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2 years ago

List the numbers of each atom in all the formulas above: H C

Chemistry

2 answers:

bezimeni [28]2 years ago

6 0

Answer:

1) H = 14; C =6.

2) H = 14; C =6.

3) H = 14; C =6.

4) H = 14; C =6.

5) H = 14; C =6.

Molecular formula of this molecules is C₆H₁₄. Those molecules are alkanes with same molecular formula and different structural formula.

General formula for alkanes is CₓH₂ₓ₊₂.

Hexane is alkane (acyclic saturated hydrocarbon, carbon-carbon bonds are single) of six carbon atoms.

d1i1m1o1n [39]2 years ago

3 0

Answer:

1) 14

2) 6

Explanation: edg 2020

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Phoenix [80]

mass of reactants = mass of products best represents the law of conservation of mass

Explanation:

This is a law of thermodynamic that applies to chemical reactions. The mass of the reactants must equal to that of the products because energy/mass cannot be destroyed or created. This is why ideally, chemical reactions should be balanced to conform to this law.

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3 years ago

PLZ HELP ASAP WILL

amid [387]

Answer:

A) Mass = 32 g of KCl

Explanation:

Given data:

Mass of potassium chloride produced = ?

Mass of potassium chlorate = 52 g

Solution:

Chemical equation:

2KClO₃ → 2KCl + 3O₂

Number of moles of KClO₃:

Number of moles = mass/molar mass

Number of moles = 52 g/ 122.55 g/mol

Number of moles = 0.424 mol

Now we will compare the moles of KClO₃ and KCl

KClO₃ : KCl

2 : 2

0.424 : 0.424

Mass of KCl:

Mass = number of moles × molar mass

Mass = 0.424 mol × 74.55 g/mol

Mass = 32 g

4 0

2 years ago

Write the balanced equation for the reaction of aqueous Pb ( ClO 3 ) 2 Pb(ClO3)2 with aqueous NaI . NaI. Include phases. chemica

FinnZ [79.3K]

Answer: The mass of precipitate (lead (II) iodide) that will form is 119.89 grams

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of NaI solution = 0.130 M

Volume of solution = 0.400 L

Putting values in above equation, we get:

$0.130M=\frac{\text{Moles of NaI}}{0.400L}\\\\\text{Moles of NaI}=(0.130mol/L\times 0.400L)=0.52mol$

The balanced chemical equation for the reaction of lead chlorate and sodium iodide follows:

$Pb(ClO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaClO_3(aq.)$

The precipitate (insoluble salt) formed is lead (II) iodide

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, 0.52 moles of NaI will produce = $\frac{1}{2}\times 0.52=0.26mol$ of lead (II) iodide

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of lead (II) iodide = 0.26 moles

Molar mass of lead (II) iodide = 461.1 g/mol

Putting values in above equation, we get:

$0.26mol=\frac{\text{Mass of lead (II) iodide}}{461.1g/mol}\\\\\text{Mass of lead (II) iodide}=(0.26mol\times 461.1g/mol)=119.89g$

Hence, the mass of precipitate (lead (II) iodide) that will form is 119.89 grams

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