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sertanlavr [38]
3 years ago
15

Which of the following is a solution? Dirt An alloy Wood Plastic

Chemistry
2 answers:
labwork [276]3 years ago
5 0
The answer would be dirt
malfutka [58]3 years ago
4 0
The answer would be an Alloy!
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A clear colorless liquid in an open beaker was heated to boiling. The liquid began to boil at 110°C, and as vapors escaped, the
Vaselesa [24]

Omitted options and they are

a) pure compound.

b. pure element.

c. pure substance.

d. homogeneous solution.

e. heterogeneous solution

Answer:d. homogeneous solution.

Explanation:

Pure substances or  elements or  compounds have a definite and sharp melting or boiling point, Any substance that is not pure is impure and will have different temperature of melting or boiling points.

To this effect, the clear colorless liquid cannot be a Pure substance, element or compound.

We can therefore say that the clear colorless liquid would be a homogeneous solution because a homogeneous solution is a mixture of constituents which completely mixes together such that each constituents cannot be seen with naked eye, When heated to boiling, each constituent in the mixture will give different boiling points.

A heterogeneous Solution, too is a mixture but contains constituents that can be seen and not a clear colourless solution.

Therefore  On the basis of this information, we can say that the material in the beaker was a Homogeneous solution

8 0
4 years ago
At a certain temperature, 0.800 mol SO 3 is placed in a 1.50 L container. 2 SO 3 ( g ) − ⇀ ↽ − 2 SO 2 ( g ) + O 2 ( g ) At equil
Elanso [62]

Answer : The value of equilibrium constant (K) is, 0.004

Explanation :

First we have to calculate the concentration of SO_3\text{ and }O_2

\text{Concentration of }SO_3=\frac{\text{Moles of }SO_3}{\text{Volume of solution}}=\frac{0.800mol}{1.50L}=1.2M

and,

\text{Concentration of }O_2=\frac{\text{Moles of }O_2}{\text{Volume of solution}}=\frac{0.150mol}{1.50L}=0.1M

Now we have to calculate the value of equilibrium constant (K).

The given chemical reaction is:

                       2SO_3(g)\rightarrow 2SO_2(g)+O_2(g)

Initial conc.       1.2             0               0

At eqm.          (1.2-2x)        2x               x

As we are given:

Concentration of O_2 at equilibrium = x = 0.1 M

The expression for equilibrium constant is:

K_c=\frac{[SO_2]^2[O_2]}{[SO_3]^2}

Now put all the given values in this expression, we get:

K_c=\frac{(2x)^2\times (x)}{(1.2-2x)^2}

K_c=\frac{(2\times 0.1)^2\times (0.1)}{(1.2-2\times 0.1)^2}

K_c=0.004

Thus, the value of equilibrium constant (K) is, 0.004

3 0
4 years ago
Need help fast !!!!!!!
Sveta_85 [38]

Answer:

Brittle

Explanation:

because thas for non metals

6 0
3 years ago
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Calculate the number of mol of solute in 8.00 × 102 mL of 0.250 M HCl.
Alenkasestr [34]

Moles = Volume*Molarity

           = 0.008*0.250

           = 0.002 mol

8 0
3 years ago
Given the following unbalanced chemical reaction: As + NaOH Na3AsO3 + H2 What would be the coefficient of the NaOH molecule in t
Lady_Fox [76]
I believe 2 would be the coefficient
6 0
3 years ago
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