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Llana [10]
3 years ago
12

Are the properties of Rubidium more similar to those of cesium or those of strontium?

Chemistry
1 answer:
fenix001 [56]3 years ago
5 0

More similar to Cesium

Explanation:

The properties of Rubidium are more similar to those of cesium compared to strontium.

Elements in the same group on the periodic table have similar chemical properties.

  • Rubidium and Cesium are located in the first group on the periodic table.
  • Other elements in this group are lithium, sodium, potassium and francium
  • Strontium belongs to the second group on the periodic table.
  • The first group have a ns¹ valence shells electronic configuration.
  • They are all referred to as alkali metals

Learn more:

Sodium brainly.com/question/6324347

Periodic table brainly.com/question/1704778

#learnwithBrainly

#learnwithBrainly

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Only someone who truly knows and understands this, please tell me if the electron configurations are right :
Setler79 [48]
Yes is 4s^23d^7=cobalt
4 0
3 years ago
Draw the best Lewis structure for NH3 by filling in the bonds, lone pairs, and formal charges. (Assign bonds, lone pairs, radica
kiruha [24]

Answer :  The Lewis-dot structure of NH_3 is shown below.

Explanation :

Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.

In the Lewis-dot structure the valance electrons are shown by 'dot'.

The given molecule is, NH_3

As we know that hydrogen has '1' valence electron and nitrogen has '5' valence electrons.

Therefore, the total number of valence electrons in NH_3 = 5 + 3(1) = 8

According to Lewis-dot structure, there are 6 number of bonding electrons and 2 number of non-bonding electrons.

Now we have to determine the formal charge for each atom.

Formula for formal charge :

\text{Formal charge}=\text{Valence electrons}-\text{Non-bonding electrons}-\frac{\text{Bonding electrons}}{2}

\text{Formal charge on N}=5-2-\frac{6}{2}=0

\text{Formal charge on }H_1=1-0-\frac{2}{2}=0

\text{Formal charge on }H_2=1-0-\frac{2}{2}=0

\text{Formal charge on }H_3=1-0-\frac{2}{2}=0

Hence, the Lewis-dot structure of NH_3 is shown below.

3 0
3 years ago
A ball sits on top of a hill. The wind blows and the ball begins rolling down the hill. What caused the ball to start moving?
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The answer should be a
4 0
3 years ago
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Determine the limiting reactant (lr) and the mass (in g) of nitrogen that can be formed from 50.0 g n2o4 and 45.0 g n2h4. some p
Licemer1 [7]
                                                   N2O4(l) + 2 N2H4(l) → 3 N2(g) + 4 H2O(g)
1) to calculate the limiting reactant you need to pass grams to moles.
<span> moles is calculated by dividing mass by molar mass
</span>
mass of N2O4: 50.0 g 
molar mass of <span>N2O4 = 92.02 g/mol
</span><span>molar mass of N2H4 = 32.05 g/mol.
</span>mass of N2H4:45.0 g

moles N2O4=50.0/92.02 g/mol= 0,54 mol of N2O4
moles N2H4= 45/32.05 g/mol= 1,40 mol of <span><span>N2H4

</span> 2)</span>
By looking at the balanced equation, you can see that 1 mol of N2O4 needs 2 moles of N2H4 to fully react . So to react  0,54 moles of N2O4, you need 2x0,54 moles of <span>N2H4 moles
</span><span>N2H4 needed = 1,08 moles.
You have more that 1,08 moles </span><span>N2H4, so this means the limiting reagent is not N2H4, it's </span>N2O4. The molecule that has molecules that are left is never the limiting reactant.

3) 1 mol of N2O4 reacting, will produce 3 mol of N2 (look at the equation)
There are 0,54 mol of N2O4 available to react, so how many moles will produce of N2?
1 mol N2O4------------3 mol of N2
0,54 mol N2O4--------x
x=1,62 mol of N2

4) the only thing left to do is convert the moles obtained, to grams.
We use the same formula as before, moles equal to mass divided by molar mass.
moles= \frac{grams}{molar mass}             (molar mass of N2= 28)
1,62 mol of N2= mass/ 28
mass of N2= 45,36 grams

4 0
3 years ago
HgS + O2 → HgO + SO2
Igoryamba

Answer:

2HgS + 3O2 → 2HgO + 2SO2

The coefficients are: 2, 3, 2, 2

Explanation:

HgS + O2 → HgO + SO2

The equation can be balance as follow:

Put 3 in front of O2 as shown below:

HgS + 3O2 → HgO + SO2

Now we can see that there are 6 atoms of O on the left side of the equation and a total of 3 atoms on the right side. It can be balance by putting 2 in front of HgO and SO2 as shown below:

HgS + 3O2 → 2HgO + 2SO2

Now we have 2 atoms of both Hg and S on the right side and 1atom each on the left. It can be balance by putting 2 in front of HgS as shown below:

2HgS + 3O2 → 2HgO + 2SO2

Now the equation is balanced.

The coefficients are: 2, 3, 2, 2

The law of conservation of mass(matter) states that matter(mass) can neither be created nor destroyed during a chemical reaction but changes from one form to another. An unbalanced equation suggests that matter has been created or destroyed. While a balanced equation proofs that matter can never be created but changes to different form. This is the more reason we have count the atoms of an element on both side of the equation to see if they are balanced irrespective of the new form they assume in the product

5 0
2 years ago
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