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Natali5045456 [20]
3 years ago
10

Atoms of elements at the top of a group on the periodic table are smaller than the atoms of elements at the bottom of the group.

How does this help explain the difference in the reactivity of metals within a group? The valence electrons of the larger atoms are farther from the nucleus and are easier to remove, so the metals near the top are more reactive than those at the bottom. The valence electrons of the larger atoms are farther from the nucleus and are harder to remove, so the metals near the bottom are more reactive than those at the top. The valence electrons of the larger atoms are farther from the nucleus and are harder to remove, so the metals near the top are more reactive than those at the bottom. The valence electrons of the larger atoms are farther from the nucleus and are easier to remove, so the metals near the bottom are more reactive than those at the top.
Chemistry
2 answers:
krok68 [10]3 years ago
8 0

Answer:

the answer is D.The valence electrons of the larger atoms are farther from the nucleus and are easier to remove, so the metals near the bottom are more reactive than those at the top.

Explanation:

Serjik [45]3 years ago
7 0

I think the correct answer would be the last one.

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Probability is a number that describes how likely it is that a certain event will occu O True O False​
lara31 [8.8K]

Answer:

Between 0 and 1

Explanation:

Hope this helps :)

6 0
3 years ago
Ice cream is made by freezing a liquid mixture that, as a first approximation, can be considered a solution of sucrose in water.
hjlf

Answer:

Freezing point is -2.81°C

Explanation:

34g/342gmol^-1 = 0.0994mol

n = m/mr

Molarity= 0.994/ 0.66 = 1.51M

◇T = -i × m ×Kf

Where ◇T is freezing depression

i= Vant Hoff factor

m = molarity

Kf = freezing content = 1.

860kgmol^-1

◇T =-1 × 1.51 × 1.860 = - 2.81°C

6 0
3 years ago
Which group of elements shares characteristics with both metals and nonmetals?
sertanlavr [38]

The answer to your question is,

Metalloids. They are a mix of elements that are both metals and non-metals in one.

-Mabel <3

4 0
3 years ago
Read 2 more answers
Find the enthalpy of neutralization of HCl and NaOH. 137 cm3 of 2.6 mol dm-3 hydrochloric acid was neutralized by 137 cm3 of 2.6
liraira [26]

Answer : The correct option is, (D) 89.39 KJ/mole

Explanation :

First we have to calculate the moles of HCl and NaOH.

\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=2.6mole/L\times 0.137L=0.3562mole

\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=2.6mole/L\times 0.137L=0.3562mole

The balanced chemical reaction will be,

HCl+NaOH\rightarrow NaCl+H_2O

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.3562 mole of HCl neutralizes by 0.3562 mole of NaOH

Thus, the number of neutralized moles = 0.3562 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = 137ml+137ml=274ml

\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 274ml=274g

Now we have to calculate the heat absorbed during the reaction.

q=m\times c\times (T_{final}-T_{initial})

where,

q = heat absorbed = ?

c = specific heat of water = 4.18J/g^oC

m = mass of water = 274 g

T_{final} = final temperature of water = 325.8 K

T_{initial} = initial temperature of metal = 298 K

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

q=274g\times 4.18J/g^oC\times (325.8-298)K

q=31839.896J=31.84KJ

Thus, the heat released during the neutralization = -31.84 KJ

Now we have to calculate the enthalpy of neutralization.

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy of neutralization = ?

q = heat released = -31.84 KJ

n = number of moles used in neutralization = 0.3562 mole

\Delta H=\frac{-31.84KJ}{0.3562mole}=-89.39KJ/mole

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 89.39 KJ/mole

3 0
4 years ago
What is a chemical equation for the second ionization for lithium
Orlov [11]
The 2nd ionization energy is removing a 2nd electron from that resulting cation: 
<span>Li+ --> Li2+ + 1e- </span>
5 0
3 years ago
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