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2 years ago

Calculate the concentration of hydronium and hydroxide ions in a 0.050 M solution of nitric acid.

Chemistry

1 answer:

Mazyrski [523]2 years ago

3 0

Answer:

[H₃O⁺] = 0.05 M & [OH⁻] = 2.0 x 10⁻¹³.

Explanation:

HNO₃ is completely ionized in water as:

HNO₃ + H₂O → H₃O⁺ + NO₃⁻.

The concentration of hydronium ion is equal to the concentration of HNO₃:

[H₃O⁺] = 0.05 M.

∵ [H₃O⁺][OH⁻] = 10⁻¹⁴.

∴ [OH⁻] = 10⁻¹⁴/[H₃O⁺] = 10⁻¹⁴/0.05 = 2.0 x 10⁻¹³.

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When benzene (c6h6) reacts with bromine (br2) bromobenzene(c6h5br) is obtained: c6h6 + br2 → c6h5br + hbr what is the theoretica

saveliy_v [14]

Answer A) : We have to calculate the number of moles of Benzene involved in the reaction,

30 g / 78 moles of benzene = 0.384 moles

For bromine it will be the same process,

65 g / 159.8 moles = 0.406 moles

By observing the reaction given above we can say that the reaction ratio of bromine and benzene is 1 : 1

We need to find the mass of bromobenzene,

which should be, 6(12) + 5 (1)+ 79.90 = 156.9 g/mol

So, the mass of bromobenzene will be 156.9 g/mol X 0.3846 mol = 60.343 g

Hence the theoretical yield will be 60.34 g

Answer B) : To calculate the actual yield we have to divide it with theoretical yield.

(56.7g / 60.343 g ) X100% = 93.96 %

Here, we can say that we got 93.96 % of actual yield.

As we know it is impossible to get 100% yield in any reaction.

3 0

3 years ago

Calculate the standard enthalpy change for the reaction at 25 ∘ 25 ∘ C. Standard enthalpy of formation values can be found in th

WINSTONCH [101]

Answer: The standard enthalpy change of the reaction is coming out to be -16.3 kJ

Explanation:

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

For the given chemical reaction:

$Mg(OH)_2(s)+2HCl(g)\rightarrow MgCl_2(s)+2H_2O(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(1\times \Delta H_f_{(MgCl_2(s))})+(2\times \Delta H_f_{(H_2O(g))})]-[(1\times \Delta H_f_{(Mg(OH)_2(s))})+(2\times \Delta H_f_{(HCl(g))})]$

We are given:

$\Delta H_f_{(Mg(OH)_2(s))}=-924.5kJ/mol\\\Delta H_f_{(HCl(g))}=-92.30kJ/mol\\\Delta H_f_{(MgCl_2(s))}=-641.8kJ/mol\\\Delta H_f_{(H_2O(g))}=-241.8kJ/mol$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(1\times (-641.8))+(2\times (-241.8))]-[(1\times (-924.5))+(2\times (-92.30))]\\\\\Delta H_{rxn}=-16.3kJ$

Hence, the standard enthalpy change of the reaction is coming out to be -16.3 kJ

6 0

3 years ago

Pbo..... = 2Pb +......

Colt1911 [192]

Answer:

It's

Explanation:

Pbo + c

= pb + co2

While balancing it becomes

2pbo + c = 2pb + co2

(I was also doing same qn)

7 0

3 years ago

What compound consists of 0.248 g magnesium, 0.655 g sulfur, and 1.31 g oxygen

Ivenika [448]

Magnesium Peroxydisulphate with formula of MgS₂O₈.

<h3>Explanation:</h3>

The question clearly stands on the concept of molarity and atomic weights.

The atomic weight of magnesium = 24.

The atomic weight of sulphur = 32.

The atomic weight of oxygen = 16.

The amount of magnesium present = 0.248 g

The amount of sulphur present = 0.665 g.

The amount of oxygen present = 1.31g.

So, moles of magnesium present = 0.01 moles.

Moles of sulphur present = 0.02 moles.

Moles of oxygen present = 0.08 moles.

So, mole ratio of the compound as magnesium : sulphur : oxygen = 1:2:8.

So the compound is Magnesium Peroxydisulphate with formula of MgS₂O₈.

3 0

3 years ago

Complete each statement to identify the form of asexual reproduction.

Mrrafil [7]

Spore formation is a form of asexual reproduction used by mushrooms and molds.

During budding, the offspring grows from the body of the parent.

Fragmentation is a form of asexual reproduction that must be followed by regeneration.

Explanation:

Asexual reproduction is the type of reproduction where the gamete formation and fusion have no relevance or existence. It functions on the process of somatic cell division via mitosis and the offsprings are identical to their parents.

The spore formation occurs in fungi through sporangia, bursting open to shed spores, forming into a new young ones. Budding occurs out as an outgrowth of the parent and attains maturity and separates. Fragmentation is the process where the parents fall apart into pieces and regeneration follows.

8 0

2 years ago

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