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padilas [110]
3 years ago
6

Is toothpaste suspension,colloid,solution?​

Chemistry
2 answers:
natima [27]3 years ago
8 0

Answer:

Collid

Explanation:

Toothpaste is a colloid, because it's part solid and part liquid. ... A colloid is a heterogeneous mixture of two substances of different phases. Shaving cream and other foams are gas dispersed in liquid. Jello, toothpaste, and other gels are liquid dispersed in solid.

vichka [17]3 years ago
5 0

Answer:

For me,its Colloid..

But i have explanation..

Explanation:

Toothpaste is neither a suspension or a solution. Toothpaste does not have a uniform composition because you can see (and feel) small particles...

But actually,I think it's COLLOID..

Well, hope it helps you..?..

Just correct me if i'm wrong or something..

(◍•ᴗ•◍)

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What volume of methane gas at 237 K and 101.33 kPa do you have when the volume is decreased to 0.50 L, with a temperature of 300
Alexxandr [17]

Answer: A volume of 0.592 L of methane gas is required at 237 K and 101.33 kPa when the volume is decreased to 0.50 L, with a temperature of 300 K and a pressure of 151.99 kPa.

Explanation:

Given: T_{1} = 237 K,   P_{1} = 101.33 kPa,      V_{1} = ?

T_{2} = 300 K,      P_{2} = 151.99 kPa,        V_{2} = 0.50 L

Formula used is as follows.

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

Substitute the values into above formula as follows.

\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\frac{101.33 kPa \times V_{1}}{237 K} = \frac{151.99 kPa \times 0.50 L}{300 K}\\V_{1} = 0.592 L

Thus, we can conclude that a volume of 0.592 L of methane gas is required at 237 K and 101.33 kPa when the volume is decreased to 0.50 L, with a temperature of 300 K and a pressure of 151.99 kPa.

8 0
3 years ago
What is the molarity of SO4^2- in a solution prepared by mixing 2.17 g of alum, KAl(SO4)2•12H2O, with 175 mL of water? The molec
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<u>Answer:</u> The concentration of sulfate ions in the solution is 0.0522 M

<u>Explanation:</u>

To calculate the the molarity of solution:, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

Given mass of alum = 2.17 g

Molar mass of alum = 478.39 g/mol

Volume of solution = 175 mL

Putting values in above equation, we get:

\text{Molarity of solution}=\frac{2.17\times 1000}{474.39\times 175}\\\\\text{Molarity of solution}=0.0261M

The chemical equation for the ionization of alum follows:

KAl(SO_4)_2.12H_2O\rightarrow K^++Al^{3+}+2SO_4^{2-}+12H_2O

1 mole of alum produces 1 mole of potassium ions, 1 mole of aluminium ions, 2 moles of sulfate ions and 12 moles of water

So, concentration of sulfate ions = (2\times 0.0261)=0.0522M

Hence, the concentration of sulfate ions in the solution is 0.0522 M

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