1.

The instruction booklet for your pressure cooker indicates that its highest setting is 12.8 psi . You know that standard atmosph

lubasha [3.4K]

Answer:At $237.57^oC$ food will cook in the pressure cooker set on high.

Explanation:

Standard atmospheric pressure $P_1$ = 14.7 Psi = 1.0001 atm (1 Psi=0.06804 atm)

Standard temperature $T_1$ = 273.15 K

Highest pressure offered by pressure cooker: 12.8 Psi = 0.8709 atm

Pressure ( $P_2$ ) inside the pressure cooker on the highest setting at temperature $T_2$ :

$P_2$ = 1.0001 atm + 0.8709 atm = 1.8710 atm

According to Gay lussac law :

$(pressure)\propto (temperature)$ (at constant Volume)

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

$T_2=\frac{P_2\times T_1}{P_1}=\frac{1.8710 atm\times 273.15 K}{1.0001 atm}=510.73 K= 237.57^oC (T(^oC)=T-273.15 K)$

At $237.57^oC$ food will cook in the pressure cooker set on high.

8 0

3 years ago

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Using the solubility of Ca(IO3)2 that you determined in pure water, calculate the value of Ksp using only concentrations, that i

Aleks04 [339]

Answer:

4.6305 * 10^-6 mol^3.L^-3

Explanation:

Firstly, we write the value for the solubility of Ca(IO3)2 in pure water. This equals 0.0105mol/L.

We proceed to write the dissociation reaction equation for Ca(IO3)2

Ca(IO3)2(s) <——->Ca2+(aq) + 2IO3-(aq)

We set up an ICE table to calculate the Ksp. ICE stands for initial, change and equilibrium. Let the concentration of the Ca(IO3)2 be x. We write the values for the ICE table as follows:

Ca2+(aq). 2IO3-(aq)

I. 0. 0.

C. +x. +2x

E. x. 2x

The solubility product Ksp = [Ca2+][IO3-]^2

Ksp = x * (2x)^2

Ksp = 4x^3

Recall, the solubility value for Ca(IO3)2 in pure water is 0.0105mol/L

We substitute this value for x

Ksp = 4(0.0105)^2 = 4 * 0.000001157625 = 4.6305 * 10^-6

5 0

3 years ago

Sodium and oxygen ionic bond diagram

stiv31 [10]

Is This it? I’m sorry if it wasn’t

8 0

3 years ago

A 6.1-kg solid sphere, made of metal whose density is 2600 kg/m3, is suspended by a cord. When the sphere is immersed in a liqui

deff fn [24]

Answer:

Density of the liquid = 1470.43 kg/m³

Explanation:

Given:

Mass of solid sphere(m) = 6.1 kg

Density of the metal = 2600 kg/m³

Thus volume of the liquid :

$Volume(V)=\frac{Mass(m)}{Density (\rho)}$

Volume of the sphere = 6.1 kg/2600 kg/m³ = 0.002346 m³

The volume of water displaced is equal to the volume of sphere (Archimedes' principle)

Volume displaced = 0.002346 m³

Buoyant force = $\rho\times gV$

Where

$\rho$ is the density of the fluid

g is the acceleration due to gravity

V is the volume displaced

The free body diagram of the sphere is shown in image.

According to image:

$mg=\rho\times gV+T$

Acceleration due to gravity = 9.81 ms⁻²

Tension force = 26 N

Applying in the equation to find the density of the liquid as:

$6.1\times 9.81=\rho\times 9.81\times 0.002346+26$

$33.841=\rho\times 9.81\times 0.002346$

$\rho=\frac{33.841}{9.81\times 0.002346}$

$\rho=1470.43 kgm^3$

<u>Thus, the density of the liquid = 1470.43 kg/m³</u>

6 0

4 years ago

Which of these is most likely to form between elements transferring electrons to

Stella [2.4K]

Answer: The answer is an ionic bond

6 0

3 years ago

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