

Answer :

The atomic number of lead is, 82. The chemical symbol of lead is, Pb.

The atomic number of silver is, 47. The chemical symbol of silver is, Ag.

The atomic number of gold is, 79. The chemical symbol of gold is, Au.

Explanation :

Atomic number : It is equal to the number of protons or electrons.

Atomic number = number of protons = number of electrons

Mass number : It is defined as the sum of number of protons and number of neutrons.

Number of neutrons = Mass number - Number of protons

As we know that:

Lead is a metal that belongs to group 14 and period 6. The atomic number of lead is, 82. The chemical symbol of lead is, Pb.

Silver is a transition metal that belongs to group 11 and period 5. The atomic number of silver is, 47. The chemical symbol of silver is, Ag.

Gold is a transition metal that belongs to group 11 and period 6. The atomic number of gold is, 79. The chemical symbol of gold is, Au.

andre [41]4 years ago
5 0
Lead-
<span>Atomic number: </span>82<span>Symbol: </span><span>Pb
Silver-
</span><span>Atomic number: </span>47<span>Symbol: </span><span>Ag
Gold-
</span><span>Atomic number: </span>79<span>Symbol: </span><span>Au</span>
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The equilibrium position shifts to the right, in accordance to the constraint principle

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Consider the following reaction:
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2. 0.58 mol

Explanation:

1.Given ΔO₂/Δt…

    2H₂O₂     ⟶      2H₂O     +     O₂

-½d[H₂O₂]/dt = +½d[H₂O]/dt = d[O₂]/dt  

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 d[H₂O]/dt =  2d[O₂]/dt =  2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ =  6.6 × 10⁻³mol·L⁻¹s⁻¹

2. Moles of O₂  

(a) Initial moles of H₂O₂

\text{Moles} = \text{1.5 L} \times \dfrac{\text{1.0 mol}}{\text{1 L}} = \text{1.5 mol }

(b) Final moles of H₂O₂

The concentration of H₂O₂ has dropped to 0.22 mol·L⁻¹.

\text{Moles} = \text{1.5 L} \times \dfrac{\text{0.22 mol}}{\text{1 L}} = \text{0.33 mol }

(c) Moles of H₂O₂ reacted

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How many molecules are in 7.62 L of CH4, at 87.5°C and 722 torr
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Explanation:

Given : Volume = 7.62 L

Temperature = 87.5^{o}C = (87.5 + 273) K = 360.5 K

Pressure = 722 torr

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Converting torr into atm as follows.

722 torr = 722 torr \times \frac{0.00131579 atm}{1 torr}\\= 0.95 atm

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PV = nRT

where,

P = pressure

V = volume

n = number of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the values into above formula as follows.

PV = nRT\\0.95 atm \times 7.62 L = n \times 0.0821 L atm/mol K \times 360.5 K\\n = \frac{0.95 atm \times 7.62 L}{0.0821 L atm/mol K \times 360.5 K}\\= \frac{7.239}{29.59705}\\= 0.244 mol

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0.244 mol \times 6.022 \times 10^{23}\\= 1.469 \times 10^{23}

Thus, we can conclude that there are 1.469 \times 10^{23} molecules present in 7.62 L of CH_4 at 87.5^{o}C and 722 torr.

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