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algol13
3 years ago
15

7. How many electrons can a single orbital hold?

Chemistry
2 answers:
Nat2105 [25]3 years ago
8 0
“It can hold two electrons”
levacccp [35]3 years ago
5 0

Answer:

2 electrons

Explanation:

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Breathing air that contains 4.0% by volume CO2 over time causes rapid breathing, throbbing headache, and nausea, among other sym
7nadin3 [17]

Answer:

1 mole of any ideal gas occupies the same volume as one mole of any other ideal gas under the same conditions of temperature and pressure.So 1 L CO2 has same number of moles as 1 L O2 and 1 L N2 etc.This means that, assuming all the gases in air are ideal gases, if CO2 is 4.0 % by volume then it is also 4.0 % by moles, because a volume of each gas has the same number of moles.

Explanation:

3 0
3 years ago
Suppose you have been given the task of distilling a mixture of hexane + toluene. Pure hexane has a refractive index of 1.375 an
tatuchka [14]

Given:

refractive index of pure hexane, \mu_{h} = 1.375

refractive index of pure hexane, \mu_{t} = 1.497

refractive index of mixture, \mu = 1.401

Formula used:

As refractive index here, behaves like a colligative prop. then using the following formula:

\mu = n_{h}\mu_{h} + n_{t}\mu_{t}                         (1)

where,

\n_{h} = hexane  mole fraction

\n_{t} = toulene  mole fraction

\n_{h} + \n_{t} = 1                                      (2)

Solution:

Now, using given formula from eqn (1)

\mu = n_{h}\mu_{h} + n_{t}\mu_{t}

1.401 = 1.375\times n_{h}+ 1.497\times n{t}              (3)

Multiply eqn (2) by 1.375, we get:

1.375\times \n_{h} + 1.375\times \n_{t} = 1.375         (4)

Now, solving eqn  (3) and (4):

\n_{t} = 0.213

Substituting the value of \n_{t} = 0.213 in eqn (1), we get:

\n_{h} = 1 - 0.213

\n_{h} = 0.787

Therefore, mole fraction of hexane in the sample is \n_{h} = 0.787

8 0
3 years ago
Science (Grade 8)
slega [8]

Answer:

1. Frictional. 2. Gravitational. 3. Concrete road. 4.0N

8 0
2 years ago
Calculate the milligrams of chlorine (35.453 g/mol) in a 3.44 g solution of 18.0 wt% CaCl2 (110.98 g/mol).
alexandr1967 [171]

Answer:

395 mg of Cl are in 3.44 g of CaCl₂

Explanation:

18.0 wt% CaCl₂ means that 18 g of salt, are contained in 100 g of solvent (assuming solvent as solution).

Let's apply a rule of three:

In 100 g of solution we have 18 g of salt

In 3.44 g of solvent we have (3.44 .18)/ 100 = 0.1692 grams of salt.

Now, let's convert this mass into moles (mass / molar mass)

0.1692 mass / 110.98 g/m = 0.00558 moles

1 mol of CaCl₂ has 1 mol of Ca and 2 moles of Cl

Then, 0.00558 moles of CaCl₂, would have the double of moles of Cl

0.00558 m .2 = 0.01116 moles of Cl.

Let's convert the moles into mass (mol . molar mass)

0.01116 moles of Cl . 35.453g/m = 0.395 grams of Cl.

Finally we convert grams to mg (x1000)

0.395 g = 395 mg

5 0
3 years ago
(PLEASE HELP, i will give brainlist, and the question is 15 points)
Mrrafil [7]
8 .6 L ndjiekebrbiekrbdk
5 0
2 years ago
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