Answer:
Explanation:
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In this case, we can divide the problem in two steps:
1. Dilution to 278 mL: here, the initial concentration and volume are 1.20 M and 52.0 mL respectively, and a final volume of 278 mL, it means that the moles remain the same so we can write:
So we solve for C2:
2. Now, since 111 mL of water is added, we compute the final volume, V3:
So, the final concentration of the 139 mL portion is:
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Answer:
Answers
1.)reactants: nitrogen and hydrogen; product: ammonia.
2.)reactants: magnesium hydroxide and nitric acid; products: magnesium nitrate and water.
3.)N 2 + 3H 2 → 2NH 3
4.)Mg(OH) 2 + 2HNO 3 → Mg(NO 3) 2 + 2H 2O.
5.)2NaClO 3 → 2NaCl + 3O 2
6.)4Al + 3O 2 → 2Al 2O 3
7.)N 2(g) + 3H 2(g) → 2NH 3(g)
Explanation:
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The equilibrium constant is found by [product]/[reactant]
If the equilibrium constant is very small, such as 4.20 * 10^-31, then that means at equilibrium there is very little product and a lot of reactant.
And likewise, if there is a lot of product formed, and very little reactant, then the K value will be very large, which tells us that it is predominantly product.
At equilibrium, for any reaction, there will always be some reactant and some product present. There cannot be zero reactant or zero product. Also keep in mind that the equilibrium constant is dependent on temperature.
At equilibrium, for your reaction, it is predominantly reactants.
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Answer:
Neutrons = ( Atomic mass – Atomic number ) ( A–Z )
Protons and Electrons are equal to the atomic number
For example Neon,
Mass number (A) = 20
Atomic Number (Z) = 10
Number of Protons = 10
Number of Electrons = 10
Number of Neutrons = ( A–Z ) = 10
Electronic distribution :
K= 2
L= 8
