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tester [92]
3 years ago
5

A student dissolves 13.8g of sodium chloride in of water in a well-insulated open cup. He then observes the temperature of the w

ater fall from 20°C to 19.1° over the course of 5.1 minutes. Use this data, and any information you need from the ALEKS Data resource, to answer the questions below about this reaction:
NaCl(s) → Na+ (aq) + Cl-(aq)

You can make any reasonable assumptions about the physical properties of the solution. Be sure answers you calculate using measured data are rounded to the correct number of significant digits. Note for advanced students: it's possible the student did not do the experiment carefully, and the values you calculate may not be the same as the known and published values for this reaction.

Is this reaction exothermic, endothermic or neither?
If you said the reaction was exothermoc or endothermic, calculate the amount of heat that was released or absorved by the reaction in this case.
Calculate the reaction enthalpy per mole.
Chemistry
1 answer:
Firdavs [7]3 years ago
3 0

Answer:

Exothermic and H=-1.55x10^6

Explanation:

Because the water is changing it is exothermic and the h is also exothermic

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As the volume of a gas at constant temperature did decreased,the pressure increases because
frosja888 [35]

Answer:

boyles law

Explanation:

volume is inversely proportional to pressure

7 0
2 years ago
The equation represents the combustion of sucrose. C12H22O11 + 12O2 12CO2 + 11H2O If there are 10.0 g of sucrose and 8.0 g of ox
Vaselesa [24]
<span>0.0292 moles of sucrose are available. First, lookup the atomic weights of all involved elements Atomic weight Carbon = 12.0107 Atomic weight Hydrogen = 1.00794 Atomic weight Oxygen = 15.999 Now calculate the molar mass of sucrose 12 * 12.0107 + 22 * 1.00794 + 11 * 15.999 = 342.29208 g/mol Divide the mass of sucrose by its molar mass 10.0 g / 342.29208 g/mol = 0.029214816 mol Finally, round the result to 3 significant figures, giving 0.0292 moles</span>
7 0
3 years ago
The atomic masses of 151eu and 153eu are 150.919860 and 152.921243 amu, respectively. The average atomic mass of europium is 151
vitfil [10]

Answer:-  The natural abundance of ^1^5^1_E_u is 0.478 or 47.8% and ^1^5^3_E_u is 0.522 or 52.2% .

Solution:- Average atomic mass of an element is calculated from the atomic masses of it's isotopes and their abundances using the formula:

Average atomic mass = mass of first isotope(abundance) + mass of second isotope(abundance)

We have been given with atomic masses for ^1^5^1_E_u and ^1^5^3_E_u as 150.919860 and 152.921243 amu, respectively.  Average atomic mass of Eu is 151.964 amu.

Sum of natural abundances of isotopes of an element is always 1. If we assume the abundance of ^1^5^1_E_u as n then the abundance of ^1^5^3_E_u would be 1-n .

Let's plug in the values in the formula:

151.964=150.919860(n)+152.921243(1-n)

151.964=150.919860n+152.921243-152.921243n

on keeping similar terms on same side:

151.964-152.921243=150.919860n-152.921243n

-0.957243=-2.001383n

negative sign is on both sides so it is canceled:

0.957243=2.001383n

n=\frac{0.957243}{2.001383}

n=0.478

The abundance of ^1^5^1_E_u is 0.478 which is 47.8%.  

The abundance of ^1^5^3_E_u is = 1-0.478

= 0.522 which is 52.2%

Hence, the natural abundance of ^1^5^1_E_u is 0.478 or 47.8% and ^1^5^3_E_u is 0.522 or 52.2% .


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interstitium

#keeponlearning

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