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stellarik [79]
3 years ago
12

Name 3 uses for mercury

Chemistry
2 answers:
notka56 [123]3 years ago
8 0

Answer:

Mercury is used in thermometers, barometers, manometers, sphygmomanometers, float valves, mercury switches, mercury relays, fluorescent lamps and other devices, though concerns about the element's toxicity have led to mercury thermometers and sphygmomanometers being largely phased out in clinical environments in favor ...

Explanation:

i know this will help because i got this right

Allushta [10]3 years ago
6 0

hydrargyrum , hg swiftplanet, quicksliver

Explanation:

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Hey can u answer this pls
Dmitry_Shevchenko [17]
The answer to your question is DNA!!!!! :)
5 0
3 years ago
Read 2 more answers
How many molecules of h2o are equal to 97.2 g h2o
Eduardwww [97]
M(H₂O) = 97,2 g.
n(H₂O) = m(H₂O) ÷ M(H₂O).
n(H₂O) = 97,2 g ÷ 18 g/mol.
n(H₂O) = 5,4 mol.
N(H₂O) = n(H₂O) · Na.
N(H₂O) = 5,4 mol · 6,023·10²³ 1/mol.
N(H₂O) = 3,25·10²⁴ molecules of water.
n - amount of substance.
Na - Avogadro number.
6 0
4 years ago
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Why is NCl3 not soluble in water?
stiks02 [169]

Answer:

Explanation:

NCl3 does not dissolve in water because it is a nonpolar molecule which is different with water. NCl3 is nonpolar due to the difference in electronegativities between 3 atoms of Cl and 1 atom if N2.

3 0
3 years ago
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In what way is sedimentary rock formed in nature with a sieve
ella [17]
Sedimentary rock is made of Brocken down rocks and living things.
4 0
3 years ago
Entalpy of vaporization of water is 41.1k/mol. if the vapor pressure of water at 373k is 101.3 kpa, what is the vapor pressure o
allsm [11]

Answer: The vapor pressure of water at 298 K is 3.565kPa.

Explanation:

The vapor pressure is determined by Clausius Clapeyron equation:

ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}(\frac{1}{T_1}-\frac{1}{T_2})

where,

P_1 = initial pressure at 298 K = ?

P_2 = final pressure at 373 K = 101.3 kPa

\Delta H_{vap} = enthalpy of vaporisation = 41.1 kJ/mol = 41100 J/mol

R = gas constant = 8.314 J/mole.K

T_1 = initial temperature = 298 K

T_2 = final temperature = 373 K

Now put all the given values in this formula, we get

\log (\frac{101.3}{P_1})=\frac{41100}{2.303\times 8.314J/mole.K}[\frac{1}{298K}-\frac{1}{373K}]

\frac{101.3}{P_1}=antilog(1.448)

P_1=3.565kPa

Therefore, the vapor pressure of water at 298 K is 3.565kPa.

3 0
4 years ago
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