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Aliun [14]
3 years ago
14

What are the balanced chemical equations for Hydrochloric acid and potassium hydroxide?

Chemistry
1 answer:
-Dominant- [34]3 years ago
7 0
HCl(aq) + KOH(s) --> KCl(aq) + H2O(l)
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Yasmin's teacher asks her to make a supersaturated saline solution. Her teacher tells her that the solubility of the salt is 360
Aleks [24]

Answer:

She can add 380 g of salt to 1 L of hot water (75 °C) and stir until all the salt dissolves. Then, she can carefully cool the solution to room temperature.  

Explanation:

A supersaturated solution contains more salt than it can normally hold at a given temperature.

A saturated solution at 25 °C contains 360 g of salt per litre, and water at 70 °C can hold more salt.

Yasmin can dissolve 380 g of salt in 1 L of water at 70 °C. Then she can carefully cool the solution to 25 °C, and she will have a supersaturated solution.

B and D are wrong. The most salt that will dissolve at 25 °C is 360 g. She will have a saturated solution.

C is wrong. Only 356 g of salt will dissolve at 5 °C, so that's what Yasmin will have in her solution at 25 °C. She will have a dilute solution.

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3 years ago
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Directions Now that the lab is complete, it is time to write your lab report. The purpose of this guide is to help you write a c
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Answer:

foot

Explanation:

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2 years ago
Knowing what you know about the nucleus and the subatomic particles in the nucleus, what do you think is the charge of the nucle
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What I think is the charge of nucleus is the proton+neutron
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The theoretical yield of a reaction is the amount of product obtained if the limiting reactant is completely converted to produc
Elis [28]

Answer: 9.9 grams

Explanation:

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

a) moles of H_2

\text{Number of moles}=\frac{8.150g}{2g/mol}=4.08moles

b) moles of C_2H_4

\text{Number of moles}=\frac{9.330g}{28g/mol}=0.33moles

H_2(g)+C_2H_4(g)\rightarrow C_2H_6(g)

According to stoichiometry :

1 mole of C_2H_4 combine with 1 mole of H_2

Thus 0.33 mole of C_2H_4 will combine with =\frac{1}{1}\times 0.33=0.33 mole of H_2

Thus C_2H_4 is the limiting reagent as it limits the formation of product.

As 1 mole of C_2H_4 give =  1 mole of C_2H_6

Thus 0.33 moles of C_2H_4 give =\frac{1}{1}\times 0.33=0.33moles  of C_2H_6

Mass of C_2H_6=moles\times {\text {Molar mass}}=0.33moles\times 30g/mol=9.9g

Thus theoretical yield (g) of C_2H_6 produced by the reaction is 9.9 grams

7 0
3 years ago
Label the
alexandr402 [8]

Answer:

0.1 m HCI

0.001m HCI

0.00001m HCI

Distilled Water

0.0001m NaOH

0.001m NaOH

0.1m NaOH

Explanation:

7 0
3 years ago
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