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nadezda [96]
3 years ago
10

How many valence electrons does Nitrogen have?​

Chemistry
1 answer:
irinina [24]3 years ago
6 0

Answer:

5 valence electrons

Nitrogen has a total of 5 valence electrons

Explanation:

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What amount of energy is required to change a spherical drop of water with a diameter of 1.80 mm to three smaller spherical drop
Gekata [30.6K]
This is a straightforward question related to the surface energy of the droplet. 

<span>You know the surface area of a sphere is 4π r² and its volume is (4/3) π r³. </span>

<span>With a diameter of 1.4 mm you have an original droplet with a radius of 0.7 mm so the surface area is roughly 6.16 mm² (0.00000616 m²) and the volume is roughly 1.438 mm³. </span>

<span>The total surface energy of the original droplet is 0.00000616 * 72 ~ 0.00044 mJ </span>

<span>The five smaller droplets need to have the same volume as the original. Therefore </span>

<span>5 V = 1.438 mm³ so the volume of one of the smaller spheres is 1.438/5 = 0.287 mm³. </span>

<span>Since this smaller volume still has the volume (4/3) π r³ then r = cube_root(0.287/(4/3) π) = cube_root(4.39) = 0.4 mm. </span>

<span>Each of the smaller droplets has a surface area of 4π r² = 2 mm² or 0.0000002 m². </span>

<span>The surface energy of the 5 smaller droplets is then 5 * 0.000002 * 72.0 = 0.00072 mJ </span>
<span>From this radius the surface energy of all smaller droplets is 0.00072 and the difference in energy is 0.00072- 0.00044 mJ = 0.00028 mJ. </span>

<span>Therefore you need roughly 0.00028 mJ or 0.28 µJ of energy to change a spherical droplet of water of diameter 1.4 mm into 5 identical smaller droplets. </span>
7 0
3 years ago
The solubility of N2 in blood can be a serious problem (the "bends") for divers breathing compressed air (78% N2 by volume) at d
OLga [1]

Answer:

The volume is 19.7 mL

Explanation:

<u>Step 1</u>: Given data

Pressure at sea level = 1.00 atm

Pressure at 50 ft = 2.47535 atm

kH for N2 in water at 25°C is 7.0 × 10−4 mol/L·atm

Molarity (M) = kH x P

<u>Step 2</u>: Calculate molarity

M at sea level:

M = 7.0*10^-4 * (1.00atm * 0.78) = 5.46*10^-4 mol/L

M at 50ft:

M = 7.0*10^-4 * (2.47535atm * 0.78) = 13.5*10^-4 mol/L

We should find the volume of N2. To find the volume whe have to find the number of moles first. This we calculate by calculating the difference between M at 50 ft and M at sea level.

13.5*10^-4 mol/L - 5.46*10^-4 mol/L = 8.04*10^-4 mol/L

Step 3: Calculate volume

P*V=nRT

with P = 1.00 atm

with V = TO BE DETERMINED

with n =  8.04*10^-4 mol/L  *1L = 8.04*10^-4

with R= 0.0821 atm * L/ mol *K

with T = 25 °C = 273+25 = 298 Kelvin

To find the volume, we re-organize the formula to: V=nRT/P

V= (8.04*10^-4 mol * 0.0821 (atm*L)/(mol*K)* 298K ) / 1.00atm = 0.0197L = 19.7ml

The volume is 19.7 mL

5 0
4 years ago
Carbon Dioxide in a 34 L container at a pressure of 387 mm Hg is transferred to a container with a pressure of 274 mm Hg. What i
AVprozaik [17]

Answer:

V2= 0.0796m3

Explanation:

We have that for a real gas the following relation is fulfilled:

PV / T = constant

To use this formula I need to pass the pressure of mmHg to KPa and the volume of L to m3, knowing that

 1Kpa = 7.50062mmHg

1L = 0.001m3

So i can say

P1V1/T1 =P2V2/T2

So

V2= P1V1T2/P2T1

V2= 51.60KPa*0.034m3*350k/36.53KPa*211k

finally

V2= 0.0796m3

7 0
3 years ago
Which statement describes the Arrhenius interpretation of acids and bases
KIM [24]
<span>it is limited to situations that involve aqueous solutions of specific compounds

I hope this helps</span>
8 0
4 years ago
Read 2 more answers
Name the ionic compound represented by the following chemical formula: CuO
Illusion [34]

Answer:

Cupric oxide, or copper (II) oxide, 

Explanation:

Correct me if im wrong tnx:<

6 0
3 years ago
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