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vova2212 [387]
3 years ago
14

How do prevailing winds affect air temperature?

Chemistry
1 answer:
GalinKa [24]3 years ago
4 0

Answer:

Explanation:

Prevailing winds bring air from one type of climate to another. For example, warm winds that travel over water tend to collect moisture as they travel; the water vapor in the air will condense as it moves into colder climates, which is why temperate coastal areas often receive heavy rainfall

Mark as brainliest for further answers :)

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hammer [34]

B. Magnesium + Hydrogen Sulfide (Reactors) ---->  Magnesium Sulfide + Hydrogen (Products)

8 0
3 years ago
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When mixtures of gaseous h2 and gaseous cl2 react, a product forms that has the same properties regardless of the relative amoun
lilavasa [31]
According to Dalton's Atomic Theory, the <em>Law of Definite Proportion is applied when a compound is always made up by a fixed fraction of its individual elements.</em> This is manifested by the balancing of the reaction.

The reaction for this problem is:

H₂ + Cl₂ → 2 HCl

1 mol of H₂ is needed for every 1 mole of Cl₂. Assuming these are ideal gases, the moles is equal to the volume. So, if equal volumes of the reactants are available, they will produce twice the given volumes of HCl.
4 0
3 years ago
How many grams are in 1.23 x 1020 atoms of arsenic?
Alina [70]

Given the data from the question, the mass of arsenic that contains 1.23×10²⁰ atoms is 0.0153 g

<h3>Avogadro's hypothesis </h3>

6.02×10²³ atoms = 1 mole of arsenic

But

1 mole of arsenic = 75 g

Thus, we can say that:

6.02×10²³ atoms = 75 g of arsenic

<h3>How to determine the mass that contains 1.23×10²⁰ atoms</h3>

6.02×10²³ atoms = 75 g of arsenic

Therefore,

1.23×10²⁰ atoms = (1.23×10²⁰ × 75) / 6.02×10²³ atoms)

1.23×10²⁰ atoms = 0.0153 g of arsenic

Thus, 1.23×10²⁰ atoms is present in 0.0153 g of arsenic

Learn more about Avogadro's number:

brainly.com/question/26141731

6 0
2 years ago
Rank the following aqueous solutions from highest to lowest freezing point: 0.1 m FeCl3, 0.30 m glucose (C6H12O6), 0.15 m CaCl2.
Tresset [83]

Answer:

Explanation: See images below for explanation

8 0
3 years ago
Find [H+] of a 0.056 M hydrofluoric acid solution. Ka = 1.45 x 10-7
brilliants [131]

Answer:  [H^+] of 0.056 M HF solution is 8.96\times 10^{-5}

Explanation:

HF\rightarrow H^+F^-

 cM              0             0

c-c\alpha        c\alpha          c\alpha  

So dissociation constant will be:

K_a=\frac{(c\alpha)^{2}}{c-c\alpha}

Give c= 0.056 M and \alpha = ?

K_a=1.45\times 10^{-7}

Putting in the values we get:

1.45\times 10^{-7}=\frac{(0.056\times \alpha)^2}{(0.056-0.056\times \alpha)}

(\alpha)=0.0016

[H^+]=c\times \alpha

[H^+]=0.056\times 0.0016=8.96\times 10^{-5}  

Thus [H^+] of 0.056 M HF solution is 8.96\times 10^{-5}

8 0
3 years ago
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