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2 years ago

5

Lanthanium found in the......... period

1 answer:

Gnesinka [82]2 years ago

5 0

I'm preatty sure it is in the periodic table xxx

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water’s molar mass is 18.01 g/mol. The molar mass of glycerol is 92.09 g/mol. At 25 celsius, glycerol is more viscous than water

Gemiola [76]

Answer is: glycerol because it is more viscous and has a larger molar mass.

Viscosity depends on intermolecular interactions.

The predominant intermolecular force in water and glycerol is hydrogen bonding.

Hydrogen bond is an electrostatic attraction between two polar groups in which one group has hydrogen atom (H) and another group has highly electronegative atom such as nitrogen (like in this molecule), oxygen (O) or fluorine (F).

4 0

3 years ago

When a lead acid car battery is recharged by the alternator, it acts essentially as an electrolytic cell in which solid lead(II)

katen-ka-za [31]

Answer:

3.81 g Pb

Explanation:

When a lead acid car battery is recharged, the following half-reactions take place:

Cathode: PbSO₄(s) + H⁺ (aq) + 2e⁻ → Pb(s) + HSO₄⁻(aq)

Anode: PbSO₄(s) + 2 H₂O(l) → PbO₂(s) + HSO₄⁻(aq) + 3H⁺ (aq) + 2e⁻

We can establish the following relations:

1 A = 1 c/s
1 mole of Pb(s) is deposited when 2 moles of e⁻ circulate.
The molar mass of Pb is 207.2 g/mol
1 mol of e⁻ has a charge of 96468 c (Faraday's constant)

Suppose a current of 96.0A is fed into a car battery for 37.0 seconds. The mass of lead deposited is:

$37.0s.\frac{96.0c}{s} .\frac{1mole^{-} }{96468c} .\frac{1molPb}{2mole^{-} } .\frac{207.2gPb}{1molPb} =3.81gPb$

6 0

3 years ago

An unknown piece of metal weighing 95.0 g is heated to 98.0°C. It is dropped into 250.0 g of water at 23.0°C. When equilibrium i

lutik1710 [3]

Answer:

$C_{metal}=126.6\frac{J}{g\°C}$

Explanation:

Hello!

In this case, when two substances at different temperature are put in contact and an equilibrium temperature is attained, we can evidence that the heat lost by the hot substance (metal) is gained by the cold substance (water) and we can write:

$Q_{metal}=-Q_{water}$

Which can be also written as:

$m_{metal}C_{metal}(T_{EQ}-T_{metal})=-m_{water}C_{water}(T_{EQ}-T_{water})$

Thus, since we need the specific heat of the metal, we solve for it as shown below:

$C_{metal}=\frac{m_{water}C_{water}(T_{EQ}-T_{water})}{-m_{metal}(T_{EQ}-T_{metal})} \\\\C_{metal}=\frac{250.0g*4.184\frac{J}{g\°C}(29.0\°C-98.0\°C)}{95.0g(29.0\°C-23.0\°C)} \\\\C_{metal}=126.6\frac{J}{g\°C}$

Best regards.

7 0

2 years ago

Balancing nuclear equations is slightly different than balancing chemical equations. The major difference is that in nuclear rea

vesna_86 [32]

Answer:

The correct answer is - 4.

Explanation:

As we known and also given that the total of the superscripts that is mass numbers, A in the reactants and products must be the same.The mass of products A can understand and calculated by this -

The sum of the product mass number of products = mass of reactant

237Np93 →233 Pa91 +AZX is the equation,

Solution:

Mass of reactants = 237

Mass of products are - Pa =233 and A = ?

233 + A = 237

A = 237 - 233

A = 4

So the equation will be:

237Np93 →233 Pa91 +4He2 (atomic number Z = 2 ∵ difference in the atomic number of reactant and products)

5 0

2 years ago

Na = ? Periodic Table

8_murik_8 [283]

Answer:

Na on the periodic table is sodium.

Hope this was useful to you

8 0

3 years ago

Read 2 more answers