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zvonat [6]
3 years ago
9

Why is more dew formed on a grass and not on polished metal surface​

Chemistry
1 answer:
Masja [62]3 years ago
8 0

Answer

Incidentally, they all form dew but because grass is closer to the soil, which maintains heat longer than the air where tree leaves are found, you will find more condensation in that region. Also, higher up the wind and sun will allow faster evaporation which makes the dew less noticable on tree leaves.

Explanation:

Preferred objects of dew formation are thus poor conducting or well isolated from the ground, and non-metallic, while shiny metal coated surfaces are poor infrared radiators. ... If the wet soil beneath is the major source of vapor, however (this type of dew formation is called distillation), wind always seems adverse.

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Human use of groundwater has _______ over time.
Harman [31]
Increased is the answer
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3 years ago
Can I please know What are the answers for #s 1,2,3,4 ?
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1 wavelength, 2 crest, 3 trough, 4 wave height <3
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Read 2 more answers
How many moles are in 213 grams of NaCl?
Nata [24]

Answer:

There are 3, 64 moles of NaCl.

Explanation:

First we calculate the mass of 1 mol of NaCl, starting from the atomic weights of Na and Cl obtained from the periodic table. Then we calculate themoles in 213 grams of NaCl, making a simple rule of three:

Weight NaCl= Weight Na + Weight Cl = 23 g + 35, 5 g= 58, 5 g/ mol

58,5 g ------1 mol NaCl

213 g---------x= (213 g x 1 mol NaCl)/ 58, 5 g= <em>3, 64 mol NaCl</em>

5 0
4 years ago
Find the empirical formula of the following compounds:
Aneli [31]

The empirical formula of the following compounds 0.903 g of phosphorus combined with 6.99 g of bromine.

<h3>What is empirical formula?</h3>

The simplest whole number ratio of atoms in a compound is the empirical formula of a chemical compound in chemistry. Sulfur monoxide's empirical formula, SO, and disulfur dioxide's empirical formula, S2O2, are two straightforward examples of this idea. As a result, both the sulfur and oxygen compounds sulfur monoxide and disulfur dioxide have the same empirical formula.

<h3>How to find the empirical formula?</h3>

Convert the given masses of phosphorus and bromine into moles by multiplying the reciprocal of their molar masses. The molar masses of phosphorus and bromine are 30.97 and 79.90 g/mol, respectively.

Moles phosphorus = 0.903 g phosphorus \frac{mol phosphorus}{ 30.97 g phosphorus}= 0.0293 mol

Moles bromine 6.99 g bromine\frac{mol bromine}{79.90 g bromine}=0.0875 mol

The preliminary formula for compound is P0.0293Bro.0875. Divide all the subscripts by the subscript with the smallest value which is 0.0293. The empirical formula is P1.00Br2.99 ≈ P₁Br3 or PBr3

To learn more about empirical formula visit:

brainly.com/question/14044066

#SPJ4

8 0
1 year ago
Pyridine (c5h5n) is a particularly foul-smelling substance used in manufacturing pesticides and plastic resins. calculate the ph
zubka84 [21]

Answer is: ph value of pyridine solution is 9.1.

Chemical reaction: C₅H₅N + H₂O → C₅H₅NH⁺ + OH⁻.<span>
c(pyridine - C</span>₅H₅N) = 0.115M.<span>
Kb(C</span>₅H₅N) = 1.4·10⁻⁹.
[C₅H₅NH⁺] = [OH⁻] = x; equilibrium concentration.<span>
[</span>C₅H₅N] = 0.115 M - x.

Kb = [C₅H₅NH⁺] · [OH⁻] / [C₅H₅N].

1.4·10⁻⁹ = x² / (0.115 M -x)

Solve quadratic equation: x = [OH⁻] = 0.0000127 M.<span>
pOH = -log(0.0000127 M) = 4.9</span>

<span>pH = 14 - 4.9 = 9.1.</span>

7 0
4 years ago
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