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2 years ago

How does a bannana flaot

2 answers:

6 0

A banana floats when put in water because bananas are less dense than water, meaning that things with a higher density naturally push it to the surface.
Root beer floats are a different matter, though ;-)
Hope that helped =)

Tatiana [17]2 years ago

6 0

Because things that float in liquids are less dense than the liquid they are put into

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Pure magnesium metal is often found as ribbons and can easily burn in the presence of oxygen. when 2.59 g of magnesium ribbon bu

stepan [7]

Burning Mg in the air and reacting with O2 forming a white powder of MnO

So the equation is going to be:
Mn + O2 ⇒ MnO (this equation is not conserved)

to make it equilibrium:
1- First we should put 2Mno to equal the O2 on both sides.
So it will be:
Mg + O2⇒ 2MgO
2- Second we should put 2Mn to equal the Mn on both sides.
2Mg + O2⇒ 2MgO (this equation is conserved)
After putting the physical states the final equilibrium equation is going to be:
Δ
2Mg(s) + O2(g)⇒ 2MgO(s)

4 0

3 years ago

-. Convert 54 m to its equivalent measurement in cm.

Sergio [31]

Answer:

5400 cm

Explanation:

$\huge \because \: 1 \: m = 100 \: cm \\ \\ \huge \therefore \: 54 \: m = 54 \times 100 \\ \\ \: \huge \therefore \: 54 \: m= 5400 \: cm$

6 0

3 years ago

PLZ HELP I WILL GIVE BRAINLISTS TO RIGHT ANSWER

galina1969 [7]

..........The answer is B

6 0

2 years ago

For each reaction, find the value of ΔSo. Report the value with the appropriate sign. (a) 3 NO2(g) + H2O(l) → 2 HNO3(l) + NO(g)

aev [14]

Answer:

ΔS° = -268.13 J/K

Explanation:

Let's consider the following balanced equation.

3 NO₂(g) + H₂O(l) → 2 HNO₃(l) + NO(g)

We can calculate the standard entropy change of a reaction (ΔS°) using the following expression:

ΔS° = ∑np.Sp° - ∑nr.Sr°

where,

ni are the moles of reactants and products

Si are the standard molar entropies of reactants and products

ΔS° = [2 mol × S°(HNO₃(l)) + 1 mol × S°(NO(g))] - [3 mol × S°(NO₂(g)) + 1 mol × S°(H₂O(l))]

ΔS° = [2 mol × 155.6 J/K.mol + 1 mol × 210.76 J/K.mol] - [3 mol × 240.06 J/K.mol + 1 mol × 69.91 J/k.mol]

ΔS° = -268.13 J/K

7 0

3 years ago

Which reactions are single displacement reactions? 2NaBr + CaF2 → 2NaF + CaBr2 SnO2 + 2H2 → Sn + 2H2O 4Na + O2 → 2Na2O 2KClO3 →

r-ruslan [8.4K]

Answer : The correct options are, $SnO_2+2H_2\rightarrow Sn+2H_2O$ and $2Al+6HCl\rightarrow 2AlCl_3+3H_2$

Explanation :

Single displacement reaction : It is a type of chemical reaction in which the more reactive element displaces the less reactive element.

Option A reaction : $2NaBr+CaF_2\rightarrow 2NaF+CaBr_2$

It is an example of double displacement reaction because in this reaction a positive cation and a negative anion of the two reactants exchange their places to form two new products.

Option B reaction : $SnO_2+2H_2\rightarrow Sn+2H_2O$

It is an example of single displacement reaction.

Option C reaction : $4Na+O_2\rightarrow 2Na_2O$

It is an example of combination reaction because in this reaction two reactants react to give a single product.

Option D reaction : $2KClO_3\rightarrow 2KCl+3O_2$

It is an example of decomposition reaction because in this reaction a single reactant decomposes into two or more products.

Option E reaction : $2Al+6HCl\rightarrow 2AlCl_3+3H_2$

It is an example of single displacement reaction because in this reaction the most react element, aluminium displaces the less reactive element, hydrogen.

Hence, the options B and E are single displacement reactions.

5 0

3 years ago