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2 years ago

how the various concentrations of acid will affect the amount of limestone that is dissolved from each jar.

1 answer:

8 0

The explanation of the how the various concentrations of acid will affect the amount of limestone has been given below.

Effects of acid rain on limestone:-

When an acid combines with a carbonate, it produces carbon dioxide as a gas and forms a salt that is soluble in the carbonate and acid's water.

There are several gases in the atmosphere that can dissolve in precipitation such as rain and snow.

Some may produce acids in rain water, such as carbonic acid, sulfuric acid, and nitric acid.

Because the concentration is modest, the rain is not highly acidic, but it is acidic enough to react with the carbonates that make up limestone.

Thus we discussed the affects of acid rain on limestones above.

Learn more about Acid Rain here:

brainly.com/question/718250

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Which of the following is NOT an example of a Heterogeneous Mixture?

yuradex [85]

Answer:

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2 years ago

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How many sodium ions are in 1.4 kg of sodium chloride, NaCl?

Whitepunk [10]

Answer:

1.44 x 10²⁵ ions of Na⁺

Explanation:

Given parameters:

Mass of NaCl = 1.4kg = 1400g

Unknown:

Number of ions of sodium = ?

Solution:

The compound NaCl in ionic form can be written as;

NaCl → Na⁺ + Cl⁻

In 1 mole of NaCl we have 1 mole of sodium ions

Now, let us find the number of moles in NaCl;

Number of moles = $\frac{mass}{molar mass}$

Molar mass of NaCl = 23 + 35.5 = 58.5g/mol

Number of moles = $\frac{1400}{58.5}$ = 23.93mol

So;

Since 1 mole of NaCl gives 1 mole of Na⁺

In 23.93 mole of NaCl will give 23.93 mole of Na⁺

1 mole of a substance = 6.02 x 10²³ ions of a substance

23.93 mole of a substance = 6.02 x 10²³ x 23.93

= 1.44 x 10²⁵ ions of Na⁺

7 0

3 years ago

The atomic theory proposed by dalton has been

Sidana [21]

What did Dalton's atomic theory contribute to science?

Dalton's atomic theory proposed that all matter was composed of atoms, indivisible and indestructible building blocks. While all atoms of an element were identical, different elements had atoms of differing size and mass.

3 0

4 years ago

CO2(g)+CCl4(g)⇌2COCl2(g) Calculate ΔG for this reaction at 25 ∘C under these conditions: PCO2PCCl4PCOCl2===0.140 atm0.185 atm0.7

padilas [110]

<u>Answer:</u> The $\Delta G$ for the reaction is 54.425 kJ/mol

<u>Explanation:</u>

For the given balanced chemical equation:

$CO_2(g)+CCl_4(g)\rightleftharpoons 2COCl_2(g)$

We are given:

$\Delta G^o_f_{CO_2}=-394.4kJ/mol\\\Delta G^o_f_{CCl_4}=-62.3kJ/mol\\\Delta G^o_f_{COCl_2}=-204.9kJ/mol$

To calculate $\Delta G^o_{rxn}$ for the reaction, we use the equation:

$\Delta G^o_{rxn}=\sum [n\times \Delta G_f(product)]-\sum [n\times \Delta G_f(reactant)]$

For the given equation:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})]$

Putting values in above equation, we get:

$\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J$

Conversion factor used = 1 kJ = 1000 J

The expression of $K_p$ for the given reaction:

$K_p=\frac{(p_{COCl_2})^2}{p_{CO_2}\times p_{CCl_4}}$

We are given:

$p_{COCl_2}=0.735atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.185atm$

Putting values in above equation, we get:

$K_p=\frac{(0.735)^2}{0.410\times 0.185}\\\\K_p=20.85$

To calculate the gibbs free energy of the reaction, we use the equation:

$\Delta G=\Delta G^o+RT\ln K_p$

where,

$\Delta G$ = Gibbs' free energy of the reaction = ?

$\Delta G^o$ = Standard gibbs' free energy change of the reaction = 46900 J

R = Gas constant = $8.314J/K mol$

T = Temperature = $25^oC=[25+273]K=298K$

$K_p$ = equilibrium constant in terms of partial pressure = 20.85

Putting values in above equation, we get:

$\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(20.85))\\\\\Delta G=54425.26J/mol=54.425kJ/mol$

Hence, the $\Delta G$ for the reaction is 54.425 kJ/mol

7 0

3 years ago

Please and thank you!

Valentin [98]

Answer:

7.5 moles of O₂.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

2KClO₃ —> 2KCl + 3O₂

From the balanced equation above,

2 moles of KClO₃ decomposed to produce 3 moles of O₂.

Finally, we shall determine the number of mole of O₂ produced by the decomposition of 5 moles of KClO₃. This can be obtained as follow:

From the balanced equation above,

2 moles of KClO₃ decomposed to produce 3 moles of O₂.

Therefore, 5 moles of KClO₃ will decompose to produce = (5 × 3)/ 2 = 7.5 moles of O₂.

Thus, 7.5 moles of O₂ were obtained from the reaction.

3 0

3 years ago