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Advocard [28]
3 years ago
7

The neutralization of H3PO4 with KOH is exothermic. H3PO4(aq)+3KOH(aq)⟶3H2O(l)+K3PO4(aq)+173.2 kJ If 60.0 mL of 0.200 M H3PO4 is

mixed with 60.0 mL of 0.600 M KOH initially at 23.43 °C, predict the final temperature of the solution, assuming its density is 1.13 g/mL and its specific heat is 3.78 J/(g·°C). Assume that the total volume is the sum of the individual volumes.
Chemistry
1 answer:
nordsb [41]3 years ago
3 0

Answer:

Final temperature of solution is 27.48^{0}\textrm{C}

Explanation:

Total volume of mixture = (60.0+60.0) mL = 120.0 mL

We know, density = (mass)/(volume)

So mass of mixture = (120.0\times 1.13)g=135.6 g

Amount of heat released per mol of H_{3}PO_{4} = \frac{(m_{mixture}\times C_{mixture}\times \Delta T_{mixture})}{n_{H_{3}PO_{4}}}

Where, m represents mass , C represents specific heat, \Delta T represents change in temperature and n is number of moles

As this reaction is an exothermic reaction therefore temperature of mixture will be higher than it's initial temperature.

Let's say final temperature of mixture is T ^{0}\textrm{C}

So, \Delta T_{mixture}=(T-23.43)^{0}\textrm{C}

Here m_{mixture}=135.6 g and C_{mixture}=3.78J/(g.^{0}\textrm{C})

Moles of H_{3}PO_{4} are added = \frac{0.200}{1000}\times 60.0moles = 0.012 moles

So, (173.2\times 10^{3})J=\frac{[(135.6g)\times (3.78J.g^{-1}.^{0}\textrm{C}^{-1})\times (T-23.43)^{0}\textrm{C}]}{0.012}

or, T = 27.48^{0}\textrm{C}

So, final temperature of solution is 27.48^{0}\textrm{C}

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Alternatively, you can go the long route and use the ideal gas law to solve for the number of moles of nitrogen given STP conditions (273 K and 1.00 atm). From PV = nRT, we can get n = PV/RT. Plugging in our values, and using 0.08206 L•atm/K•mol as our gas constant, R, we get n = (1.00)(846)/(0.08206)(273) = 37.8 moles, which confirms our answer.
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Calculate the concentration (in mol/l) of 33% by weight (33 g naoh per 100 g of solution) naoh solution. (the density of the 33%
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Molarity is measured in moles per Liter. If there are 1.35 g/mL, find out how many grams there are in a liter of solution.

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3 years ago
Ultra Clorox bleach is 6.68% sodium hypochlorite in water. How much of each is needed to make an 800.mL bottle?
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Match the prefixes. 1. 1 hex- 2. 2 eth- 3. 3 prop- 4. 4 hept- 5. 5 non- 6. 6 dec- 7. 7 pent- 8. 8 but- 9. 9 meth- 10. 10 oct-
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Hi!

All these prefixes are commonly employed in the nomenclature in organic chemistry. These are associated with organic compounds, such as alkanes, alkenes, alcohols and carboxylic acids -and these prefixes are known as word root.

<h3>The answers would be:</h3><h3>a) 1 - meth</h3>

Meth is associated with the number one, in that it signifies the presence of one carbon in the compound in concern. For instance, a one carbon alkane would be known as methane -with <em>ane </em>being indicative of the compound being an alkane, and meth denoting the presence of one carbon

<h3>b) 2 - eth</h3>

Eth is the associated prefix with the number 2 as it denotes the presence of 2 carbon atoms in the compound. For instance, the compound ethene gets it name for having 2 carbon atoms (denoted by eth). The eth acts as a prefix to the characteristic <em>ene</em> of alkenes.

<h3>c) 3 - prop</h3>

Prop is the prefix that signifies the presence of 3 carbons in the compound in concern. For instance, propanol is a three carbon alcohol, which we can tell by the name. Prop indicating three carbons in the compound, and <em>anol</em> being indicative of the compound being an alcohol.

<h3>d) 4 - but</h3>

But is associated with the number one, in that it denotes the presence of 4 carbons in the compound. For instance, a 4 carbon carboxylic acid would be known as butanoic acid -with anoic<em> </em>being indicative of the compound being a carboxylic acid, and but signifying the presence of four carbons.

<h3>e) 5 - pent</h3>

Pent is the associated prefix with the number five as it denotes the presence of five carbon atoms in the compound. For instance, the compound pentene gets it name for having five carbon atoms (denoted by pent). The pent acts as a prefix to the characteristic <em>ene</em> of alkenes.

<h3>f) 6 - hex</h3>

Hex is the prefix that signifies the presence of six carbons in the compound in question. For instance, a six carbon alkane would be known as hexane -with <em>ane </em>being indicative of the compound being an alkane, and hex denoting the presence of one carbon

<h3>g) 7 - hept</h3>

Hept is the associated prefix with the number seven as it denotes the presence of seven carbon atoms in the compound. For instance, the compound heptene gets it name for having seven carbon atoms (denoted by hept). The hept acts as a prefix to the characteristic <em>ene</em> of alkenes.

<h3>h) 8 - oct</h3>

Oct is the prefix that is associated with the number 8. This is because it denotes the presence of eight carbon atoms in the compound. For instance, the compound octane gets it name for having eight carbon atoms (denoted by oct). The oct acts as a prefix to the characteristic <em>ane</em> of alkanes.

<h3>i) 9 - non</h3>

Non is the prefix that is associated with the number nine, and thus denotes the presence of nine carbon atoms in the compound. For instance, for alkanes, a nonane would be a nine carbon atom alkane, with non denoting the presence of nine carbon atoms, and <em>ane</em> being indicative that the compound it an alkane.

<h3>j) 10- dec</h3>

Dec is the prefix that is associated with the number nine, and thus denotes the presence of nine carbon atoms in the compound. For instance, for alkanes, a decane would be an alkane with ten carbon atoms, with dec denoting the presence of ten carbon atoms, and <em>ane</em> being indicative that the compound it an alkane.

<h3>Hope this helps!</h3>
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