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Pepsi [2]
2 years ago
7

Alkanes are hydrocarbons containing only single bonds. Acyclic alkanes have carbon atoms arranged in chains, whereas cycloalkane

s have carbon atoms arranged in one or more rings. How many hydrogen atoms are in an acyclic alkane with 7 7 carbon atoms?
Chemistry
1 answer:
sashaice [31]2 years ago
4 0

Answer:

156 Hydrogen atoms

Explanation:

<u>Any acyclic alkane has a molecular formula that can be expressed as</u>:

CₙH₂ₙ₊₂

Where <em>n</em> is any integer and the number of carbon atoms. For example, Propane has 3 carbon atoms, this means it would have [2*3+2] 8 hydrogen atoms, resulting with a formula of C₃H₈.

An acyclic alkane with 77 carbon atoms would thus have:

2*77 + 2 = 156 hydrogen atoms

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On a cool morning (12"C), a balloon is filled with 1.5 L of helium. By mid afternoon, the temperature has soared to 32°C. What i
ddd [48]

Answer:

1.6 L

Explanation:

Using Charle's law  

\frac {V_1}{T_1}=\frac {V_2}{T_2}

Given ,  

V₁ = 1.5 L

V₂ = ?

T₁ = 12 °C

T₂ = 32 °C  

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (12 + 273.15) K = 285.15 K  

T₂ = (32 + 273.15) K = 305.15 K  

Using above equation as:

\frac{1.5}{285.15}=\frac{V_2}{305.15}

V_2=\frac{1.5\cdot \:305.15}{285.15}

New volume = 1.6 L

8 0
2 years ago
A student determines measures the mass of one mole of carbon and finds it to be 12.22 grams. if the accepted value is 12.11 gram
Gre4nikov [31]

Answer:-

0.91% is the students % of error

Explanation: -

Accepted value= 12.11 grams

Measured value = 12.22 grams

Error = 12.22-12.11 = 0.11 grams

Percentage error = \frac{0.11 grams}{12.11 grams}x100

                           = 0.91 %

Thus 0.91% is the students % of error

5 0
2 years ago
A 50-gram sample has a half-life of 12 days. How much material will remain after 12 days?
Elis [28]
<span>A 50-gram sample with a half-life of 12 days will have a remaining mass of 25 grams after its 12-day half-life. Every cycle of a half-life, the sample will lose half of its mass, so if the half-life, itself, is 12 days and the time period passing is 12 days, one half-life has passed and the material will be halved.</span>
8 0
1 year ago
Read 2 more answers
A) To generate the spectrum above a source capable of producing electromagnetic radiation with an energy of 7 * 10 ^ 4 * k * J p
marin [14]

Answer:

Gamma

Explanation:

I'm not sure how to do it without calculations but:

E=hv

7*10^7 J/mol=6.626*10^34 Js * v

v=1*10^41

Gamma rays.

More here: https://www.hasd.org/faculty/AndrewSchweitzer/spectroscopy.pdf

3 0
2 years ago
Read 2 more answers
A compound that is composed of carbon, hydrogen, and oxygen contains 70.6% C, 5.9% H, and 23.5% O by mass. The molecular weight
zhannawk [14.2K]

Answer: The molecular formula will be C_8H_8O_2

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C= 70.6 g

Mass of H = 5.9 g

Mass of O = 23.5 g

Step 1 : convert given masses into moles.

Moles of C =\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{70.6g}{12g/mole}=5.9moles

Moles of H =\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{5.9g}{1g/mole}=5.9moles

Moles of O =\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{23.5g}{16g/mole}=1.5moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = \frac{5.9}{1.5}=4

For H = \frac{5.9}{1.5}=4

For O =\frac{1.5}{1.5}=1

The ratio of C : H: O= 4: 4:1

Hence the empirical formula is C_4H_4O

The empirical weight of C_4H_4O = 4(12)+4(1)+1(16)= 68g.

The molecular weight = 136 g/mole

Now we have to calculate the molecular formula.

n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{136}{68}=2

The molecular formula will be=2\times C_4H_4O=C_8H_8O_2

4 0
2 years ago
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