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3 years ago

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

Chemistry

1 answer:

NNADVOKAT [17]3 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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How do the metals in group 1 compare with the transition metals?

Nimfa-mama [501]

Answer:

Group 1 metals and transition metals are different from each other, mainly based on the colour of the chemical compounds that they form. The key difference between group 1 metals and transition metals is that the group 1 metals form colourless compounds, whereas the transition metals form colourful compounds.

6 0

3 years ago

Read 2 more answers

Quinine, an antimalarial drug, is 8.63% nitrogen. There are two nitrogen atoms per molecule. What is the molecular weight of qui

Furkat [3]

Answer:

324.18 g/mol

Explanation:

Let the molecular mass of the antimalarial drug, Quinine is x g/mol

According to question,

Nitrogen present in the drug is 8.63% of x

So, mass of nitrogen = $\frac {8.63}{100}\times x$

Also, according to the question,

2 atoms are present in 1 molecule of the drug.

Mass of nitrogen = 14.01 amu = 14.01 g/mol (grams for 1 mole)

So, mass of nitrogen = 14.01×2 = 28.02

These 2 must be equal so,

$\frac {8.63}{100}\times x=28.02$

solving for x, we get:

6 0

3 years ago

A 20.00-mL sample of a weak base is titrated with 0.0568 M HCl. At the endpoint, it is found that 17.88 mL of titrant was used.

nata0808 [166]

Answer: 0.0508mL

Explanation: Using the basic formula that states: C acid * V acid = C base * V base. we have:0.568 * 17.88 = 20 * C base.

therefore concentration of the base is 1.0156/20 = 0.0508 mL

7 0

3 years ago

Consider the reaction CH4(g) 2O2(g)CO2(g) 2H2O(g) Using standard thermodynamic data at 298K, calculate the entropy change for th

zvonat [6]

Answer:

the entropy change for the surroundings when 1.62 moles of CH4(g) react at standard conditions is −8.343 J/K

Explanation:

The balanced chemical equation of the reaction in the question given is:

$CH_{4(g)} + 2O_{2(g)} \to CO_{2(g)} + 2 H_2O _{(g)}$

Using standard thermodynamic data at 298K.

The entropy of each compound above are listed as follows in a respective order.

Entropy of (CH4(g)) = 186.264 J/mol.K

Entropy of (O2(g)) = 205.138 J/mol.K

Entropy of (CO2(g)) = 213.74 J/mol.K

Entropy of (H2O(g)) = 188.825 J/mol.K

The change in Entropy (S) of the reaction is therefore calculated as follows:

$=1*S(CO2(g)) + 2*S(H2O(g)) - 1*S( CH4(g)) - 2*S(O2(g))$

$=1*(213.74) + 2*(188.825) - 1*(186.264) - 2*(205.138)$

= -5.15 J/mol.K

Given that :

the number of moles = 1.62 of CH4(g) react at standard conditions.

Then;

The change in entropy of the rxn $= 1.62 \ mol * -5.15 \ J/mol.K$

= −8.343 J/K

6 0

3 years ago

Ep-15 why is carbon monoxide especially dangerous?

Rzqust [24]

Carbon monoxide is dangerous because it binds with hemoglobin in the blood.

Hemoglobin is made up of proteins that bind to iron atoms. The structure of the protein facilitates loose binding of oxygen. On other hand, Carbon monoxide binds very strongly to the iron in hemoglobin. Once carbon monoxide is bonded to hemoglobin, it is very difficult to release. This, eventually results in blood losing it its ability to transport oxygen. Hence, the person will suffocate. Due to this, CO is dangerous.

8 0

3 years ago