Los fenomenos fisicos son cambios permamentes en la materia?

Where Q is the amount of energy transferred (J), m is the mass of the substance (kg), c is the specific heat (J g⁻¹ °C⁻¹) and ΔT is the temperature difference (°C).

Density = mass / volume

The density of water = 0.997 g/mL

Hence mass of 1.25 L (1250 mL) of water = 0.997 g/mL x 1250 mL

= 1246.25 g

Specific heat capacity of water = 4.186 J/ g °C.

Let's assume that there is no heat loss to the surrounding and the final temperature is T.

By applying the equation,

5430 J = 1246.25 g x 4.186 J/ g °C x (T - 23) °C
(T - 23) °C = 5430 J / 1246.25 g x 4.186 J/ g °C
(T - 23) °C = 1.04 °C
T = 1.04 °C + 23 °C
T = 24.04 °C

Hence, the final temperature of the water is 24.04 °C.

4 0

3 years ago

Calculate the Ka of your acetic acid solution. Discuss this calculation. Based on the value of Ka, is acetic acid a strong acid

lord [1]

Based on our knowledge of strong and weak acids, we can confirm that the Ka value for acetic acid will be relatively low since it is a weak acid.

Acids can be strong or weak. This is determined by its tendency to break apart into ions or stay together to form molecules. Although somewhat counter-intuitive, strong acids are those that are most likely to break apart and therefore contain a high number of ions within their solutions.

Weak acids, on the other hand, are those that tend to stay together in the form of molecules and therefore possess very low ion counts in their solutions. The acid dissociation constant, Kₐ, is used to measure whether an acid is weak or strong and how much so. In the case of Acetic acid, the ka measurement will offer a low value, indicating a weak acid.

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4 0

2 years ago

Strontium chloride and sodium fluoride react to form strontium fluoride and sodium chloride, according to the reaction shown.

grin007 [14]

Answer:

4.07L of a 0.110M NaF are needed

Explanation:

Based on the reaction:

SrCl₂(aq)+2NaF(aq)⟶SrF₂(s)+2NaCl(aq)

1 mole of strontium chloride react with 2 moles of NaF

361mL of 0.620M SrCl₂ solution has:

0.361L ₓ (0.620mol / L) = 0.22382 moles SrCl₂.

Moles of NaF for a complete reaction must be:

0.22382 moles SrCl₂ ₓ (2 mol NaF / 1 mol SrCl₂) = 0.44764 moles of NaF

If you have a solution of 0.110M NaF, the moles of NaF needed are:

0.44764 moles of NaF ₓ (1L / 0.110mol NaF) = 4.07L of a 0.110M NaF are needed

6 0

3 years ago

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