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kolbaska11 [484]
3 years ago
5

Which word equation represents a neutralization reaction

Chemistry
2 answers:
Vlad1618 [11]3 years ago
8 0
Base+salt > acid +alkali > neutralization i think this is the reaction
Arisa [49]3 years ago
8 0

<u>Answer:</u> The word equation for neutralization reaction is given below.

<u>Explanation:</u>

Neutralization reaction is defined as the reaction in which acid reacts with base to produce a salt and water as products. The chemical equation for this reaction follows:

HX+BOH\rightarrow BX+H_2O

where,

HX is a acid

BOH is a base

BX is the salt

H_2O is the water molecule.

The word equation for the above reaction follows:

\text{Acid }+\text{ Base}\rightarrow \text{Salt }+H_2O

Hence, the word equation for the neutralization reaction is given above.

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The table shows data for two groups of plants one grown with fertilizer and the other without fetalizer was the mean height of t
Dmitrij [34]

Answer:

D is the correct answer.

Explanation:

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8 0
2 years ago
How many moles of oxygen gas (O2) are required to completely react with 10 moles of hydrogen gas (H2)?
Brums [2.3K]

Answer:

5 moles of oxygen are required.

Explanation:

Given data:

Moles of O₂ required = ?

Moles of H₂ present = 10 mol

Solution:

Chemical equation:

O₂ + 2H₂       →     2H₂O

Now we will compare the moles of oxygen and hydrogen.

                      H₂        :        O₂

                        2         :         1

                       10         :     1/2×10 = 5 mol

5 moles of oxygen are required.

3 0
3 years ago
What are the characteristics of the two systems of government in this region? Explain in detail.
Anna11 [10]

Answer:

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Explanation:

7 0
3 years ago
Equation for the for formation of ethanoic acid<br>​
vladimir2022 [97]

Answer:

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Explanation:

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5 0
3 years ago
What is the excess reactant in the combustion of 23 g of methane in the open atmosphere?
Marta_Voda [28]

Answer : The excess reactant in the combustion of methane in opem atmosphere is O_{2} molecule.

Solution : Given,

Mass of methane = 23 g

Molar mass of methane = 16.04 g/mole

The Net balanced chemical reaction for combustion of methane is,

CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)

First we have to calculate the moles of methane.

\text{ Moles of methane}=\frac{\text{ Given mass of methane}}{\text{ Molar mass of methane}} = \frac{23g}{16.04g/mole} = 1.434 moles

From the above chemical reaction, we conclude that

1 mole of methane react with the 2 moles of oxygen

and 1.434 moles of methane react to give \frac{2moles\times 1.434moles}{1moles} moles of oxygen

The Moles of oxygen = 2.868 moles

Now we conclude that the moles of oxygen are more than the moles of methane.

Therefore, the excess reactant in the combustion of methane in open atmosphere is O_{2} molecule.


6 0
3 years ago
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