Answer:
Given
mass of H2O (m) =35.6g
molarmass (mr) = H2O ), 1x2+16=18g/mol
moles of H2O (n) =?
sln
n=m/mr
n=35.6g/18g/mol
n=1.978moles
the moles of H2O are 1.978moles
D. The final substance in Beaker A is a mixture and in Beaker B is a pure substance.
Explanation:
Cations are positively charged ions with fewer electrons than protons. To partake in reaction, metals lose electrons to achieve a stable octet configuration. It has lost valence electrons and wold have a positive charge associated with it.
Potassium is a group 1 element. A metal.
Potassium has 1 valence electron so it loses the valence electron to form a stable octet.
A potassium ion has a positive charge and therefore cannot be an anion but is a cation.
The group number pretty much denotes the number of valence electrons.
Group 1 = 1
Group 2 = 2
Group 17 = 7
Group 18 = 8
Answer: Because it's a combination of chemicals, vodka doesn't freeze at the same temperature as either water or alcohol. Of course, vodka will freeze, but not at the temperature of an ordinary freezer. This is because vodka contains enough alcohol to lower the freezing point of water below the -17°C of your typical freezer.
Explanation: .......
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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