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3 years ago

Why are group 1 elements more reactive than group 2 elements?

Chemistry

2 answers:

sashaice [31]3 years ago

4 0

Answer:

its because in group 2 elements,outermost electron is difficult to be removed than group 1 elements

Hence group 2 elements are less reactive than group 1

kaheart [24]3 years ago

3 0

Less reactive than Group I elements. The reasoning for this is because it is more difficult to lose two electrons compared to losing just one electron. They mostly React with water to form alkaline solutions. ...Now This is because the smaller an atom the closer the outer electrons are to the nucleus.

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In a cell, what is the function of the Cell Membrane?

joja [24]

The answer is C :

It controls the entry and exit of substances.

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3 years ago

A cooling curve has two flat lines, or plateaus. What does the plateau at the higher temperature represent?

blagie [28]

Answer:

It is in the state of "thermal arrest"

Explanation:

The temperature stays constant during the phase change process . This is because the matter has more internal energy and heat has to be taken away for the solidification process to begin. The energy that is required for a phase change is know as latent heat (which is the energy released or absorbed by a body during a thermodynamic process).

5 0

3 years ago

Read 2 more answers

A hypothetical element has an atomic weight of 48.68 amu. It consists of three isotopes having masses of 47.00 amu, 48.00 amu, a

Morgarella [4.7K]

Answer : The percent abundance of the heaviest isotope is, 78 %

Explanation :

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$

As we are given that,

Average atomic mass = 48.68 amu

Mass of heaviest-weight isotope = 49.00 amu

Let the percentage abundance of heaviest-weight isotope = x %

Fractional abundance of heaviest-weight isotope = $\frac{x}{100}$

Mass of lightest-weight isotope = 47.00 amu

Percentage abundance of lightest-weight isotope = 10 %

Fractional abundance of lightest-weight isotope = $\frac{10}{100}$

Mass of middle-weight isotope = 48.00 amu

Percentage abundance of middle-weight isotope = [100 - (x + 10)] % = (90 - x) %

Fractional abundance of middle-weight isotope = $\frac{(90-x)}{100}$

Now put all the given values in above formula, we get:

$48.68=[(47.0\times \frac{10}{100})+(48.0\times \frac{(90-x)}{100})+(49.0\times \frac{x}{100})]$

$x=78\%$

Therefore, the percent abundance of the heaviest isotope is, 78 %

5 0

3 years ago

Read 2 more answers

One mole of an ideal gas with a volume of 1.0 L and a pressure of 5.0 atm is allowed to expand isothermally into an evacuated bu

Deffense [45]

Answer:

w= - 1.7173 kJ, q= 1.7173 kJ, q(rev) = 1717.3 J = 1.7173 kJ.

Explanation:

Okay, from the question we are given the information below;

Number of moles, n= 1 mole; initial volume, v(1) = 1.0 litres (L); pressure (p) = 5atm, final volume(v2) = 2.0 Litres(L) ; the workdone, w= not given; the heat, q and q(rev)= not given and the gas was said to expand isothermally.

So, this question is a question from the part of chemistry known as thermodynamics. Therefore, grip yourself we are delving into thermodynamics 'waters' now.

For expansion isothermally; the workdone, w= -nRT ln v2/v1.

Where T= temperature= 25° C = 298 k and R= gas constant.

Therefore; workdone, w = - 1 × 8.314 × 298 × ln(2/1).

Workdone,w= - 1717.32204643. =

- 1717.3 Joules (J).

==> Workdone,w= - 1.7173 kJ.

Then, we are to find q. q can be solved by using the first law of thermodynamics, which by mathematical representation is:

∆U= q + w. Where ∆U= change in internal enegy. Since the question is dealing with isothermal expansion, there is this rule that says for an isothermal expansion ∆U = 0.

Hence, 0 =q + [- 1717.3 Joules (J)].

q=1717.3 J = 1.7173 kJ.

Finally, the q(rev) which is= nRT ln (v2/V1).

q(rev) = 1 × 8.314 × 298 ln (2/1).

q(rev) = 1717.3 J = 1.7173 kJ.

PS: please note the negative signs in the workdone and the positive sign in the q(rev).

7 0

3 years ago

7.02 x 10^23 molecules of X2, a ssubstance consisting of diatomic molecules, has a mass of 296 grams. Determine the atomic weigh

Fiesta28 [93]

Answer:

d. 127 g/mol.

Explanation:

Hello!

In this case, since we have the amount of molecules of this this compound, we are able to compute the moles out there by using the Avogadro's number:

$mol=7.02x10^{23}molec*\frac{1mol}{6.022x10^{23}molec}=1.17mol$

Which correspond to the moles of X2. Then, by using the mass we are able to compute the molar mass of X2:

$MM=\frac{296g}{1.17mol}\\\\MM=254g/mol$

It means that the atomic mass of X halves the molar mass of X2, which is then d. 127 g/mol.

Best regards!

4 0

3 years ago