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Sindrei [870]
3 years ago
11

HELP PLEASE (If you cant answer all, that okay! First to answer all will be mark BRAINLIEST!)

Chemistry
1 answer:
Yakvenalex [24]3 years ago
4 0

Answer:

<em>Hope this helps :)</em>

Explanation:

Antoine Lavoisier- used patterns to predict undiscovered elements

John Newlands- Arranged elements according to their atomic

Johann Wolfgang- created groups of three elements, each based on Johann Wolfgang Dobereiner

Dmitri Mendeleev- divided elements into four categories

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The reaction CO2(g) → C(s) + O2(g) is a decomposition reaction.<br><br> True<br> False
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  The reaction CO₂(g)  → C(s) +O₂(g)  is  a decomposition  reaction is true

Explanation

 A Decomposition reaction is a reaction in which involve breaking down  a compound into   two or more  simpler  substances.

 The reaction above is a decomposition reaction since CO₂    molecule is broken down  into C (an  element) and O₂  (a  molecule).

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Name Formula: Lead (II) Oxide<br> F) Pb2O<br><br> G) PbO2<br><br> H) PbO<br><br> I) Pb2O2
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How will the price of a product impact on the profits of the entrepreneur​
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The higher your price, the less volume you have to produce for a given dollar amount of profit! Even a small price increase can generate significant additional profit. ... When a business comes out with a new product or service and they are the first to market, they may be able to charge high prices initially.

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If the ph of a solution is decreased from ph 8 to ph 6 it means that the
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Explanation:

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Therefore, the solution becomes basic in nature.

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Given the following data:
bagirrra123 [75]

176.0 \; \text{kJ} \cdot \text{mol}^{-1}

As long as the equation in question can be expressed as the sum of the three equations with known enthalpy change, its \Delta H can be determined with the Hess's Law. The key is to find the appropriate coefficient for each of the given equations.

Let the three equations with \Delta H given be denoted as (1), (2), (3), and the last equation (4). Let a, b, and c be letters such that a \times (1) + b \times (2) + c \times (3) = (4). This relationship shall hold for all chemicals involved.

There are three unknowns; it would thus take at least three equations to find their values. Species present on both sides of the equation would cancel out. Thus, let coefficients on the reactant side be positive and those on the product side be negative, such that duplicates would cancel out arithmetically. For instance, 3 + (-1) = 2 shall resemble the number of \text{H}_2 left on the product side when the second equation is directly added to the third. Similarly

  • \text{NH}_4 \text{Cl} \; (s): -2 \; a = 1
  • \text{NH}_3\; (g): -2 \; b = -1
  • \text{HCl} \; (g): 2 \; c = -1

Thus

a = -1/2\\b = 1/2\\c = -1/2 and

-\frac{1}{2} \times (1) + \frac{1}{2} \times (2) - \frac{1}{2} \times (3)= (4)

Verify this conclusion against a fourth species involved- \text{N}_2 \; (g) for instance. Nitrogen isn't present in the net equation. The sum of its coefficient shall, therefore, be zero.

a + b = -1/2 + 1/2 = 0

Apply the Hess's Law based on the coefficients to find the enthalpy change of the last equation.

\Delta H _{(4)} = -\frac{1}{2} \; \Delta H _{(1)} + \frac{1}{2} \; \Delta H _{(2)} - \frac{1}{2} \; \Delta H _{(3)}\\\phantom{\Delta H _{(4)}} = -\frac{1}{2} \times (-628.9)+ \frac{1}{2} \times (-92.2) - \frac{1}{2} \times (184.7) \\\phantom{\Delta H _{(4)}} = 176.0 \; \text{kJ} \cdot \text{mol}^{-1}

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3 years ago
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