Answer:
B
Explanation:
catches fire easily, there is a chemical in the substance that reacts to the flame, and a key to it being a chemical property is when exposed to other substances, there is a chemical in the other substances that makes it chemically react
Explanation:
Copper (II) sulfate is usually present as a hydrous state, which is of the form CuSO4 * nH2O, where n is a whole number.
Mass of sample (CuSO4 * nH2O)
= 152.00g - 128.10g = 23.90g.
Mass of water loss during heating
= 152.00g - 147.60g = 4.40g.
Molar mass of H2O = 18g/mol
Moles of H2O in sample
= 4.40g / (18g/mol) = 0.244mol.
Mass of anhydrous sample (CuSO4)
= 23.90g - 4.40g = 19.50g
Molar mass of CuSO4 = 159.61g/mol
Moles of CuSO4 in sample
= 19.50g / (159.61g/mol) = 0.122mol.
Since mole ratio of CuSO4 to H2O
= 0.122mol : 0.244mol = 1:2, n = 2.
Hence we have CuSO4 * 2H2O.
Answer:
Velocity = 22.11 m/s
Explanation:
Using the equation;
λ = h / mv;
where
λ = wavelength ( 3.46 x 10^-33 )
m = mass of the marble ( 8.66 g )
v = velocity ( we gotta find out )
h = planck's constant ( 6.626 * 10^-34 )
Making v the subject of the formula;
V = h /mλ
=(6.626 x 10^-34 ) / 0.00866 ( 3.46 x 10^-33 )
= 22.1135
≈ 22.11 m/s
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
Answer:
=> 572.83 K (299.83°C).
=> 95.86 m^2.
Explanation:
Parameters given are; Water flowing= 13.85 kg/s, temperature of water entering = 54.5°C and the temperature of water going out = 87.8°C, gas flow rate 54,430 kg/h(15.11 kg/s). Temperature of gas coming in = 427°C = 700K, specific heat capacity of hot gas and water = 1.005 kJ/ kg.K and 4.187 KJ/kg. K, overall heat transfer coefficient = Uo = 69.1 W/m^2.K.
Hence;
Mass of hot gas × specific heat capacity of hot gas × change in temperature = mass of water × specific heat capacity of water × change in temperature.
15.11 × 1.005(700K - x ) = 13.85 × 4.187(33.3).
If we solve for x, we will get the value of x to be;
x = 572.83 K (2.99.83°C).
x is the temperature of the exit gas that is 572.83 K(299.83°C).
(b). ∆T = 339.2 - 245.33/ln (339.2/245.33).
∆T = 93.87/ln 1.38.
∆T = 291.521K.
Heat transfer rate= 15.11 × 1.005 × 10^3 (700 - 572.83) = 1931146.394.
heat-transfer area = 1931146.394/69.1 × 291.521.
heat-transfer area= 95.86 m^2.
