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timofeeve [1]
3 years ago
9

What does it mean if two objects are in thermal equilibrium?

Chemistry
1 answer:
galina1969 [7]3 years ago
6 0

Answer:

ang hirap nixbsxbaaaaksbs

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Chemistry Lab Question: Filtration
tatyana61 [14]

For the first question, salt is soluble while sand is insoluble or not dissolvable in water. The salt should have vanished or melted, but the sand stayed noticeable or visible, making a dark brown solution probably with some sand particles caught on the walls of the container when the boiling water was put in to the mixture of salt and sand. The solubility of a chemical can be disturbed by temperature, and in the case of salt in water, the hot temperature of the boiling water enhanced the salt's capability to melt in it.

For the second question, the melted or dissolved salt should have easily made its way through the filter paper and into the second container, while the undissolved and muddy sand particles is caught on the filter paper. The size of the pores of the filter paper didn’t change. On the contrary, the size of the salt became smaller because it has been dissolved which is also the reason why it was able to go through the filter paper, while the size of the sand may have doubled or even tripled which made it harder to pass through.

6 0
3 years ago
2. Why do you think electroplated jewelleries are in demand?<br> Chemistry plsss
Ivan

Explanation:

Electroplated jeweleries are in demand because firstly they are as shiny and attractive as real jeweleries, they are also light weighted and cost effective.

8 0
2 years ago
If 40.5 J of heat is added to a 15.4 g sample of silver, what will the change in temperature be? (Specific heat of silver is 0.2
crimeas [40]

Answer:

623.7

Explanation:

40.5*15.4=623.7

4 0
3 years ago
The ph of a 0.55 m aqueous solution of hypobromous acid, hbro, at 25.0 °c is 4.48. what is the value of ka for hbro?
gladu [14]
Hbro  dissociate  as  follows
HBro--->  H+  +  BrO-
  Ka=  (H+)(BrO-)  /  HBro
 PH  =  -log (H+)
therefore (H+)  =  10^-4.48= 3.31  x10^-5
ka is  therefore= ( 3.31  x 10^-5)^2/0.55=1.99  x10^-9
6 0
3 years ago
How many milliliters of a 1.5 m h2so4 are needed to neutralize 35ml sample of a 1.5 m solution?
DochEvi [55]

Answer:

1) 17.5 mL

Explanation:

Hello,

In this case, the reaction between sulfuric acid and potassium hydroxide is:

H_2SO_4+2KOH\rightarrow K_2SO_4+2H_2O

In such a way, we notice a 1:2 molar ratio between the acid and the base, therefore, at the equivalence point we have:

2*n_{acid}=n_{base}

And in terms of concentrations and volumes:

2*M_{acid}V_{acid}=M_{base}V_{base}

Thus, we solve for the volume of acid:

V_{acid}=\frac{M_{base}V_{base}}{2*M_{acid}} =\frac{35mL*1.5M}{2*1.5M} \\\\V_{acid}=17.5mL

Best regards.

5 0
3 years ago
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