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BlackZzzverrR [31]

3 years ago

5

Match each element with the element that is most similar to it. If you don’t know the answer, use this periodic table to help yo

u make your best guess.
Drag the tiles to the correct boxes to complete the pairs...

1 answer:

yan [13]3 years ago

5 0

Answer:

Lithium - Potassium

Carbon - Silicon

Fluorine - Bromine

Oxygen - Sulfur

Explanation:

I hope this helps!! :3

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A laboratory analysis of a sample finds it is composed of 38.8% carbon, 16.2% hydrogen, and 45.1% nitrogen. What is its empirica

Sladkaya [172]

Answer: The empirical formula for the given compound is $CH_5N$

Explanation : Given,

Percentage of C = 38.8 %

Percentage of H = 16.2 %

Percentage of N = 45.1 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 38.8 g

Mass of H = 16.2 g

Mass of N = 45.4 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{38.8g}{12g/mole}=3.23moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{16.2g}{1g/mole}=16.2moles$

Moles of Nitrogen = $\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{45.4g}{14g/mole}=3.24moles$

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 3.23 moles.

For Carbon = $\frac{3.23}{3.23}=1$

For Hydrogen = $\frac{16.2}{3.23}=5.01\approx 5$

For Oxygen = $\frac{3.24}{3.23}=1.00\approx 1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N = 1 : 5 : 1

Hence, the empirical formula for the given compound is $C_1H_5N_1=CH_5N$

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3 years ago

What is the chemical name for K2Se

Oduvanchick [21]

Answer:

Potassium selenide

Explanation:

Potassium selenide (K2Se)

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Hank's Garage has an air compressor with a holding tank that contains 200L of compressed air at 5200 torr. One day a hose ruptur

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Hank's Garage has an air compressor with a holding tank that contains 200L of compressed air at 5200 torr. One day a hose ruptured and all the compressed air was released to a volume of 1370 L at atmospheric pressure.

Hank's Garage has an air compressor with a holding tank that contains a volume of 200L (V₁) of compressed air at a pressure of 5200 torr (P₁).

One day a hose ruptured and all the compressed air was released. The final pressure was the atmospheric pressure (1 atm = 760 torr) (P₂).

We can calculate the new volume (V₂) in these conditions using Boyle's law, which states there is an inverse relationship between the volume and the pressure of an ideal gas.

$P_1 \times V_1 = P_2 \times V_2\\\\V_2 = \frac{P_1 \times V_1}{P_2} = \frac{5200 torr \times 200L}{760torr} = 1370 L$

Hank's Garage has an air compressor with a holding tank that contains 200L of compressed air at 5200 torr. One day a hose ruptured and all the compressed air was released to a volume of 1370 L at atmospheric pressure.

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2 years ago

What coefficients are needed to balance the equation for the complete combustion of methane? enter the coefficients in the order

Art [367]

The coefficients in the order CH₄, O₂, CO₂, and H₂O, are 1,2,1,2

The equation becomes:

CH₄ (g) + 2O₂ (g) -> CO₂ (g) + 2H₂0 (g)

<h3><em>Further explanation</em></h3>

Complete combustion of hydrocarbon with oxygen will be obtained by CO₂ and H₂O compounds.

If O is insufficient there will be incomplete combustion produced by CO and H and O

Hydrocarbon combustion reactions (specifically alkanes)

$\large {\boxed {\bold {C_nH _ (_2_n _ + _ 2_) + \frac {3n + 1} {2} O_2 ----> nCO_2 + (n + 1) H_2O}}}$

Equalization of chemical reaction equations can be done using variables.

Steps in equalizing the reaction equation:

1. gives a coefficient on substances involved in the equation of reaction such as a, b, or c etc.

2. make an equation based on the similarity of the number of atoms where the number of atoms = coefficient × index between reactant and product

3. Select the coefficient of the substance with the most complex chemical formula equal to 1

For gas combustion reaction which is a reaction of hydrocarbons with oxygen produces CO₂ and H₂O (water vapor). can use steps:

Balancing C atoms, H and last atoms O atoms

Methana combustion reaction

CH₄ (g) + O₂ (g) -> CO₂ (g) + H₂0 (g)

We give the most complex compounds, namely Methane, we give the number 1

So the reaction becomes

CH₄ (g) + aO₂ (g) -> bCO₂ (g) + cH₂0 (g)

C atom on the left 1, right b, so b = 1

H atom on the left 4, right 2c, so 2c = 4 ---> c = 2

Atom O on the left 2a, right 2b + c, so 2a = 2b + c

2a = 2.1 + 2

2a = 4

a = 2

The equation becomes:

CH₄ (g) + 2O₂ (g) -> CO₂ (g) + 2H₂0 (g)

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