All categories

2 years ago

The no.of radial nodes of 3s and 2p orbital are respectively

Chemistry

1 answer:

vesna_86 [32]2 years ago

7 0

Is this incomplete, because this makes no sense?

You might be interested in

In the Gabriel synthesis of primary amines, N-potassiophthalimide is used as a source of the nitrogen atom. Complete the synthes

Amiraneli [1.4K]

Reaction is attached:

4 0

2 years ago

I need help Type th temperature (in Kelvin) in the left column and the volumes in the right column, being sure to keep pairs of

e-lub [12.9K]

Answer:

sorry nahi pata tha muje

7 0

2 years ago

Read 2 more answers

If 25.0 g of NH₃ and 45.0g of O₂ react in the following reaction, what is the mass in grams of NO that will be formed? 4 NH₃ (g)

Allisa [31]

Answer:

The correct answer would be : 33.8 g

Explanation:

Molar mass of ammonia,

Molar Mass = 1* Molar Mass(N) + 3* Molar Mass (H)

= 1*14.01 + 3*1.008 = 17.034 g/mol

mass(NH3)= 25.0 g (given)

number of mol of NH3,

n = mass of NH3/molar mass of NH3

=(25.0 g)/(17.034 g/mol)

= 1.468 mol

Now,

Molar mass of O2

= 32 g/mol

mass(O2)= 45.0 g

similar as ammonia

n (O2)=(45.0 g)/(32 g/mol)

= 1.406 mol

Balanced chemical equation is:

4 NH3 + 5 O2 ---> 4 NO + 6 H2O

1.83456 mol of O2 is required for 1.46765 mol of NH3

by the calculation we have only 1.40625 mol of O2

Thus, the limiting agent will be - O2

now the Molar mass of NO,

= 1*14.01 + 1*16.0

= 30.01 g/mol (similar formula used for NH3)

Balanced equation :

mol of NO formed = (4/5)* moles of O2

= (4/5)×1.40625 (from above calculation)

= 1.125 mol

mass of NO = number of moles × molar mass

= 1.125*30.01

= 33.8 g

Thus, the correct answer would be : 33.8 g

5 0

2 years ago

Electroplating is a way to coat a complex metal object with a very thin (and hence inexpensive) layer of a precious metal, such

8_murik_8 [283]

Answer:

0.0164 g

Explanation:

Let's consider the reduction of silver (I) to silver that occurs in the cathode during the electroplating.

Ag⁺(aq) + 1 e⁻ → Ag(s)

We can establish the following relations.

1 A = 1 C/s
The charge of 1 mole of electrons is 96,468 C (Faraday's constant)
1 mole of Ag(s) is deposited when 1 mole of electrons circulate.
The molar mass of silver is 107.87 g/mol

The mass of silver deposited when a current of 0.770 A circulates during 19.0 seconds is:

$19.0s \times \frac{0.770c}{s} \times \frac{1mole^{-} }{96,468C} \times \frac{1molAg}{1mole^{-}} \times \frac{107.87g}{1molAg} = 0.0164 g$

5 0

2 years ago

kondor19780726 [428]

Explanation:

It is known that 1 gram contains 1000 milligrams. And, mathematically we can represent it as follows.

$\frac{1 g}{1000 mg}$ or $\frac{1000 mg}{1 g}$

So, when we have to convert grams into milligrams then we simply multiply the digit with 1000. And, if we have to convert a digit from milligrams to grams then we simply divide it by 1000.

4 0

2 years ago