All categories

3 years ago

PLEASE HELP. ONLY IF YOU KNOW THE ANSWER

Chemistry

1 answer:

Nezavi [6.7K]3 years ago

8 0

Answer:

24. B

25. B

26. B

Explanation:

24. As depicted in the image attached, all living organisms including animals undergo cellular respiration, which is the process whereby sugar molecule is broken down to release energy (ATP). In this process, CARBON DIOXIDE (CO2) is released as a product. The equation is as follows:

C6H12O6 + 6O2 → 6CO2 + 6H2O + energy

On the other hand, plants undergo a unique process called photosynthesis in order to synthesize their food (sugars) and hereby release oxygen gas in the process. Based on the diagram in option B, plants release oxygen using CO2, while animals release CO2 using O2.

25. The structure depicted in the image of this question is an organelle called MITOCHONDRIA. The mitochondrion is the site of cellular respiration (Krebs cycle and ETC) in eukaryotic cells.

26. Photosynthesis is a metabolic process unique to the cells of autotrophic organisms like green plants. It is the process where they synthesize their food in form of organic molecules using SUNLIGHT as a source of energy.

6CO2 + 6H2O + sunlight → C6H12O6 + 6O2

You might be interested in

Beta particles will: 1. bend toward the North Pole of a magnet. 2 bend toward the South Pole of the magnet 3 not be affected by

Setler79 [48]

Beta particles will bend toward the South Pole of the magnet.

Hope this helped!

8 0

3 years ago

Read 2 more answers

Why OF2 is a polar molecule. Explain in detail ?

ale4655 [162]

Explanation:

Hybridization of $OF_2$ is $sp^3$

Bond pair in $OF_2$ = 2

lone pair in $OF_2$ = 2

Because of presence of two lone pairs of electron shape is bent.

F is more electronegative than O, therefore two O-F bons are polar and hence, two dipole exists. The resultant diople moment in a molecule is the vector sum of individual existing dipoles (shown in the attachment).

because of the bent shape of $OF_2$ two dipole moments add to each other. this result in the existence of permanent dipole moment in the molecule and thus $OF_2$ is a polar molecule.

6 0

4 years ago

Why is potassium sulfate K2SO4, but strontium sulfate SrSO4?

Oksi-84 [34.3K]

The sulphate ion has a charge of 2-, and potassium has a charge of 1+. To fully neutralise the charge on a sulphate ion, you need two potassium ions.

However, strontium has a charge of 2+. To neutralise the sulphate ion's 2- charge, one strontium ion is enough, so only one bonds to the sulphate.

Hope I helped! xx

7 0

3 years ago

Select the two variables that are held constant when testing Boyle's law in a manometer.

MrRissso [65]

Start studying Pressure - Volume Relationships in Gases (Boyle's Law). Learn vocabulary, terms ...Select all that apply. V2 = k/P2 V2 = P1V1/P2 ... What two variables are held constant when testing Boyle's Law in a manometer? Temperature hope this helps

4 0

3 years ago

Read 2 more answers

A) Combustion analysis of toluene, a common organic solvent, gives 5.86 mg of CO2 and 1.37 mg of H2O. If the compound contains o

Mumz [18]

Answer:

For a: The empirical formula for the given compound is $CH$

For b: The empirical and molecular formula for the given organic compound are $C_{10}H_{20}O$

Explanation:

For a:

The chemical equation for the combustion of hydrocarbon follows:

$C_xH_y+O_2\rightarrow CO_2+H_2O$

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

We are given:

Conversion factor used: 1 g = 1000 mg

Mass of $CO_2=5.86mg=5.86\times 10^{-3}g$

Mass of $H_2O=1.37mg=1.37\times 10^{-3}g$

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 $5.86\times 10^{-3}g$ of carbon dioxide, $\frac{12}{44}\times 5.86\times 10^{-3}=1.60\times 10^{-3}g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in $1.37\times 10^{-3}g$ of water, $\frac{2}{18}\times 1.37\times 10^{-3}=0.152\times 10^{-3}g$ of hydrogen will be contained.

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{1.60\times 10^{-3}g}{12g/mole}=0.133\times 10^{-3}moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.152\times 10^{-3}g}{1g/mole}=0.152\times 10^{-3}moles$

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 $0.133\times 10^{-3}$ moles.

For Carbon = $\frac{0.133\times 10^{-3}}{0.133\times 10^{-3}}=1$

For Hydrogen = $\frac{0.152\times 10^{-3}}{0.133\times 10^{-3}}=1.14\approx 1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H = 1 : 1

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

For b:

The chemical equation for the combustion of menthol 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 $CO_2$ = 0.2829 g

Mass of $H_2O$ = 0.1159 g

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 0.2829 g of carbon dioxide, $\frac{12}{44}\times 0.2829=0.077g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 0.1159 g of water, $\frac{2}{18}\times 0.1159=0.013g$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.1005) - (0.077 + 0.013) = 0.105 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.077g}{12g/mole}=0.0064moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.013g}{1g/mole}=0.013moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.0105g}{16g/mole}=0.00065moles$

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 0.00065 moles.

For Carbon = $\frac{0.0064}{0.00065}=9.84\approx 10$

For Hydrogen = $\frac{0.013}{0.00065}=20$

For Oxygen = $\frac{0.00065}{0.00065}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 10 : 20 : 1

The empirical formula for the given compound is $C_{10}H_{20}O$

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 = 156 g/mol

Mass of empirical formula = 156 g/mol

Putting values in above equation, we get:

$n=\frac{156g/mol}{156g/mol}=1$

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

$C_{(1\times 10)}H_{(1\times 20)}O_{(1\times 1)}=C_{10}H_{20}O$

Hence, the empirical and molecular formula for the given organic compound are $C_{10}H_{20}O$

3 0

4 years ago