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LekaFEV [45]
3 years ago
10

If a barometer were built using ethanol () instead of mercury (), would the column of ethanol be higher than, lower than, or the

same as the column of mercury at 1.00 atm?
Chemistry
1 answer:
Semenov [28]3 years ago
6 0

Answer:

The column of ethanol will be higher than the column of mercury.

Explanation:

The pressure (P) of a column of liquid can be calculated using the following expression.

P = ρ . g . h

where,

ρ: density of the liquid

g: gravity

h: height of the column

h = P / ρ . g

We can see that the height of the column is inversely proportional to the density of the liquid. The density of ethanol (0.789 g/mL) is lower than the density of mercury (13.6 g/mL). Then, under the same conditions, the column of ethanol will be higher than the column of mercury.

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B; Seismometer would be the answer.
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3 years ago
A beaker containing 80 grams of lead(ii) nitrate, pb(no3)2, in 100 grams of water has a temperature of 30 ºc. approximately how
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Answer:

14 g.

Explanation:

  • From the figure attached:

<em>the solubility of lead(II) nitrate, Pb(NO₃)₂, in 100 grams of water has a temperature of 30ºC is </em><em>(66 g).</em>

When beaker containing 80 grams of lead(II) nitrate, Pb(NO₃)₂, in 100 grams of water has a temperature of 30ºC.

<em>∴ The grams of the salt are undissolved, on the bottom of the beaker are </em><em>(14 g).</em>

4 0
3 years ago
A 14.4-gg sample of granite initially at 86.0 ∘C∘C is immersed into 24.0 gg of water initially at 25.0 ∘C∘C. What is the final t
kari74 [83]

Answer:

The final temperature of both substances when they reach thermal equilibrium is 31.2 °C

Explanation:

Step 1: Data given

Mass of sample granite = 14.4 grams

Initial temperature = 86.0 °C

Mass of water = 24.0 grams

The initial temperature of water = 25.0 °C

The specific heat of water = 4.18 J/g°C

The specific heat of granite = 0.790 J/g°C

Step 2: Calculate the final temperature

Heat lost = heat gained

Qgranite = - Qwater

Q = m*c*ΔT

m(granite)*c(granite)*ΔT(granite) = -m(water)*c(water)*ΔT(water)

⇒with m(granite) = the mass of granite = 14.4 grams

⇒with c(granite) = The specific heat of granite = 0.790 J/g°C

⇒with ΔT⇒(granite) = the change of temperature of granite = T2 - T1 = T2 - 86.0 °C

⇒with m(water) = the mass of water = 24.0 grams

⇒with c(water) = The specific heat of water = 4.18 J/g°C

⇒with ΔT(water) = the change of temperature of granite = T2 - T1 = T2 -25.0°C

14.4 grams * 0.790 * (T2 - 86.0°C) = -24.0 *4.18 * (T2 - 25.0°C)

11.376T2 - 978.336 = -100.32T2 + 2508

111.696 T2 = 3486.336

T2 = 31.2 °C

The final temperature of both substances when they reach thermal equilibrium is 31.2 °C

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Answer:

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Explanation:

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