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lubasha [3.4K]
3 years ago
10

A student decreases the temperature of 556 cm^3 balloon from 278 K to 231 K. Assuming constant pressure, what should the new vol

ume of the balloon be?
Chemistry
1 answer:
julia-pushkina [17]3 years ago
6 0
In this problem, Charles' law is applicable since constant pressure with varying volume and temperature are given. Charles' law derived from ideal gas law is expressed as \frac{ V_{1} }{ T_{1} }=  \frac{ V_{2} }{ T_{2} }. Subsituting V1=556 cm3, T1=278 K and T2=231 K, V2 or the new volume is 462 cm3. 
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A solution of H2SO4(aq) with a molal concentration of 5.25 m has a density of 1.266 g/mL. What is the molar concentration of thi
Igoryamba
<span>There are a number of ways to express concentration of a solution. This includes molarity and molality. Molarity is expressed as the number of moles of solute per volume of the solution.  MOlality is expressed as moles per kg solution. 

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5 0
4 years ago
3Fe + 4H2O &lt;=&gt; Fe3O4 + 4H2
Maslowich
1)
4 molecules of Hygdrogen is produced.


2)
4 atoma of Oxygen is produced.

3)
1 molecule of Fe3O4 is formed.

4)
1mole of 3 Fe : mole of 4 H2O

3 :4
=3/4


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6 0
4 years ago
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What is the concentration of barium ions in a 1.00 M BaCl2 solution?
Luden [163]
Concentration of Ba2+ is 1.00 mol/dm3 or 1M
5 0
3 years ago
How many hydrogen molecules are in 1.2 moles of hydrogen?
nalin [4]

Answer:

7.22 x 10²³molecules

Explanation:

Given parameters:

Number of moles of hydrogen  = 1.2moles

Unknown:

Number of molecules of hydrogen  = ?

Solution:

From the concept of moles, a mole of a substance contains the Avogadro's number of particles.

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So;  1.2 moles of hydrogen  = 1.2 x  6.02 x 10²³ molecules;

                                               = 7.22 x 10²³molecules

6 0
3 years ago
How many atoms are in 1.75 mol CHCL3
vaieri [72.5K]

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So you have 1.75 mol CHC1₃ x (6.022x10²³) = 1.05385 x 10²⁴ atoms of CHCl₃

But now you have to round because of the rules of significant figures so you get 1.05 x 10²⁴ atoms of CHCl₃

6 0
3 years ago
Read 2 more answers
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