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Vinil7 [7]
3 years ago
13

Electron dot diagram for four elements are shown. Which element would be found in Group 13?

Chemistry
2 answers:
gladu [14]3 years ago
8 0

Answer:

D. 1

Explanation:

Just took it :)

ANTONII [103]3 years ago
6 0

Answer:

D: 1

Explanation:

A PEX

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for the reaction HCI+NaOH➡️NaCi+H2O, how many moles of hydrochloric acid are required to produce 150 g. of water?
Ne4ueva [31]

Answer: 8.33 mol of HCl (Hydrochloric Acid)

Explanation:

150 g H2O  x __1 mol__ x __1 mol HCl__  =  8.33 mol of HCl

                         18.016 g        1 mol H2O

6 0
3 years ago
How do I find density
skad [1K]
Divide mass by the volume to find density.
7 0
3 years ago
Which physical change is endothermic?
Helen [10]

Answer: The correct option is 1.

Explanation: Endothermic reactions are the reactions in which heat is provided to break down the reactant molecules.

In option 1:

The stronger intermolecular forces between the particles in solid molecule are broken down to convert into gaseous form. Hence, some energy in the form of heat is provided to move them far apart. Therefore, it is considered as an endothermic reaction.

In option 2, 3 and 4:

All the other processes involves the formation of bonds and thus there is no need to provide any energy.

5 0
3 years ago
Read 2 more answers
A 118-ml flask is evacuated and found to have a mass of 97.129 g. when the flask is filled with 768 torr of helium gas at 35 ?c,
Inessa05 [86]
The full question asks to decide whether the gas was a specific gas. That part is missing in your question. You need to decide whether the gas in the flask is pure helium.

To decide it you can find the molar mass of the gas in the flask, using the ideal gas equation pV = nRT, and then compare with the molar mass of the He.

From pV = nRT you can find n, after that using the mass of gass in the flask you use MM = mass/moles.

1) From pV = nRT, n = pV / RT

Data:
V = 118 ml = 0.118 liter
R = 0.082 atm*liter/mol*K
p = 768 torr * 1 atm / 760 torr = 1.0105 atm
T = 35 + 273.15 = 308.15 K

n = 1.015 atm * 0.118 liter / [ 0.082 atm*liter/K*mol  * 308.15K] =0.00472 mol

mass of gas = mass of the fask with the gas - mass of the flasl evacuated = 97.171 g - 97.129 g = 0.042

=> MM =  mass/n = 0.042 / 0.00472 = 8.90 g/mol

Now from a periodic table or a table you get that the molar mass of He is 4g/mol

So the numbers say that this gas is not pure helium , because its molar mass is more than double of the molar mass of helium gas.
7 0
3 years ago
The heats of combustion of ethane (C2H6) and butane (C4H10) are 52 kJ/g and 49 kJ/g, respectively. We need to produce 1.000 x 10
LekaFEV [45]

Answer :

(1) The number of grams needed of each fuel (C_2H_6)\text{ and }(C_4H_{10}) are 19.23 g and 20.41 g respectively.

(2) The number of moles of each fuel (C_2H_6)\text{ and }(C_4H_{10}) are 0.641 moles and 0.352 moles respectively.

(3) The balanced chemical equation for the combustion of the fuels.

C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O

C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O

(4) The number of moles of CO_2 produced by burning each fuel is 1.28 mole and 1.41 mole respectively.

The fuel that emitting least amount of CO_2 is C_2H_6

Explanation :

<u>Part 1 :</u>

First we have to calculate the number of grams needed of each fuel (C_2H_6)\text{ and }(C_4H_{10}).

As, 52 kJ energy required amount of C_2H_6 = 1 g

So, 1000 kJ energy required amount of C_2H_6 = \frac{1000}{52}=19.23g

and,

As, 49 kJ energy required amount of C_4H_{10} = 1 g

So, 1000 kJ energy required amount of C_4H_{10} = \frac{1000}{49}=20.41g

<u>Part 2 :</u>

Now we have to calculate the number of moles of each fuel (C_2H_6)\text{ and }(C_4H_{10}).

Molar mass of C_2H_6 = 30 g/mole

Molar mass of C_4H_{10} = 58 g/mole

\text{ Moles of }C_2H_6=\frac{\text{ Mass of }C_2H_6}{\text{ Molar mass of }C_2H_6}=\frac{19.23g}{30g/mole}=0.641moles

and,

\text{ Moles of }C_4H_{10}=\frac{\text{ Mass of }C_4H_{10}}{\text{ Molar mass of }C_4H_{10}}=\frac{20.41g}{58g/mole}=0.352moles

<u>Part 3 :</u>

Now we have to write down the balanced chemical equation for the combustion of the fuels.

The balanced chemical reaction for combustion of C_2H_6 is:

C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O

and,

The balanced chemical reaction for combustion of C_4H_{10} is:

C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O

<u>Part 4 :</u>

Now we have to calculate the number of moles of CO_2 produced by burning each fuel to produce 1000 kJ.

C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O

From this we conclude that,

As, 1 mole of C_2H_6 react to produce 2 moles of CO_2

As, 0.641 mole of C_2H_6 react to produce 0.641\times 2=1.28 moles of CO_2

and,

C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O

From this we conclude that,

As, 1 mole of C_4H_{10} react to produce 4 moles of CO_2

As, 0.352 mole of C_4H_{10} react to produce 0.352\times 4=1.41 moles of CO_2

So, the fuel that emitting least amount of CO_2 is C_2H_6

5 0
3 years ago
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