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Living things respire to stay alive. In the mitochondria, sugar is combined with oxygen to produce energy, water and carbon diox

olya-2409 [2.1K]

Answer:

C6H12O6 + 6O2 → 6CO2 + 6H2O

Explanation:

Cellular respiration occurs in the mitochondria of cells. It is a process in which sugar is combined with oxygen to produce energy, water and carbon dioxide. This is the major process by which energy is released in living organisms.

Aerobic respiration involves a series of chemical reactions. These reactions commence with sugar and oxygen then it produces carbon dioxide and water according to the reaction equation; C6H12O6 + 6O2 → 6CO2 + 6H2O.

8 0

3 years ago

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Crime scene investigators keep a wide variety of compounds on hand to help with identifying unknown substances they find in the

almond37 [142]

The powder could be acetaminophen, analgesic having chemical formula $C_{17}H_{21}NO_{4}$

<h3>What is an empirical formula?</h3>

A chemical formula showing the simplest ratio of elements in a compound rather than the total number of atoms in the molecule.

We are given:

Percentage of C = 67.31 %

Percentage of H = 6.978 %

Percentage of N = 4.617 %

Percentage of O = 21.10 %

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

Mass of C = 67.31 g

Mass of H = 6.978 g

Mass of N = 4.617 g

Mass of O = 21.10 g

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

Step 1: Converting the given masses into moles.

Moles of carbon = $\frac{mass}{molar \;mass}$

Moles of carbon = $\frac{67.31g}{12g/mole}$

=5.60 moles

Moles of hydrogen = $\frac{mass}{molar \;mass}$

Moles of hydrogen = $\frac{6.978 g}{1 g/mole}$

=6.978 moles

Moles of nitrogen = $\frac{mass}{molar \;mass}$

Moles of nitrogen = $\frac{4.617 g}{14 g/mole}$

=0.329 moles

Moles of oxygen = $\frac{mass}{molar \;mass}$

Moles of oxygen = $\frac{21.10 g}{16 g/mole}$

=1.31 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 0.329 moles.

We get the ratio of C : H : N : O = 17 : 21 : 1 : 4

The empirical formula for the given compound is $C_{17}H_{21}NO_{4}$ .

Learn more about the empirical formula here:

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3 0

2 years ago

How many moles of hydrogen gas can be produced if 0.78 moles of hydrochloric acid reacts with excess solid zinc according to the

White raven [17]

Answer:

The answer to your question is It will be formed 0.39 moles of H₂

Explanation:

Data

moles of H₂ = ?

moles of HCl = 0.78

moles of Zinc = excess

Balanced chemical reaction

2 HCl + Zn ⇒ 1 H₂ + ZnCl₂

Process

1.- Use proportions to solve this problem. Consider the coefficients of the balanced reaction.

2 moles of HCl ---------------------- 1 mol of H₂

0.78 moles of HCl ----------------- x

x = (0.78 x 1) / 2

- Simplification

x = 0.78 / 2

- Result

x = 0.39 moles of H₂

3 0

3 years ago

n the diagram shown, when an object ‘X’ is brought near the ring shaped magnet, the magnet moves away from it. Four friends are

kati45 [8]

Answer:

Akash

Explanation:

it could be a magnet with the same poles facing eachoher

6 0

3 years ago

Calculate the number of hydrogen atoms present in 40g of urea, (NH2)2CO

tekilochka [14]

Answer: There are $16.14 \times 10^{23}$ atoms of hydrogen are present in 40g of urea, $(NH_{2})_{2}CO$ .

Explanation:

Given: Mass of urea = 40 g

Number of moles is the mass of substance divided by its molar mass.

First, moles of urea (molar mass = 60 g/mol) are calculated as follows.

$Moles = \frac{mass}{molar mass}\\= \frac{40 g}{60 g/mol}\\= 0.67 mol$

According to the mole concept, 1 mole of every substance contains $6.022 \times 10^{23}$ atoms.

So, the number of atoms present in 0.67 moles are as follows.

$0.67 mol \times 6.022 \times 10^{23} atoms/mol\\= 4.035 \times 10^{23} atoms$

In a molecule of urea there are 4 hydrogen atoms. Hence, number of hydrogen atoms present in 40 g of urea is as follows.

$4 \times 4.035 \times 10^{23} atoms\\= 16.14 \times 10^{23} atoms$

Thus, we can conclude that there are $16.14 \times 10^{23}$ atoms of hydrogen are present in 40g of urea, $(NH_{2})_{2}CO$ .

7 0

2 years ago

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