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3 years ago

A flask is filled with 50.0 ml of 0.15 M HCl acid. Exactly 0.054 grams of CaCo3 are placed in the acid,

Chemistry

1 answer:

ira [324]3 years ago

8 0

Answer:

See explanation

Explanation:

2HCl(aq) + CaCO3(aq) ------->CaCl2(aq) + CO2(g) + H2O(l)

Number of moles of acid present = 50/1000 * 0.15 = 0.0075 moles

Number of moles of calcium carbonate = 0.054g/100 g/mol = 0.00054 moles

2 moles of HCl reacts with 1 mole of calcium carbonate

x moles of HCl reacts with 0.00054 moles of calcium carbonate

x = 2 * 0.00054/1

x = 0.00108 moles of HCl

Amount of acid left = 0.0075 moles - 0.0075 moles = 0.00642 moles

Reaction of HCl and NaOH

HCl(aq) + NaOH(aq) ------> NaCl(aq) + H2O(l)

Since the reaction is in the mole ratio of 1:1

0.00642 moles of HCl is neutralized by 0.00642 moles of NaOH

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2 years ago

g a package of aluminum foil contains 50 ft2 of foil, which weighs approximately 8.0 oz. aluminum has a density of 2.70 g/cm3. w

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Answer:

1.8×10^-2mm or 0.018mm

Explanation:

Density = mass/Volume

But volume= area×thickness

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2 years ago

A certain second-order reaction (B→products) has a rate constant of 1.30×10−3 M−1⋅s−1 at 27 ∘C and an initial half-life of 224 s

soldier1979 [14.2K]

Answer:

$\large\boxed{\large\boxed{0.291M}}$

Explanation:

By definition one half-life is the time to reduce the initial concentration to half.

For a second order reaction the rate law equations are:

$\dfrac{d[B]}{dt}=-k[B]^2$

$\dfrac{1}{[B]}=\dfrac{1}{[B]_0}+kt$

The half-life equation is:

$t_{1/2}=\dfrac{1}{k[A]_0}$

Thus, substitute the rate constant $1.30\times 10^{-3}M^{-1}\cdot s^{-1}$ and the half-life time 224s to find [A]₀:

$224s=\dfrac{1}{1.30\times10^{-3}M^{-1}\cdot s^{-1}[A]_0}$

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