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andreev551 [17]
3 years ago
11

When a hydrochloric acid solution is combined with a potassium hydroxide solution, an acid-base reaction occurs Write a balanced

molecular equation for this reaction. Express your answer as a chemical equation. Identify all of the phases in your answer..
Chemistry
1 answer:
Svetach [21]3 years ago
7 0

Answer:

HCl(aq) + KOH(aq) ⇒ KCl(aq) + H₂O(l)

Explanation:

Hydrochloric acid is an acid because it releases H⁺ in an aqueous solution.

Potassium hydroxide is a base because it releases OH⁻ in an aqueous solution.

When an acid reacts with a base they form a salt and water. This is a neutralization reaction. The neutralization reaction between hydrochloric acid and potassium hydroxide is:

HCl(aq) + KOH(aq) ⇒ KCl(aq) + H₂O(l)

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What occurs in endothermic reactions?
kondaur [170]

Answer:

a) Heat energy is absorbed.

Explanation:

What occurs in endothermic reactions?

a) Heat energy is absorbed. YES. This is the definition of an endothermic reaction.

b) Water is produced. NO. Endothermic refers to the heat absorbed and not to the products formed.

c) Oxygen is produced. NO. Endothermic refers to the heat absorbed and not to the products formed.

d) Heat energy is released. NO. This is known as an exothermic reaction.

8 0
2 years ago
You are provided with a compound fertilizer, 40-15-10. Calculate the quantity of fertilizer to add to a one hectare field to sup
Kryger [21]

Answer:

The correct answer is a) 300 Kg b) 225 Kg c) 400 Kg and d) 600 Kg.

Explanation:

Based on the given value, that is. 40-15-10 shows the compositions of nitrogen, phosphorus, and potassium found in the compound fertilizer in the form of percentage.  

41 percent nitrogen is equal to 40/100 = 0.4, 15 percent phosphorus is equal to 15/100 = 0.15, and 10 percent potassium is equal to 10/100 = 0.1.  

To find the amount of fertilizer, which is needed per hectare to supply the different nutrients per hectare, there is a need to divide the quantity to be supplied by the percentage composition, now the values comes out as:  

a) At 120 Kg per ha, the required nitrogen will be 120/0.4 = 300 Kilogram of fertilizer.  

b) At 90 kg per ha, the required nitrogen will be 90/0.4 = 225 Kilograms of fertilizer.  

c) At 60 kg per ha, the required phosphorus will be 60/0.15 = 400 Kilograms of fertilizer.  

d) At 60 Kg per ha, the required potassium will be 60/0.1 = 600 Kilograms of fertilizer.  

5 0
2 years ago
Energía necesaria para arrancar un electrón de un átomo neutro​
zalisa [80]
La energía de ionización o potencial de ionización, es la energía necesaria para extraer un electrón de un átomo neutro. Por ejemplo, cuando un átomo de cloro neutro en forma gaseosa recoge un electrón para formar un ion Cl-, libera una energía de 349 kJ/mol o 3,6 eV/átomo.

Nomas ponlo En tus palabras
5 0
3 years ago
The primary gas in a volcano is: water vapor carbon dioxide sulfur dioxide nitrogen
Snezhnost [94]

Answer:

Water vapor

Explanation:

The magma consist of dissolved gases when these gases produce the force the volcanic eruption take place. The volcanic gases comes out and their volume is increased tremendously. The gases present in volcano are listed below:

The volcanic gases consist of water vapors, carbon dioxide and sulfur.

These three re the primary gases but the water is present in higher amount.

The percentage of water is 60%.

The carbon dioxide present in 10-40%.

Other gases present in valcano are nitrogen, argon, helium, neon methane and hydrogen.

7 0
2 years ago
Read 2 more answers
13.50 grams of Pb(NO3)4 are dissolved in enough water to make 250 mL of solution. What is the molar its of the resulting solutio
likoan [24]
Data:
M (molarity) = ? (M or Mol/L)
m (mass) = 13.50 g
V (volume) = 250 mL → 0.25 L
MM (Molar Mass) of Lead(IV) Nitrate Pb(NO_3)_4
Pb = 1*207 = 207 amu
N = (1*14)*4 = 14*4 = 56 amu
O = (3*16)*4 = 48*4 = 192 amu
------------------------------------
MM of Pb(NO_3)_4 = 207+56+192 = 455 g/mol

Formula:
M =  \frac{m}{MM*V}

Solving:
M = \frac{m}{MM*V}
M =  \frac{13.50}{455*0.25}
M =  \frac{13.50}{113.75}
M = 0.118681318...\:\:\to\:\:\boxed{\boxed{M \approx 0.119\:Mol/L}}\end{array}}\qquad\quad\checkmark

Answer:
<span>B. 0.119 M</span>
5 0
3 years ago
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