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3 years ago

Balance the equation : CH3COOH(aq) + NaHCO3(s) -> CO2(g) + H2O(l) + Na+(aq) + CH3COO-(aq) and provide an explanation

Chemistry

1 answer:

mihalych1998 [28]3 years ago

3 0

Answer:

Explanation:

CH₃COOH(aq) + NaHCO₃(s) -> CO₂(g) + H₂O(l) + Na+(aq) + CH₃COO-(aq)

The equation is already balanced .

In the reaction acetic acid reacts with sodium bicarbonate to produce carbon dioxide gas , water and sodium acetate . Sodium acetate will be in ionic form ie in the form of Na⁺ ( sodium ion ) and CH₃COO ⁻ ion ( acetate ion ) .

This is an acid base neutralisation reaction . one mole of acetic acid reacts with one mole of sodium bicarbonate to produce one mole of carbon dioxide gas , one mole of water and one mole of sodium acetate in ionic form .

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Sterling silver is 92.5% silver and 7.5% copper. Its density is 10.25 g/cm3. A sterling silver pendant is added to a graduated c

Mariulka [41]

From the information given:

The volume of the graduated cylinder = 50.0 mL
when a sterling silver pendant is added, the volume increases to = 61.3 mL

∴

The volume of the sterling silver pendant is:

= 61.3 mL - 50.0 mL

= 11.3 mL

Since, 1 mL = 1cm³

Then;

11.3 mL = 11.3 cm³

the density of the sterling silver = 10.25 g/cm³

Using the relation for Density; i.e.

$\mathbf{Density = \dfrac{mass}{volume}}$

$\mathbf{10.25 \ g/cm^3= \dfrac{mass}{11.3 \ cm^3}}$

mass = 10.25 g/cm³× 11.3 cm³

mass of the sterling silver = 115.825 grams

Recall that sterling silver has:

92.5% silver and;
7.5% copper

∴

The mass of the copper contained in the sterling silver pendant can be calculated as:

$\mathbf{= \dfrac{115.825 \ g \times 7.5}{100}}$

= 8.687 grams

Therefore, we can conclude that the mass of the copper contained in the sterling silver pendant is 8.687 grams

Learn more about the relation between Density, Mass, and Volume here:

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2 years ago

Symptoms of a mental disorder are ways of

kkurt [141]

Answer: thinking

Explanation:

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3 years ago

Two unknown molecular compounds were being studied. A solution containing 5.00 g of compound A in 100. g of water froze at a low

LenaWriter [7]

Answer:

Compound B has greater molar mass.

Explanation:

The depression in freezing point is given by ;

$\Delta T_f=i\times k_f\times m$ ..[1]

$m=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Mass of solvent in kg}}$

Where:

i = van't Hoff factor

$k_f$ = Molal depression constant

m = molality of the solution

According to question , solution with 5.00 g of A in 100.0 grams of water froze at at lower temperature than solution with 5.00 g of B in 100.0 grams of water.

The depression in freezing point of solution with A solute: $\Delta T_{f,A}$