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Arlecino [84]
4 years ago
7

How many moles are in 3.84 grams of calcium chloride?

Chemistry
1 answer:
iris [78.8K]4 years ago
5 0

Answer:

How many grams CaCl2 in 1 mol? The answer is 110.984. We assume you are converting between grams CaCl2 and mole. You can view more details on each measurement unit: molecular weight of CaCl2 or mol This compound is also known as Calcium Chloride

Explanation:

i hope i helped =)

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Convert the following mass of a compound into moles.<br> 23.5g of NaCl
krok68 [10]

Answer:

0.402 moles

Explanation:

1 mole NaCl/58.44g Nacl ×23.5g NaCl

grams get cancelled out and you are left with moles

5 0
3 years ago
The temperature of a 500 ml sample of gas increases from 150 k to 300 k. what is the final volume of the sample of gas, if the p
seropon [69]
Upon a constant pressure (P), volume (V) of a gas will vary in direct proportion to changes in temperature (T). So V1/T1 = V2/T2
V2 = V1T2/T1 = (500)(300)/150
V2 = 150000/150 = 1000 mL
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Using the thermodynamic information in the ALEKS Data tab, calculate the boiling point of titanium tetrachloride . Round your an
ddd [48]

Answer:

The boiling point is 308.27 K (35.27°C)

Explanation:

The chemical reaction for the boiling of titanium tetrachloride is shown below:

TiCl_{4(l)} ⇒ TiCl_{4(g)}

ΔH°_{f} (TiCl_{4(l)}) = -804.2 kJ/mol

ΔH°_{f} (TiCl_{4(g)}) = -763.2 kJ/mol

Therefore,

ΔH°_{f} = ΔH°_{f} (TiCl_{4(g)}) - ΔH°_{f} (TiCl_{4(l)}) = -763.2 - (-804.2) = 41 kJ/mol = 41000 J/mol

Similarly,

s°(TiCl_{4(l)}) = 221.9 J/(mol*K)

s°(TiCl_{4(g)}) = 354.9 J/(mol*K)

Therefore,

s° = s° (TiCl_{4(g)}) - s°(TiCl_{4(l)}) = 354.9 - 221.9 = 133 J/(mol*K)

Thus, T = ΔH°_{f} /s° = [41000 J/mol]/[133 J/(mol*K)] = 308. 27 K or 35.27°C

Therefore, the boiling point of titanium tetrachloride is 308.27 K or 35.27°C.

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3 years ago
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The answer I got was C.
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What were the first 10 elements of the periodic table?
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H. Hydrogen
He. Helium
Li. Lithium
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B. Boron
C. Carbon
N. Nitrogen
O. Oxygen
F. Fluorine
N. Neon

5 0
3 years ago
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