basically all of the molecules in the can of soda started to stop moving due to the compression and change of temperature. the ice had frozen the molecules and which then made the can of soda nice and cold lol
Answer:
m = 998 g
Explanation:
Hello there!
In this case, according to the definition of the molar mass as the mass of one mole of the compound, it is possible to state the 1 mole of C8H18 has a mass of 114.26 grams; therefore, the mass in 8.65 moles turn out to be:

In agreement to the notation requirement.
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Answer:
- <em>Hydration number:</em> 4
Explanation:
<u>1) Mass of water in the hydrated compound</u>
Mass of water = Mass of the hydrated sample - mass of the dehydrated compound
Mass of water = 30.7 g - 22.9 g = 7.8 g
<u>2) Number of moles of water</u>
- Number of moles = mass in grams / molar mass
- molar mass of H₂O = 2×1.008 g/mol + 15.999 g*mol = 18.015 g/mol
- Number of moles of H₂O = 7.9 g / 18.015 g/mol = 0.439 mol
<u>3) Number of moles of Strontium nitrate dehydrated, Sr (NO₃)₂</u>
- The mass of strontium nitrate dehydrated is the constant mass obtained after heating = 22.9 g
- Molar mass of Sr (NO₃)₂ : 211.63 g/mol (you can obtain it from a internet or calculate using the atomic masses of each element from a periodic table).
- Number of moles of Sr (NO₃)₂ = 22.9 g / 211.63 g/mol = 0.108 mol
<u>4) Ratio</u>
- 0.439 mol H₂O / 0.108 mol Sr(NO₃)₂ ≈ 4 mol H₂O : 1 mol Sr (NO₃)₂
Which means that the hydration number is 4.
Answer:
16.89g of PbBr2
Explanation:
First, let us calculate the number of mole of Pb(NO3)2. This is illustrated below:
Molarity of Pb(NO3)2 = 0.595M
Volume = 77mL = 77/1000 = 0.077L
Mole =?
Molarity = mole/Volume
Mole = Molarity x Volume
Mole of Pb(NO3)2 = 0.595x0.077
Mole of Pb(NO3)2 = 0.046mol
Convert 0.046mol of Pb(NO3)2 to grams as shown below:
Molar Mass of Pb(NO3)2 =
207 + 2[ 14 + (16x3)]
= 207 + 2[14 + 48]
= 207 + 2[62] = 207 +124 = 331g/mol
Mass of Pb(NO3)2 = number of mole x molar Mass = 0.046 x 331 = 15.23g
Molar Mass of PbBr2 = 207 + (2x80) = 207 + 160 = 367g/mol
Equation for the reaction is given below:
Pb(NO3)2 + CuBr2 —> PbBr2 + Cu(NO3)2
From the equation above,
331g of Pb(NO3)2 precipitated 367g of PbBr2
Therefore, 15.23g of Pb(NO3)2 will precipitate = (15.23x367)/331 = 16.89g of PbBr2