Consider the diagram of the basic structure of a bacterium. A diagram of a bacterium is labeled. Part A is the cell membrane, pa

Utilize the ideal gas law:

n = PV / RT

P = 100kPa = 100 x 1000 x (9.8 x 10^{-6}) = 0.98 atm
Convert kPa to atm, where 1 Pa = 9.8 x 10^{-6} atm.
T = 293 K
V = 6.8 L
R = 1/12
Substituting all values leads to:
n = 0.272

Answer:

The temperature difference is 293.15 Kelvin.

Explanation:

The provided information:

The temperature difference between the two matters is 20°C

We need to determine this difference in Kelvin =?

To solve this;

Using the formula:

0°C +273.15

Now substituting the values in place of 0.

20°C + 273.15 = 293.15 K

Hence, the temperature differential between the two samples is 293.15 K.

Response:

The specific heat of the alloy

Clarification:

Weight of the alloy = 25 gm

Initial temperature = 100°c = 373 K

Weight of the water = 90 gm

Initial temperature of water = 25.32 °c = 298.32 K

Final temperature = 27.18 °c = 300.18 K

Using the energy balance equation,

Heat released by the alloy = Heat absorbed by the water

[[ - ] = ( - )

25 × × ( 373 - 300.18 ) = 90 × 4.2 (300.18 - 298.32)

This gives us the specific heat of the alloy.

Answer:

Explanation:

Given data:

Initial temperature T₁ = 25.2°C = 298.2K

Initial pressure P₁ = 0.6atm

Final temperature = 72.4°C = 345.4K

What we need to find:

Final pressure = ?

To determine this, we apply a modified version of the combined gas law with constant volume. This simplifies our calculations to:

Here, P and T signify pressure and temperatures, 1 refers to initial and 2 to final temperatures.

Now we can substitute the known variables:

P₂ = 0.7atm

The pH level is 1.39. To explain, we start with the given information: the concentration of HClO is 0.15 M, with an acid dissociation constant of 2.9 × 10-8. The objective is to calculate the pH of the solution. Through the process, we find that the equilibrium concentration after applying the formula yields 0.04069 M for H3O⁺, leading us to a pH of 1.39.