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MrRissso [65]
4 years ago
12

brainliest!!!!!!!!!!!!!!!! asap please What is Kepler-186f? a. telescope used to investigate outer space b.star in another solar

system c. recently discovered moon of Saturn d.Earth-like planet in another solar system
Chemistry
1 answer:
Anika [276]3 years ago
6 0

D-It was discovered a while ago and it's about 500 light years away from earth

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Which of these properties would be possessed by both acids and bases? note: we do not mean mixing both in the same solution. we
Firdavs [7]
Answer is: <span>3. they conduct electricity.
1) acids taste sour, bases taste bitter.
2) acid </span><span>would react with metals and produce hydrogen gas:
Ca(s) + 2HCl(aq) </span>→ CaCl₂(aq) + H₂(g).
3) Acids and bases give ions in water solutions, so <span>they conduct electricity.
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An Arrhenius base is a substance that dissociates in water to form hydroxide ions (OH⁻<span>). 
In this example lithium hydroxide is an Arrhenius base:</span>

LiOH(aq) → Li⁺(aq) + OH⁻(aq).

An Arrhenius acid is a substance that dissociates in water to form hydrogen ions or protons (H⁺<span>). 
For example hydrochloric acid: HCl(aq) → H</span>⁺(aq) + Cl⁻(aq).

4) red litmus turn blue whe base is drop on it and blue litmus turn red when acid is drop on it.

5) bases have pH greater than 7, acids have pH less than 7.

5 0
3 years ago
A compound with the empirical formula, ch2, has a molar mass equal to 84.2 g/mol. what is the molecular formula for the compound
MatroZZZ [7]
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7 0
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Why is the cellulose part of human diet​
Luden [163]

Answer:

Cellulose stimulates the work of intestines and increases their discharge.

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What is the name of the compound CaCo3​
ddd [48]
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4 0
3 years ago
Express the van der Waals equation of state as a virial expansion in powers of 1/Vm and obtain expressions for B and C in terms
nirvana33 [79]

Answer:

PV_{m} = RT[1 + (b-\frac{a}{RT})\frac{1}{V_{m} } + \frac{b^{2} }{V^{2} _{m} } + ...]

B = b -a/RT

C = b^2

a = 1.263 atm*L^2/mol^2

b = 0.03464 L/mol

Explanation:

In the given question, we need to express the van der Waals equation of state as a virial expansion in powers of 1/Vm and obtain expressions for B and C in terms of the parameters a and b. Therefore:

Using the van deer Waals equation of state:

P = \frac{RT}{V_{m}-b } - \frac{a}{V_{m} ^{2} }

With further simplification, we have:

P = RT[\frac{1}{V_{m}-b } - \frac{a}{RTV_{m} ^{2} }]

Then, we have:

P = \frac{RT}{V_{m} } [\frac{1}{1-\frac{b}{V_{m} } } - \frac{a}{RTV_{m} }]

Therefore,

PV_{m} = RT[(1-\frac{b}{V_{m} }) ^{-1} - \frac{a}{RTV_{m} }]

Using the expansion:

(1-x)^{-1} = 1 + x + x^{2} + ....

Therefore,

PV_{m} = RT[1+\frac{b}{V_{m} }+\frac{b^{2} }{V_{m} ^{2} } + ... -\frac{a}{RTV_{m} }]

Thus:

PV_{m} = RT[1 + (b-\frac{a}{RT})\frac{1}{V_{m} } + \frac{b^{2} }{V^{2} _{m} } + ...]           equation (1)

Using the virial equation of state:

P = RT[\frac{1}{V_{m} }+ \frac{B}{V_{m} ^{2}}+\frac{C}{V_{m} ^{3} }+ ...]

Thus:

PV_{m} = RT[1+ \frac{B}{V_{m} }+ \frac{C}{V_{m} ^{2} } + ...]     equation (2)

Comparing equations (1) and (2), we have:

B = b -a/RT

C = b^2

Using the measurements on argon gave B = −21.7 cm3 mol−1 and C = 1200 cm6 mol−2 for the virial coefficients at 273 K.

b = \sqrt{C} = \sqrt{1200} = 34.64[tex]cm^{3}/mol[/tex] = 0.03464 L/mol

a = (b-B)*RT = (34.64+21.7)*(1L/1000cm^3)*(0.0821)*(273) = 1.263 atm*L^2/mol^2

3 0
3 years ago
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