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lbvjy [14]
3 years ago
14

Which group of elements ,in the periodic table, would most likely be used for electrical wires?

Chemistry
1 answer:
mash [69]3 years ago
3 0
Most electrical wiring uses copper.
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1. In an equilibrium experiment, acetic acid (which is a weak acid) is mixed with sodium acetate (a soluble salt), with methyl o
koban [17]

Explanation:

It is known that acetic acid is a weak acid. It's equilibrium of dissociation will be represented as follows.

          CH_{3}COOH(aq) + H_{2}O(l) \rightleftharpoons CH_{3}COO^{-}(aq) + H_{3}O^{+}(aq)

On the other hand, sodium acetate (CH_{3}COONa) is a salt of weak acid, that is, CH_{3}COOH and strong base, that is, NaOH. Therefore, aqueous solution of sodium acetate will be basic in nature.

Since, acetic acid is a weak acid but still it is an acid. So, when methyl orange is added in a solution of acetic acid then it given a reddish-orange color because of its acidity.

When sodium acetate is mixed into this solution then it will dissociate as follows.

            CH_{3}COO^{-}Na^{+}(aq) \rightleftharpoons CH_{3}COO^{-}(aq) + Na^{+}(aq)

As both solutions are liberating acetate ion upon dissociation. Hence, it is the common ion.

So, when more acetate ions will increase from dissociation of sodium acetate the according to Le Chatelier's principle the equilibrium will shift on left side.

As a result, there will be decrease in the concentration of hydronium ions. As a result, there will be increase in the pH of the system.

Hence, color of methyl orange will change from reddish orange to yellow. This shift in equilibrium is due to the common ion which is CH_{3}COO^{-} ion.

4 0
3 years ago
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8 0
3 years ago
Name any two<br> chemical fertilizers rich in potassium​
lbvjy [14]

Explanation:

Fertilizers are chemically synthesized plant nutrients.

Nitrogen (N), Phosphorus (P) and Potassium (K) are macronutrients and are required in large amounts by plants. So, farmers use fertilizers in order to supply these nutrients. NPK 15:15:15 , NPK 20:20:20, NPK 15:30:15 are examples of fertilizers used to supply N, P, K to crops.

3 0
3 years ago
Read 2 more answers
Which equation is used to help form the combined gas law?<br> eP, V, P, V, т.
alexdok [17]

Answer:

The combined gas law is formulated from PV/T =K.

Explanation:

The combined gas law comprises of Boyle's law, Charles's law and Gay lusaac's law. This laws were not discovered but simply put together considering other cases of ideal gas law. It states that if the amount of gas is left unchanged, the ratio between the pressure, volume, and temperature is constant.

7 0
3 years ago
Find percent yield:
saveliy_v [14]

<u>Answer:</u> The percent yield of the reaction is 91.8 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For B_5H_9 :</u>

Given mass of B_5H_9 = 4.0 g

Molar mass of B_5H_9 = 63.12 g/mol

Putting values in equation 1, we get:

\text{Moles of }B_5H_9=\frac{4g}{63.12g/mol}=0.0634mol

  • <u>For oxygen gas:</u>

Given mass of oxygen gas = 10.0 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

\text{Moles of oxygen gas}=\frac{10g}{32g/mol}=0.3125mol

The chemical equation for the reaction of B_5H_9 and oxygen gas follows:

2B_5H_9+12O_2\rightarrow 5B_2O_3+9H_2O

By Stoichiometry of the reaction:

12 moles of oxygen gas reacts with 2 moles of B_2H_5

So, 0.3125 moles of oxygen gas will react with = \frac{2}{12}\times 0.3125=0.052mol of B_2H_5

As, given amount of B_2H_5 is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

12 moles of oxygen gas produces 5 moles of B_2O_3

So, 0.3125 moles of oxygen gas will produce = \frac{5}{12}\times 0.3125=0.130moles of water

Now, calculating the mass of B_2O_3 from equation 1, we get:

Molar mass of B_2O_3 = 69.93 g/mol

Moles of B_2O_3 = 0.130 moles

Putting values in equation 1, we get:

0.130mol=\frac{\text{Mass of }B_2O_3}{69.63g/mol}\\\\\text{Mass of }B_2O_3=(0.130mol\times 69.63g/mol)=9.052g

To calculate the percentage yield of B_2O_3, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of B_2O_3 = 8.32 g

Theoretical yield of B_2O_3 = 9.052 g

Putting values in above equation, we get:

\%\text{ yield of }B_2O_3=\frac{8.32g}{9.052g}\times 100\\\\\% \text{yield of }B_2O_3=91.8\%

Hence, the percent yield of the reaction is 91.8 %

6 0
4 years ago
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