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KIM [24]
3 years ago
8

In neutralization reaction _______ and ________ is produced.

Chemistry
2 answers:
dezoksy [38]3 years ago
4 0

Answer:

In neutralization reaction water and a salt is produced.

Oxana [17]3 years ago
4 0

salt and water

to explain i can use an example

this is the reaction between an acid and a base

NaOH+HCl-->NaCl+H2O

we get NaCl which is a salt and H2O which is water

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How does water affect the growth of a plant?
irina1246 [14]

Answer:

Water transports important nutrients throughout the plant.

Explanation:

Much like blood in an animal, water is necessary for a plants survival, as it transports important nutrients from the soil all through the plant's stems and leaves.

3 0
4 years ago
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Write IUPAC names of the following compound : (1) CH3-CH- CH - CH, (2) CH3- C- CH2 - CH - COOH OH CH,​
lawyer [7]

\large \displaystyle\sf \prod^{ \infty }_{n = 1} \bigg( \frac{25}{777} \bigg)^{ {( - 1)}^{n} \bigg( \frac{ \tan^{4n} (x) + 1}{ \tan ^{2n} (x) } \bigg) }

5 0
3 years ago
PLEASE HELP!!!
poizon [28]

The net ionic equation of the reaction is:

  • Pb²⁺ (aq) +2 I⁻ (aq) → PbI2(s)

<h3>What are net ionic equations?</h3>

Net ionic equations are equations where ions which remain in solution known as spectator ions are not shown in the equation. Only ions involved in formation of product are shown.

In the given equation, sodium and nitrate ions are spectator ions.

The net ionic equation of the given reaction is as follows:

  • Pb²⁺ (aq) +2 I⁻ (aq) → PbI2(s)

In conclusion, spectator ions are not shown in net ionic equation.

Learn more about net ionic equations at: brainly.com/question/19705645
#SPJ1

6 0
2 years ago
If 0.25 liters of solution contains 0.030 moles of KOH what is the molarity of the solution?
sdas [7]

Answer:

The molarity of the solution is 0,12 M.

Explanation:

We calculate the molarity, which is a concentration measure that indicates the moles of solute (in this case KOH) in 1000ml of solution (1 liter):

0,25 L solution----- 0,030 moles of KOH

1 L solution----x= (1 L solution x 0,030 moles of KOH)/0,25 L solution

x= 0,12 moles of KOH ---> <em>The solution is 0,12 M</em>

7 0
3 years ago
For the following reaction, 5.22 grams of aluminum oxide are mixed with excess sulfuric acid. The reaction yields 12.9 grams of
Mamont248 [21]

Answer:

m_{Al_2(SO_4)_3}=17.5gAl_2(SO_4)_3

Explanation:

Hello,

In this case, the undergoing balanced chemical reaction is:

Al_2O_3(s)+3H_2SO_4(aq)\rightarrow Al_2(SO_4)_3(aq)+3H_2O(l)

Thus, as 5.22 grams of aluminium oxide reacts, the required yielded amount of aluminium sulfate results:

m_{Al_2(SO_4)_3}=5.22gAl_2O_3*\frac{1molAl_2O_3}{102gAl_2O_3}*\frac{1molAl_2(SO_4)_3}{1molAl_2O_3}*\frac{342gAl_2(SO_4)_3}{1molAl_2(SO_4)_3} \\\\m_{Al_2(SO_4)_3}=17.5gAl_2(SO_4)_3

Moreover, the percent yield is:

Y=\frac{12.9g}{17.5g} *100\%=73.7\%

Best regards.

6 0
4 years ago
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