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3 years ago

9.86 x 10²⁸ O-atoms require what volume (L) N₂O₂ at STP?

1 answer:

8 0

First, we use avogadro's number to convert atoms into moles. Then, relate the number of moles from elemental to the compound. Lastly, we use conditions at STP to calculate the volume. We do as follows:

9.86 x 10²⁸ O-atoms ( 1 mol / 6.022x10^23 atoms O) ( 1 mol N2O2 / 2 mol O ) ( 22.4 L / 1 mol ) = 1833809.37 L needed

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I need help with number 3. is the answer 1,2,3,or4

lubasha [3.4K]

Answer:

1 and 4

Explanation:

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3 years ago

2. How many grams of zinc will be formed if 32.0 g of copper reacts with zinc nitrate? Copper(I)nitrate is the other product.

Yakvenalex [24]

Answer:

16.46 g.

Explanation:

It is a stichiometry problem.

We should write the balance equation of the mentioned chemical reaction:

2Cu + Zn(NO₃)₂ → Zn + 2Cu(NO₃).

It is clear that 2.0 moles of Cu reacts with 1.0 mole of Zn(NO₃)₂ to produce 1.0 mole of Zn and 2.0 moles of Cu(NO₃).

We need to calculate the number of moles of the reacted Cu (32.0 g) using the relation:

n = mass / molar mass

The no. of moles of Cu = mass / atomic mass = (32.0 g) / (63.546 g/mol) = 0.503 mol.

Using cross multiplication:

2.0 moles of Cu produces → 1.0 mole of Zn, from the stichiometry.

0.503 mole of Cu produces → ??? mole of Zn.

The no. of moles of Zn produced = (1.0 mol)(0.503 mol) / (2.0 mol) = 0.2517 mol.

∴ The grams of Zn produced = no. of moles x atomic mass of Zn = (0.2517 mol)(65.38 g/mol) = 16.46 g.

3 0

3 years ago

Using the reaction C12H22O11 + 12 O2 ->12CO2 + 11H2O (e). How many grams of carbon dioxide can be produced from 456.7g sucros

klio [65]

Answer:

704.6 g CO2

Explanation:

MM sucrose = 342.3 g/mol

MM CO2 = 44.01 g/mol

g CO2 = 456.7 g sucrose x (1 mol sucrose/MM sucrose) x (12 moles CO2/1 mol sucrose) x (MM CO2/1mol CO2) = 704.6 g CO2

7 0

2 years ago

A certain skin lotion is a fine mixture of water and various oils. This lotion is cloudy and cannot be separated into oil and wa

timofeeve [1]

The answer you are looking for would be a colliod. Hope this helps have a great day!!

6 0

3 years ago

Read 2 more answers

Burning 12.00 g of an oxoacid produces 17.95 g of carbon dioxide and 4.87 g of water. Consider that 0.25

Veseljchak [2.6K]

Answer: The molecular formula will be $C_6H_6O_6$

Explanation:

Mass of $CO_2$ = 17.95 g

Mass of $H_2O$ = 4.87 g

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 17.95 g of carbon dioxide, = $\frac{12}{44}\times 17.95=4.89g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 4.87 g of water, = $\frac{2}{18}\times 4.87=0.541g$ of hydrogen will be contained.

Mass of oxygen in the compound = (12.00) - (4.89+0.541) = 6.57 g

Mass of C = 4.89 g

Mass of H = 0.541 g

Mass of O = 6.57 g

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{4.89g}{12g/mole}=0.407moles$

Moles of H= $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.541g}{1g/mole}=0.541moles$

Moles of O= $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{6.57g}{16g/mole}=0.410moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{0.407}{0.407}=1$

For H = $\frac{0.541}{0.407}=1$

For O = $\frac{0.410}{0.407}=1$

The ratio of C : H : O = 1: 1 : 1

Hence the empirical formula is $CHO$ .

Hence the empirical formula is $CHO$

The empirical weight of $CHO$ = 1(12)+1(1)+1(16)= 29 g.

If 0.25 moles has mass of 44.0 g

Thus 1 mole has mass of = $\frac{44.0}{0.25}\times 1=176g$

Thus molecular mass is 176 g

Now we have to calculate the molecular formula.

$n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{176g}{29g}=6$

The molecular formula will be= $6\times CHO=C_6H_6O_6$

5 0

3 years ago