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2 years ago

Which bond is the strongest CO CF CN CC?

Chemistry

1 answer:

romanna [79]2 years ago

3 0

Answer:

The CN double bond and triple bond are stronger than the corresponding CC bonds

Explanation:

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What is the empirical formula of an oxide of nitrogen containing 63.61% by mass of nitrogen and 36.69% by mass of oxygen?

katen-ka-za [31]

Answer:

D. N₂O

Explanation:

Let's assume we have 100 g of the compound. That means it consists of 63.61 grams of nitrogen and 36.69 grams of oxygen.

Converting masses to moles:

63.61 g N × (1 mol N / 14.01 g N) = 4.540 mol N

36.69 g O × (1 mol O / 16.00 g O) = 2.293 mol O

Normalize by dividing by the smallest:

4.540 / 2.293 = 1.980 mol N

2.293 / 2.293 = 1.000 mol O

So there is approximately twice as many N atoms as O atoms. The empirical formula is therefore N₂O.

8 0

3 years ago

Number of grams in four moles of BaCO3

Irina-Kira [14]

This compound is also known as Barium Carbonate. 1 mole is equal to 1 moles BaCO3, or 197.3359<span> grams.</span>

5 0

2 years ago

How many moles are 8.8 grams of CO2?

Lemur [1.5K]

The answer is .2grams

8 0

3 years ago

How many liters of hydrogen gas will be produced at STP from the reaction of 7.179×10^23 atoms of magnesium with 54.219g of phos

Alexeev081 [22]

Answer: The volume of hydrogen gas produced will be, 12.4 L

Explanation : Given,

Mass of $H_3PO_4$ = 54.219 g

Number of atoms of $Mg$ = $7.179\times 10^{23}$

Molar mass of $H_3PO_4$ = 98 g/mol

First we have to calculate the moles of $H_3PO_4$ and $Mg$ .

$\text{Moles of }H_3PO_4=\frac{\text{Given mass }H_3PO_4}{\text{Molar mass }H_3PO_4}$

$\text{Moles of }H_3PO_4=\frac{54.219g}{98g/mol}=0.553mol$

and,

$\text{Moles of }Mg=\frac{7.179\times 10^{23}}{6.022\times 10^{23}}=1.19mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

$3Mg+2H_3PO_4\rightarrow Mg(PO_4)_2+3H_2$

From the balanced reaction we conclude that

As, 3 mole of $Mg$ react with 2 mole of $H_3PO_4$

So, 0.553 moles of $Mg$ react with $\frac{2}{3}\times 0.553=0.369$ moles of $H_3PO_4$

From this we conclude that, $H_3PO_4$ is an excess reagent because the given moles are greater than the required moles and $Mg$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $H_2$

From the reaction, we conclude that

As, 3 mole of $Mg$ react to give 3 mole of $H_2$

So, 0.553 mole of $Mg$ react to give 0.553 mole of $H_2$

Now we have to calculate the volume of $H_2$ gas at STP.

As we know that, 1 mole of substance occupies 22.4 L volume of gas.

As, 1 mole of hydrogen gas occupies 22.4 L volume of hydrogen gas

So, 0.553 mole of hydrogen gas occupies $0.553\times 22.4=12.4L$ volume of hydrogen gas

Therefore, the volume of hydrogen gas produced will be, 12.4 L

4 0

3 years ago

3 & 4 Help pleaseeeee

kvasek [131]

3 is C i believe, and as for 4 minerals are solid, natural.. that's about all I can say...

4 0

3 years ago