____Cl2+____H2=___

What process is occurring as iodine solid is transformed into iodine vapor?

Luba_88 [7]

The process that is occurring as iodine solid is transformed into iodine vapour is SUBLIMATION.
Sublimation is the process by which a substance transformed from a solid state straight into a gaseous state without passing through the liquid state. Iodine is capable of undergoing sublimation process and this property is usually used to separate iodine from mixtures.<span />

5 0

3 years ago

Which ionic compound would you predict to have a higher boiling point: NaBr or

nasty-shy [4]

Answer:

The boiling point of NaBr is higher than that of Na2S

Explanation:

Bromine has higher electronegativity as compared to the sulphur. The ionic strength of the bond formed with Sodium is higher in the NaBr compound. The higher the ionic bond strength the higher will be the boiling point

Also the size of bromine is large as compared to that of sulphur. Large molecules consists of more electron and hence they create der Waals attractive forces due to which the boiling point of compound increases.

Hence, the boiling point of NaBr is higher than that of Na2S

8 0

3 years ago

In 2.00 min, 29.7 mL of He effuse through a small hole. Under the same conditions of pressure and temperature, 9.28 mL of a mixt

sergiy2304 [10]

Answer : The percent composition by volume of mixture of $CO$ and $CO_2$ are, 18.94 % and 81.06 % respectively.

Solution :

According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.

$R\propto \sqrt{\frac{1}{M}}$

And the relation between the rate of effusion and volume is :

$R=\frac{V}{t}$

or, from the above we conclude that,

$(\frac{V_1}{V_2})^2=\frac{M_2}{M_1}$ ..........(1)

where,

$V_1$ = volume of helium gas = 29.7 ml

$V_2$ = volume of mixture = 9.28 ml

$M_1$ = molar mass of helium gas = 4 g/mole

$M_2$ = molar mass of mixture = ?

Now put all the given values in the above formula 1, we get the molar mass of mixture.

$(\frac{29.8ml}{9.28ml})^2=\frac{M_2}{4g/mole}$

$M_2=40.97g/mole$

The average molar mass of mixture = 40.97 g/mole

Now we have to calculate the percent composition by volume of the mixture.

Let the mole fraction of $CO$ be, 'x' and the mole fraction of $CO_2$ will be, (1 - x).

As we know that,

$\text{Average molar mass of mixture}=\text{Mole fraction of }CO$

$\text{Average molar mass of mixture}=(\text{Mole fraction of }CO\times \text{Molar mass of } CO)+(\text{Mole fraction of }CO_2\times \text{Molar mass of } CO_2)$

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

$40.94g/mole=((x)\times 28g/mole)+((1-x)\times 44g/mole)$

$x=0.1894$

The mole fraction of $CO$ = x = 0.1894

The mole fraction of $CO_2$ = 1 - x = 1 - 0.1894 = 0.8106

The percent composition by volume of mixture of $CO$ = $0.1894\times 100=18.94\%$

The percent composition by volume of mixture of $CO_2$ = $0.8106\times 100=81.06\%$

Therefore, the percent composition by volume of mixture of $CO$ and $CO_2$ are, 18.94 % and 81.06 % respectively.

6 0

3 years ago

Marian claims that Food A has slightly more than three times the Calories as Food B. Johan claims that Food B has more Calories.

Liula [17]

Explanation:

In order to justify Marian's statement we have to look at the '' Amount per serving calories'' ⇒

In food label A we can see that this value is 160 calories

In food label B we can see that this value is 50 calories

⇒ 160 calories is slightly more than three times 50 calories

Otherwise If we want to justify Johan statement we need to look at the '' serving size '' ⇒

In food label A we can see that the serving size is 1 cup (237 mL)

In food label B we can see that the serving size is $\frac{1}{4}$ cup (56g)

Working with the information of the food label A we can write the following expression :

$\frac{1Cup}{160calories}=\frac{\frac{1}{4}Cup}{x}$ ⇒ Looking at the value of '' $x$ '' ⇒

$x=160calories(\frac{1}{4})=40calories$

$x=40calories$

If we look at the same amount of portion volume :

In $\frac{1}{4}$ cup of food A we have 40 calories

In $\frac{1}{4}$ cup of food B we have 50 calories

We could conclude that Food B has more calories.

That's how both claims could both be justified.

8 0

4 years ago

What is non stoichiometric reaction ?

yanalaym [24]

Nonstoichiometric compounds are chemical compounds deviated from stoichiometry, namely, their elemental composition cannot be represented by a ratio of well-defined natural numbers, and therefore they violate the law of definite proportions.

6 0

4 years ago

____Cl2+____H2=____HCl

Cl2+H2=HCl