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2 years ago

What is the vapor pressure of a solution with a benzene to octane molar ratio of 2 1?

1 answer:

8 0

At 50 °C the vapor pressure of Benzene is 280 mmHg while that of Octane is 400 mmHg.

Mole Fraction of Benzene;

2 / 2 + 1 = 0.666

So, Vapor pressure of Benzene is,

0.666 × 280 mmHg = 186.66 mmHg

Mole Fraction of Octane;

1 / 2 + 1 = 0.333

So, Vapor pressure of Benzene is,

0.333 × 400 mmHg = 133.33 mmHg

Vapor Pressure of Solution;

Vapor pressure of Benzene + Vapor pressure of Octane

186.66 mmHg + 133.33 mmHg = 319.99 mmHg

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in collecting the precipitate, why would it be inappropriate to heat the reacted mixture and evaporate off the water?

djverab [1.8K]

In collecting the precipitate, it is inappropriate to heat <span>the reacted mixture and evaporate off the water because it is possible that the mixture contains other substances that precipitates as well when the mixture is being heated so you will not be able to collect what you want.</span>

6 0

2 years ago

A beach has a supply of sand grains composed of calcite, ferromagnesian silicate minerals, and non-ferromagnesian silicate miner

bezimeni [28]

Ferromagnesian silicate minerals (i looked it up)

4 0

2 years ago

Calculate the radius ratio for NaBr if the ionic radii of Na + and Br − are 102 pm and 196 pm , respectively. radius ratio: Base

Fudgin [204]

Answer : The expected coordination number of NaBr is, 6.

Explanation :

Cation-anion radius ratio : It is defined as the ratio of the ionic radius of the cation to the ionic radius of the anion in a cation-anion compound.

This is represented by,

$\frac{r_{cation}}{r_{anion}}$

When the radius ratio is greater than 0.155, then the compound will be stable.

Now we have to determine the radius ration for NaBr.

Given:

Radius of cation, $Na^+$ = 102 pm

Radius of cation, $Br^-$ = 196 pm

$\frac{r_{cation}}{r_{anion}}=\frac{102}{196}=0.520$

As per question, the radius of cation-anion ratio is between 0.414-0.732. So, the coordination number of NaBr will be, 6.

The relation between radius ratio and coordination number are shown below.

Therefore, the expected coordination number of NaBr is, 6.

8 0

2 years ago

Pure platinum is too soft to be used in jewelry because it scratches easily. To increase the hardness so that it can be used in

Zina [86]

Answer:

percentage mass of platinum in the alloy ≈ 90.60 %

Explanation:

The alloy is 8.528 g sample of an alloy containing platinum and cobalt . The alloy react with excess nitric acid to form cobalt(ii) nitrate . Platinum is resistant to acid so it will definitely not react with the acid only the cobalt metal in the alloy will react with the acid.

The chemical reaction can be represented as follows:

Co (s) + HNO₃ (aq) → Co(NO₃)₂ (aq) + NO₂ (l) + H₂O (l)

The balanced equation

Co (s) + 4HNO₃ (aq) → Co(NO₃)₂(aq) + 2NO₂ (l) + 2H₂O (l)

Cobalt is the limiting reactant

atomic mass of cobalt = 58.933 g/mol

Molar mass of Co(NO₃)₂ = 58.933 + 14 × 2 + 16 × 6 = 58.933 + 28 + 96 = 182.933 g

58.933 g of cobalt produce 182.933 g of Co(NO₃)₂

? gram of cobalt will produce 2.49 g of Co(NO₃)₂

cross multiply

grams of cobalt that will react = (58.933 × 2.49)/182.933

grams of cobalt that will react = 146.74317000/182.933

grams of cobalt that will react= 0.8021689362 g

grams of cobalt that will react = 0.802 g

mass of platinum in the alloy = 8.528 g - 0.802 g = 7.726 g

percentage mass of platinum in the alloy = 7.726/8.528 × 100 = 772.600/8.528 = 90 .595 %

percentage mass of platinum in the alloy ≈ 90.60 %

6 0

2 years ago

How many moles of sodium bicarbonate is needed to neutralize 0.8 ml of sulphuric acid?

svet-max [94.6K]

Answer:

n NaHCO3 = 9.6 E-3 mol

Explanation:

balanced reaction:

2 NaHCO3(s) + H2SO4(ac) ↔ Na2SO4(ac) + 2 CO2(g) + 2 H2O(l)

assuming a concentration of H2SO4 6M....normally worked in the lab

⇒ n H2SO4 = 8 E-4 L * 6 mol/L = 4.8 E-3 mol H2SO4

according to balanced reaction, we have that for every mol of H2SO4 there are two mol of NaHCO3 ( sodium bicarbonate)

⇒ mol NaHCO3 = 4.8 E-3 mol H2SO4 * ( 2 mol NaHCO3 / mol H2SO4 )

⇒ ,mol NaHCO3 = 9.6 E-3 mol

So 9.6 E-3 mol NaHCO3, are the minimun moles necessary to neutralize the acid.

6 0

2 years ago