Answer:
Anhydride, any chemical compound obtained, either in practice or in principle, by the elimination of water from another compound. Examples of inorganic anhydrides are sulfur trioxide, SO3, which is derived from sulfuric acid, and calcium oxide, CaO, derived from calcium hydroxide
Explanation:
<h3>
<em><u>examples</u></em><em><u>.</u></em></h3>
1)acid anhydride.
2)basic anhydrides.
<h3>
<em><u>reactions</u></em><em><u>. </u></em></h3>
1)reaction with water
(CH3CO)2O + H2O → 2 CH3CO2H.
Explanation:
Since, the given reaction is as follows.

Initial: 36.1 atm 0 0
Change: 2x x x
Equilibrium: (36.1 - 2x) x x
Now, expression for
of this reaction is as follows.
![K_{p} = \frac{[N_{2}][O_{2}]}{[NO]^{2}}](https://tex.z-dn.net/?f=K_%7Bp%7D%20%3D%20%5Cfrac%7B%5BN_%7B2%7D%5D%5BO_%7B2%7D%5D%7D%7B%5BNO%5D%5E%7B2%7D%7D)
As the initial pressure of NO is 36.1 atm. Hence, partial pressure of
at equilibrium will be calculated as follows.
![K_{p} = \frac{[N_{2}][O_{2}]}{[NO]^{2}}](https://tex.z-dn.net/?f=K_%7Bp%7D%20%3D%20%5Cfrac%7B%5BN_%7B2%7D%5D%5BO_%7B2%7D%5D%7D%7B%5BNO%5D%5E%7B2%7D%7D)
x = 18.1 atm
Thus, we can conclude that partial pressure of
at equilibrium is 18.1 atm.
Physicist Ernest Rutherford<span> established the nuclear theory of the atom with his </span>gold-foil experiment<span>. When he shot a beam of alpha particles at a sheet of </span>gold foil<span>, a few of the particles were deflected. He concluded that a tiny, dense nucleus was causing the deflections.</span>
Answer:
70000 cm in 7km
Explanation:
There are 1000 m and 100 cm in 1 m in 1 km, so to find how many cm are in 7 km you need to do 10000 x 7. You get 70000
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Answer:
1.1 × 10² g
Explanation:
First, we will convert 1.0 L to cubic centimeters.
1.0 L × (10³ mL/1 L) × (1 cm³/ 1 mL) = 1.0 × 10³ cm³
The density of water is 1.0 g/cm³. The mass corresponding to 1.0 × 10³ cm³ is:
1.0 × 10³ cm³ × (1.0 g/cm³) = 1.0 × 10³ g
1 mole of water (H₂O) has a mass of 18 g, consisting of 2 g of H and 16 g of O. The mass of Hydrogen in 1.0 × 10³ g of water is:
1.0 × 10³ g H₂O × (2 g H/18 g H₂O) = 1.1 × 10² g