Answer:
A meteor is an asteroid or other object that burns and vaporizes upon entry into the Earth's atmosphere; meteors are commonly known as "shooting stars." If a meteor survives the plunge through the atmosphere and lands on the surface, it's known as a meteorite.
Explanation:
Answer:
Chemical reaction are irreversible. Some of the example of chemical reaction are cooking, rusting, and burning. During a chemical reaction, the composition of substances changes and the particles rearrange to form a new substance. The new substance formed after chemical reaction of substance has different physical and chemical properties.
When a chemical reaction occur, the atoms or molecules of the substances change its physical and chemical properties such as while cooking of vegetable, the molecules of vegetable undergo changes in their properties and form a new substance which is different from the earlier.
Answer:
The concentration c is equal to Ka
Explanation:
The acid will ionize as observed in the following reaction:
HA = H+ + A-
H+ is the proton of the acid and A- is the conjugate base
. The equation to calculate the Ka is as follows:
Ka = ([H+]*[A -])/[HA]
Initially we have to:
[H+] = 0
[A-] = 0
[HA] = c
During the change we have:
[H+] = +x
[A-] = +x
[HA] = -x
During balance we have:
[H+] = 0 + x
[A-] = 0 + x
[HA] = c - x
Substituting the Ka equation we have:
Ka = ([H+]*[A-])/[HA]
Ka = (x * x)/(c-x)
x^2 + Kax - (c * Ka) = 0
We must find c, having as [H+] = 1/2c. Replacing we have:
(1/2c)^2 + (Ka * 1/2 * c) - (c * Ka) = 0
(c^2)/2 + Ka(c / 2 - c) = 0
(c^2)/2 + (-Ka * c/2) = 0
c^2 -(c*Ka) = 0
c-Ka = 0
Ka = c
Answer:
Explanation:
<u>1) Mass of carbon (C) in 2.190 g of carbon dioxide (CO₂)</u>
- atomic mass of C: 12.0107 g/mol
- molar mass of CO₂: 44.01 g/mol
- Set a proportion: 12.0107 g of C / 44.01 g of CO₂ = x / 2.190 g of CO₂
x = (12.0107 g of C / 44.01 g of CO₂ ) × 2.190 g of CO₂ = 0.59767 g of C
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<u>2) Mass of hydrogen (H) in 0.930 g of water (H₂O)</u>
- atomic mass of H: 1.00784 g/mol
- molar mass of H₂O: 18.01528 g/mol
- proportion: 2 × 1.00784 g of H / 18.01528 g of H₂O = x / 0.930 g of H₂O
x = ( 2 × 1.00784 g of H / 18.01528 g of H₂O) × 0.930 g of H₂O = 0.10406 g of H
<u>3) Mass of oxygen (O) in 1.0857 g of pure sample</u>
- Mass of O = mass of pure sample - mass of C - mass of H
- Mass of O = 1.0857 g - 0.59767 g - 0.10406 = 0.38397 g O
Round to four decimals: Mass of O = 0.3840 g
<u>4) Mole calculations</u>
Divide the mass in grams of each element by its atomic mass:
- C: 0.59767 g / 12.0107 g/mol = 0.04976 mol
- H: 0.10406 g / 1.00784 g/mol = 0.10325 mol
- O: 0.3840 g / 15.999 g/mol = 0.02400 mol
<u>5) Divide every amount by the smallest value (to find the mole ratios)</u>
- C: 0.04976 mol / 0.02400 mol = 2.07 ≈ 2
- H: 0.10325 mol / 0.02400 mol = 4.3 ≈ 4
- O: 0.02400 mol / 0.02400 mol = 1
Thus the mole ratio is 2 : 4 : 1, and the empirical formula is:
