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ddd [48]
3 years ago
10

Коя от следни те течности има най- ниско парно налягане при 250 С? -- Вода → tк ═ 1000С -диетилов етер → tк ═ 34,60С -пентан → t

к ═ 360С - етанол → tк ═ 780С
Chemistry
1 answer:
BaLLatris [955]3 years ago
7 0

Answer:

sorry i dont understand

Explanation:

i am russian but i do school in english so i dont know what some of these elements are exept for water

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Significant Figures For 7.06 × 10^5 ÷ 5.3 × 10^-2
Kay [80]

Answer:

\boxed{\text{two}}

Explanation:

In multiplication  and division problems, your answer can have no more significant figures than the number with the fewest significant figures.

\dfrac{7.06 \times 10^{5}}{5.3 \times 10^{-2}}= 1.332 075 472 \times 10^{7} (by my calculator)  

There are three significant figures in 7.06 and two in 2.3.

You must round to \boxed{\textbf{two}} significant figures and report the answer as 1.3 × 1.0⁷.

4 0
3 years ago
Does somebody know the answer ?
Alex777 [14]
H2O is the missing reactant.

Just a caveat: this equation isn’t balanced.
3 0
3 years ago
What is the definition of a physical property of matter?
Bezzdna [24]
<span>Physical properties are any properties of matter which can be perceived or observed without changing the chemical identity of the sample.</span>
7 0
3 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
4. In the lab activity Making Connections, an experiment was designed to test the effect of exercise on the ability to squeeze a
zvonat [6]

Answer:

The independent variable

Explanation:

This is because, we are looking for what changed which that immediately crosses out control variable and hypothesis

4 0
3 years ago
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