Answer:
2 and 3 because there are the protons and neutrons which have a greater mass than the electrons which are found in the locations 1 and 4. The atom contains a nucleus, which is made from protons and neutrons, and electrons which are found around the nucleus. The mass of the atoms is concentrated in the very tiny space represented by the nucleus. Of course the electrons have a mass too, but is very small compared to the protons and neutrons, and we usually neglect its mass.
Answer:
The temperature will decrease
Explanation:
the particles spread out while releasing heat
Answer:
<em>The correct option is the structure in the option D. </em>
Explanation:
Ammonia contains a single nitrogen atom bonded to three hydrogen atoms.
The structure present in option A shows a single nitrogen atom bonded with a single hydrogen atom hence it is not correct.
The structure in option B shows a nitrogen atom bonded to four hydrogen atoms hence it is also not correct.
The structure in option C shows a nitrogen atom bonded to two hydrogen atoms hence it is also not correct.
The structure in option D shows a nitrogen atom bonded to three hydrogen atoms which is the correct structure of ammonia (NH3).
Answer:
1. ![K_eq = [Ca^{2+][OH^-]^2 = K_{sp}](https://tex.z-dn.net/?f=K_eq%20%3D%20%5BCa%5E%7B2%2B%5D%5BOH%5E-%5D%5E2%20%3D%20K_%7Bsp%7D)
2. a. No effect;
b. Products;
c. Products;
d. Reactants
Explanation:
1. Equilibrium constant might be written using standard guidelines:
- only aqueous species and gases are included in the equilibrium constant excluding solids and liquids;
- the constant involves two parts: in the numerator of a fraction we include the product of the concentrations of products;
- the denominator includes the product of the concentrations of reactants;
- the concentrations are raised to the power of the coefficients in the balanced chemical equation.
Based on the guidelines, we have two ions on the product side, a solid on the left side. Thus, the equilibrium constant has the following expression:
![K_eq = [Ca^{2+][OH^-]^2 = K_{sp}](https://tex.z-dn.net/?f=K_eq%20%3D%20%5BCa%5E%7B2%2B%5D%5BOH%5E-%5D%5E2%20%3D%20K_%7Bsp%7D)
2. a. In the following problems, we'll be considering the common ion effect. According to the principle of Le Chatelier, an increase in concentration of any of the ions would shift the equilibrium towards the formation of our precipitate.
In this problem, we're adding calcium carbonate. It is insoluble, so it wouldn't have any effect on the equilibrium.
b. Sodium carbonate is completely soluble, it would release carbonate ions. The carbonate ions would combine with calcium cations and more precipitate would dissolve. This would shift the equilibrium towards formation of the products to reproduce the amount of calcium cations.
c. HCl would neutralize calcium hydroxide to produce calcium chloride and water, so the amount of calcium ions would increase, therefore, the products are favored.
d. NaOH contains hydroxide anions, so we'd have a common ion. An increase in hydroxide would produce more precipitate, so our reactants are favored.
Answer:
1.23 g
Explanation:
<em>A chemist adds 1.55 L of a 0.00582 M calcium sulfate solution to a reaction flask. Calculate the mass in grams of calcium sulfate the chemist has added to the flask.</em>
Step 1: Given data
- Volume of the solution (V): 1.55 L
- Molar concentration of the solution (C): 0.00582 M (0.00582 mol/L)
Step 2: Calculate the moles (n) of calcium sulfate added
We will use the following expression.
n = C × V
n = 0.00582 mol/L × 1.55 L
n = 0.00902 mol
Step 3: Calculate the mass corresponding to 0.00902 moles of calcium sulfate
The molar mass of calcium sulfate is 136.14 g/mol.
0.00902 mol × 136.14 g/mol = 1.23 g