Answer:
*sodium chloride
*potassium chloride
*calcium chloride dihydrate
*magnesium chloride hexahydrate
*sodium acetate trihydrate
*sodium citrate dihydrate
*sodium hydroxide
Explanation:
Are the components of salt
Answer:
Rubidium (Rb).
Explanation:
From the question given above, the following data were obtained:
Electronic configuration => 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s¹
Name of element =?
To know the name of the element with the above electronic configuration, we shall determine the atomic number of the element. This can be obtained as follow:
Number of electrons = 2 + 2 + 6 + 2 + 6 + 2 + 10 + 6 + 1
Number of electrons = 37
Next, we shall determine the number of protons. This can be obtained as follow:
From the question given above, the element has no charge. Hence the element contains equal numbers of protons and electrons.
Number of electrons = 37
Number of protons = number of electrons = 37
Next, we shall determine the atomic number. This can be obtained as follow:
The atomic number of an element is simply defined as the number of protons present in the atom of the element. Thus,
Atomic number = proton number
Proton is = 37
Therefore,
Atomic number = 37
Finally, we shall determine the name of the element by comparing the atomic number of those in the periodic table.
Thus, the element with the above electronic configuration is Rubidium (Rb) since no two elements have the same atomic number
An enzyme speeds up chemical reactions as it acts as a catalyst.
Answer:
The Percent composition of isopropanol in the mixture is 29.07 %
Explanation:
Step 1: Data given
Number of moles isopropanol (C3H8O) = 2.52 moles
Mass of the solution = 521 grams
Molar mass of isopropanol (C3H8O) = 60.1 g/mol
Step 2: Calculate mass of isopropanol
Mass isopropanol = moles isopropanol * molar mass isopropanol
Mass isopropanol = 2.52 moles * 60.1 g/mol
Mass isopropanol = 151.45 grams
Step 3: Calculate the percent composition of isopropanol in the mixture
Percent composition of isopropanol = (mass isopropanol / total mass of mixture) * 100 %
Percent composition of isopropanol = (151.45 grams / 521 grams ) * 100 %
Percent composition of isopropanol = 29.07 %
The Percent composition of isopropanol in the mixture is 29.07 %
