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ikadub [295]
2 years ago
13

How many mL of 0.506 M HClare needed to dissolve 9.85 g of BaCO3?

Chemistry
1 answer:
Blababa [14]2 years ago
8 0

Answer:

197mL of 0,506M HCl

Explanation:

The reaction of HCl + BaCO₃ is:

BaCO₃(s) + 2HCl → BaCl₂(aq) + CO₂ + H₂O.

The moles of BaCO₃ in 9,85 g are:

9,85 g of BaCO₃ × \frac{1mol}{197,34 g} = <em>0,0499 moles of BaCO₃</em>

As 1 mol of BaCO₃ reacts with two moles of HCl, for a complete reaction of BaCO₃ to dissolve this compound in water you need:

0,0499 moles of BaCO₃ × \frac{2molHCl}{1molBaCO_{3}} =<em> 0,0998 moles of HCl</em>

If you have a 0,506M HCl, you need to add:

0,0998 moles of HCl× \frac{1L}{0,506moles} = 0,197 L ≡ 197mL

I hope it helps!

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Answer:

+ 636 KJ

Explanation:

We want to arrive to the equation

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by manipulating algebraically the first four  given equations.

We notice the first one has our product H2CO(g) as a reactant. This indicates we must take the inverse of that equation. Also we need 6 mol of H2CO(g), thus it also needs to be multiplied by 6

6 CO2(g) + 6 H2O(g)  ------->6H2CO(g) + 6O2(g)  ΔH °comb = + 572.9 KJ/x 6

Now we want C6H12O6(s) as a reactant and it  is a product in the second one, therefore lets reverse it

C6H12O6(s)  -------> 6 C(s) + 6 H2(g) + 3 O2(g)   ΔH ° f = + 1274.4 KJ/mol

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6 H2(g) + 3 O2(g) -----------> 6H2O(g) ΔH ° f = - 285.8 KJ/mol x 6

then lets add them to get ΔH ° rxn:

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+ C6H12O6(s)  -------> 6 C(s) + 6 H2(g) + 3 O2(g)      ΔH ° f = + 1274.4 KJ

+ 6C(s) + 6O2(g) ---------> 6 CO2(g)                            ΔH ° f = - 2361.0 KJ

+6 H2(g) + 3 O2(g) -----------> 6H2O(g)                      ΔHº f  = - 1714.8 KJ

<u>                                                                                                                            </u>

C6H12O6(s) ---------> 6 H2CO(g)  

ΔH ° rxn =  3437.4 + 1274.4 - 2361.0 - 1714.8 =  636 KJ

8 0
3 years ago
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The general electronic configuration of transition element is (n−1)d1−10ns1−2.

<h3>What are the transition element in period 6?</h3>

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