Answer:
Well they didn't transfer any energy when they weren’t touching and it did t produce any energy if it didn’t move. Since they are on top of each other they are causing momentum on each other creating kinetic energy
Explanation:
Answer:
volume in L = 0.25 L
Explanation:
Given data:
Mass of Cu(NO₃)₂ = 2.43 g
Volume of KI = ?
Solution:
Balanced chemical equation:
2Cu(NO₃)₂ + 4KI → 2CuI + I₂ + 4KNO₃
Moles of Cu(NO₃)₂:
Number of moles = mass/ molar mass
Number of moles = 2.43 g/ 187.56 g/mol
Number of moles = 0.013 mol
Now we will compare the moles of Cu(NO₃)₂ with KI.
Cu(NO₃)₂ : KI
2 : 4
0.013 : 4 × 0.013=0.052 mol
Volume of KI:
<em>Molarity = moles of solute / volume in L</em>
volume in L = moles of solute /Molarity
volume in L = 0.052 mol / 0.209 mol/L
volume in L = 0.25 L
Answer:
C₄F₈
Explanation:
Using their mole ratio to compute their mass
molar mass of carbon = 12.0107 g/mol
molar mass of fluorine gas = 37.99681
let x = mass of carbon
given mass of fluorine = 1.70 g
x / 12.01067 = 1.70 / 37.99687
cross multiply
x = ( 1.70 × 12) / 37.99687 = 20.4 / 37.99687 = 0.53688 g
mass of one mole of CF₂ = 0.53688 + 1.70 = 2.23688 g
number of mole of CF₂ = 8.93 g / 2.23688 = 3.992 approx 4
molecular formula of CF₂ = 4 (CF₂) = C₄F₈
Answer:
<span>23.6
g carbon dioxide comes from 8.6 g of CH4 or 10.7 g carbon dioxide comes from
15.6 g O that means the 15.6 g of oxygen is still the limiting reactant because
it gets used up and only makes 10.7 g of CO2. </span>
Explanation:
1) Balanced chemical equation:
CH₄ + 2O₂ → CO₂ + 2H₂O
2) mole ratios:
1 mol CH₄ : 2mol O₂ : 1 mol CO₂ : 2 mol H₂O
3) molar masses
CH₄: 16.04 g/mol
O₂: 32.0 g/mol
CO₂: 44.01 g/mol
4) Convert the reactant masses to number of moles, using the formula
number of moles = mass in grams / molar mass
CH₄: 8.6g / 16.04 g/mol = 0.5362 moles
<span />
O₂: 15.6 g / 32.0 g/mol = 0.4875 moles
5) If the whole 0.5632 moles of CH₄ reacted that yields to the same number of moles of CO₂ and that is a mass of:
mass of CO₂ = number of moles x molar mass = 23.60 g of CO₂
Which is what the first part of the answer says.
6) If the whole 0.4875 moles of O₂ reacted that would yield 0.4875 / 2 = 0.24375 moles of CO₂, and that is a mass of:
mass of CO₂ = 0.4875 grams x 44.01 g/mol = 10.7 grams of CO₂.
Which is what the second part of the answer says.
7) From the mole ratio you know infere that 0.5362 moles of CH₄ needs more twice number of moles of O₂, that is 1.0724 moles of O₂, and since there are only 0.4875 moles of O₂, this is the limiting reactant.
Which is what the chosen answer says.
8) From the mole ratios 0.4875 moles of O₂ produce 0.4875 / 2 moles of CO₂, and that is:
0.4875 / 2 mols x 44.01 g/mol = 10.7 g of CO₂, which is the last part of the answer.
1) Balanced chemical reaction:
2H2 + O2 -> 2H20
Sotoichiometry: 2 moles H2: 1 mol O2 : 2 moles H2O
2) Reactant quantities converted to moles
H2: 5.00 g / 2 g/mol = 2.5 mol
O2: 50.0 g / 32 g/mol = 1.5625 mol
Limitant reactant: H2 (because as per the stoichiometry it will be consumed with 1.25 mol of O2).
3) Products
H2 totally consumed -> 0 mol at the end
O2 = 1.25 mol consumed -> 1.5625 mol - 1.25 mol = 0.3125 mol at the end
H2O: 2.5 mol H2 produces 2.5 mol H2O -> 2.5 mol at the end.
Total number of moles: 0.3125mol + 2.5 mol = 2.8125 mol
4) Pressure
Use pV = nRT
n = 2.8125
V= 9 liters
R = 0.082 atm*lit/K*mol
T = 35 C + 273.15 = 308.15K
p = nRT/V = 7.9 atm
