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Ede4ka [16]
2 years ago
6

What is the concentration of hydronium ions after a 100ml solution of 0.02m magnesium hydroxide and 100ml of 0.05m nitric acid?

Chemistry
1 answer:
Svetllana [295]2 years ago
4 0
Mg(OH)_2 + 2HNO_3 ---\ \textgreater \   Mg(NO_3)_2 + 2H_2O

That is the balanced chemical equation.

Before reaction,
milli-moles of Mg(OH)2 = 100 x 0.02 = 2
milli-moles of HNO3 = 100 x 0.05 = 5

After reaction:
Mg(OH)2 is limiting reagent. It will react completely leaving some HNO3 behind.
milli-moles of HNO3 react = 2 x coefficient of HNO3 = 2 x 2 = 4

milli-moles of HNO3 left = 5 -4 = 1

Now, total volume = 100+ 100 = 200 mL

It is asked for hydronium ion concentration, which is basically hydrogen ion conc.
 1 HNO3 will dissociate to give 1 H+ ion.

so, milli-moles of H+ = 1 

So, concentration of H+  or hydronium(H3O+) = 1 / 200 = 0.005

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The recommended daily intake of potassium ( K ) is 4.725 g . The average raisin contains 3.513 mg K . Fill in the denominators o
kondor19780726 [428]

Explanation:

It is known that 1 gram contains 1000 milligrams. And, mathematically we can represent it as follows.

             \frac{1 g}{1000 mg} or \frac{1000 mg}{1 g}

So, when we have to convert grams into milligrams then we simply multiply the digit with 1000. And, if we have to convert a digit from milligrams to grams then we simply divide it by 1000.

4 0
2 years ago
A mixture of gases A2 and B2 are introduced to a slender metal cylinder that has one end closed and the other fitted with a pist
REY [17]

Answer:

q < 0, w > 0, the sign of ΔE cannot be determined from the information given

Explanation:

Determination of sign of q

Temperature of the water bath before the reaction = 25 °C

Temperature of the water bath after the completion of the reaction = 28 °C

After the completion of the reaction, temperature of the water bath is increased that means heat is released during the reaction and flows out of the system.

If heat is absorbed by the system, then q is indicated by positive sign and if heat is released by the system, then q is indicated by negative sign.

As in the given case, heat is released by the system, so sign of q is negative, or q < 0

Determination of sign of w

After the completion of the reaction, piston moved downward, that means volume of the system decreases or compression occur. During the compression, work is done on the system.

if work is done on the system, sign of w is positive.

If work is done by the system, sign of w is negative.

In the given case, work is done on the system, therefore sign of w is positive, or w > 0

Determination of sign of ΔE

Relationship between ΔE, q and w is given by first law of thermodynamics:

ΔE = q + w

In this case, q is positive and w is negative, so the sign of ΔE depends of magnitude of q and w. As magnitude of w and q cannot be determined in this case, thus, the given information is insufficient for the determination of sign of ΔE.

So, among the given option, option c is correct.

q < 0, w > 0, the sign of ΔE cannot be determined from the information given

3 0
2 years ago
Octane (C8H18) undergoes combustion according to the following thermochemical equation. 2C8H18(l) + 25O2(g) → 16CO2(g) + 18H2O(l
Zepler [3.9K]

Answer: The standard enthalpy of formation of liquid octane is -250.2 kJ/mol

Explanation:

The given balanced chemical reaction is,

2C_8H_{18}(l)+25O_2(g)\rightarrow 16CO_2(g)+18H_2O(l)

First we have to calculate the enthalpy of reaction (\Delta H^o).

\Delta H^o=H_f_{product}-H_f_{reactant}

\Delta H^o=[n_{O_2}\times \Delta H_f^0_{(O_2)}+n_{H_2O}\times \Delta H_f^0_{(H_2O)}]-[n_{C_8H_{18}}\times \Delta H_f^0_{(C_8H_{18})+n_{O_2}\times \Delta H_f^0_{(O_2)}]

where,

We are given:

\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(C_8H_{18}(l))}=?kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-285.8kJ/mol

Putting values in above equation, we get:

-1.0940\times 10^4=[(16\times -393.5)+(18\times -285.8)]-[(25\times 0)+(2\times \Delat H_f{C_8H_{18}(l)}]

\Delta H^o_f_{(C_8H_{18}(l))}=-250.2kJ/mol

Thus the standard enthalpy of formation of liquid octane is -250.2 kJ/mol

4 0
2 years ago
Imagine you needed to identify if an object has undergone a physical change or a chemical change. What information would you nee
pishuonlain [190]
How it looks. basically the thing that tells you how it change. for example if an ice cube was melted (heat), it only changed physically not chemically as the h20 molecules are still there. however lets say you burn woos— you cant get that would back. its ash now and it has changed chemically.
7 0
1 year ago
Which equation represents the total ionic equation for the reaction of HNO3 and NaOH? Upper H superscript plus, plus upper O upp
DIA [1.3K]

Answer:

Total Ionic equation:

H⁺(aq) + NO₃⁻ (aq) + Na⁺(aq) + OH⁻(aq) → H₂O(l) + Na⁺(aq) + NO₃⁻ (aq)

Explanation:

Chemical equation:

HNO₃ + NaOH →  NaNO₃ + H₂O

Balanced chemical equation:

HNO₃(aq) + NaOH(aq) →  NaNO₃(aq) + H₂O(l)

Total Ionic equation:

H⁺(aq) + NO₃⁻ (aq) + Na⁺(aq) + OH⁻(aq) → H₂O(l) + Na⁺(aq) + NO₃⁻ (aq)

Net ionic equation:

H⁺(aq) + OH⁻(aq) → H₂O(l)  

The NO₃⁻ (aq)  and Na⁺ (aq) are spectator ions that's why these are not written in net ionic equation. The water can not be splitted into ions because it is present in liquid form.

Spectator ions:

These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation

6 0
2 years ago
Read 2 more answers
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