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2 years ago

5

Give IUPAC name

1 answer:

Diano4ka-milaya [45]2 years ago

7 0

Answer:

The IUPAC name of the given compound is Pent-1-ene.

Explanation:

Longest chain has 5 carbons and the double bond is at C-1 carbon

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3. Could a molecule<br> contain only one atom?

Blababa [14]

A molecule is defined as two or more atoms of the same element different element that are bound together. A molecule may be homonuclear, which means, it consists of atoms of one chemical element, as with oxygen (O2); or it may be heteronuclear, a chemical compound composed of more than one element, as with water (H2O).

8 0

2 years ago

Which compound has the highest percent composition by mass of strontium

8090 [49]

So if the compound has the smallest gram formula mass it has the highest percentage composition by mass of strontium

3 0

2 years ago

Harim placed 5mL of ethanol into a container that weighs 1 gram using a dropper. He already knew the density of ethanol is 0.78

Aleks04 [339]

Answer:

there it is fella tried on ma own consciousness

8 0

2 years ago

Air is compressed from an inlet condition of 100 kPa, 300 K to an exit pressure of 1000 kPa by an internally reversible compress

ElenaW [278]

Answer:

(a) $W_{isoentropic}=8.125\frac{kJ}{mol}$

(b) $W_{polytropic}=7.579\frac{kJ}{mol}$

(c) $W_{isothermal}=5.743\frac{kJ}{mol}$

Explanation:

Hello,

(a) In this case, since entropy remains unchanged, the constant $k$ should be computed for air as an ideal gas by:

$\frac{R}{Cp_{air}}=1-\frac{1}{k} \\\\\frac{8.314}{29.11} =1-\frac{1}{k}\\$

$0.2856=1-\frac{1}{k}\\\\k=1.4$

Next, we compute the final temperature:

$T_2=T_1(\frac{p_2}{p_1} )^{1-1/k}=300K(\frac{1000kPa}{100kPa} )^{1-1/1.4}=579.21K$

Thus, the work is computed by:

$W_{isoentropic}=\frac{kR(T_2-T_1)}{k-1} =\frac{1.4*8.314\frac{J}{mol*K}(579.21K-300K)}{1.4-1}\\\\W_{isoentropic}=8.125\frac{kJ}{mol}$

(b) In this case, since $n$ is given, we compute the final temperature as well:

$T_2=T_1(\frac{p_2}{p_1} )^{1-1/n}=300K(\frac{1000kPa}{100kPa} )^{1-1/1.3}=510.38K$

And the isentropic work:

$W_{polytropic}=\frac{nR(T_2-T_1)}{n-1} =\frac{1.3*8.314\frac{J}{mol*K}(510.38-300K)}{1.3-1}\\\\W_{polytropic}=7.579\frac{kJ}{mol}$

(c) Finally, for isothermal, final temperature is not required as it could be computed as:

$W_{isothermal}=RTln(\frac{p_2}{p_1} )=8.314\frac{J}{mol*K}*300K*ln(\frac{1000kPa}{100kPa} ) \\\\W_{isothermal}=5.743\frac{kJ}{mol}$

Regards.

8 0

2 years ago

C12H22O11 + 11 H2SO4 12 C + 11 H2SO4 + 11 H2O

Vladimir [108]

Answer:

B: The sulfuric acid is not consumed or react with the reactant.

4 0

2 years ago