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3 years ago

When Earth is between the Moon and the Sun, how is the Moon seen?

2 answers:

7 0

When the Earth is between the moon and the sun, how is the moon seen, in a solar eclipse on Earth. The answer should be A. Full moon. Hope it helped you, and have a great day.

AVprozaik [17]3 years ago

6 0

It is B, because it is an eclipse.

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If 2 CH3OH + 3 O2 -> 2CO2 + 4 H2O was carried out in the laboratory and 219 g of water was produced, what would the percent y

ANEK [815]

Answer:

Percent yield = 84.5 %

Explanation:

Given data:

Mass of methanol = 229 g

Actual yield of water = 219 g

Percent yield of water = ?

Solution:

Chemical equation:

2CH₃OH + 3O₂ → 2CO₂ + 4H₂O

Number of moles of methanol:

Number of moles = mass/ molar mass

Number of moles = 229 g/ 32 g/mol

Number of moles = 7.2 mol

Now we will compare the moles of water with methanol.

CH₃OH : H₂O

2 : 4

7.2 : 4/2×7.2 = 14.4 mol

Mass of water:

Mass = number of moles × molae mass

Mass = 14.4 mol × 18 g/mol

Mass = 259.2 g

Percent yield:

Percent yield = actual yield / theoretical yield × 100

Percent yield = 219 g / 259.2 g × 100

Percent yield = 84.5 %

7 0

3 years ago

When 0.42 g of a compound containing C, H, and O is burned completely, the products are 1.03 g CO2 and 0.14 g H2O. The molecular

Luda [366]

Answer: The empirical and molecular formula for the given organic compound is $C_2H_2O_4$ and $C_6H_4O_2$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=1.03g$

Mass of $H_2O=0.14g$

We know that:

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 1.03 g of carbon dioxide, $\frac{12}{44}\times 1.03=0.28g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 0.14 g of water, $\frac{2}{18}\times 0.14=0.016g$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.42) - (0.28 + 0.016) = 0.124 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.28g}{12g/mole}=0.023moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.016g}{1g/mole}=0.016moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.124g}{16g/mole}=0.00775moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.00775 moles.

For Carbon = $\frac{0.023}{0.00775}=2.96\approx 3$

For Hydrogen = $\frac{0.016}{0.00775}=2.06\approx 2$

For Oxygen = $\frac{0.00775}{0.00775}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 3 : 2 : 1

Hence, the empirical formula for the given compound is $C_3H_{2}O_1=C_3H_2O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{molecular mass}}{\text{empirical mass}}$

We are given:

Mass of molecular formula = 108.10 g/mol

Mass of empirical formula = 54 g/mol

Putting values in above equation, we get:

$n=\frac{108.10g/mol}{54g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(3\times 2)}H_{(2\times 2)}O_{(1\times 2)}=C_6H_4O_2$

Thus, the empirical and molecular formula for the given organic compound is $C_3H_2O$ and $C_6H_4O_2$

6 0

3 years ago

The study of the composition of matter is?

NemiM [27]

Chemistry. More specifically, analytical chemistry.

6 0

3 years ago

Approximately how many particles are in 3 moles

mash [69]

Answer:

18.066 × 10²³ particles

Explanation:

Given data:

Number of moles = 3 moles

Number of particles = ?

Solution:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

For example,

18 g of water = 1 mole = 6.022 × 10²³ molecules of water

1.008 g of hydrogen = 1 mole = 6.022 × 10²³ atoms of hydrogen

For 3 moles of substance:

One mole = 6.022 × 10²³ particles

3 mol × 6.022 × 10²³ particles/ 1 mol = 18.066 × 10²³ particles

8 0

3 years ago

Which of the following breaks and forms bonds?

svp [43]

Your answer would be "Chemical Change"

7 0

3 years ago