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Vesna [10]
3 years ago
15

1) An unlabeled laboratory solution is found to contain a H+ ion concentration of 1 x 10-4

Chemistry
1 answer:
marishachu [46]3 years ago
7 0

Answer:

C) 1 x 10-10 M

Explanation:

To solve this question we must use the equation:

Kw = [H+] [OH-]

<em>Where Kw is the equilibrium dissociation of water = 1x10-14</em>

<em>[H+] is the molar concentration of hydronium ion = 1x10-4M</em>

<em>[OH-] is the molar concentration of hydroxyl ion</em>

<em />

Replacing:

1x10-14= 1x10-4 [OH-]

<em>[OH-] = 1x10-14 / 1x10-4M</em>

<em>[OH-] = 1x10-10 M</em>

Right option is:

<h3>C) 1 x 10-10 M </h3>

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adelina 88 [10]
Ionic bonds are made up of a metal and a nonmetal elements while covalent bonds are are made up of two metal. So, you can just look at the periodic table and identify which compounds have two metals and which have one metal and one nonmetal. Also, those compounds with high electronegativity difference are very likely to be ionic while the opposite is covalent. HOWEVER, some compounds can be both covalent and ionic. For instance, if HCI gas is at higher temperatures, then it is ionic while it would be covalent at room temp. 

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6 0
3 years ago
370 cm3 of water at 80°C is mixed with 120 cm3 of water at 4°C. Calculate the final equilibrium temperature, assuming no heat is
NikAS [45]

Answer : The final temperature of the mixture is 61.4^oC

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

And as we know that,

Mass = Density × Volume

Thus, the formula becomes,

(\rho_1\times V_1)\times c_1\times (T_f-T_1)=-(\rho_2\times V_2)\times c_2\times (T_f-T_2)

where,

c_1 = c_2 = specific heat of water = same

m_1 = m_2 = mass of water  =  same

\rho_1 = \rho_2 = density of water = 1.0 g/mL

V_1 = volume of water at 80.0^oC  = 370cm^3=370mL

V_2 = volume of water at 4^oC  = 120cm^3=120mL

T_f = final temperature of mixture = ?

T_1 = initial temperature of water = 80.0^oC

T_2 = initial temperature of water = 4^oC

Now put all the given values in the above formula, we get:

(\rho_1\times V_1)\times (T_f-T_1)=-(\rho_2\times V_2)\times (T_f-T_2)

(1.0g/mL\times 370mL)\times (T_f-80.0)^oC=-(1.0g/mL\times 120mL)\times (T_f-4)^oC

T_f=61.4^oC

Therefore, the final temperature of the mixture is 61.4^oC

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