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Delvig [45]
3 years ago
7

2 HCI → H2 + Cl2 Endothermic or exothermic

Chemistry
1 answer:
Oksana_A [137]3 years ago
6 0

Answer:

Exothermic

Explanation:

I think it is exothermic because heat is being released

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Indicate the number of unpaired electrons for following: [noble gas]ns2np5
Fantom [35]
1 unpaired electron. 
4 0
3 years ago
Why were the Spanish interested in conquering Peru?
mr_godi [17]

Answer:

531–1532 – Pizarro's third voyage to Peru. Spaniards form a bond with the Natives (Huancas, Chankas, Cañaris and Chachapoyas) who were under the oppression of the Inca Empire, and Pizarro includes them among his troops to face the Incas. Atahualpa is captured by Spanish.

Explanation:  Is this right ?

if it is please give thanks

6 0
2 years ago
Write the scientific term.
Reil [10]

Answer:  Solubility.

Explanation:

Solubility is defined as the maximum amount of solute dissolved per 100 g of the solvent at a certain fixed temperature to form a saturated solution.

STP condition is Standard Temperature and Pressure condition which is  temperature of 273 K and pressure of 1 atm.

Thus the scientific term for "the number of grams of solute dissolved in 100 g of the solvent to form a saturated solution at STP​" is called as Solubility.

6 0
2 years ago
Consider the following reaction:
adell [148]

Answer:

1. d[H₂O₂]/dt = -6.6 × 10⁻³ mol·L⁻¹s⁻¹; d[H₂O]/dt = 6.6 × 10⁻³ mol·L⁻¹s⁻¹

2. 0.58 mol

Explanation:

1.Given ΔO₂/Δt…

    2H₂O₂     ⟶      2H₂O     +     O₂

-½d[H₂O₂]/dt = +½d[H₂O]/dt = d[O₂]/dt  

d[H₂O₂]/dt = -2d[O₂]/dt = -2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = -6.6 × 10⁻³mol·L⁻¹s⁻¹

 d[H₂O]/dt =  2d[O₂]/dt =  2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ =  6.6 × 10⁻³mol·L⁻¹s⁻¹

2. Moles of O₂  

(a) Initial moles of H₂O₂

\text{Moles} = \text{1.5 L} \times \dfrac{\text{1.0 mol}}{\text{1 L}} = \text{1.5 mol }

(b) Final moles of H₂O₂

The concentration of H₂O₂ has dropped to 0.22 mol·L⁻¹.

\text{Moles} = \text{1.5 L} \times \dfrac{\text{0.22 mol}}{\text{1 L}} = \text{0.33 mol }

(c) Moles of H₂O₂ reacted

Moles reacted = 1.5 mol - 0.33 mol = 1.17 mol

(d) Moles of O₂ formed

\text{Moles of O}_{2} = \text{1.33 mol H$_{2}$O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{2 mol H$_{2}$O}_{2}} = \textbf{0.58 mol O}_{2}\\\\\text{The amount of oxygen formed is $\large \boxed{\textbf{0.58 mol}}$}

8 0
3 years ago
Suppose the flask is already at equilibrium but then the volume of the reaction flask is then reduced what will happen
sveta [45]

1) Answer is: c) The reaction will proceed right.

Balanced chemical reaction: N₂(g) + 3H₂(g) ⇄ 2NH₃(g) ΔH = +92 kJ.

Reducing the volume of the system increase the partial pressures of the products and reactants.

With a pressure increase due to a decrease in volume, the side of the equilibrium with fewer moles is more favorable, there are 4 moles at the left side (three moles of hydrogen and one mole of nitrogen) and 2 moles (ammonia) at the right side of the reaction.

2) Answer is: d) The partial pressure of ammonia will increase.

This reaction is endothermic (enthalpy is higher than zero), which means that heat is added.

According to Le Chatelier's principle when the reaction is endothermic heat is included as a reactant and when the temperature increased, the heat of the system increase, so the system consume some of that heat by shifting the equilibrium to the right,  producing more ammonia.

8 0
3 years ago
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