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3 years ago

Determine the mass of 18.7ml of rubbing alcohol with a density of 0.79 g/ml?

Chemistry

2 answers:

ycow [4]3 years ago

7 0

23.67 g unless you're rounding to significant figures than it's 24 g

alexandr1967 [171]3 years ago

5 0

Answer:

The mass of the rubbing alcohol is 14.77 g

Explanation:

The formula to be used here is that of density which is shown below

Density = Mass ÷ volume

The density of the rubbing alcohol (also known as isopropyl alcohol) is given in the question as 0.79 g/ml while the volume of same is given as 18.7ml. The mass of this rubbing alcohol is the unknown (which we can derive from the formula given earlier).

Hence,

Mass = Density × volume

Mass = 0.79 × 18.7

Mass = 14.77 g

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HACTEHA [7]

Not sure gets hotter prolly

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3 years ago

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Alice and Bob are experimenting with two moles of neon, a monatomic gas, that starts out at conditions of standard temperature a

Archy [21]

Answer:

29273.178 joules have been added to the gas for the entire process.

Explanation:

The specific heats of monoatomic gases, measured in joules per mol-Kelvin, are represented by the following expressions:

Isochoric (Constant volume)

$c_{v} = \frac{3}{2}\cdot R_{u}$ (1)

Isobaric (Constant pressure)

$c_{p} = \frac{5}{2}\cdot R_{u}$ (2)

Where $R_{u}$ is the ideal gas constant, measured in pascal-cubic meters per mol-Kelvin.

Under the assumption of ideal gas, we notice the following relationships:

1) Temperature is directly proportional to pressure.

2) Temperature is directly proportional to volume.

Now we proceed to find all required temperatures below:

(i) Alice heats the gas at constant volume until its pressure is doubled:

$\frac{T_{2}}{T_{1}} = \frac{P_{2}}{P_{1}}$ (3)

( $\frac{P_{2}}{P_{1}} = 2$ , $T_{1} = 273.15\,K$ )

$T_{2} = \frac{P_{2}}{P_{1}} \times T_{1}$

$T_{2} = 2\times 273.15\,K$

$T_{2} = 546.3\,K$

(ii) Bob further heats the gas at constant pressure until its volume is doubled:

$\frac{T_{3}}{T_{2}} =\frac{V_{3}}{V_{2}}$ (4)

( $\frac{V_{3}}{V_{2}} = 2$ , $T_{2} = 546.3\,K$ )

$T_{3} = \frac{V_{3}}{V_{2}}\times T_{2}$

$T_{3} = 2\times 546.3\,K$

$T_{3} = 1092.6\,K$

Finally, the heat added to the gas ( $Q$ ), measured in joules, for the entire process is:

$Q = n\cdot [c_{v}\cdot (T_{2}-T_{1})+c_{p}\cdot (T_{3}-T_{2})]$ (5)

If we know that $R_{u} = 8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K}$ , $n = 2\,mol$ , $T_{1} = 273.15\,K$ , $T_{2} = 546.3\,K$ and $T_{3} = 1092.6\,K$ , the heat added to the gas for the entire process is:

$c_{v} = \frac{3}{2}\cdot \left(8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K} \right)$

$c_{v} = 12.471\,\frac{J}{mol\cdot K}$

$c_{p} = \frac{5}{2}\cdot \left(8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K} \right)$

$c_{p} = 20.785\,\frac{J}{mol\cdot K}$

$Q = (2\,mol)\cdot \left[\left(12.471\,\frac{J}{mol\cdot K} \right)\cdot (536.3\,K-273.15\,K)+\left(20.785\,\frac{J}{mol\cdot K} \right)\cdot (1092.6\,K-546.3\,K )\right]$

$Q = 29273.178\,J$

29273.178 joules have been added to the gas for the entire process.

6 0

3 years ago

Running water can transport weathered pieces of rocks downstream what will happen to these weathered pieces of rocks and sedimen

OleMash [197]

Answer:

those small sediments begin to precipitate, condense on the ground and finally settle.

Explanation:

Sedimentation is a natural process, which occurs spontaneously in nature and promotes the formation of sedimentary rocks.

8 0

3 years ago

Question 3 (1 point)

larisa86 [58]

Answer:

1. Formulae of hydrate is Na₂CO₃.10H₂O

2. Formulae of hydrate is CaSO₄.2H₂O

Explanation:

Q1. Mass of hydrated salt = 8.50 g mass

mass of anhydrous salt = 7.26 grams

mass of water lost = 1.24

Formula of hydrated salt = Na₂CO₃.xH₂O

Formula of anhydrous salt = Na₂CO₃

Molar mass of anhydrous salt is obtained as below where Na = 23, C = 12, O = 16, H = 1

Molar mass of Na₂CO₃ = (23 *2 + 12 + 16 * 3) = 106 g

molar mass of water = (1 *2 + 16) = 18 g

Mole ratio of Na₂CO₃ to H₂O = 7.26/106 : 1.24/18

Mole ratio of Na₂CO₃ to H₂O = 0.0068 : 0.068

Mole ratio of Na₂CO₃ to H₂O = 1 :10

Therefore, formulae of hydrate is Na₂CO₃.10H₂O

Q2. Mass of hydrated salt = 60.98 g mass

mass of anhydrous salt = 48.23 grams

mass of water lost = 12.75

Formula of hydrated salt = CaSO₄.xH₂O

Formula of anhydrous salt = CaSO₄

Molar mass of anhydrous salt is obtained as below where Ca = 40, S = 32, O = 16, H = 1

Molar mass of CaSO₄= (40 + 32 + 16 * 4) = 136 g

molar mass of water = (1 *2 + 16) = 18 g

Mole ratio of CaSO₄ to H₂O = 48.23/136 : 12.75/18

Mole ratio of CaSO₄ to H₂O = 0.35 : 0.70

Mole ratio of CaSO₄ to H₂O = 1 : 2

Therefore, formulae of hydrate is CaSO₄.2H₂O

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3 years ago

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During blank,two divisions of the nucleus and the cytoplasm occur?

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Cytokinensis is what I found.

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