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4 years ago

Determine the oh of a solution that is 0.220 m in hco3

Chemistry

1 answer:

ICE Princess25 [194]4 years ago

6 0

Answer : The hydroxide ion concentration of a solution is, $5\times 10^{-14}$

Explanation :

As we know that $HCO_3^-$ dissociates in water to give hydrogen ion $(H^+)$ and carbonate ion $(CO_3^{2-})$ .

As, 1 mole of $HCO_3^-$ dissociates to give 1 mole of hydrogen ion $(H^+)$

Or, 1 M of $HCO_3^-$ dissociates to give 1 M of hydrogen ion $(H^+)$

So, 0.200 M of $HCO_3^-$ dissociates to give 0.200 M of hydrogen ion $(H^+)$

Now we have to calculate the hydroxide ion concentration.

As we know that:

$[H^+][OH^-]=1\times 10^{-14}$

$0.200\times [OH^-]=1\times 10^{-14}$

$[OH^-]=5\times 10^{-14}$

Therefore, the hydroxide ion concentration of a solution is, $5\times 10^{-14}$

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How many calories of heat are necessary to raise the temperature of 319.5 g of water from 35.7 °C

rjkz [21]

20600Cal

Explanation:

Given parameters:

Mass of water = 319.5g

Initial temperature = 35.7°C

Final temperature = 100°C

Unknown:

Calories needed to heat the water = ?

Solution:

The calories is the amount of heat added to the water. This can be determined using;

H = m c Ф

c = specific heat capacity of water = 4.186J/g°C

H is the amount of heat

Ф is the change in temperature

H = m c (Ф₂ - Ф₁)

H = 319.5 x 4.186 x (100 - 35.7) = 85996.56J

Now;

1kilocalorie = 4184J

85996.56J to kCal; $\frac{85996.56}{4184}$ = 20.6kCal = 20600Cal

learn more:

Specific heat brainly.com/question/3032746

#learnwithBrainly

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3 years ago

Why does naphthalene have a higher melting point than biphenyl?

AlexFokin [52]

Naphthalene has a higher melting point than biphenyl because naphthalene is a polar compound while biphenyl is a non-polar compound. Studies show that polar compounds have higher melting and boiling points than nonpolar compounds. It is because polar compounds have strong intermolecular forces.

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3 years ago

What is the mass in grams of a single formula unit of silver chloride, AgCI? A) 4.21 x 1021 g B) 8.61 x 10258 C) 1.66 x 10-248 D

bekas [8.4K]

Answer: D) $2.38*10^-^2^2$ g

Explanation: The question asks to convert formula unit to grams. It is a unit conversion problem.

1 mole equals to Avogadro number of formula units. So, to convert the given number of formula units to moles, we need to divide by the Avogadro number. After this, we do moles to grams conversion and for this the moles are multiplied by the molar mass of the compound. Molar mass of AgCl is 143.32 gram per mol.

$1FormulaUnitAgCl(\frac{1mol}{6.022*10^2^3formulaUnits})(\frac{143.32g}{1mol})$

= $2.38*10^-^2^2$ g

So, the correct option is D) $2.38*10^-^2^2$ g

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3 years ago

Read 2 more answers

A cheetah can run 112km/h over a 100-m distance. what is this speed in meters per second

Naya [18.7K]

First convert the 112 km/hr ratio into m/s (meters per second). To do this you multiply 112 km with 1000 m/km (since there's 1000 m in one km). You get 112000 m. Then multiply 1 hr with 60 min/hr (since there's 60 min in one hr. You get 60 min, but you want seconds, so multiply 60 min with 60 s/min to get 3600 s. There you go! Your answer is the speed of 112000m/3600s, but you can simplify that to 31.11m/s (since the answer should be in ? meters per 1 second.

Also, the "100-m-distance" part of the question is just to throw you off, because one particular speed obviously stays constant over any distance. Hope that helps :)

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4 years ago

In which of the following reactions does a decrease in the volume of the reaction vessel at constant

Black_prince [1.1K]

Answer:

The correct option is: A) 2H₂(g) + O₂(g) → 2H₂O(g)

Explanation:

According to the Le Chatelier's principle, change in the volume of the reaction system causes equilibrium to shift in the direction that reduces the effect of the volume change.

When the volume decreases then the pressure of the reaction vessel increases, then the equilibrium shifts towards the reaction side that produces less number of moles of gas.

A) 2H₂(g) + O₂(g) → 2H₂O(g)

The number of moles of reactant is 3 and number of moles of product is 2.

Therefore, when volume decreases, the equilibrium shifts towards the product side, thereby favoring the formation of products.

B) NO₂(g) + CO(g) → NO(g) + CO₂(g)

The number of moles of reactant and product both is 2.

Therefore, when the volume decreases, the equilibrium does not shift in any direction.

C) H₂(g) + I₂(g) → 2HI(g)

The number of moles of reactant and product both is 2.

Therefore, when the volume decreases, the equilibrium does not shift in any direction.

D) 2O₃(g) → 3O₂(g)

The number of moles of reactant is 2 and number of moles of product is 3.

Therefore, when volume decreases, the equilibrium shifts towards the reactant side, thereby favoring the formation of reactants.

E) MgCO₃(s) → MgO(s) + CO₂(g)

The number of moles of reactant is 1 and number of moles of product is 2.

Therefore, when volume decreases, the equilibrium shifts towards the reactant side, thereby favoring the formation of reactants.

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3 years ago