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Assoli18 [71]
3 years ago
15

A forward reaction in which adding heat decreases product formation is _________, while a forward reaction in which adding heat

increases product formation is_______.
a.) exothermic, endothermic
b.) irreversible, reversible
c.) endothermic, exothermic
d.) none of the above
Chemistry
1 answer:
ELEN [110]3 years ago
3 0
I believe the correct answer from the choices listed above is option A. <span>A forward reaction in which adding heat decreases product formation is exothermic, while a forward reaction in which adding heat increases product formation is endothermic. Exothermic would mean that heat is being released by the process while the opposite is called endothermic in which it absorbs heat.</span>
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If there was a drink that could make you younger, how much would you drink from the bottle? If for one drink you become 1 year y
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if you drink the water your not 1year old you can drink just little.

Explanation:

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6 0
2 years ago
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A compounds empirical formula is H.O. If the formula mass is 34 amu what is the molecular formula?
BigorU [14]

Answer:

C

Explanation:

1. The molecule has to have a ratio of 1 hydrogen to 1 oxygen.

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damaskus [11]

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3 0
2 years ago
A 3.452 g sample containing an unknown amount of a Ce(IV) salt is dissolved in 250.0-mL of 1 M H2SO4. A 25.00 mL aliquot is anal
SOVA2 [1]

Answer:

1,812 wt%

Explanation:

The reactions for this titration are:

2Ce⁴⁺ + 3I⁻ → 2Ce³⁺ + I₃⁻

I₃⁻ + 2S₂O₃⁻ → 3I⁻ + S₄O₆²⁻

The moles in the end point of S₂O₃⁻ are:

0,01302L×0,03428M Na₂S₂O₃ = 4,463x10⁻⁴ moles of S₂O₃⁻. As 2 moles of S₂O₃⁻ react with 1 mole of I₃⁻, the moles of I₃⁻ are:

4,463x10⁻⁴ moles of S₂O₃⁻×\frac{1molI_{3}^-}{2molS_{2}O_{3}^-} = 2,2315x10⁻⁴ moles of I₃⁻

As 2 moles of Ce⁴⁺ produce 1 mole of I₃⁻, the moles of Ce⁴⁺ are:

2,2315x10⁻⁴ moles of I₃⁻×\frac{2molCe^{4+}}{1molI_{3}^-} = 4,463x10⁻⁴ moles of Ce(IV). These moles are:

4,463x10⁻⁴ moles of Ce(IV)×\frac{140,116g}{1mol} = <em>0,0625 g of Ce(IV)</em>

As the sample has a 3,452g, the weight percent is:

0,0625g of Ce(IV) / 3,452g × 100 = <em>1,812 wt%</em>

I hope it helps!

5 0
3 years ago
An electron in a hydrogen atom relaxes to the n=4 level, emitting light of 74 THz.
Solnce55 [7]
Delta E = Ef - Ei
E = energy , h = plank constant  , v = frequency
h= 6.626 * 10 ^-34 j*s  ,  T = 10 ^ 12  , v = 74 * 10 ^12 Hz  ,  Hz = s^-1 

E = ( 6.626 * 10^ -34 j*s) ( 74 * 10 ^ 12 s^ -1 )  =   4.90 * 10 ^ -20 J
Delta E  = Ef  -  Ei
-4.90 * 10 ^ -20 J =  -2.18 * 10 ^ -18J ( 1/4 ^2 - 1/x ^2)
0.0225 = 0.0625 -  ( 1/x ^ 2)
0.225 - 0.0625 =  - 1/ x ^ 2 

- 0.0400 = - 1/x ^2    =   -1 / - 0.0400    =   x^2
25   =  x^2 
     x = 5 




6 0
3 years ago
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