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3 years ago

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The chemical formula of glucose isC6H12O6. What are the names of the elements glucose, and how many atoms of each element are pr

esent in a glucose molecule?

1 answer:

Vlad [161]3 years ago

6 0

Answer:

A glucose molecule is made up of 6 Carbon atoms, 12 Hydrogen atoms, and 6 Oxygen atoms. This is a 1:2:1 ratio. Hopefully, this helps!

Explanation:

You might be interested in

Gallium is a metallic element in Group III. It has similar properties to aluminium.

Liula [17]

Answer:

Gallium has Orthorhombic and forms ionic bonds with metals

Explanation:

The structure of gallium is Orthorhombic and it has a high neutron capture cross section. It is very stable in dry air but oxidizes in moist air. It reacts with oxygen to form Ga2O3. It has a valence of III. Its electronic configuration is

Ar 3d10 4s2 4p1

It is found in the 3A group of periodic table and it mostly forms ionic bonds with metal.

3 0

3 years ago

A solution is prepared by dissolving 0.5636 g oxalic acid (H2C2O4) in enough water to make 100.0 mL of solution. A 10.00-mL aliq

zepelin [54]

Answer:

The final molarity of the diluted oxalic acid solution is 0.002504 M

Explanation:

Step 1: Data given

Mass of oxalic acid (H2C2O4) = 0.5636 grams

Volume of the solution = 100.0 mL = 0.100 L

A 10 ml of this solution diluted in 250 ml of solution.

Molecular weight of H2C2O4 = 90.03 g/mol

Step 2: Calculate initial moles of H2C2O4

Moles H2C2O4 = mass / molar mass

Moles H2C2O4 = 0.5636 grams / 90.03 g/mol

Moles H2C2O4 = 0.00626 moles

Step 3: Calculate molarity of the solution

Molarity = moles / volume

Molarity = 0.00626 moles / 0.100 L

Molarity = 0.0626 M

Step 4: Calculate moles of a 10.00 mL aliquot

Moles = 0.0626 M * 0.010 L

Moles = 0.000626 moles

Step 5: Calculate the new molarity

Molarity = 0.000626 moles / 0.250 L

Molarity = 0.002504 M

The final molarity of the diluted oxalic acid solution is 0.002504 M

6 0

3 years ago

Calculate the concentration of so42− ions in a 0.010 m aqueous solution of sulfuric acid. express your answer to four decimal pl

dimulka [17.4K]

<span>Answer: 0.00649M

The question is incomplete,
</span>

<span>You are told that the first ionization of the sulfuric acid is complete and the second ionization of the sulfuric acid has a constant Ka₂ = 0.012
</span>
<span>
With that you can solve the question following these steps"
</span>

<span>1) First ionization:
</span>
<span>
H₂SO₄(aq) --> H⁺ (aq) + HSO₄⁻ (aq)

Under the fully ionization assumption the concentration of HSO4- is the same of the acid = 0.01 M

2) Second ionization
</span>

<span>HSO₄⁻ (aq) ⇄ H⁺ + SO₄²⁻ with a Ka₂ = 0.012
</span>

<span>Do the mass balance:
</span>

<span><span> HSO₄⁻ (aq) H⁺ SO₄²⁻</span>
</span>
<span /><span /><span> 0.01 M - x x x

</span><span>Ka₂ = [H⁺] [SO₄²⁻] / [HSO₄⁻]</span>
<span /><span>
=> Ka₂ = (x²) / (0.01 - x) = 0.012
</span><span />

<span>3) Solve the equation:

</span><span>x² = 0.012(0.01 - x) = 0.00012 - 0.012x</span>
<span /><span>
x² + 0.012x - 0.0012 = 0
</span><span />

<span>Using the quadratic formula: x = 0.00649
</span><span />

<span>So, the requested concentratioN is [SO₄²⁻] = 0.00649M</span>

4 0

3 years ago

Read 2 more answers

Calcutale Grxn for the following equation at 25°C: <br><br> 4KClO3(s) → 3KClO4(s) KCl(s)

Step2247 [10]

Answer:

-133.2 kJ

Explanation:

Let's consider the following balanced equation.

4 KClO₃(s) → 3 KClO₄(s) + KCl(s)

We can calculate the standard Gibbs free energy of the reaction (ΔG°rxn) using the following expression.

ΔG°rxn = 3 mol × ΔG°f(KClO₄(s)) + 1 mol × ΔG°f(KCl(s)) - 4 mol × ΔG°f(KClO₃(s))

ΔG°rxn = 3 mol × (-303.1 kJ/mol) + 1 mol × (-409.1 kJ/mol) - 4 mol × (-296.3 kJ/mol)

ΔG°rxn = -133.2 kJ

5 0

3 years ago

For the following reaction, identify the element that was oxidized, the element that was reduced, and the reducing agent. Give a

Volgvan

<span>For this reaction, oxidation number of Carbon in CO would be +2 while oxidation number of carbon in CO2 would be +4 and so this means that carbon has oxidized. Oxidation number of nitrogen in NO is +2. While oxidation number of nitrogen in N2 is 0 so this means that nitrogen had reduced. The reducing agent is the one which provides electrons by oxidizing itself so in this case; CO is the reducing agent while the C in CO oxidized to produce electrons. </span><span>I am hoping that this answer has satisfied your query about and it will be able to help you, and if you’d like, feel free to ask another question.</span>

6 0

3 years ago