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N76 [4]
3 years ago
15

Cuanto se demora un clavo en un envase con vinagre en oxidarse?

Chemistry
1 answer:
Alexxandr [17]3 years ago
6 0

Answer:

Verter lentamente partes iguales de vinagre y peróxido de hidrógeno para formar la solución. Mezclar estos dos ingredientes crea una pequeña cantidad de ácido peracético que oxidará el metal en la uña y creará óxido. La solución debe empezar a inflamarse y ponerse roja en los próximos cinco minutos.

Explanation:

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The pressure of a sample of helium in a 200. ml. container is 2.0 atm. If the 5 points
just olya [345]

The pressure of the gas = 40 atm

<h3>Further explanation</h3>

Given

200 ml container

P = 2 atm

final volume = 10 ml

Required

Final pressure

Solution

Boyle's Law  

At a fixed temperature, the gas volume is inversely proportional to the pressure applied  

\tt \rm p_1V_1=p_2.V_2\\\\\dfrac{p_1}{p_2}=\dfrac{V_2}{V_1}

Input the value :

P₂ = P₁V₁/V₂

P₂ = 2 x 200 / 10

P₂ = 40 atm

3 0
3 years ago
Using the data below, calculate the enthalpy for the combustion of C to CO
Leona [35]

Answer:

ΔH3 = -110.5 kJ.

Explanation:

Hello!

In this case, by using the Hess Law, we can manipulate the given equation to obtain the combustion of C to CO as shown below:

C(s) + 1/2O2(g) --> CO(g)

Thus, by letting the first reaction to be unchanged:

C(s) + O2(g)--> CO2 (g) ; ΔH1 = -393.5 kJ

And the second one inverted:

CO2(g) --> CO(g) + 1/2O2(g) ; ΔH2= 283.0kJ

If we add them, we obtain:

C(s) + O2(g) + CO2(g) --> CO(g) + CO2 (g) + 1/2O2(g)

Whereas CO2 can be cancelled out and O2 subtracted:

C(s) + 1/2O2(g)  --> CO(g)

Therefore, the required enthalpy of reaction is:

ΔH3 = -393.5 kJ + 283.0kJ

ΔH3 = -110.5 kJ

Best regards!

6 0
3 years ago
If element X has 25 protons, how many electrons does it have?
igor_vitrenko [27]
Im pretty sure it would also be 25
7 0
4 years ago
Read 2 more answers
How many atoms are in 4.4 mol Zn?
nirvana33 [79]
1 mol zn ------------- 6.02x10²³ atoms
4.4 mol -------------- ??

4.4 x ( 6.02x10²³ ) / 1 => 2.648x10²⁴ atoms of zn
6 0
3 years ago
How much work is needed to assemble an atomic nucleus containing three protons (such as Be) if we model it as an equilateral tri
klio [65]

Answer:

2.16 MeV

Explanation:

To determine the amount of work done that is needed to assemble the atomic mass; we need to apply the equation;

U = \frac{3}{4 \pi E_o} (\frac{e^2}{r})

where:

\frac{1}{4 \pi E_o} = proportionality constant = (9*10^9N.m^2/C^2)

e = magnitude of the charge of each electron = (1.6*10^{-19} C)^2

r = length of each side of the vertex = (2.00*10^{-15}m)

So; replacing our values into above equation; we have:

U = 3\frac{(9*10^9N.m^2/C^2)(1.6*10^{-19} C)^2}{(2.00*10^{-15}m)}

U = 3.456 × 10 ⁻¹³ J

If we have to convert our unit from J to Mev; then we are going to have:

U = 3.456 × 10 ⁻¹³ J (\frac{1 MeV}{1.602*10^{-13}J})

U = 2.16 MeV

Therefore, the amount of work done needed to assemble an atomic nucleus =  2.16 MeV

8 0
3 years ago
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