Observation on which fruit has more acidity: lemons, watermelon and oranges

Which element has properties of both metals and nonmetals?

PilotLPTM [1.2K]

Metalloids have the properties of metals and nonmetals.

8 0

3 years ago

How many liters of nitrogen gas are needed to make 25 mol of nitrogen trifluoride

ikadub [295]

The reaction between N₂ and F₂ gives Nitrogen trifluoride as the product. The balanced equation is;

N₂ + 3F₂ → 2NF₃

The stoichiometric ratio between N₂ and NF₃ is 1 : 2
Hence,
moles of N₂ / moles of F₂ = 1 / 2
moles of N₂ / 25 mol = 0.5
moles of N₂ = 0.5 x 25 mol = 12.5 mol

Hence N₂ moles needed = 12.5 mol

At STP (273 K and 1 atm) 1 mol of gas = 22.4 L

Hence needed N₂ volume = 22.4 L mol⁻¹ x 12.5 mol
= 280 L

6 0

3 years ago

Quick question: Are water molecules attracted to each other because of the polar nature of water?

Colt1911 [192]

Answer:

Because the opposite charge between them

Explanation:

4 0

3 years ago

Re-order each list in the table below, if necessary, so that the atoms or ions in it are listed in order of decreasing size.

Serjik [45]

Answer: Na, S, Cl

Explanation:

Atomic size decreases as one moves from left to right on the periodic table with elements in the same period. This is as a result of the electrons increasing in the outer circle and thus being drawn to the protons in the nucleus which will lead to the outer shell area decreasing.

Sodium (Na) comes before Sulfur (S) which comes before Chlorine (S) so this is the decreasing order as they are all in the same period.

7 0

2 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 :

Part 1 :

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 $C_4H_{10}$ = $\frac{1000}{49}=20.41g$

Part 2 :

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,

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

Part 3 :

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:

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

Part 4 :

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