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Mariulka [41]
2 years ago
15

3. Round the number 10.4656 to the tenths place * 10 10.4 10.5 or 10.6

Chemistry
1 answer:
motikmotik2 years ago
5 0
Answer: 10.5 hope this helps!
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Which pair of elements do you expect to be most similar in their chemical properties?
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Isotopes of carbon or any element
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What is the molarity of 122.5 g of AlCl3 in 1.0 L of solution? (MM = 133 g/mol)
yawa3891 [41]
Answer: Molarity is defined as moles of solute per liter of solution. So, find the moles of solute and divide by the liters of solution.
molar mass AlCl3 = 133g/mole
moles AlCl3 = 127 g x 1 mole/133 g = 0.955 moles
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4 0
3 years ago
The vapor pressure of ethanol is 30°C at 98.5 mmHg and the heat of vaporization is 39.3 kJ/mol. Determine the normal boiling poi
Gelneren [198K]

Answer : The normal boiling point of ethanol will be, 348.67K or 75.67^oC

Explanation :

The Clausius- Clapeyron equation is :

\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})

where,

P_1 = vapor pressure of ethanol at 30^oC = 98.5 mmHg

P_2 = vapor pressure of ethanol at normal boiling point = 1 atm = 760 mmHg

T_1 = temperature of ethanol = 30^oC=273+30=303K

T_2 = normal boiling point of ethanol = ?

\Delta H_{vap} = heat of vaporization = 39.3 kJ/mole = 39300 J/mole

R = universal constant = 8.314 J/K.mole

Now put all the given values in the above formula, we get:

\ln (\frac{760mmHg}{98.5mmHg})=\frac{39300J/mole}{8.314J/K.mole}\times (\frac{1}{303K}-\frac{1}{T_2})

T_2=348.67K=348.67-273=75.67^oC

Hence, the normal boiling point of ethanol will be, 348.67K or 75.67^oC

3 0
3 years ago
List the first 20 elements chemistry
timama [110]

Answer:

h- Hydrogen

He - Helium

Li - Lithium

Be - Beryllium

B - Boron

C - Carbon

N - Nitrogen

O - Oxygen

Explanation:

Happy to help:)

8 0
3 years ago
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