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rodikova [14]
3 years ago
11

What layer of earths atmosphere is the layer next to the earth

Chemistry
1 answer:
Pavlova-9 [17]3 years ago
5 0

Answer:

troposphere

Explanation:

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HELP PLZ
Vaselesa [24]

Answer:

Scientific Explanation of weather change and cause storms is described below in detail.

Explanation:

The weather is just the nature of the environment at any time, including items such as warmth, rainfall, air pressure, and cloud shelter. Daily variations in the weather are due to storms and winds. Seasonal variations are due to the Earth spinning nearby the sun. Warm beginnings often produce stormy weather as the heated air mass at the exterior rises above the cool air mass, creating clouds and storms.

4 0
3 years ago
Fill in the coefficients that will balance the following reaction: a0Ca + a1CO2 + a2O2 → a3CaCO3
VashaNatasha [74]
The chemical equation without coefficients is:


Ca + CO2 + O2 --------> Ca CO3


You can balance that equation by trial an error.


This is the chemical equation balanced:


2Ca + 2CO2 + O2 --------> 2Ca CO3


Count the atoms on each side to check the balance


Atom    Left side         right side

Ca         2                     2 

C           2                     2

O          2*2 + 2 = 6       2*3 = 6


Then those are the coefficients:

a0 = 2

a1 = 2

a2 = 1

a3 = 2
6 0
3 years ago
A 0.5438 g of a C.H.O. compound was combusted in air to make 1.039 g of CO2 and 0.6369 g H20. What is the empirical formula? Bal
goblinko [34]

Answer:

C₃H₅O₂

4C₃H₅O₂ + 13O₂ → 12CO₂ + 10H₂O

Explanation:

The reaction can be expressed as:

CₓHₓOₓ + nO₂ → CO₂ + H₂O

Under the assumption that there was a total combustion, all of the carbon in the reactant was combusted into CO₂, so <u>the mass of C contained in the C.H.O. compound is the same mass of C contained in 1.039 g of CO₂</u>:

1.039gCO_{2}*\frac{1molCO_{2}}{44gCO_{2}} *\frac{1molC}{1molCO_{2}} *\frac{12gC}{1molC} =0.2834gC

All of the hydrogens atoms in the compound ended up becoming H₂O, so <u>the mass of H contained in the C.H.O. compound is the same mass of H contained in 0.6369 g of H₂O</u>:

0.6369g*\frac{1molH_{2}O}{18gH_{2}O} *\frac{1molH}{1molH_{2}O} *\frac{1gH}{1molH} =0.0354gH

Because the compound is composed only by C, H and O, <u>the mass of O in the compound can be calculated by substraction</u>:

0.5438 g Compound - 0.2834 g C - 0.0354 g H = 0.2250 g O

In order to determine the empirical formula, we calculate the moles of each component:

  • mol C = 0.2834 g C ÷ 12 g/mol = 0.0236 mol C
  • mol H = 0.0354 g H ÷ 1 g/mol = 0.0354 mol H
  • mol O = 0.2250 g O ÷ 16 g/mol = 0.0141 mol O

Then we divide those values by the lowest one:

0.0236 mol C ÷ 0.0141 = 1.67

0.0354 mol H ÷ 0.0141 = 2.51

0.0141 mol O ÷ 0.0141 = 1

If we multiply those values by 2, we're left with the empirical formula C₃H₅O₂.

  • The reaction is:

4C₃H₅O₂ + 13O₂ → 12CO₂ + 10H₂O

8 0
3 years ago
All factors that cause an increase in the rate of dissolving have what in common?
Nikitich [7]

Answer:

They increase the attractive forces between the solute and solvent particles.

Explanation:

The dissolution of a solute in a solvent depends on interaction between the solute and the solvent. The more the attractive force and interaction between solute and solvent, the greater the greater the rate of dissolution of the solute in the solvent.

The absence of interaction between solute and solvent molecules means that the substance can not dissolve in that particular solvent. Hence, any factor that enhances solute-solvent interaction will enhance dissolution of a solute in a particular solvent.

3 0
3 years ago
Asdfjkl;asdfjkl;asdfjkl;asdfjkl;asdfjkl;asdfjkl;asdfjkl;
horsena [70]

Answer: ok ill just take my points in dip

Explanation:

3 0
3 years ago
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