Answer:
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The given solution of Mn²⁺ is 0.60 mg/mL.
Hence mass of Mn²⁺ in 5 mL of solution = 0.60 mg/mL x 5 mL = 3 mg
Molar mass of Mn = 54.9 g/mol
Hence, moles of Mn²⁺ = 3 x 10⁻³ g / 54.9 g/mol = 5.46 x 10⁻⁵ mol
The balanced equation for the reaction is,
2Mn²⁺ + 5KIO₄ + 3H₂O → 2MnO₄⁻ + 5KIO₃ + 6H⁺
The stoichiometric ratio between Mn²⁺ and KIO₄ is 2 : 5
Hence, moles of KIO₄ reacted = 5.46 x 10⁻⁵ mol x (5 / 2)
= 13.65 x 10⁻⁵ mol
Molar mass of KIO₄ = 230 g/mol
Hence needed mass of KIO₄ = 13.65 x 10⁻⁵ mol x 230 g/mol
= 0.031395 g
= 31.395 mg
≈ 31.4 mg
Data Given:
Pressure = P = ?
Volume = V = 3.0 L
Temperature = T = 115 °C + 273 = 388 K
Mass = m = 75.0 g
M.mass = M = 44 g/mol
Solution:
Let suppose the Gas is acting Ideally. Then According to Ideal Gas Equation,
P V = n R T
Solving for P,
P = n R T / V ------ (1)
Calculating Moles,
n = m / M
n = 75.0 g / 44 g.mol⁻¹
n = 1.704 mol
Putting Values in Eq. 1,
P = (1.704 mol × 0.08205 atm.L.mol⁻¹.K⁻¹ × 388 K) ÷ 3.0 L
P = 18.08 atm
Answer:
last choice
Explanation:
Specific heat of a metal is j/kg-c
It says nothing about melting point or heat of fusion
A = twice B = 2 * j/kg-c B = j/kg-c
if j is constant and mass is the same we can write this more like
A = 2 /C and B = 1/C
or A ~ C/2 and B~ C
now you can see that the temp change in A will be 1/2 that of B
