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zalisa [80]
3 years ago
10

Each of the following values was read on an instrument of measuring device. In each case the last digit was estimated. Tell what

divisions would have been made on the scale or measuring device.
example: 16.00 The scale read to tenths; hundredths were estimated

16 g

16.00 mL

160 cm

1600 g

160.0 g

For this question, you must find the divisions made for each of the numbers above
Chemistry
1 answer:
Drupady [299]3 years ago
6 0

Answer:

<h3>160 cm</h3>

Explanation:

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A rigid, closed vessel of volume V =648 liters maintained at constant temperature T=365 K is loaded with 112 mol of n-hexane, 15
Marrrta [24]

Answer:

P = 2.92 atm

Explanation:

With the three assumptions in mind, the system consists of:

  • A liquid phase containing n-hexane and n-heptane, and
  • A gaseous phase containing n-hexane vapor, n-heptane vapor, and nitrogen gas.

First we use PV=nRT to calculate the moles of n-hexane and n-heptane in the gaseous phase:

  • n-hexane:

P = 0.199 MPa ⇒ 0.199 * 1.869 = 1.964 atm

  • 1.964 atm * 648 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 365 K
  • n = 42.52 moles
  • n-heptane:

P = 0.083 MPa ⇒ 0.083 * 1.869 = 0.155 atm

  • 0.155 atm * 648 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 365 K
  • n = 3.358 moles

So <u>the gaseous phase consists of 42.52 moles of n-hexane, 3.358 moles of n-heptane, and 14 mol of nitrogen</u>.

For the liquid phase, we <u>calculate the remaining moles of n-hexane and n-heptane</u>. Then we<u> convert to liters</u>, using their molar volumes:

  • n-hexane:
  • n = 112 mol - 42.52 mol = 69.48 mol
  • 69.48 mol * 0.146 L/mol = 10.14 L
  • n-heptane:
  • n = 155 mol - 3.358 mol = 151.642 mol
  • 151.642 mol * 0.162 L/mol = 24.57 L

So the liquid phase occupies (10.14+24.57) = 34.71 L, and <u>contains 69.48 mol of n-hexane and 151.64 mol of n-heptane</u>.

Finally, to<u> calculate the pressure in the vessel</u>, we use PV=nRT:

P = ?

V = 648 - 34.71 = 613.29 L

n = 42.52 mol hexane + 3.35 mol heptane + 14 mol nitrogen = 59.87 mol

T = 365 K

  • P * 613.29 L = 59.87 mol *  0.082 atm·L·mol⁻¹·K⁻¹ * 365 K
  • P = 2.92 atm
6 0
3 years ago
Compare and contrast atoms elements compounds and molecules
N76 [4]

atoms = smallest unit of an element

elements = contain only one type of atom

compounds = two or more elements that are chemically bound together

molecules = a group of atoms bonded together

6 0
3 years ago
14)
MAXImum [283]

Answer:

the vapor pressure of this solution would increase if some of the water were allowed to evaporate

4 0
3 years ago
A gold-colored ring has a mass of 17.5 grams and a volume of 0.82 mL. What is the density of this ring?
max2010maxim [7]

Answer:

21 g/mL

Explanation:

To solve this problem, first look at the density equation, which is D=M/V, which D stands for density, M stands for mass, and V stands for volume. When you substitute in the variables, you get D=17.5/.82, which is equivalent to 21.34. However, since we need to pay attention to the sig fig rules for multiplying, we need to have the same amount of sig figs as the value with the least amount of sig figs, which is the number .82. .82 has two sig figs, so you round down. Your answer will be 21 g/mL.

4 0
3 years ago
Using spectroscopic notation write the complete electron configuration for the fluorine atom. Using noble gas notation write the
prohojiy [21]

Explanation:

Flourine has atomic number of 9 and hence 9 electrons in its neutral state. The full electronic configuration is given as;

1s2 2s2 2p5

Carbon has atomic number of 6 and hence 6 electrons in it's neutral state. The noble gas notation as the following format;

[closest noble gas before the element] remaining electrons

The nearest noble gas to carbon is Helium, the noble gas notation is given as;

[He] 2s4

3 0
3 years ago
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