Using charles law
v1/t1=v2/t2
v1=49ml
v2=74
t1=7+273=280k
t2=?
49/280=74/t2
0.175=74/t2 cross multiply
0.175t2=74
t2=74/0.175
t2=422k or 149celcius
Answer:
The new volume will be 42, 7 L.
Explanation:
We use the gas formula, which results from the combination of the Boyle, Charles and Gay-Lussac laws. According to which at a constant mass, temperature, pressure and volume vary, keeping constant PV / T. The conditions STP are: 1 atm of pressure and 273 K of temperature.
P1xV1/T1 =P2xV2/T2
1 atmx 22,4 L/273K = 0,5atmx V2/260K
V2=((1 atmx 22,4 L/273K )x 260K)/0,5 atm= 42, 67L
Answer:
Explanation:
sp² hybridization is found in those compounds having double bond .
Out of the given compounds only C₂H₂Cl₂ has double bond so this compound contains carbon with sp² hybridization .
Rest have sp³ hybridization because they are saturated compounds .
<u>Given:</u>
Mass of calcium nitrate (Ca(NO3)2) = 96.1 g
<u>To determine:</u>
Theoretical yield of calcium phosphate, Ca3(PO4)2
<u>Explanation:</u>
Balanced Chemical reaction-
3Ca(NO3)2 + 2Na3PO4 → 6NaNO3 + Ca3(PO4)2
Based on the reaction stoichiometry:
3 moles of Ca(NO3)2 produces 1 mole of Ca3(PO4)2
Now,
Given mass of Ca(NO3)2 = 96.1 g
Molar mass of Ca(NO3)2 = 164 g/mol
# moles of ca(NO3)2 = 96.1/164 = 0.5859 moles
Therefore, # moles of Ca3(PO4)2 produced = 0.0589 * 1/3 = 0.0196 moles
Molar mass of Ca3(PO4)2 = 310 g/mol
Mass of Ca3(PO4)2 produced = 0.0196 * 310 = 6.076 g
Ans: Theoretical yield of Ca3(PO4)2 = 6.08 g
Mass of KNO₃ : = 40.643 g
<h3>Further explanation</h3>
Given
28.5 g of K₃PO₄
Required
Mass of KNO₃
Solution
Reaction(Balanced equation) :
2K₃PO₄ + 3 Ca(NO₃)₂ = Ca₃(PO₄)₂ + 6 KNO₃
mol K₃PO₄(MW=212,27 g/mol) :
= mass : MW
= 28.5 : 212,27 g/mol
= 0.134
Mol ratio of K₃PO₄ : KNO₃ = 2 : 6, so mol KNO₃ :
= 6/2 x mol K₃PO₄
= 6/2 x 0.134
= 0.402
Mass of KNO₃ :
= mol x MW KNO₃
= 0.402 x 101,1032 g/mol
= 40.643 g