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Montano1993 [528]
4 years ago
11

In the periodic table below, shade all the elements for which the neutral atom has a valence electron configuration of ns2np3 ,

where n stands for an integer.
Chemistry
1 answer:
Lorico [155]4 years ago
5 0
Answer is: chemical elements<span> in </span>group 15<span> of the </span>periodic table have <span>neutral atom with a valence electron configuration of ns</span>²<span>np</span>³<span>.
</span>Nitrogen: [He] 2s² 2p³.
Phosphorus: [Ne] 3s² 3p³.
Arsenic: <span>[Ar] 3d</span>¹⁰ 4s² 4p³.
Bismuth: <span>[Xe] 4f</span>¹⁴ 5d¹⁰ 6s² 6p³.
Moscovium: predicted [Rn] 5f¹⁴<span> 6d</span>¹⁰<span> 7s</span>² 7p³.
These elements have 5 valence electrons.

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If you combine 230.0 mL of water at 25.00 ∘ C and 140.0 mL of water at 95.00 ∘ C, what is the final temperature of the mixture?
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<u>Answer:</u> The final temperature of the mixture is 51.49°C

<u>Explanation:</u>

When two samples of water are mixed, the heat released by the water at high temperature will be equal to the amount of heat absorbed by water at low temperature

Heat_{\text{absorbed}}=Heat_{\text{released}}

The equation used to calculate heat released or absorbed follows:

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

m_1\times c\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]       ......(1)

where,

q = heat absorbed or released

m_1 = mass of water at high temperature = 140 g     (Density of water = 1.00 g/mL)

m_2 = mass of water at low temperature = 230 g

T_{final} = final temperature = ?°C

T_1 = initial temperature of water at high temperature = 95.00°C

T_2 = initial temperature of water at low temperature = 25.00°C

c = specific heat of water= 4.186 J/g°C

Putting values in equation 1, we get:

140\times 4.186\times (T_{final}-95)=-[230\times 4.186\times (T_{final}-25)]

T_{final}=51.49^oC

Hence, the final temperature of the mixture is 51.49°C

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