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kap26 [50]
3 years ago
8

2,2-dimethyl-4-propyloctane has how many secondary carbons? view available hint(s) 2,2-dimethyl-4-propyloctane has how many seco

ndary carbons? five nine six seven
Chemistry
1 answer:
Citrus2011 [14]3 years ago
7 0
Structure of <span>2,2-dimethyl-4-propyloctane is in the Word document below.
Answer is: six </span><span>secondary carbons.
</span>Secondary carbon (2°)<span> is attached to two other carbons. Secundary carbons are third, fifth, sixth and seventh in octane and first and second in propyl.
</span>Primary carbon (1°) is <span> attached to only </span>one <span>carbon.
</span>Tertiary carbon (3°)<span> is attached to three other carbons.
</span>Quaternary carbon (4°)<span> is attached to </span>four<span> other carbons.</span>
Download docx
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In physical changes, the atoms or molecules that compose the matter do not change their identity, even though the matter may cha
slavikrds [6]

Answer:

True

Explanation:

when a physical change occurs in a substance, the appearance might change but the chemical identity remains same. For example, if liquid water is heated, the water changes to gaseous form .This means the appearance of the water has changed, but the chemical identity remains same as both liquid water and steam has both oxygen and hydrogen as the component molecules or atoms.

3 0
4 years ago
Hello, everyone!
marusya05 [52]

Answer: 27.09 ppm and 0.003 %.

First, <u>for air pollutants, ppm refers to parts of steam or gas per million parts of contaminated air, which can be expressed as cm³ / m³. </u>Therefore, we must find the volume of CO that represents 35 mg of this gas at a temperature of -30 ° C and a pressure of 0.92 atm.

Note: we consider 35 mg since this is the acceptable hourly average concentration of CO per cubic meter m³ of contaminated air established in the "National Ambient Air Quality Objectives". The volume of these 35 mg of gas will change according to the atmospheric conditions in which they are.

So, according to the <em>law of ideal gases,</em>  

PV = nRT

where P, V, n and T are the pressure, volume, moles and temperature of the gas in question while R is the constant gas (0.082057 atm L / mol K)

The moles of CO will be,

n = 35 mg x \frac{1 g}{1000 mg} x \frac{1 mol}{28.01 g}

→ n = 0.00125 mol

We clear V from the equation and substitute P = 0.92 atm and

T = -30 ° C + 273.15 K = 243.15 K

V =  \frac{0.00125 mol x 0.082057 \frac{atm L}{mol K}  x 243 K}{0.92 atm}

→ V = 0.0271 L

As 1000 cm³ = 1 L then,

V = 0.0271 L x \frac{1000 cm^{3} }{1 L} = 27.09 cm³

<u>Then the acceptable concentration </u><u>c</u><u> of CO in ppm is,</u>

c = 27 cm³ / m³ = 27 ppm

<u>To express this concentration in percent by volume </u>we must consider that 1 000 000 cm³ = 1 m³ to convert 27.09 cm³ in m³ and multiply the result by 100%:

c = 27.09 \frac{cm^{3} }{m^{3} } x \frac{1 m^{3} }{1 000 000 cm^{3} } x 100%

c = 0.003 %

So, <u>the acceptable concentration of CO if the temperature is -30 °C and pressure is 0.92 atm in ppm and as a percent by volume is </u>27.09 ppm and 0.003 %.

5 0
3 years ago
How many sulfur atoms would be present in 41.8 moles of
Ede4ka [16]

Answer: 2.52*10^25

Explanation:

6.02214076*10^23 in 1 mole

3 0
3 years ago
Veronica is taking an average of 16 height measurements in a brand-new experiment. Which term is this the best example of?
Varvara68 [4.7K]

Answer:

Repetition

Explanation:

The best term to describe this example is repetition and not replication.

Most times, to obtain an accurate and precise reading which is reliable, scientist makes several repeated measurements. The average gives the most reliable representation of the phenomenon being tested.

This process is called repetition.

On the other hand, duplicating an experiment is replication. Experiments are replicated to proof their validity.

  • Since this is a brand new experiment where 16 height measurements are reported, we are dealing with a repetition situation.
4 0
3 years ago
Read 2 more answers
Hydrogen sulfide decomposes according to the following reaction: 2H2S(g) ⇋ 2H2(g) + S2(g) Kc=9.30x10-8 at 700.°C.If 0.45 mol of
Natalija [7]

Answer:

[H₂] = 1.61x10⁻³ M

Explanation:

2H₂S(g) ⇋ 2H₂(g) + S₂(g)

Kc = 9.30x10⁻⁸ = \frac{[H_{2}]^2[S_{2}]}{[H_{2}S]^2}

First we <u>calculate the initial concentration</u>:

0.45 molH₂S / 3.0L = 0.15 M

The concentrations at equilibrium would be:

[H₂S] = 0.15 - 2x

[H₂] = 2x

[S₂] = x

We <u>put the data in the Kc expression and solve for x</u>:

\frac{(2x^2) * x}{(0.15-2x)^2}=9.30x10^{-8}

\frac{4x^3}{0.0225-4x^2}=9.30*10^{-8}

We make a simplification because x<<< 0.0225:

\frac{4x^3}{0.0225} =9.30*10^{-8}

x = 8.058x10⁻⁴

[H₂] = 2*x = 1.61x10⁻³ M

5 0
4 years ago
Read 2 more answers
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