When more than one variable changes during a scientific experiment,

Who would most likely be required to work with sodium hydroxide?

maw [93]

Answer:

i think plumber

4 0

4 years ago

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M= 2 kg, a = 2 m/s^2. <br><br> Find F

saul85 [17]

Answer:

Explanation:

8 0

3 years ago

Consider the equation below. CaCO3(s) <—>CaO(s) + CO2(g) what is the equilibrium constant expression for the given reactio

hichkok12 [17]

Answer:

Answer is: Keq = [CO₂].

Explanation:

Balanced chemical reaction: CaCO3(s) ⇄ CaO(s) + CO₂(g).

The equilibrium constant (Keq) is a ratio of the concentration of the products to the concentration of the reactants.

Pure liquids (shown in chemical reactions by appending (l) to the chemical formula) and solids (shown in chemical equations by appending (s) to the chemical formula) not go in to he equilibrium constant expression, only gas state (shown in chemical reactions by appending (g) to the chemical formula) reactants and products go in to the equilibrium constant expression

8 0

3 years ago

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A student did not read the direction to the experiment properly and mixed up where to place the NaOH solution and the vinegar. H

Mariana [72]

Answer:

At end point there will a transition from pink to colorless.

Explanation:

As the student put the vinegar in the titrator and NaOH in the beaker, it means that he has poured phenolphthalein in the NaOH solution.

The pH range of phenolphthalein is 8.3-10 (approx), it means it will show pink color in basic medium.

So on addition of phenolphthalein in NaOH the solution will become pink in color.

When we start pouring vinegar from titrator neutralization of NaOH will begin.

On complete neutralization , on addition of single drop of vinegar the solution will become acidic and there will be complete disappearance of pink color solution in the beaker.

3 0

4 years ago

At 298 K, what is the Gibbs free energy change (ΔG) for the following reaction?

9966 [12]

Answer:

(a). The Gibbs free energy change is 2.895 kJ and its positive.

(b). The Gibbs free energy change is 34.59 J/mole

(c). The pressure is 14924 atm.

(d). The Gibbs free energy of diamond relative to graphite is 4912 J.

Explanation:

Given that,

Temperature = 298 K

Suppose, density of graphite is 2.25 g/cm³ and density of diamond is 3.51 g/cm³.

$\Delta H\ for\ diamond = 1.897 kJ/mol$

$\Delta H\ for\ graphite = 0 kJ/mol$

$\Delta S\ for\ diamond = 2.38 J/(K mol)$

$\Delta S\ for\ graphite = 5.73 J/(K mol)$

(a) We need to calculate the value of $\Delta G$ for diamond

Using formula of Gibbs free energy change

$\Delta G=\Delta H-T\Delta S$

Put the value into the formula

$\Delta G= (1897-0)-298\times(2.38-5.73)$

$\Delta G=2895.3$

$\Delta G=2.895\ kJ$

The Gibbs free energy change is positive.

(b). When it is compressed isothermally from 1 atm to 1000 atm

We need to calculate the change of Gibbs free energy of diamond

Using formula of gibbs free energy

$\Delta S=V\times\Delta P$

$\Delta S=\dfrac{m}{\rho}\times\Delta P$

Put the value into the formula

$\Delta S=\dfrac{12\times10^{-6}}{3.51}\times999\times10130$

$\Delta S=34.59\ J/mole$

(c). Assuming that graphite and diamond are incompressible

We need to calculate the pressure

Using formula of Gibbs free energy

$\beta= \Delta G_{g}+\Delta G+\Delta G_{d}$

$\beta=V(-\Delta P_{g})+\Delta G+V\Delta P_{d}$

$\beta=\Delta P(V_{d}-V_{g})+\Delta G$

Put the value into the formula

$0=\Delta P(\dfrac{12\times10^{-6}}{3.51}-\dfrac{12\times10^{-6}}{2.25})\times10130+2895.3$

$0=-0.0194\Delta P+2895.3$

$\Delta P=\dfrac{2895.3}{0.0194}$

$\Delta P=14924\ atm$

(d). Here, $C_{p}=0$

So, The value of $\Delta H$ and $\Delta S$ at 900 k will be equal at 298 K

We need to calculate the Gibbs free energy of diamond relative to graphite

Using formula of Gibbs free energy

$\Delta G=\Delta H-T\Delta S$

Put the value into the formula

$\Delta G=(1897-0)-900\times(2.38-5.73)$

$\Delta G=4912\ J$

Hence, (a). The Gibbs free energy change is 2.895 kJ and its positive.

(b). The Gibbs free energy change is 34.59 J/mole

(c). The pressure is 14924 atm.

(d). The Gibbs free energy of diamond relative to graphite is 4912 J.

7 0

3 years ago