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2 years ago

Use the successive ionization energies for this unknown element to identify the family it belongs to.

Chemistry

1 answer:

NNADVOKAT [17]2 years ago

4 0

Answer: Belongs to the group 2A

Explanation:

As you can see, the first two ionization energies are close and low, meaning that this element ionizates easily.

Not only loses easily the first electron, but the second too

To remove the third electron you requiered a huge amount of energy

Now, elements easily ionizable are the ones from group IA, group 2A and transition metals.

The last ones have mixed characteristics in matter of how many electrons you can remove from them, so they are not a family.

Now the question: group I or group II ?

The elements of group I have low ionization energies for the first electron but high energies for the second ones.

Being all that said, the unknown element belongs to the Group 2A

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3 years ago

One mole of which of these compounds contains two moles of hydrogen atoms? f NaOH g H2S h CH4 j NH3

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3 years ago

What is the molality of a solution if 100.0 g of glucose (C&Hi20e) were dissolved into 750. mL of water?

hodyreva [135]

<u>Answer:</u> The molality of solution is 0.740 m.

<u>Explanation:</u>

To calculate the mass of solvent (water), we use the equation:

$Density=\frac{Mass}{Volume}$

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Density of water = 1 g/mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{750mL}\\\\\text{Mass of water}=750g$

To calculate the molality of solution, we use the equation:

$Molarity=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

Where,

$m_{solute}$ = Given mass of solute $(C_6H_{12}O_6)$ = 100.0 g

$M_{solute}$ = Molar mass of solute $(C_6H_{12}O_6)$ = 180 g/mol

$W_{solvent}$ = Mass of solvent (water) = 750 g

Putting values in above equation, we get:

$\text{Molality of }C_6H_{12}O_6=\frac{100\times 1000}{180\times 750}\\\\\text{Molality of }C_6H_{12}O_6=0.740m$

Hence, the molality of solution is 0.740 m.

6 0

3 years ago

How do I solve the following word problem: The half-life of Phosphorus - 32 is 14.3 days. It is used to study a plant's use of f

BaLLatris [955]

Half life is the time that it takes for half of the original value of some amount of a radioactive element to decay.

We have the following equation representing the half-life decay:

$A=A_o\times2^{(-\frac{t}{t_{half}})_{}_{}}$

A is the resulting amount after t time

Ao is the initial amount = 50 mg

t= Elapsed time

t half is the half-life of the substance = 14.3 days

We replace the know values into the equation to have an exponential decay function for a 50mg sample

$A=\text{ 50 }\times2^{\frac{-t}{14.3}}$

That would be the answer for a)

To know the P-32 remaining after 84 days we have to replace this value in the equation:

$\begin{gathered} A=\text{ 50 }\times2^{\frac{-84}{14.3}} \\ A=0.85\text{ mg} \end{gathered}$

So, after 84 days the P-32 remaining will be 0.85 mg

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1 year ago