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murzikaleks [220]
3 years ago
5

Which two elements have chemical properties that are most similar?

Chemistry
1 answer:
masya89 [10]3 years ago
5 0

Answer:

Li and Na

Explanation:

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Which phrase best describes activation energy? (1 point)
iren [92.7K]

Answer: the minimum amount of energy required to break bonds and start a chemical reaction

Explanation: got a 100% on the quick check

8 0
2 years ago
Does anyone know what Hydrogen's oxidation state would be if it was acting as an anion non-metal?
Ilia_Sergeevich [38]

Answer:

yeah,The oxidation state of an atom does not represent the "real" charge on that atom, or any other actual atomic property.Hydrogen has OS = +1, but adopts −1 when bonded as a hydride to metals or metalloids. Oxygen in compounds has OS = −2. This set of postulates covers .

Explanation:

3 0
3 years ago
I need an explanation
Rina8888 [55]
<span>The Kelvin scale is an absolute temperature scale, while the Celsius scale is not. When you convert a temperature from Celsius to Kelvin, you add 273 degrees to the temperature. However, when you calculate a temperature change, you get the same number, whether you use the Celsius or the Kelvin scale.</span>
8 0
3 years ago
The normal boiling point of bromine is 58.8°C, and its enthalpy of vaporization is 30.91 kJ/mol. What is the approximate vapor p
saul85 [17]

Answer : The vapor pressure of bromine at 10.0^oC is 0.1448 atm.

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 bromine at 10.0^oC = ?

P_2 = vapor pressure of propane at normal boiling point = 1 atm

T_1 = temperature of propane = 10.0^oC=273+10.0=283.0K

T_2 = normal boiling point of bromine = 58.8^oC=273+58.8=331.8K

\Delta H_{vap} = heat of vaporization = 30.91 kJ/mole = 30910 J/mole

R = universal constant = 8.314 J/K.mole

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

\ln (\frac{1atm}{P_1})=\frac{30910J/mole}{8.314J/K.mole}\times (\frac{1}{283.0K}-\frac{1}{331.8K})

P_1=0.1448atm

Hence, the vapor pressure of bromine at 10.0^oC is 0.1448 atm.

4 0
3 years ago
Cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc
MAXImum [283]

<3333333333333333 >>:)))

Explanation:

c:

5 0
2 years ago
Read 2 more answers
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