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3 years ago

7

What does this chemical reaction describe? water → hydrogen + oxygen respiration decomposition of water combustion of hydrogen f

uel chemical reaction in a battery

1 answer:

Lubov Fominskaja [6]3 years ago

4 0

I think it describes a chemical reaction in a battery involving the decomposition of water into hydrogen plus oxygen gas by electrolysis due to an electric current passing through the water and causing the dissociation of the two constituent elements. This is why water in wet cell battery can be seen to be bubbling during this process. Wet cell batteries must have water to function properly. and the water must be filled up regularly as it gets consumed.

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Sandra is having difficulty with her reading assignment because she does not fully understand the language. Which online tool wo

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Answer:

The answer is a TRANSLATION TOOL or D

Explanation:

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3 years ago

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A particle of mass 4.00 kg is attached to a spring with a force constant of 100 N/m. It is oscillating on a frictionless, horizo

zloy xaker [14]

Solution :

Given :

Mass attached to the spring = 4 kg

Mass dropped = 6 kg

Force constant = 100 N/m

Initial amplitude = 2 m

Therefore,

a). $v_{initial} = A w$

$= 2 \times \sqrt{\frac{100}{4}}$

= 10 m/s

Final velocity, v at equilibrium position, v = 5 m/s

Now, $\frac{1}{2}(4+4)5^2 = \frac{1}{2} kA'$

A' = amplitude = 1.4142 m

b). $T=2 \pi \sqrt{\frac{m}{k}}$

m' = 2m

Hence, $T'=\sqrt2 T$

c). $\frac{\frac{1}{2}(4+4)5^2 + \frac{1}{2}\times 4 \times 10^2}{\frac{1}{2} \times 4 \times 10^2}$

$=\frac{1}{2}$

Therefore, factor $=\frac{1}{2}$

Thus, the energy will change half times as the result of the collision.

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3 years ago

View the chart on the characteristics of electromagnetic and mechanical waves and answer the question.

Elza [17]

I think the answer is c

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3 years ago

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a chamber with a fixed volume of 1.0 meters cubed contains a monatomic gas at 3.00 *10^K. The chamber is heated to a temperature

igomit [66]

Answer:

Explanation:

Given

Volume of fixed chamber $V=1 m^3$

Initial Temperature $T_1=300 K$

Final Temperature $T_2=400 K$

Heat Supplied $Q=10 J$

From First law of thermodynamics

Change in internal energy of the system is equal to heat added minus work done by the system

$\Delta U=Q-W$

as the volume is fixed therefore work

$W=\int PdV=0$

thus $\Delta U=mc_v\Delta T=Q$

$c_v$ for mono-atomic gas is $12.471 J/K-mol$

$n\times 12.471\times (400-300)=10$

$n=0.008018 mol$

and 1 mole contains $6.022\times 10^{23} molecules$

thus No of molecules $=0.008018\times 6.022\times 10^{23}$

No of molecules $=4.82\times 10^{21} molecules$

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4 years ago

The energy from 0.015 moles of octane was used to heat 250 grams of water. The temperature of the water rose from 293.0 K to 371

arsen [322]

Answer : The correct option is, (B) -5448 kJ/mol

Explanation :

First we have to calculate the heat required by water.

$q=m\times c\times (T_2-T_1)$

where,

q = heat required by water = ?

m = mass of water = 250 g

c = specific heat capacity of water = $4.18J/g.K$

$T_1$ = initial temperature of water = 293.0 K

$T_2$ = final temperature of water = 371.2 K

Now put all the given values in the above formula, we get:

$q=250g\times 4.18J/g.K\times (371.2-293.0)K$

$q=81719J$

Now we have to calculate the enthalpy of combustion of octane.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of combustion of octane = ?

q = heat released = -81719 J

n = moles of octane = 0.015 moles

Now put all the given values in the above formula, we get:

$\Delta H=\frac{-81719J}{0.015mole}$

$\Delta H=-5447933.333J/mol=-5447.9kJ/mol\approx -5448kJ/mol$

Therefore, the enthalpy of combustion of octane is -5448 kJ/mol.

5 0

4 years ago