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MrRissso [65]
3 years ago
5

Which of the following descriptions of the storage requirements for alkali metals is the most accurate? Select the correct answe

r below: A. They must be kept separate from moisture and oxygen. B. They must be kept separate from other alkali metals. C. They must be kept separate from non-alkali metals. D. They must be kept separate from mineral oil.
Chemistry
1 answer:
elena55 [62]3 years ago
7 0

Answer:

A. They must be kept separate from moisture and oxygen

Explanation:

Alkali metals refers to a group of chemical elements in the periodic table. It is the common name given to Group 1 elements (excluding hydrogen). They are generally metallic in nature with physical properties like shining lustre, silvery in appearance etc. Alkali metals include lithium, Pottasium, Sodium, Rubidium etc.

One important chemical property of alkali metals is their high reactivity i.e. they tend to react quickly with other substances. For example, alkali metals react vigorously with water to form hydroxides. They also react with oxygen to form oxides. Due to this highly reactive properties, they are kept or stored away from moisture (water) and oxygen (present in air).

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Is backbonding is possible in BF4- ? If no then why?
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What property can be used to determine if a sample is a pure substance or a mixture?
Yuliya22 [10]

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calculate the difference in slope of the chemical potential against temperature on either side of the normal freezing point of w
kipiarov [429]

Answer:

(a) The normal freezing point of water (J·K−1·mol−1) is -22Jmole^-^1k^-^1

(b) The normal boiling point of water (J·K−1·mol−1) is -109Jmole^-^1K^-^1

(c) the chemical potential of water supercooled to −5.0°C exceed that of ice at that temperature is  109J/mole

Explanation:

Lets calculate

(a) - General equation -

      (\frac{d\mu(\beta )}{dt})p-(\frac{d\mu(\alpha) }{dt})_p = -5_m(\beta )+5_m(\alpha ) =  -\frac{\Delta H}{T}

 \alpha ,\beta → phases

ΔH → enthalpy of transition

T → temperature transition

 (\frac{d\mu(l)}{dT})_p -(\frac{d\mu(s)}{dT})_p == -\frac{\Delta_fH}{T_f}

            = \frac{-6.008kJ/mole}{273.15K} ( \Delta_fH is the enthalpy of fusion of water)

           = -22Jmole^-^1k^-^1

(b) (\frac{d\mu(g)}{dT})_p-(\frac{d\mu(l)}{dT})_p= -\frac{\Delta_v_a_p_o_u_rH}{T_v_a_p_o_u_r}

                                  = \frac{40.656kJ/mole}{373.15K} (\Delta_v_a_p_o_u_rH is the enthalpy of vaporization)

                               = -109Jmole^-^1K^-^1

(c) \Delta\mu =\Delta\mu(l)-\Delta\mu(s) =-S_m\DeltaT

[\mu(l-5°C)-\mu(l,0°C)] =  [\mu(s-5°C)-\mu(s,0°C)]=-S_mΔT

\mu(l,-5°C)-\mu(s,-5°C)=-Sm\DeltaT [\mu(l,0

\Delta\mu=(21.995Jmole^-^1K^-^1)\times (-5K)

     = 109J/mole

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The answer is A I’m pretty sure lol
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