Combination
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Answer:
- Question 19: the three are molecular compounds.
Explanation:
<em>Question 19.</em>
All of them are the combination of two kinds of different atoms in fixed proportions.
- C₂H₄: two carbon atoms per four hydrogen atoms
- HF: one hydrogen atom per one fluorine atom
- H₂O₂: two hydrogen atoms per two oxygent atoms
Thus, they all meet the definition of compund: a pure substance formed by two or more different elements with a definite composition.
Molecular compounds are formed by covalent bonds and ionic compounds are formed by ionic bonds.
Two non-metal elements, like H-F, C - C, C - H, H-O, H - H, and O - O will share electrons forming covalent bonds to complete their valence shell. Thus, the three compounds are molecular and not ionic.
<em>Question 20. </em>Formula of copper(II) sulfate hydrate with 36.0% water.
Copper(II) sulfate is CuSO₄. Its molar mass is 159.609g/mol
Water is H₂O. Its molar mass is 18.015g/mol
Calling x the number of water molecules in the hydrate, the percentage of water is:

From which we can solve for x:

Thus, there are 5 molecules of water per each unit of CuSO₄, and the formula is:
The grams of carbon dioxide that are in 35.6 liters of Co2 is calculates as below
calculate the number of moles of CO2
At STP 1 mole = 22.4 L
what about 35.6 liters
= 1mole x 35.6 liters/ 22.4 liters = 1.589 moles
mass of CO2 = moles x molar mass of CO2
= 1.589 mol x 44 g/mol = 69.92 grams
<u>Given:</u>
Mass of MgBr2 = 0.500 g
<u>To determine:</u>
Number of anions in 0.500 g MgBr2
<u>Explanation:</u>
Molar mass of MgBr2 = 24 + 2 (80) = 184 g/mol
Moles of MgBr2 = 0.500 g/184 g.mol-1 = 0.00271 moles
Based on stoichiometry-
1 mole of MgBr2 has 1 mole of Mg2+ cations and 2 moles of Br- anions
Therefore, 0.00271 moles of MgBr2 will have: 2 * 0.00271 = 0.00542 moles of Br-
Now,
1 mole of Br- contains 6.023 * 10²³ anions
0.00542 moles of Br- contain: 0.00542 * 6.023*10²³ = 3.264*10²¹ anions
Ans: There are 3.264*10²¹ anions in 0.5 g of MgBr2
Volume of Hydrogen V1 = 351mL
Temperature T1 = 20 = 20 + 273 = 293 K
Temperature T2 = 38 = 38 + 273 = 311 K
We have V1 x T2 = V2 x T1
So V2 = (V1 x T2) / T1 = (351 x 311) / 293 = 372.56
Volume at 38 C = 373 ml