1. ionic compound , aqueous cations and aqueous anions
2. covalent compound aqueous covalent compound
<u>Explanation:</u>
1. A(n) <u>ionic compound </u> dissolves in water , H₂O(l), will produce <u>aqueous cations </u> and <u>aqueous anions </u>in solution.
When NaCl dissolves in water it will produce Na⁺ and Cl⁻ ions in solution
2. A(n) <u>covalent compound </u> dissolves in water , H₂O(l), will produce <u>aqueous covalent compound </u>in solution.
When Ammonia (NH₃) dissolves in water it forms aqueous ammonia, NH₃(aq)
Organic compounds, like carbohydrates, proteins, nucleic acids, and lipids, are all good examples of covalent compounds.
Answer:
18.45 g of C
Explanation:
This is a problem of rules of three:
1 mol of C₃H₈ contains 3 moles of C and 8 moles of H
If 8 moles of H are contained in 1 mol of propane
4.10 moles of H are contained in (4.1 . 1) /8 = 0.5125 moles
Now, If 1 mol of propane contains 3 moles of C
0.5125 moles of propane may contain (0.5125 . 3) / 1 = 1.5375 moles of C
Let's convert the moles to mass:
1.5375 mol . 12 g /mol = 18.45 g
Answer:
There are 255, 85 grams of Strontium in 2,92 moles.
Explanation:
We perform a simple rule of three to calculate the number of grams, knowing that one mole of Sr weights 87, 62 grams:
1 mol Sr---------87, 62 grams
2,92 mol Sr-----x=(2,92 mol Sr x-87, 62 grams)/1 mol Sr= 255,8504 grams
It will be number 1.
NO loses oxygen ie its being reduced to N2
NH3 gains oxygen atom ie its being oxidised to H2O
Answer:
2.2 °C/m
Explanation:
It seems the question is incomplete. However, this problem has been found in a web search, with values as follow:
" A certain substance X melts at a temperature of -9.9 °C. But if a 350 g sample of X is prepared with 31.8 g of urea (CH₄N₂O) dissolved in it, the sample is found to have a melting point of -13.2°C instead. Calculate the molal freezing point depression constant of X. Round your answer to 2 significant digits. "
So we use the formula for <em>freezing point depression</em>:
In this case, ΔTf = 13.2 - 9.9 = 3.3°C
m is the molality (moles solute/kg solvent)
- 350 g X ⇒ 350/1000 = 0.35 kg X
- 31.8 g Urea ÷ 60 g/mol = 0.53 mol Urea
Molality = 0.53 / 0.35 = 1.51 m
So now we have all the required data to <u>solve for Kf</u>:
