decameters - meters: multiply by 10
meters to meters: multiply by 1
centimeters to meters: divide by 100
millimeters to meters: divide by 1000
For the rows at the bottom:
hectometer row: 100, multiply by 100, 4500
decameter row: 10, multiply by 10, 450
meter row: 1, multiply by 1, 45
decimeter row: 0.1, divide by 10, 4.5
centimeter row: 0.01, divide by 100, 0.45
im guessing theres a millimeter row at the bottom:
millimeter row: 0.001, divide by 1000, 0.045
hope this helps!
Answer:
1.0 *10^(-4) mol
Explanation:
For gases:
n1/n2 = V1/V2
n1/3.8*10^(-4) mol = 230 mL/ 860 mL
n1 = 3.8*10^(-4)*230/860 = 1.0 *10^(-4) mol
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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Answer:
1 has the highest density because it has the most amount of circles in the least amount of space- it is the most densely filled with circles; it is the most dense.
Answer:
Everything to the right of the arrow is a product.
Explanation:
