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Oduvanchick [21]
2 years ago
14

Explain why on a hot sunny day it is cooler in the shade​

Chemistry
1 answer:
kotykmax [81]2 years ago
5 0

Shade only feels cooler because you are avoiding solar radiation. In reality, the temperature in the sun is the same as the temperature in the shade. ... When in the shade, your skin is not being “heated” by the sun's rays, so your skin and your body feel a more comfortable temperature.

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Plz help me : Copper (I) chloride CuCl is made up of:
TEA [102]
C. Metal and Nonmetal
8 0
2 years ago
The reduction of nitrogen monoxide is described by the following chemical equation: 2H2 (g) +2NO (g) 2H20 ()N2 (g Suppose a two-
Ipatiy [6.2K]

Answer:

Reasonable Second step- N_{2}O(g)+H_{2}(g)\rightarrow N_{2}(g)+H_{2}O(g)

Explanation:

The given single step chemical reaction is as follows.

2H_{2}(g)+2NO \rightarrow 2H_{2}O(g)+N_{2}(g)

Suppose a two-step mechanism is proposed for this reaction,

The reaction occured in two steps they are as follows.

Step -1:H_{2}(g)+2NO \rightarrow N_{2}O(g)+H_{2}O(g)

Step-2:N_{2}O(g)+H_{2}(g)\rightarrow N_{2}(g)+H_{2}O(g)

8 0
3 years ago
How many moles of helium are needed to fill a balloon to a volume of 5.3 L at 22 ℃ and 632 mmHg?
son4ous [18]

Answer:

0.18 moles

Explanation:

Applying,

PV = nRT................... Equation 1

Where P = pressure, V = volume, n = number of moles, R = molar gas constant, T = temperature.

make n the subject of the equation

n = PV/RT............... Equation 2

Given: V = 5.3 L, T = 22 °C = (22+272) K = 295 K, P = 632 mmHg = (0.00131579×632) = 0.8316 atm,  R = 0.083 L.atm/K.mol

Substitute these values into equation 2

n = (0.8316×5.3)/(0.083×295)

n = 0.18 moles

6 0
3 years ago
A student wishes to calculate the experimental value of Ksp for AgI. S/he follows the procedure in Part 3 and finds Ecell to be
Ymorist [56]

Answer:

a)    [Ag+]dilute = 6.363  × 10⁻¹⁶ M  

b)    1.273 × 10⁻¹⁶

c)    2.629×10⁻¹⁹ M Thus; the value for  [Ag+ ]dilute will be too low

Explanation:

In an Ag | Ag+ concentration cell ,

The  anode reaction can be written as :

Ag ----> Ag+(dilute) + e-    &:

The  cathode reaction can be written as:

Ag+(concentrated) + e- ----> Ag

The  Overall Reaction : is

Ag+(concentrated) -----> Ag+(dilute)

However, the Standard Reduction potential of cell = E°cell = 0

( since both cathode and anode have same Ag+║Ag )

Also , given that the theoretical slope is - 0.0591 V

Therefore; the reduction potential of cell ; i.e

Ecell = E°cell - 0.0591 V × log ( [Ag+]dilute / [Ag+]concentrated )

0.839 V = 0 - 0.0591 V × log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) )  

log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) ) = - 14.1963  

[Ag+]dilute = \mathbf{10^{-14.1963} } × 1.0 × 10⁻¹ M

[Ag+]dilute = 6.363  × 10⁻¹⁶ M  

b)

AgI ----> Ag + (dilute) + I⁻

So , Solubility product = Ksp = [Ag⁺]dilute × [I⁻]  

= 6.363 × 10⁻¹⁶ M × 0.20 M  

= 1.273 × 10⁻¹⁶

c) If s/he mistakenly uses 1.039 V as Ecell; then the value for [Ag+]dilute will be :

Ecell = E°cell - 0.0591 V × log ( [Ag+]dilute / [Ag+]concentrated )

1.039 V = 0 - 0.0591 V × log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) )  

log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) ) = - 17.5804  

[Ag+]dilute = \mathbf{10^{-17.5804} } × 1.0 × 10⁻¹ M

[Ag+]dilute = 2.629×10⁻¹⁹ M

Thus, the value for  [Ag+ ]dilute will be too low

5 0
3 years ago
In a neutral atom, the number of protons _______ the number of electrons.
sergejj [24]

Answer:

equal the number of electrons.

Explanation:

8 0
3 years ago
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