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slamgirl [31]
3 years ago
11

HCHO2 + NaOH = NaCHO2 + H2O complete ionic

Chemistry
1 answer:
Llana [10]3 years ago
4 0

Answer:

HCHO₂(aq) + Na⁺(aq) + OH⁻(aq) ⟶ Na⁺(aq) + CHO₂⁻(aq) + H₂O(ℓ)  

Explanation:

An ionic equation uses the symbols (aq) [aqueous] to indicate molecules and ions that are soluble in water, (s) [solid] to indicate insoluble solids, and (ℓ) to indicate substances (usually water) in the liquid state.

In this reaction, aqueous sodium hydroxide reacts with aqueous formic acid to form sodium formate and water.

1. Molecular equation

HCHO₂(aq) + NaOH(aq) ⟶ NaCHO₂(aq) + H₂O(ℓ)  

2. Ionic equation

Sodium hydroxide and sodium formate are soluble ionic compounds, so we write them as hydrated ions.

HCHO₂(aq) + Na⁺(aq) + OH⁻(aq) ⟶ Na⁺(aq) + CHO₂⁻(aq) + H₂O(ℓ)  

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4 years ago
Most Bic lighters hold 5.0ml of liquified butane (density = 0.60 g/ml). Calculate the minimum size container you would need to "
Hatshy [7]

Answer:

Volume of container = 0.0012 m³ or 1.2 L or 1200 ml

Explanation:

Volume of butane = 5.0 ml

density = 0.60 g/ml

Room temperature (T) = 293.15 K

Normal pressure (P) = 1 atm = 101,325 pa

Ideal gas constant (R) = 8.3145 J/mole.K)

volume of container V = ?

Solution

To find out the volume of container we use ideal gas equation

PV = nRT

P = pressure

V = volume

n = number of moles

R = gas constant

T = temperature

First we find out number of moles

<em>As Mass = density × volume</em>

mass of butane = 0.60 g/ml ×5.0 ml

mass of butane = 3 g

now find out number of moles (n)

n = mass / molar mass

n = 3 g / 58.12 g/mol

n = 0.05 mol

Now put all values in ideal gas equation

<em>PV = nRt</em>

<em>V = nRT/P</em>

V = (0.05 mol × 8.3145 J/mol.K × 293.15 K) ÷ 101,325 pa

V = 121.87 ÷ 101,325 pa

V = 0.0012 m³ OR 1.2 L OR 1200 ml

8 0
4 years ago
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