In a certain compound of iron and oxygen, FexOy, it is found that a sample of this compound weighing 6.285 g contains 4.396 g of

The new pressure of the gas is calculated to be 40.7 kPa. Using the principle that P1 • V1 = P2 • V2, we can set 98.8 kPa (P1) multiplied by 21.7 mL (V1) equal to P2 (unknown pressure) multiplied by 52.7 mL (V2). To isolate P2, we rearrange the equation to P2 = (98.8 kPa • 21.7 mL) / 52.7 mL, resulting in P2 equal to 40.7 kPa.

The specific heat of titanium metal is 0.524 J/g°C. Given that Q = 1.68 kJ, which equates to 1680 Joules, with a mass of 126 grams and initial and final temperatures of 20°C and 45.4°C respectively, the specific heat is computed using the formula Q = (mass)(ΔT)(Cp), where ΔT = T₂ - T₁ = 25.4°C. Plugging in the numbers leads us to Cp = 0.524 J/g°C.

Density is calculated as mass divided by volume.
  Step one:
Convert m³ to ml.
1 m³ = 1,000,000 ml
0.250 m³ x 1,000,000 = 250,000 ml
  Step two: Convert mg to g.
1 mg = 0.001 g, hence 4.25 x 10^8 mg equals 0.459 g.
Consequently, the density comes out to be 0.459 g/250,000 = 1.836 x 10^-6 g/ml.

The connection between Gibb's free energy and temperature can be described as follows. The value for is specified as 1.7 kJ/mol. Additionally, we know that k =

= . Since k has a value of 0.50357 at a temperature of 298 K, we can conclude that

= 0.50357, leading to + 1 = 1.50357, which equals = 0.665. Therefore, we can deduce that 0.665 of the total fructose in the solution exists as fructopyranose.

Answer:

Explanation:

AgNO3 + NaCl --> AgCl + NaNO3

Moles

of AgNO3

= molarity * volume

= 1 * 0.01

= 0.01 mol

for NaCl

= 0.01 * 1

= 0.01 mol.

According to stoichiometry, one mole of silver nitrate corresponds to one mole of NaCl reacted. Hence,

Moles of AgCl generated = 0.01 × 1

= 0.01 mol AgCl produced.

Heat gained by the solution as precipitation occurs:

Solution mass = density × volume

= 1 × 20

= 20 g.

Using q = m * Cp * (T2 - T1)

= 20 * 4.18 * (32.6 - 25.0)

= 635 J

The absorbed heat of 635 J indicates the reaction released -635 J

Thus, Delta H = -635 J/0.01 mol

= -63500 J/mol

= -63.5 kJ/mol.