Answer:
FeCl₃
Explanation:
4FeCl₃ + 3O₂ => 2Fe₂O₃+ 6Cl₂
Given => 7moles 9moles
A simple way to determine which reagent is the limiting reactant is to convert all given data to moles then divide by the respective coefficients of the balanced equation. The smaller value will be the limiting reactant.
4FeCl₃ + 3O₂ => 2Fe₂O₃+ 6Cl₂
Given => 7/4 = 1.75* 9/3 = 3
*Smaller value => FeCl₃ is limiting reactant.
NOTE: However, when working problems, one must use original mole values given.
Condensation is a chemical reaction in which two molecules are combined together and as a product formed a large molecule. There is also loss of small molecules in the condensation reaction. And mostly the functional groups combined in the reaction.
The other name for a condensation reaction is dehydration. As there is loss of molecule in condensation reaction, when that molecule is water it is said to be dehydration.
Answer:
atom is the answer I think
First we have to find moles of C:
Molar mass of CO2:
12*1+16*2 = 44g/mol
(18.8 g CO2) / (44.00964 g CO2/mol) x (1 mol C/ 1 mol CO2) =0.427 mol C
Molar mass of H2O:
2*1+16 = 18g/mol
As there is 2 moles of H in H2O,
So,
<span>(6.75 g H2O) / (18.01532 g H2O/mol) x (2 mol H / 1 mol H2O) = 0.74mol H </span>
<span>Divide both number of moles by the smaller number of moles: </span>
<span>As Smaaler no moles is 0.427:
So,
Dividing both number os moles by 0.427 :
(0.427 mol C) / 0.427 = 1.000 </span>
<span>(0.74 mol H) / 0.427 = 1.733 </span>
<span>To achieve integer coefficients, multiply by 2, then round to the nearest whole numbers to find the empirical formula:
C = 1 * 2 = 2
H = 1.733 * 2 =3.466
So , the empirical formula is C2H3</span>
Answer:
The equilibrium temperature of the coffee is 72.4 °C
Explanation:
Step 1: Data given
Mass of cream = 15.0 grams
Temperature of the cream = 10.0°C
Mass of the coffee = 150.0 grams
Temperature of the coffee = 78.6 °C
C = respective specific heat of the substances( same as water) = 4.184 J/g°C
Step 2: Calculate the equilibrium temperature
m(cream)*C*(T2-T1) = -m(coffee)*c*(T2-T1)
15.0 g* 4.184 J/g°C *(T2 - 10.0°C) = -150.0g *4.184 J/g°C*(T2-78.6°C)
62.76T2 - 627.6 = -627.6T2 + 49329.36
690.36T2 = 49956.96
T2 = 72.4 °C
The equilibrium temperature of the coffee is 72.4 °C
