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2 years ago

An element forms an oxide, E₂O₃, and a fluoride, EF₃.(a) Of which two groups might E be a member?

Chemistry

1 answer:

Fittoniya [83]2 years ago

8 0

a) The E might belong to group 13.

As the formula of a chemical compound is derived by the cross multiplication of the valency of the atoms. As formula of the given oxide is and valency of O atom is -2, therefore valency of element E must be +3 in order to obtain E2O3.

Also, in EF3, the valency of E will be +3 because there are three atoms of fluorine who has an individual valency of -1. Thus, e will have the valency of +3.

The Group 13 is the boron group which has the following elements:

Boron
Aluminium
Gallium
Indium
Thallium

All these elements have the valency of +3.

To know more about Valency, refer to this link:

brainly.com/question/12717954

#SPJ4

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What is the final temperature, in degrees Celsius, of the gas in the bubble outside the volcano if the final volume of the bubbl

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The question is incomplete, here is the complete question:

A 127 mL bubble of hot gases at 213°C and 1.89 atm is emitted from an active volcano.

What is the final temperature, in degrees Celsius, of the gas in the bubble outside the volcano if the final volume of the bubble is 136 mL and the pressure is 0.850 atm , if the amount of gas does not change?

<u>Answer:</u> The temperature of the gas is -40°C

<u>Explanation:</u>

To calculate the volume when temperature and pressure has changed, we use the equation given by combined gas law.

The equation follows:

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1,V_1\text{ and }T_1$ are the initial pressure, volume and temperature of the gas

$P_2,V_2\text{ and }T_2$ are the final pressure, volume and temperature of the gas

We are given:

$P_1=1.89atm\\V_1=127mL\\T_1=213^oC=[213+273]K=484K\\P_2=0.850atm\\V_2=136mL\\T_2=?$

Putting values in above equation, we get:

$\frac{1.89atm\times 127mL}{484K}=\frac{0.850atm\times 136mL}{T_2}\\\\T_2=\frac{0.850\times 136\times 484}{1.89\times 127}=233K$

Converting the temperature from kelvins to degree Celsius, by using the conversion factor:

$T(K)=T(^oC)+273$

$233=T(^oC)+273\\\\T(^oC)=-40^oC$

Hence, the temperature of the gas is -40°C

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Explanation:

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4) 0.75 mol S × (1 mol CS2/4 mol S) = 0.19 mol CS2

5) 3 mol H2S × (1 mol CH4/2 mol H2S) = 1.5 mol CH4

Fe2O3 + 2Al ---> 2Fe + Al2O3

6) 25 g FeO3 × (1 mol Fe2O3/159.69 g Fe2O3) = 0.16 mol Fe2O3

0.16 mol Fe2O3 × (2 mol Al/1 mol Fe2O3) = 0.32 mol Al

0.32 mol Al × (26.98 g Al/1 mol Al) = 8.6 g Al

7) Given:

45 g Al × (1 mol Al/26.98 g Al) = 1.6 mol Al

85 g Fe2O3 ×(1 mol Fe2O3/159.69 g Fe2O3)

= 0.53 mol Fe2O3

Let's look at how much Fe each reactant will produce:

1.6 mol Al × (2 mol Fe/2 mol Al) = 1.6 mol Fe

0.53 mol Fe2O3 × (2 mol Fe/1 mol Fe2O3) = 1.1 mol Fe

Note that the given amount of Fe2O3 will give us fewer Fe. Therefore, Fe2O3 is the limiting reactant.

8) Al will produce 1.6 mol Fe × (55.845 g Fe/1 mol Fe)

= 89 g Fe

Fe2O3 will produce 1.1 mol Fe × (55.845 Fr/1 mol Fe)

= 61 g Fe

9) Since Fe2O3 is the limiting reactant, the ideal yield of Fe for the reaction is 61 g. If the actual reaction only gave us 25 g Fe. then the percent yield of Fe is

%yield = (25 g Fe/61 g Fe) × 100% = 41%

10) If we only got 25 g Fe, then the amount of Al actually used in the reaction is

25 g Fe × (1 mol Fe/55.845 g Fe) = 0.45 mol Fe

0.45 mol Fe × (2 mol Al/2 mol Fe) = 0.45 mol Al

0.45 mol Al × (26.98 g Al/1 mol Al) = 12 g

Therefore, the leftover amount of Al is

25 g Al - 12 g Al = 13 g Al

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<h2>ANSWER:</h2>

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