How can rice and flour mixture could be separated?

pantera1 [17]

Depositional landforms are the visible evidence of processes that have deposited sediments or rocks after they were transported by flowing ice or water, wind or gravity. Examples include beaches, deltas, glacial moraines, sand dunes and salt domes

3 0

3 years ago

The individual dipole moments in ammonia (NH3) do not cancel each other

kozerog [31]

Answer:

The strongest force that exists between molecules of Ammonia is <em>Hydrogen Bonding</em>.

Explanation:

Hydrogen Bond Interactions are those interactions which are formed between a partial positive hydrogen atom bonded directly to most electronegative atoms (i.e. F, O and N) of one molecule interacts with the partial negative most electronegative atom of another molecule.

Hence, in ammonia the nitrogen atom being more electronegative element than Hydrogen will be having partial negative charge and making the hydrogen atom partial positive. Therefore, the attraction between these partials charges will be the main force of interaction between ammonia molecules.

Other than Hydrogen bonding interactions ammonia will also experience dipole-dipole attraction and London dispersion forces.

4 0

3 years ago

The chemical combination of two or more different atoms in fixed amounts is called a(n)?

raketka [301]

A .compound is the combination of two or more atoms

3 0

3 years ago

A 0.1014 g sample of a purified compound containing C, H, and, O was burned in a combustion apparatus and produced 0.1486 g CO2

Alina [70]

Answer: The empirical formula for the given compound is $CH_2O$

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 $CO_2=0.1486g$

Mass of $H_2O=0.0609g$

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 0.1486 g of carbon dioxide, $\frac{12}{44}\times 0.1486=0.0405g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 0.0609 g of water, $\frac{2}{18}\times 0.0609=0.00677$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.1014) - (0.0405 + 0.00677) = 0.054 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.0405g}{12g/mole}=0.003375moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.00677g}{1g/mole}=0.00677moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.054g}{16g/mole}=0.003375moles$

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 0.485 moles.

For Carbon = $\frac{0.003375}{0.003375}=1$

For Hydrogen = $\frac{0.00677}{0.003375}=2.00\times 2$

For Oxygen = $\frac{0.003375}{0.003375}=1$

Step 3: Taking the mole ratio as their subscripts.

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

The empirical formula for the given compound is $C_1H_2O_1=CH_2O$

Thus, the empirical formula for the given compound is $CH_2O$

8 0

3 years ago

Please help!

lakkis [162]

Answer:B

Explanation:

5 0

2 years ago