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Masja [62]
3 years ago
13

Imagine that you wanted to make a time capsule in which you would seal important artifacts from your life (pictures, poems, a lo

ck of your baby hair, etc.), to be opened by your heirs 1,000 years from now. To prevent these artifacts from decaying, you want to fill the capsule with a gas that would be least reactive. Which of these gases would you choose: oxygen gas (O2) carbon dioxide (CO2), argon gas (Ar), or hydrogen gas (H2)? (The atomic numbers of the atoms in these molecules are as follows: oxygen = 8; carbon = 6; argon = 18; and hydrogen = 2.)
Chemistry
1 answer:
Elza [17]3 years ago
5 0

Answer:

Argon gas

Explanation:

Between the options given, argon gas is the least reactive. Argon is known as a <em>noble gas</em>, this means that it is located in the group 18 in the periodic table. These gases are known for their low reactivity, under most circumstances.

The <u>electronic configuration</u> of argon helps us see why that is the case: With an atomic number of 18, it has an electronic configuration of 1s² 2s² 2p⁶ 3s² 3p⁶. This means the most external orbital is filled with electrons, meaning that there are not unpaired electrons able to react with other substances.

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Which noble gas is closest to sulfur on the periodic table
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How many moles of water are in a beaker with 50 mL?
tatiyna

Answer:

Number of moles = 2.8 mol

Explanation:

Given data:

Number of moles of water = ?

Volume of water = 50 mL

Density of water = 1.00 g/cm³

Solution:

1 cm³ =  1 mL

Density = mass/ volume

1.00 g/mL = mass/ 50 mL

Mass = 1.00 g/mL× 50 mL

Mass = 50 g

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5 0
3 years ago
How much heat is lost when 575 grams molten iron at 1825 k becomes solid iron at 293 k? The melting point or iron is 1811 k.​
Galina-37 [17]

Answer:

146 kJ  

Explanation:

There are two heat flows in this question.  

Heat lost on cooling + heat lost on solidifying = 0  

                 q₁              +                 q₂                   = 0  

              mCΔT          +             nΔHsol              = 0  

Data:  

       m = 575 g  

       C = 0.449 J·K⁻¹g⁻¹  

    T_i = 1825 K  

    T_f = 1811 K  

ΔHsol = -13.8 kJ·mol⁻¹  

Calculations:  

(a) Heat lost on cooling  

ΔT = T_f - T_i = 1811 K - 1825 K = -14 K  

q₁ = mCΔT = 575 g × 0.449 J·K⁻¹g⁻¹ × (-14 K) = -361 J = -3.61 kJ  

(b) Heat lost on solidifying  

n = \text{575 g} \times \dfrac{\text{1 mol}}{\text{55.84 g}} = \text{10.30 mol}\\\\q_{2} = n\Delta_{\text{sol}}H = \text{10.30 mol} \times \dfrac{\text{-13.8 kJ}}{\text{1 mol}}= \text{-142.1 kJ}

(c) Total heat lost  

q = q₁ + q₂ = -3.61 kJ - 142.1 kJ = -146 kJ  

The heat lost was 146 kJ.

 

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