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3 years ago

Which of the following is an example of a saturated hydrocarbon?

Chemistry

2 answers:

MakcuM [25]3 years ago

4 0

C. Ethyne. Hope this helped :)

KIM [24]3 years ago

3 0

Answer: B. Heptane

Explanation:

Saturated hydrocarbons are defined as the hydrocarbons in which a single bond is present between carbon and carbon atoms. they are known as alkanes. The general formula for these hydrocarbons is $C_nH_{2n+2}$

Unsaturated hydrocarbons are defined as the hydrocarbons which have double or triple covalent C-C bonds. They are known as alkenes and alkynes respectively. The general formula for these hydrocarbons is $C_nH_{2n}$ and $C_nH_{2n-2}$

A. Nonyne is an alkyne with 9 carbon atoms with molecular formula of $C_9H_{16}$

B. Heptane is an alkane with 7 carbon atoms with molecular formula of $C_7H_{16}$

C. Ethyne is an alkyne with 2 carbon atoms with molecular formula of $C_2H_{2}$ .

D. Hexene is an alkene with 6 carbon atoms with molecular formula of $C_6H_{12}$

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At 1023 K and 1 atm, a 3.00 gram sample of Snoz(s) (gram-formula mass = 151 g/mol) reacts with hydrogen gas to produce tin and w

Blizzard [7]

Answer:

2 moles of Sn are produced when 4 moles of H2(g) are consumed completely

Explanation:

to determine the number of moles of sn (l) produced when 4.0 moles of H2 (g) is consumed completely.

First, find the number of moles of H2 consumed by taking this as limiting reagent.

$n = \frac{g}{M.W (g/mol)}$

Then find the moles of Sn (l) taking into account the stoichiometric relationship between H2(g) and Sn(l). 2:1

$SnO_{2}$ (s) + 2 $H_{2}$ (g) ⇒ Sn(l) + 2 $H_{2}O$ (g)

$mol Sn(l) = \frac{1mol Sn}{2mol H_{2} } . 4 mol H_{2} = 2 mol$

∴2 moles of Sn are produced when 4 moles of H2(g) are consumed completely.

4 0

3 years ago

A solution contains 3.1 mM Zn(NO3)2 and 4.2 mM Ca(NO3)2. The p-function for Zn2+ is _____, and the p-function for NO3- is _____.

seropon [69]

Answer: The p-function of $Zn^{2+}$ and $NO_3^{-}$ ions are 2.51 and 2.14 respectively.

Explanation:

p-function is defined as the negative logarithm of any concentration.

We are given:

Millimolar concentration of zinc nitrate = 3.1 mM

Millimolar concentration of calcium nitrate = 4.2 mM

Converting this into molar concentration, we use the conversion factor:

1 M = 1000 mM

Concentration of zinc nitrate = 0.0031 M = 0.0031 mol/L

1 mole of zinc nitrate produces 1 mole of zinc ions and 2 moles of nitrate ions

Concentration of zinc ions = 0.0031 M

Concentration of nitrate ions in zinc nitrate, $M_1=(2\times 0.0031)=0.0062M$

Concentration of calcium nitrate = 0.0042 M = 0.0042 mol/L

1 mole of calcium nitrate produces 1 mole of calcium ions and 2 moles of nitrate ions

Concentration of calcium ions = 0.0042 M

Concentration of nitrate ions in calcium nitrate, $M_2=(2\times 0.0042)=0.0084M$

To calculate the concentration of nitrate ions in the solution, we use the equation:

$M=\frac{M_1V_1+M_2V_2}{V_1+V_2}$

Putting values in above equation, we get:

$M=\frac{(0.0062\times 1)+(0.0084\times 1)}{1+1}\\\\M=0.0073M$

Calculating the p-function of zinc ions and nitrate ions in the solution:

For zinc ions:

$\text{p-function of }Zn^{2+}\text{ ions}=-\log[Zn^{2+}]$

$\text{p-function of }Zn^{2+}\text{ ions}=-\log(0.0031)\\\\\text{p-function of }Zn^{2+}\text{ ions}=2.51$

For nitrate ions:

$\text{p-function of }NO_3^{-}\text{ ions}=-\log[NO_3^{-}]$

$\text{p-function of }NO_3^{-}\text{ ions}=-\log(0.0073)\\\\\text{p-function of }NO_3^{-}\text{ ions}=2.14$

Hence, the p-function of $Zn^{2+}$ and $NO_3^{-}$ ions are 2.51 and 2.14 respectively.

5 0

3 years ago

A small amount wet of hydrogen gas can be prepared by the reaction of zinc with excess hydrochloric acid and trapping the gas pr

dybincka [34]

Answer : The partial pressure of the hydrogen is, 705.9 mmHg

Explanation :

According to the Dalton's law of partial pressure,

$P_T=P_{H_2}+P_{H_2O}$

where,

$P_T$ = total pressure of the gas = 729.7 mmHg

$P_{H_2}$ = partial pressure of the hydrogen gas = ?

$P_{H_2O}$ = partial pressure of the water = 23.8 mmHg (standard value)

Now put all the given values in the above expression, we get:

$729.7mmHg=P_{H_2}+23.8mmHg$

$P_{H_2}=705.9mmHg$

Therefore, the partial pressure of the hydrogen is, 705.9 mmHg

4 0

4 years ago

____H2 + ____O2 ---> ____H20 2, 1, 1 2, 2, 1 2, 1, 2 1, 2, 2

Tems11 [23]

2H2+O2----->2H2O. I hope this helps

6 0

3 years ago

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Calculate the molar solubility of copper(II) arsenate (Cu3(AsO4)2) in water. Use 7.6 x 10^-36 as the solubility product constant

Molodets [167]

Molar solubility is the number of moles of a substance (the solute) that can be dissolved per liter of solution before the solution becomes saturated. We calculate as follows:

3Cu2+ + 2(AsO4)3- = Cu3(AsO4)2

7.6 x 10^-36 = (3x^3)(2x^2)
x = 6.62 x 10^-8 M

8 0

3 years ago

Read 2 more answers