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Yakvenalex [24]
3 years ago
12

Which chemical reaction involving water is a decomposition reaction (breaking up) and which is a synthesis (building up) reactio

n?
Physics
2 answers:
vichka [17]3 years ago
5 0
Water breaking down into hydrogen and oxygen is a decomposition reaction.
<span>Hydrogen and oxygen combining to form water is a synthesis reaction.</span>
tino4ka555 [31]3 years ago
3 0

A decomposition reaction would be the electrolysis.

Usually it is separated in two equations, one for oxidation and another for reduction, which are called half reactions:

2 H+ + 2e− → H2

2 H2O → O2 + 4 H+ + 4e−

Then you should add them up. In order for the equation to be balanced, the first equation has to be added twice:

2 H+ + 2e− + 2 H+ + 2e−+2 H2O → H2+ H2 + O2 + 4 H+ + 4e−

4 H+ + 4e− +2 H2O → 2H2 + O2 + 4 H+ + 4e−

Then you should cancel the substances which are both in reactives and in products, and also the electrons:

2 H2O → 2H2 + O2

And there you have the overall equation for electrolysis.

Electrolysis does not happen naturally – you have to provide electric energy for it to happen.

A synthesis reaction would be the opposite reaction, which would be called: electrosyntesis.

2H2 + O2 → 2 H2O


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3 years ago
Consider an electron with charge −e and mass m orbiting in a circle around a hydrogen nucleus (a single proton) with charge +e.
alexandr1967 [171]

Answer:

v=\sqrt{k\frac{e^2}{m_e r}}, 2.18\cdot 10^6 m/s

Explanation:

The magnitude of the electromagnetic force between the electron and the proton in the nucleus is equal to the centripetal force:

k\frac{(e)(e)}{r^2}=m_e \frac{v^2}{r}

where

k is the Coulomb constant

e is the magnitude of the charge of the electron

e is the magnitude of the charge of the proton in the nucleus

r is the distance between the electron and the nucleus

v is the speed of the electron

m_e is the mass of the electron

Solving for v, we find

v=\sqrt{k\frac{e^2}{m_e r}}

Inside an atom of hydrogen, the distance between the electron and the nucleus is approximately

r=5.3\cdot 10^{-11}m

while the electron mass is

m_e = 9.11\cdot 10^{-31}kg

and the charge is

e=1.6\cdot 10^{-19} C

Substituting into the formula, we find

v=\sqrt{(9\cdot 10^9 m/s) \frac{(1.6\cdot 10^{-19} C)^2}{(9.11\cdot 10^{-31} kg)(5.3\cdot 10^{-11} m)}}=2.18\cdot 10^6 m/s

7 0
3 years ago
A residential subdivision encompasses 1100 acres with a housing density of four houses per acre. Assume that a high-value reside
aleksandrvk [35]

Answer:

(a). The average daily demand of this subdivision is 2444.44 gallon/min.

(b). The design-demand used to design the distribution system is 2444.44 gallon/min.

Explanation:

Given that,

Area = 1100 acres

Number of house in 1 acres = 4

\text{Number of house in 1100 acres} = 4\times1100

\text{Number of house in 1100 acres} = 4400

Per house water demand = 800 g/day/house

(a). We need to calculate the average daily demand of this subdivision

Using formula for average daily demand

\text{average daily demand}=house\times\text{Per house water demand}

\text{average daily demand}=4400\times800\ gallon/day

\text{average daily demand}=3520000\ gallon/day

\text{average daily demand}=\dfrac{3520000}{24\times60}\ gallon/min

\text{average daily demand}=2444.44\ gallon/min

The average daily demand of this subdivision is 2444.44 gallon/min.

(b). We need to calculate the design-demand used to design the distribution system

Using formula for the design-demand

\text{design demand}=(Q_{max})daily\times\text{fire flow}

\text{design demand}=1.64\times2444.4\times1000

\text{design demand}=4008816\ gallon/m

Hence, (a). The average daily demand of this subdivision is 2444.44 gallon/min.

(b). The design-demand used to design the distribution system is 2444.44 gallon/min.

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2 years ago
Can anyone help with this physics question
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physical science, Earth science and life science

i know if its right but hope this helps :)

3 0
3 years ago
Read 2 more answers
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