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3 years ago

Using the graph and particle diagram below, create an LOL diagram that demonstrates the change in

Chemistry

1 answer:

Usimov [2.4K]3 years ago

3 0

Answer:

first bring craft paper then make 12345678 910 then make lines how many box you want to make then then draw write down their your answer bye-bye

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Choose the molecular compound among the substances listed below.

agasfer [191]

Answer: Among the listed substances $XeCl_{4}$ is the molecular compound.

Explanation:

A chemical compound formed by the chemical combination of two or more non-metals is called a molecular compound or covalent compound.

For example, Xe and Cl are non-metals. The compound formed by them is $XeCl_{4}$ which is a molecular compound.

A molecular compound is formed by sharing of atoms between the combining atoms.

Whereas NaF, $MnCl_{2}$ and CaO are all ionic compounds as they are formed by chemical combination of a metal and a non-metal.

Thus, we can conclude that among the listed substances $XeCl_{4}$ is the molecular compound.

4 0

3 years ago

This question deals with waste disposal in the Solutions and Spectroscopy experiment. What should be done to waste solutions con

Korvikt [17]

Answer:

b. It should be dumped in a beaker labeled "waste copper" on one's bench during the experiment.

d. It should be disposed of in the bottle for waste copper ion when work is completed.

Explanation:

Solutions containing copper ion should never be disposed of by dumping them in a sink or in common trash cans, because this will cause pollution in rivers, lakes and seas, being a contaminating agent to both human beings and animals. They should be placed in appropriate compatible containers that can be hermetically sealed. The sealed containers must be labeled with the name and class of hazardous substance they contain and the date they were generated.

It never should be returned to the bottle containing the solution, since it can contaminate the solution of the bottle.

In the Solutions and Spectroscopy experiments there is always wastes.

3 0

3 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

Which element is less electronegative than silicon (Si)? A. Sulfur (S) B. Magnesium (Mg) C. Carbon (C) D. Oxygen (0)

Luden [163]

Answer:

C. Carbon

Explanation:

Carbon has an electronegativity of 2.55, followed by Tin at 1.96, Silicon at 1.90 and the least electronegative would be Lead at 1.87.

7 0

3 years ago

At 25°C and constant pressure, carbon monoxide gas combines with oxygen gas to give carbon dioxide gas with the evolution of 10.

stealth61 [152]

Answer : The value of $\Delta H$ for the reaction is, -565.6 kJ

Explanation :

First we have to calculate the molar mass of CO.

Molar mass CO = Atomic mass of C + Atomic mass of O = 12 + 16 = 28 g/mole

Now we have to calculate the moles of CO.

$\text{Moles of }CO=\frac{\text{Mass of }CO}{\text{Molar mass of }CO}=\frac{1g}{28g/mole}=\frac{1}{28}mole$

Now we have to calculate the value of $\Delta H$ for the reaction.

The balanced equation will be,

$2CO(g)+O_2(g)\rightarrow 2CO_2(g)$

From the balanced chemical reaction we conclude that,

As, $\frac{1}{28}mole$ of CO release heat = 10.1 kJ

So, 2 mole of CO release heat = $2\times 28\times 10.1=565.6kJ$

Therefore, the value of $\Delta H$ for the reaction is, -565.6 kJ (The negative sign indicates the amount of energy is released)

4 0

3 years ago