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3 years ago

Which of the following atoms or ions does not have a filled outer subshell?CaSZn2+S2Ca2+

Chemistry

1 answer:

ipn [44]3 years ago

4 0

S and S²⁻ do not have the outer subshell fully filled with electrons.

Explanation:

We look at electronic configurations:

Ca 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² - the outer subshell 4s² is fully-filled with electrons

S 1s² 2s² 2p⁶ 3s² 3p⁴ - the outer subshell 3p⁴ is not fully-filled with electrons

Zn²⁺ 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s⁰ - here the 4s subshell is higher in energy than 3d subshell so will consider 3d¹⁰ the out subshell which is fully-filled with electrons

S²⁻ 1s² 2s² 2p⁶ 3s² 3p² - the outer subshell 3p² is not fully-filled with electrons

Ca²⁺ 1s² 2s² 2p⁶ 3s² 3p⁶ - the outer subshell 3p⁶ is fully-filled with electrons

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electron configurations

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What current (in a) is required to plate out 1. 22 g of nickel from a solution of ni2 in 0. 50 hour?

Mashcka [7]

The current required to accumulate the 1.22 grams of nickel in 0.5 hours is 2.23 A.

<h3>What is current?</h3>

The current is given as the product of the charge with time. In the electrochemical analysis of the nickel, there will be a reduction of the nickel ion to nickel. The formation is given as:

$\rm Ni^{2+}+2e^-\;\rightleftharpoons Ni$

There is the deposition of 1 mole of Ni with 2 electrons transfer. The transfer of charge for 1 mole that is 58.7 grams Nickel is:

$\rm 58.7\;g=2\;\times\;96487\;C\\58.7\;g=192974\;C$

The mass of Ni to be deposited is 1.22 grams. The charge required is given as:

$\rm 58.7\;grams\;Ni=192974\;C\\\\1.22\;grams\;Ni=\dfrac{192974}{58.7}\;\times\;1.22\;C\\\\1.22\;grams\;Ni=4010.7\;C$

The current required to transfer 4010.7 C of charge in 1800 seconds is given as:

$\rm Charge=Current\;\times\;Time\\4010.7\;C=Current\;\times\;1800\;sec\\Current=2.23\;A$

Thus, the current required to accumulate the 1.22 grams of nickel in 0.5 hours is 2.23 A.

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1 year ago

Temperature is used to measure the movement of particles such as atoms and molecules. As temperature increases, what is true of

ch4aika [34]

This is thermodynamics. When you increase the temperature of an object, the particles gain on kinethic energy ergo the move faster. When you decrease it, they slow down.

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2 years ago

Read 2 more answers

It takes 167 s for an unknown gas to effuse through a porous wall and 99 s for the same volume of n2 gas to effuse at the same t

irakobra [83]

Answer : The molar mass of the unknown gas will be 79.7 g/mol

Explanation : To solve this question we can use graham's law;

Now we can use nitrogen as the gas number 2, which travels faster than gas 1;

So, 167 / 99 = 1.687 So the nitrogen gas is 1.687 times faster that the unknown gas 1

We can compare the rates of both the gases;

So here, Rate of gas 2 / Rate of gas 1 = $\sqrt{(molar mass 1 / molar mass 2)}$

Now, 1.687 = square root [ $\sqrt{(molar mass 1) / (28.01 g/mol N_{2})}$ ]

When we square both the sides we get;

2.845 = (molar mass 1) / (28.01 g/mol N2)

On rearranging, we get,

2.845 X (28.01 g/mol N2) = Molar mass 1

So the molar mass of unknown gas will be = 79.7 g/mol

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3 years ago

A 0.150-kg sample of a metal alloy is heated at 540 Celsius an then plunged into a 0.400-kg of water at 10.0 Celsius, which is c

Zarrin [17]

Answer:

$C_{alloy}=0.497\frac{J}{g\°C}$

Explanation:

Hello there!

In this case, according to this calorimetry problem on equilibrium temperature, it is possible for us to infer that the heat released by the metal allow is absorbed by the water for us to write:

$Q_{allow}=-(Q_{water}+Q_{Al})$

Thus, by writing the aforementioned in terms of mass, specific heat and temperature, we have:

$m_{alloy}C_{alloy}(T_{eq}-T_{alloy})=-(m_{water}C_{water}(T_{eq}-T_{water})+m_{Al}C_{Al}(T_{eq}-T_{Al})$

Then, we solve for specific heat of the metallic alloy to obtain:

$C_{alloy}=\frac{-(m_{water}C_{water}(T_{eq}-T_{water})+m_{Al}C_{Al}(T_{eq}-T_{Al})}{m_{alloy}(T_{eq}-T_{alloy})}$

Thereby, we plug in the given data to obtain:

$C_{alloy}=\frac{-(400g*4.184\frac{J}{g\°C} (30.5\°C-10.0\°C)+200g*0.900\frac{J}{g\°C}(30.5\°C-10.0\°C)}{150g(30.5\°C-540\°C)} \\\\C_{alloy}=0.497\frac{J}{g\°C}$

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The mass of the ice cubes and the water will be equal because the same amount of matter is in the beaker.

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