Definition: This is an alternative form of a gene.

Two moles of magnesium and five moles of oxygen are placed in a vessel. When magnesium is ignited ca two moles of magnesium and

Greeley [361]

Limiting reactant in this experiment would be Magnesium since it will run out first

4 0

3 years ago

Please help thank you (15 points)

Eva8 [605]

Answer:

D. The sun has less gravitational pull on Pluto because it is farther away.

6 0

2 years ago

A 13.30 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 13.00 grams of CO2 and 2

a_sh-v [17]

Answer: The empirical and molecular formula for the given organic compound is $CHO_2$ and $C_2H_2O_4$

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 $CO_2=13.00g$

Mass of $H_2O=2.662g$

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

So, in 13.00 g of carbon dioxide, $\frac{12}{44}\times 13.00=3.54g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 2.662 g of water, $\frac{2}{18}\times 2.662=0.296g$ of hydrogen will be contained.

Mass of oxygen in the compound = (13.30) - (3.54 + 0.296) = 9.464 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{3.54g}{12g/mole}=0.295moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.296g}{1g/mole}=0.296moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{9.465g}{16g/mole}=0.603moles$

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 0.295 moles.

For Carbon = $\frac{0.295}{0.295}=1$

For Hydrogen = $\frac{0.296}{0.295}=1$

For Oxygen = $\frac{0.603}{0.295}=2.044\approx 2$

Step 3: Taking the mole ratio as their subscripts.

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

Hence, the empirical formula for the given compound is $CHO_2$

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 = 90.04 g/mol

Mass of empirical formula = 45 g/mol

Putting values in above equation, we get:

$n=\frac{90.04g/mol}{45g/mol}=2$

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

$C_{(1\times 2)}H_{(1\times 2)}O_{(2\times 2)}=C_2H_2O_4$

Hence, the empirical and molecular formula for the given organic compound is $CHO_2$ and $C_2H_2O_4$

3 0

3 years ago

Determine the mass in grams of 3.75 x 10^21 atoms of zinc. (the mass of one mole of zinc is 65.39 g)

Nastasia [14]

Answer: The mass in $3.75 \times 10^{21}$ atoms of zinc is 0.405 g.

Explanation:

Given: Atoms of zinc = $3.75 \times 10^{21}$

It is known that 1 mole of every substance contains $6.022 \times 10^{23}$ atoms. So, the number of moles in given number of atoms is as follows.

$Moles = \frac{3.75 \times 10^{21}}{6.022 \times 10^{23}}\\= 0.622 \times 10^{-2}\\= 0.0062 mol$

As moles is the mass of a substance divided by its molar mass. So, mass of zinc (molar mass = 65.39 g/mol) is calculated as follows.

$Moles = \frac{mass}{molar mass}\\0.0062 mol = \frac{mass}{65.39 g}\\mass = 0.405 g$

Thus, we can conclude that the mass in $3.75 \times 10^{21}$ atoms of zinc is 0.405 g.

6 0

3 years ago

at standard pressure how do the boiling point and the freezing point of NaCl compare to the boiling point and freezing point of

Ede4ka [16]

Usually in this context you would be referring to the boiling and freezing point of a NaCl solution (saltwater) compared to pure H_{2}O. Sematics would be different for NaCl compound itself, you would say melting and boiling point for a solid substance- and the temperatures would be very, very radical (high).
The boiling point of pure water is 100 degrees C (212 F), and the freezing/melting point is below 0 degrees C (32 F). For a salt water solution, the boiling point is raised and the melting point is lowered. This means that water will stay liquid for an increased range of temperature. Depending on the amount of NaCl solute in the water, the boiling and melting points may change a few degrees.

7 0

3 years ago