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1 year ago

All weather is energy in motion. Where does the energy come from

Chemistry

2 answers:

alekssr [168]1 year ago

8 0

Answer:

The Sun radiates huge amounts of energy. Only a small portion of that energy hits the Earth, but it is enough to light our days, heat our air and land, and create weather systems over the oceans. Most of the energy you will learn about comes from the Sun. The Earth also gives off energy.

Explanation:

expeople1 [14]1 year ago

3 0

Answer:

"The Sun radiates huge amounts of energy. Only a small portion of that energy hits the Earth, but it is enough to light our days, heat our air and land, and create weather systems over the oceans. Most of the energy you will learn about comes from the Sun. The Earth also gives off energy."

By the way I got this off thee internet and copy and pasted it so don't write it just look and write in your own words please. Please do not copy and paste or write what i wrote because this is someone else's explanation or answer

Hope this helps!! Also please rank thank you.

Explanation:

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svlad2 [7]

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2 years ago

What is the smallest part of a tree that can do all the things needed for life??

nalin [4]

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2 years ago

Problem PageQuestion Aqueous sulfuric acid will react with solid sodium hydroxide to produce aqueous sodium sulfate and liquid w

quester [9]

Answer:

0.72g

Explanation:

Step 1:

We'll begin by writing a balanced equation for the reaction. This is illustrated below:

H2SO4 + 2NaOH —> Na2SO4 + 2H2O

Step 2:

Determination of the mass of sulphuric acid (H2SO4) and the mass of sodium hydroxide (NaOH) that reacted from the balanced equation. This is illustrated below:

Molar Mass of H2SO4 = (2x1) + 32 +(16x4) = 2 + 32 + 64 = 98g/mol

Molar Mass of NaOH = 23 + 16 + 1 = 40g/mol

Mass of NaOH from the balanced equation = 2 x 40 = 80g

Step 3

Determination of the limiting reactant. To do this, we need to know which of the reactant is excess.

Now let us consider using all of the mass of NaOH given to see if there will be left over for H2SO4. This is illustrated below:

From the balanced equation above,

98g of H2SO4 required 80g of NaOH.

Therefore, Xg of H2SO4 will require 1.6g of NaOH i.e

Xg of H2SO4 = (98x1.6)/80

Xg of H2SO4 = 1.96g

Now comparing the mass of H2SO4 that reacted ( i.e 1.96g) and the mass of H2SO4 given ( i.e 2.94g), we can see clearly that there are left over ( i.e 2.94 - 1.96 = 0.98g) of H2SO4. Therefore, H2SO4 is the excess reactant and NaOH is the limiting reactant.

Step 4:

Determination of the mass of water produced from the reaction. This is illustrated below:

The balanced equation for the reaction is given below:

H2SO4 + 2NaOH —> Na2SO4 + 2H2O

Molar Mass of H2O = (2x1) + 16 = 2 + 16 = 18g/mol

Mass of H2O from the balanced equation = 2 x 18 = 36g

From the balanced equation above,

80g of NaOH reacted to produced 36g of H2O.

Therefore, 1.6g of NaOH will react to produce = (1.6 x 36)/80 = 0.72g of H2O.

Therefore, the maximum mass of water (H2O) produced by the chemical reaction of aqueous sulfuric acid with solid sodium hydroxide is 0.72g

4 0

2 years ago

Which of the following series of isoelectronic ions (Mg²⁺, N³⁻, F⁻, Si⁴⁺) has the ionic radii in order of largest to smallest? A

Rom4ik [11]

Answer:

Option (d) is correct

N³⁻ > F⁻ > Mg²⁺ > Si⁴⁺

Explanation:

Total electrons for all the species = 10

So these are iso electronic with each other.

We know

Ionic radii ∝ $\frac{Magnitude of Negative Charge}{Magnitude of Positive Charge}$

Si⁴⁺ has 14 protons and 10 electrons
Mg²⁺ has 12 protons and 10 electrons
N³⁻ has 7 protons and 10 electrons
F⁻ has 9 protons and 10 electrons

Iso electronic species with greatest number of protons have small size and vice versa.
So Si⁺⁴ have smallest size and N³⁻ have largest in size

4 0

2 years ago

Consider the first-order reaction described by the equation At a certain temperature, the rate constant for this reaction is 5.8

zubka84 [21]

Answer: The half life of the reaction is 1190.7 seconds

Explanation:

The equation used to calculate rate constant from given half life for first order kinetics:

$t_{1/2}=\frac{0.693}{k}$

where,

k = rate constant of the reaction = $5.82\times 10^{-4}s^{-1}$

$t_{1/2}$ = half life of the reaction = ?

Putting values in above equation, we get:

$t_{1/2}=\frac{0.693}{5.82\times 10^{-4}s^{-1}}=1190.7s$

Hence, the half life of the reaction is 1190.7 seconds

8 0

2 years ago