Answer : The concentration of and at equilibrium is, 0.0031 M and 0.0741 M respectively.
Explanation : Given,
Moles of = 0.166 mol
Volume of solution = 2.15 L
First we have to calculate the concentration of
Now we have to calculate the concentration of and at equilibrium.
Initial conc. 0.0772 0 0
At eqm. 0.0772-x x x
The expression for equilibrium constant is:
By solving the term, we get the value of 'x'.
x = 0.0741
Thus,
The concentration of at equilibrium = (0.0772-x) = (0.0772-0.0741) = 0.0031 M
The concentration of at equilibrium = x = 0.0741 M
<span>decomposition of SrCO3 to SrO and CO2 =change in mass
moles of CO2 =(1.850 g - 1.445 g).
</span>Mass of <span>C<span>O2</span></span><span> in mixture: 1.850-1.445 = 0.405g
</span>0.405g/44.01 g/mol <span>C<span>O2</span></span><span> = 0.0092 moles </span><span>C<span>O2</span></span><span>.
</span>ratio of <span>C<span>O2</span></span><span> to SrO in Sr</span><span>C<span>O3</span></span><span> is 1:1
</span><span> mass ratio = 1.358/1.850 = 0.7341, </span>
or 73.41% Sr<span>C<span>O3</span></span><span>.
</span>hope this helps
694,563,239 rounded to the nearest thousand is 694,563.
It's because the first digit from the right is for ones, second for tens, third for hundreds and fourth for thousands and that's the one that we should take a closer look at. You can round it either to 3 or 4, depends on the digit of hundreds. In this case 3239 is clearly closer to 3000 than 4000, that's why we round it to 694,563, not 694,564.
The answer is c .................