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3 years ago

Not able to be separated by physical means but is chemical what is it

1 answer:

8 0

My science text book said that it was either diamond or gold. Gold may not be right, but I am pretty sure diamond is.

Sorry if I got this wrong.

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What is the reaction which a metal element displaces the metal atom of a compound?

Marina86 [1]

A single replacement reaction, sometimes called a single displacement reaction, is a reaction in which one element is substituted for another element in a compound. The starting materials are always pure elements, such as a pure zinc metal or hydrogen gas, plus an aqueous compound.

4 0

3 years ago

What kind of properties can only be observed when a substance changes into a different substance?

Shkiper50 [21]

<h3>Answer:</h3><h2>Chemical properties</h2><h3>Explanation:</h3>

By its very definition, a chemical property is one which is exhibited as a result of a chemical reaction. This may happen during or after the reaction. This is because in a chemical reaction there is a transformation in the physical composition of the components and this directly affects its chemical properties.

4 0

3 years ago

“Kids and grown-ups love it so, the happy world of HARIBO!” Who does not like the dark brown liquorice and its lovely taste?! Ta

KiRa [710]

Answer:

Any Questions?

So i can give you the answer

3 0

2 years ago

1.) The process for converting ammonia to nitric acid involves the conversion of NH3 to

Firdavs [7]

Answer:

a) 1.39 g ; b) O₂ is limiting reactant, NH₃ is excess reactant; c) 0.7 g

Explanation:

We have the masses of two reactants, so this is a limiting reactant problem.

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.

MM: 17.03 32.00 30.01

4NH₃ + 5O₂ ⟶ 4NO + 6H₂O

Mass/g: 1.5 1.85

2. Calculate the moles of each reactant

$\text{moles of NH}_{3} = \text{1.5 g NH}_{3} \times \dfrac{\text{1 mol NH}_{3}}{\text{17.03 g NH}_{3}} = \text{0.0881 mol NH}_{3}\\\\\text{moles of O}_{2} = \text{1.85 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.057 81 mol O}_{2}$

3. Calculate the moles of NO we can obtain from each reactant

From NH₃:

The molar ratio is 4 mol NO:4 mol NH₃

$\text{Moles of NO} = \text{0.0881 mol NH}_{3} \times \dfrac{\text{4 mol NO}}{\text{4 mol NH}_{3}} = \text{0.0881 mol NO}$

From O₂:

The molar ratio is 4 mol NO:5 mol O₂

$\text{Moles of NO} = \text{0.057 81 mol O}_{2}\times \dfrac{\text{4 mol NO}}{\text{5 mol O}_{2}} = \text{0.046 25 mol NO}$

4. Identify the limiting and excess reactants

The limiting reactant is O₂ because it gives the smaller amount of NO.

The excess reactant is NH₃.

5. Calculate the mass of NO formed

$\text{Mass of NO} = \text{0.046 25 mol NO}\times \dfrac{\text{30.01 g NO}}{\text{1 mol NO}} = \textbf{1.39 g NO}$

6. Calculate the moles of NH₃ reacted

The molar ratio is 4 mol NH₃:5 mol O₂

$\text{Moles reacted} = \text{0.057 81 mol O}_{2} \times \dfrac{\text{4 mol NH}_{3}}{\text{5 mol O}_{2}} = \text{0.046 25 mol NH}_{3}$

7. Calculate the mass of NH₃ reacted

$\text{Mass reacted} = \text{0.046 25 mol NH}_{3} \times \dfrac{\text{17.03 g NH}_{3}}{\text{1 mol NH}_{3}} = \text{0.7876 g NH}_{3}$

8. Calculate the mass of NH₃ remaining

Mass remaining = original mass – mass reacted = (1.5 - 0.7876) g = 0.7 g NH₃

8 0

3 years ago

Calculate the ratio of the mass ratio of SS to OO in SOSO to the mass ratio of SS to OO in SO2SO2. Consider a sample of SOSO in

Veronika [31]

Answer:

The ratio of the mass ratio of S to O; in SO, to the mass ratio of S to O; in SO₂, is 2:1

Explanation:

According to the consideration, let us first find the ratio of S and O in both the compounds

For SO:

$\frac{m_{S} }{m_{O} }= \frac{32}{16}\\\\ \frac{m_{S} }{m_{O} }= 2$

Let us express it as

$SO_{\frac{m_{S} }{m_{O} }} = 2$

For SO₂,

Due to two oxygen atoms in the molecule, the mass of oxygen will be taken two times

$\frac{m_{S} }{m_{O} }= \frac{32}{(2)(16)}\\\\ \frac{m_{S} }{m_{O} }= 1$

Let us express it as

$SO_2_{\frac{m_{S} }{m_{O} }}= 1$

Now, for the ratio of both the above-calculated ratios,

$\frac{SO_{\frac{m_{S} }{m_{O} }}}{SO_2_{\frac{m_{S} }{m_{O} }}}=\frac{2}{1}$

The required ratio is 2:1

3 0

3 years ago