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3 years ago

Determine the [OH⁻] concentration in a 0.344 M Ca(OH)₂ solution.

Chemistry

1 answer:

slavikrds [6]3 years ago

8 0

Answer:

$0.688M$

Explanation:

Hello,

In this case, it is widely acknowledged that strong bases usually correspond to those formed with metals in groups IA and IIA which have relatively high activity and reactivity, therefore, when they are dissolved in water the following dissociation reaction occurs (for calcium hydroxide):

$Ca(OH)_2\rightarrow Ca^{2+}+2OH^-$

In such a way, for the same volume, we can compute the concentration of hydroxyl ions by simple stoichiometry (1:2 molar ratio):

$0.344\frac{molCa(OH)_2}{L}*\frac{2molOH^-}{1molCa(OH)_2} \\\\0.688\frac{mol OH^-}{L}$

Or simply:

$0.688M$

Regards.

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3 years ago

Read 2 more answers

Be sure to answer all parts. Consider the reaction N2(g) + 3H2(g) → 2NH3(g)ΔH o rxn = −92.6 kJ/mol If 4.0 moles of N2 react with

Sphinxa [80]

Answer : The work done is, $1.98\times 10^4J$

Explanation :

The given balanced chemical reaction is:

$N_2(g)+3H_2(g)\rightarrow 2NH_3(g)$

When 4 moles of $N_2$ react with 12 moles of $H_2$ then it gives 8 moles of $NH_3$

First we have to calculate the change in moles of gas.

Moles on reactant side = Moles of $N_2$ + Moles of $H_2$

Moles on reactant side = 4 + 12 = 16 moles

Moles on product side = Moles of $NH_3$

Moles on reactant side = 8 moles

Change in moles of gas = 16 - 8 = 8 moles

Now we have to calculate the change in volume of gas.

Using ideal gas equation:

$PV=nRT$

where,

P = Pressure of gas = 1.0 atm

V = Volume of gas = ?

n = number of moles of gas = 8 mole

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of gas = $25^oC=273+25=298K$

Putting values in above equation, we get:

$1.0atm\times V=8mole\times (0.0821L.atm/mol.K)\times 298K$

$V=195.7L$

As the number of moles of gas decreased. So, the volume also deceased. Thus, the volume of gas will be, -195.7 L

Now we have to calculate the work done.

Formula used :

$w=-p\Delta V$

where,

w = work done

p = pressure of the gas = 1.0 atm

$\Delta V$ = change in volume = -195.7 L

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

$w=-p\Delta V$

$w=-(1.0atm)\times (-195.7L)$

$w=195.7L.artm=195.7\times 101.3J=19824.41J=1.98\times 10^4J$

conversion used : (1 L.atm = 101.3 J)

Thus, the work done is, $1.98\times 10^4J$

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3 years ago

When a 3.00 grams sample of a compound containing only c, h, and o was completely burned, 1.17 grams of h2o and 2.87 grams of co

SVEN [57.7K]

Answer:- $CH_2O_2$

Solution:- From given masses of carbon dioxide and water we could calculate the moles that helps to calculate the moles of C and H.

Molar mass of carbon dioxide = 44 gram per mol

molar mass of water = 18.02 gram per mol

From given info, combustion of compound gives 1.17 grams of water and 2.87 grams of carbon dioxide. Let's calculate the moles of these:

$1.17gH_2O(\frac{1mol}{18.02g})$

= $0.0649molH_2O$

Similarly, $2.87gCO_2(\frac{1mol}{44g})$

= $0.0652molCO_2$

One mol of water has two moles of H. So, the moles of H would be two times the moles of water as calculated above.

So, moles of H = 2* 0.0649 = 0.1298 mol

One mol of carbon dioxide contains one mol of C. So, the moles of C would be equal to the moles of carbon dioxide calculated above.

moles of C = 0.0652 mol

Let's convert the moles of H and C to grams so that we could calculate the amount of oxygen present in the sample as:

grams of H in sample = 1.008 x 0.1298 = 0.1308 g

grams of C in sample = 12*0.0652 = 0.7824 g

If we subtract the sum of the masses of C and H from sample mass then it would give as the mass of oxygen since the sample has only C, H and O.

mass of O in sample = 3.00g - (0.1308 g + 0.7824 g)

= 3.00 g - 0.9132 g

= 2.0868 g

Let's convert these grams of oxygen to moles on dividing by it's atomic mass as:

$2.0868gO(\frac{1mol}{15.999g})$

= 0.130 mol O

Now, we have the moles of all the three atoms and we know that an empirical formula is the simplest whole number ratio of the moles of atoms. So, let's calculate the ratio. For this, we divide the moles of each by the least one of them.Looking at the moles, the least value is for carbon. So, let's divide the moles of each by the moles of C as:

C = $\frac{0.0652}{0.0652}$ = 1

H = $\frac{0.1298}{0.0652}$ = 2

O = $\frac{0.130}{0.0652}$ = 2

The ratio of C, H and O is 1:2:2. So, the simplest formula of the compound is $CH_2O_2$ .

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3 years ago

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