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vagabundo [1.1K]
2 years ago
11

Why is NaNO3 a homogenous mixture

Chemistry
1 answer:
wlad13 [49]2 years ago
6 0
Every pure compound is homogeneous...
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Which of the following pairs of elements could possibly be in the same group? X has a 1+ ion; Y has a 1- ion. X tends to form a
exis [7]
Let's eliminate these one by one.
The first pair would not be the same, as X would most likely be in group IA, and Y would be in group VIIA, because of their tendency to gain and lose electrons.
The second pair would also violate the same rule, but X would most likely be in group IIA, and Y would most likely be in group VIA.
The third pair would not be the same, as X is most likely in group VIIA, and since Y has eight valence electrons, it is most likely a noble gas.
The final pair has X with atomic number 15, making it phosphorous. Phosphorous wants to gain 3 electrons to have a full octet of 8 outer "valence" electrons, and Y would also like to gain 3 electrons. This means it is possible that the final pair would be in the same group.
5 0
3 years ago
Read 2 more answers
A rectangular block of solid carbon (graphite) floats at the interface of two immiscible liquids.
Sergio039 [100]

Explanation:

Let us take the volume of block is x.

Since, the block is floating this means that it is in equilibrium. Formula to calculate net force will be as follows.

                F_{net} = Buoyancy force(F_{b}) - weight force(w)

Also, buoyancy force (F_{b}) = (volume submerged in water × density of water) + (volume in oil × density of oil)

          (F_{b}) = (0.592 V \times \rho) + (1 - 0.592)V \times 1000 g          

                      = (0.592 V \times \rho + 408 V) g

As,   W = V × density of graphite × g

It is given that density of graphite is 2.16 g/cm^{3} or 2160 kg/m^{3}.

So, W = 2160 V g

F_{net} = (0.592 V \rho + 408 V) g - 2160 V g = 0

            0.592 \rho = 1752

     \rho = 2959.46 kg/m^{3} or 2.959 g/cm^{3} is the density of oil.

It is given that mass of flask is 124.8 g.

Mass of 35.3 cm^{3} oil = 35.3 \times 2.959 104.7 g

Hence, in second weighing total mass will be calculated as follows.

                       (124.8 + 104.7) g

                       = 229.27 g

Thus, we can conclude that in the second weighing mass is 229.27 g.

5 0
2 years ago
How much heat is needed to raise the temperature of 55.0 g sample of water by 65.0 oC.
sveticcg [70]

Answer: 14943.5 J

Explanation:

The quantity of heat energy (Q) required to raise the temperature of a substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

Thus, Q = MCΦ

Given that,

Q = ?

Mass of water = 55.0g

C = 4.18 J/g°C

Φ = 65.0°C

Then, Q = MCΦ

Q = 55.0g x 4.18 J/g°C x 65.0°C

Q = 14943.5 J

Thus, 14943.5 joules of heat is needed to raise the temperature of water.

3 0
3 years ago
Read 2 more answers
This is the equation for the combustion of propane.
yan [13]

Answer:

B

Explanation:

Reactants are the compounds which reaction produce the products. In general terms this can be expressed symbolically as follows:

reactants -> products

Other phenomena like heat are omitted because are not always present, that is, only compounds are included. Therefore, in this reaction the reactants are C3H8 (propane) and O2 ( oxygen) and the products are CO2 (carbon dioxide) and H2O (water)

8 0
2 years ago
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A 0.10 M solution of Na2HPO4 could be made a buffer solution with all of the following EXCEPT ________.NaH2PO4K3PO4Na3PO4H3PO4
Kipish [7]

The answer is: H₃PO₄.

A phosphoric acid is three protic acid, which means that in water release tree protons.

Phosphoric acid ionizes in three steps in water.

First step: H₃PO₄(aq) ⇄ H₂PO₄⁻(aq) + H⁺(aq).

Second step: H₂PO₄⁻(aq)⇄ HPO₄²⁻(aq) + H⁺(aq).

Third step: HPO₄²⁻(aq) ⇄ PO₄³⁻(aq) + H⁺(aq).

Species that are present: H₃PO₄, H₂PO₄⁻, HPO₄²⁻, PO₄³⁻ and H⁺.

A buffer can be defined as a substance that prevents the pH of a solution from changing by either releasing or absorbing H⁺ in a solution.

Buffer is a solution that can resist pH change upon the addition of an acidic or basic components and it is able to neutralize small amounts of added acid or base, pH of the solution is relatively stable.

7 0
3 years ago
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