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Ber [7]
2 years ago
6

Write insoluble or soluble solution for the following mixtores12345​

Chemistry
1 answer:
dimaraw [331]2 years ago
6 0

<em>Gasoline</em><em> </em><em>in</em><em> </em><em>water</em><em> </em><em>=</em><em> </em><em>insoluble</em><em> </em>

<em>acetone</em><em> </em><em>in</em><em> </em><em>nail</em><em> </em><em>polished</em><em> </em><em>=</em><em> </em><em>soluble</em><em> </em>

<em>salt</em><em> </em><em>in</em><em> </em><em>alcohol</em><em> </em><em>=</em><em> </em><em>soluble</em><em> </em>

<em>oil</em><em> </em><em>in</em><em> </em><em>vinegar</em><em> </em><em>=</em><em> </em><em>insoluble</em><em> </em>

<em>tawas</em><em> </em><em>in</em><em> </em><em>water</em><em> </em><em>=</em><em> </em><em>soluble</em><em>.</em><em>.</em><em>.</em>

<em>Sorry</em><em> </em><em>if</em><em> </em><em>i</em><em> </em><em>am</em><em> </em><em>incorrect</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em>

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Which would be expected to have the higher boiling point, t-butyl alcohol (2-methyl-2-propanol) or n-butyl alcohol (1-butanol)?
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D=853g<br> 310 cm² <br> find the density
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In the electrolysis of molten libr, which product forms at the cathode?
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A student was assigned the task of determining the identity of an unknown liquid. The student weighed a clean, dry 250-mL Erlenm
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Answer:

248.4 mL

Explanation:

Erlenmeyer = 78.649 g

Erlenmeyer + Water = 327.039 g

Water = (Erlenmeyer + Water) - Erlenmeyer

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4 0
3 years ago
You have a red ballooe that has a volume of 20 liters, a pressure of 1.5 atmospheres and a temperature of 28 C. What is the volu
ki77a [65]

Explanation:

The given data is as follows.

 P_{1} = 1.5 atm,     V_{1} = 20 L,  T_{1} = (28 + 273) K = 301 K

   P_{2} = 5 atm,     V_{2} = ?,  T_{2} = (50 + 273) K = 323 K

Formula to calculate the volume will be as follows.

         \frac{P_{1} \times V_{1}}{T_{1}} = \frac{P_{2} \times V_{2}}{T_{2}}

Putting the given values into the above formula as follows.

        \frac{P_{1} \times V_{1}}{T_{1}} = \frac{P_{2} \times V_{2}}{T_{2}}  

        \frac{1.5 atm \times 20 L}{301 K} = \frac{5 atm \times V_{2}}{323 K}  

                 V_{2} = 0.64 L

Thus, we can conclude that the change in volume of the balloon will be 0.64 L.

6 0
3 years ago
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