What is the molality of a solution made by dissolving 137.9g of sucrose in 414.1g of water? The density of the solution is 1.104g/ml.

Answer:

mass of platinum = 2526.12 g

Explanation:

Given data:

Mass of water = 125 g

Initial temperature of water= 100.0°C

Initial temperature of Pt = 20.0°C

Final temperature = 235°C

Specific heat of Pt = 0.13 j/g°C

Specific heat of water = 4.184 j/g°C

Mass of platinum = ?

Solution:

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 = T2 - T1

Q(w) = Q(Pt)

m.c.  (T2 - T1)   =   m.c.   (T2 - T1)

125 g × 4.184 j/g°C ×  (235°C - 100.0°C)  = m × 0.13 j/g°C ×  (235°C - 20°C)

125 g × 4.184 j/g°C × 135°C  = m × 0.13 j/g°C × 215°C

70605 j = m×27.95 j/g

m = 70605 j /27.95 j/g

m = 2526.12 g

To find the mass of platinum dropped into hot water, you would apply the conservation of energy principle and use the specific heat values for both substances. However, the provided temperatures suggest an error in the question, as platinum would not heat to a temperature higher than the water's temperature.

The question involves finding the mass of platinum, which was dropped into hot water, by using the concept of heat transfer between the metal and water. According to the law of conservation of energy, the heat lost by the water is equal to the heat gained by the platinum. To solve for the mass of the platinum, we use the formula q = mcΔT, where 'q' is the heat transfer, 'm' is the mass, 'c' is the specific heat, and 'ΔT' is the change in temperature.

However, the given information seems to contain a mistake, as platinum will not naturally heat up to 235.0°C when dropped into water that is at 100.0°C. There must be an error in the given temperatures. Assuming we had the correct temperature details and using the given specific heats for water (4.184 J/g°C) and platinum (0.13 J/g°C), we could set up the heat transfer equations and solve for the unknown mass of platinum.

Answer: Solution attached.

Each equation is now balanced.

Explanation:

Answer:

3.5 mol·L⁻¹  

Explanation:

1. Set up an ICE table.

[tex]\begin{array}{cccccc}\text{2NO} & + & \text{I}_{2} &\, \rightleftharpoons \, & \text{2NOI} & & \\ 2.0 & & 4.0 & & 1.0 & & \\ -2x & & -x & & +2x & & \\ 2.0-2x & & 4.0-x & & 1.0+2x & & \\\end{array}[/tex]

2. Solve for x

The equilibrium concentration of NO is 1.0 mol·L⁻¹, so

       1.0 = 2.0 - 2x

2x + 1.0 = 2.0

        2x =  1.0

          x = 0.5

3. Calculate the equilibrium concentration of I₂

[I₂] = 4.0 - x = 4.0 - 0.5 = 3.5 mol·L⁻¹

The concentration of I₂ at equilibrium is calculated to be 3.5 M by using the initial concentration of NO to determine the stoichiometric change in I₂ concentration based on the reaction 2NO(g) + I₂(g) → 2NOI(g).

To find the concentration of I₂ at equilibrium for the reaction 2NO(g) + I₂(g) → 2NOI(g), we use the initial and equilibrium concentrations of NO to determine the change in concentration of I₂. Given the stoichiometry of the reaction, for every 1 mole decrease in NO, there is a 0.5 mole decrease in I₂. The initial concentration of NO is 2.0 M, and at equilibrium, it is 1.0 M, which means there has been a 1.0 M decrease (2.0 M - 1.0 M). The I₂ concentration at equilibrium can be found by subtracting half of this change from the initial I₂ concentration. Since initially the concentration of I₂ is 4.0 M, the equilibrium concentration is calculated as 4.0 M - (1.0 M / 2) = 3.5 M.

Final answer:

Each SI unit measures a specific fundamental quantity: kilograms (kg) for mass, kelvin (K) for temperature, seconds (s) for time, and amperes (A) for electric current.

Explanation:

The student has asked to match each SI unit to the quantity it measures among mass, temperature, time, and electric current. Here are the matches:

These four units are part of the metric system, which uses powers of 10 to relate quantities over various ranges of nature. All other physical quantities, such as force and charge, are derived from these fundamental units.