Which of the following has the same empirical and molecular formula?

The pressure exerted by 1.5 mol of gas in a 13 L flask at 22 °C is ____ kPa

sukhopar [10]

Answer:

282.7KPa

Explanation:

Step 1:

Data obtained from the question.

Number of mole of (n) = 1.5 mole

Volume (V) = 13L

Temperature (T) = 22°C = 22 + 273°C = 295K

Pressure (P) =..?

Gas constant (R) = 0.082atm.L/Kmol

Step 2:

Determination of the pressure exerted by the gas.

This can be obtained by using the ideal gas equation as follow:

PV = nRT

P = nRT /V

P = 1.5 x 0.082 x 295 / 13

P = 2.79atm.

Step 3:

Conversion of 2.79atm to KPa.

This is illustrated below:

1 atm = 101.325KPa

Therefore, 2.79atm = 2.79 x 101.325 = 282.7KPa

Therefore, the pressure exerted by the gas in KPa is 282.7KPa

8 0

2 years ago

Answer all parts of the following questions on Particle Stoichiometry.

Ugo [173]

<span>1. Translate, predict the products, and balance the equation above.

Li + Cu(NO3)2 = Li(NO3)2 + Cu

2. How many particles of lithium are needed to produce 125 g of copper?

125 g Cu ( 1 mol / 63.55 g ) (1 mol Li / 1 mol Cu ) ( 6.022 x 10^23 particles / 1 mol ) = 1.18x10^24 Li particles

3. How many grams of lithium nitrate are produced from 4.83E24 particles of copper (II) nitrate?

</span>4.83E24 particles of copper (II) nitrate ( 1 mol / 6.022x10^23 particles ) (1 mol Li(NO3)2 / 1 mol Cu(NO3)2 ) ( 130.95 g / 1 mol ) = 1043.77 grams Li(NO3)2

3 0

3 years ago

Please help!!!! Best answer will get brainliest.

Yuri [45]

Answer:

ionic

Explanation:

I don't know if this correct

3 0

2 years ago

Two aqueous solutions are prepared: 1.00 m Na2CO3 and 1.00 m LiCl. Which of the following statements is true?

svlad2 [7]

Answer:

The $Na_{2}CO_{3}$ solution has a higher osmotic pressure and higher boiling point than LiCl solution.

Explanation:

As concentrations of two aqueous solutions are same therefore we can write:

$\Delta P\propto i$ , $\Delta T_{b}\propto i$ and $\pi \propto i$

where $\Delta P$ , $\Delta T_{b}$ and $\pi$ are lowering of vapor pressure, elevation in boiling point and osmotic pressure of solution respectively. $i$ is van't hoff factor.

$i$ = total number of ions generated from dissolution of one molecule of a substance (for strong electrolyte).

Here both $Na_{2}CO_{3}$ and LiCl are strong electrolytes.

So, $i(Na_{2}CO_{3})=3$ and $i(LiCl)=2$

Hence, lowering of vapor pressure, elevation in boiling point and osmotic pressure will be higher for $Na_{2}CO_{3}$ solution.

Therefore the $Na_{2}CO_{3}$ solution has a higher osmotic pressure and higher boiling point than LiCl solution.

7 0

3 years ago

Electrophilic addition of hbr to 3-methyl-2-hexene creates an asymmetric center at c-2. what is the product of this reaction?

AleksandrR [38]

This reaction would give rise to two products.

2-bromo-3-methylhexane, and
3-bromo-3-methylhexane.

However, 2-bromo-3-methylhexane would be more common than 3-bromo-3-methylhexane among the products.

The hydrogen atom in a hydrogen bromide molecule carries a partial positive charge. It is attracted to the double bond region with a high electron density. The hydrogen-bromine bond breaks when HBr gets too close to a double bond to produces a proton $\text{H}^{+}$ and a bromide ion $\text{Br}^{-}$ .

The proton would attack the double bond to produce a carbocation. It could attach itself to either the second or the third carbon atom.

$\phantom{\text{H}_3\text{C}-\text{CH}-\;}+\\\text{H}_3\text{C}-\text{CH}-\text{CH}-\text{CH}_2-\text{CH}_2-\text{CH}_3\\\phantom{\text{H}_3\text{C}-\;\;}|\phantom{\text{CH}-}\;}|\\\phantom{\text{H}_3\text{C}-\;} \text{H}\phantom{\text{CH}}\;\;}\text{CH}_3$
$\phantom{\text{H}_3\text{C}-\text{CH}-}\;\;\text{H}\\\phantom{\text{H}_3\text{C}-\;}+\phantom{\text{H}-\;}|\\\text{H}_3\text{C}-\text{CH}-\text{CH}-\text{CH}_2-\text{CH}_2-\text{CH}_3\\\phantom{\text{H}_3\text{C}-\;\;\;}\phantom{\text{CH}-}\;}|\\\phantom{\text{H}_3\text{C}-\;} \phantom{\text{H}}\phantom{\text{CH}}\;\;}\text{CH}_3$

Carbocations are unstable and might decompose over time. The first carbocation is more stable than the second for having three alkyl groups- i.e., straight carbon chains- attached to the center of the positive charge. Alkyl groups have stabilizing positive induction effect on positively-charged carbon. The second carbocation has only two, and is therefore not as stable. The first carbocation thus has the greatest chance to react with a bromide ion to produce a stable halocarbon.

Bromide ions are negatively charged. They attach themselves to carbocations at the center of positive charge. Adding a bromide ion to the first carbocation would produce 3-bromo-3-methylhexane whereas adding to the second produces 2-bromo-3-methylhexane.

The <em>most likely</em> product of this reaction is therefore 3-bromo-3-methylhexane.

6 0

2 years ago