You have to provide options, otherwise people will not reply to your question.
Ti^2+(g)-->Ti^3+(g)+-3rd IP=2652.5
Answer:
During nuclear fission and fusion matter that seems to disappear but is actually converted into energy. The amount of energy (E) produced in such a reaction can be calculated using Einstein's formula for the equivalence of mass and energy: E = mc^2.
Explanation:
Answer:
The answer to your question is: 101.2 g of CO2
Explanation:
C = 27.6 g
O₂ = 86.5 g remained 12.9 g
O₂ that reacted = 86.5 - 12.9 = 73.6 g
C + O₂ ⇒ CO₂ The equation is balanced
27.6 73.6 ?
MW 12 32 44
Rule of three
12 g of C------------------ 44 g CO2
27.6 g C ------------------ x
x = 27.6(44)/12 = 101.2 g of CO2
32 g of O2 --------------- 44 g of CO2
73.6 g of O2 ------------ x
x = 73.6(44)/32 = 101.2 g of CO2
Answer:
41.45 mL
Explanation:
Applying the general gas equation,
PV/T = P'V'/T'............... Equation 1
Where P = Initial pressure of hydrogen, V = Initial volume of hydrogen, T= Initial Temperature of hydrogen, P' = Final pressure of hydrogen, V' = Final Volume of Hydrogen, T' = Final Temperature.
make V' the subject of the equation
V' = PVT'/TP'................ Equation 2
Given: P = 718 torr = (718×133.322) N/m² = 95725.196 N/m², V = 47.9 mL = 0.0479 dm³, T = 26 °C = (26+273) = 299 K, T' = 273 K, P' = 101000 N/m²
Substitute these values into equation 2
V' = ( 95725.196×0.0479×273)/(299×101000)
V' = 0.04145 dm³
V' = 41.45 mL
