Answer:
13.2 g of gold
Explanation:
We'll begin by converting 5.25 L to ft³.
This can be obtained as follow:
Recall:
1 L = 0.0353 ft³
Therefore,
5.25 L = 5.25 × 0.0353
5.25 L = 1.85×10¯¹ ft³
From the question given above,
2.45 g of gold is present in 5.25 L ( i.e 1.85×10¯¹ ft³) of soil.
Therefore, Xg of gold will be present in 1 ft³ of soil i.e
Xg of gold = 2.45/1.85×10¯¹
Xg of gold = 13.2 g
Therefore, 13.2 g of gold is present in 1 ft³ of the soil.
Answer:
0.66atm
Explanation:
Based on the reaction:
NH₄HS(s) ↔ NH₃(g) + H₂S(g)
The equilibrium constant, Kp, is defined as:
Kp = 0.11 = 
As moles of gas produced for NH₃(g) and H₂S(g) are the same, it is possible to write:
That means pressure of NH₃(g) is 0.33atm and H₂S(g) is, also, 0.33atm. Thus, total pressure is:
0.33atm×2 = <em>0.66atm</em>
Solids' particles move a lot faster and closer together. Liquids' are more spread apart and move slower. So there is more energy in the solid
The answer is: " 56 g CaCl₂ " .
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Explanation:
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2.0 M CaCl₂ = 2.0 mol CaCl₂ / L ;
Since: "M" = "Molarity" (measurement of concentration);
= moles of solute per L {"Liter"} of solution.
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Note the exact conversion: 1000 mL = 1 L .
Given: 250 mL ;
250 mL = ? L ? ;
250 mL * (1 L / 1000 L) = (250/1000) L = 0.25 L .
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(2.0 mol CaCl₂ / L ) * (0.25L) = (2.0) * (0.25) mol = 0.50 mol CaCl₂ ;
We have: 0.50 mol CaCl₂ ; Convert to "g" (grams):
→ 0.50 mol CaCl₂ .
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1 mol CaCl₂ = ? g ?
From the Periodic Table of Elements:
1 mol Ca = 40.08 g
1 mol Cl = <span>35.45 g .
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There are 2 atoms of Cl in " CaCl₂ " ;
→ Note the subscript, "2", in the " Cl₂ " ;
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So, to calculate the molar mass of "CaCl₂" :
40.08 g + 2(35.45 g) =
40.08 g + 70.90 g = 110.98 g ; round to 4 significant figures;
→ round to 111 g/mol .
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So:
→ 0.50 mol CaCl₂ = ? g CaCl₂ ? ;
→ 0.50 mol CaCl₂ * (111 g CaCl₂ / mol CaCl₂) ;
= (0.50) * (111 g) CaCl₂ ;
= 55.5 g CaCl₂ ;
→ round to 2 significant figures;
→ 56 g CaCl₂ .
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The answer is: " 56 g CaCl₂ " .
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D: The atomic mass number
