Answer:
c = 0.528 J/g.°C
Explanation:
Given data:
Mass of titanium = 43.56 g
Heat absorbed = 0.476 KJ = 476 j
Initial temperature = 20.5°C
Final temperature = 41.2°C
Specific heat capacity = ?
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 41.2°C - 20.5°C
ΔT = 20.7 °C
476 J = 43.56 g × c × 20.7 °C
476 J = 901.692 g.°C × c
c = 476 J / 901.692 g.°C
c = 0.528 J/g.°C
Answer:
You need 375 mL of BaCl2 solution.
Explanation:
M1V1=M2V2
Dilution formula. Substitute known values and solve for V1.
M1 = 2.0 M
M2 = 1.50 M
V2 = 500 mL
(2.0 M)(V1) = (1.50 M)(500 mL)
V1 = (1.50 M)(500 mL) / (2.0 M)
V1 = 375 mL
1) ideal gas law: p·V = n·R·T.
p - pressure of gas.
V -volume of gas.
n - amount of substance.
R - universal gas constant.
T - temperature of gas.
n₁ = 0,04 mol, V₁ = 0,06 l.
n₂ = 0,07 mol, V₂ = 0,06 · 0,07 ÷ 0,04 = 0,105 l.
2) V₁ = 0,06 l, T₁ = 240,00 K.
T₂ = 340,00 K, V₂ = 340 · 0,06 ÷ 240 = 0,05 l.
C3H8.gas reacts with 5L of O2 at STP
Add 7 water atom to the right hand side to adjust the quantity of oxygen. Increase Cr(+3) by two to adjust the quantity of Cr. Duplicate Cl-by two to adjust the quantity of chlorine molecules.
Cr2O7[2-](aq) +2 Cl[-](aq) < - >2 Cr[3+] (aq) + Cl2(g)+7H2O
Presently adjust that charges.
you have - 4 charges on the left hand side, while +18 charges on the right hand side, there for include 14H+ the left hand side to adjust the charges
Cr2O7[2-](aq) +2 Cl[-](aq)+14H+ < - >2 Cr[3+] (aq) + Cl2(g)+7H2O
take note of that the oxidation number of hydrogen in water is +1
