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2 years ago

Which of the following is an empirical formula? H6O2 H402 H2O H2O2

Chemistry

2 answers:

Jobisdone [24]2 years ago

6 0

Answer:

hola responderé

Explanation:

el h2o es agua y el agua es mojada gracias por su atención

topjm [15]2 years ago

5 0

${\qquad\qquad\huge\underline{{\sf Answer}}}$

<h3>What does an empirical formula mean ? </h3>

- Empirical formula represents a molecule with its elements combined in simplest ratio, it's not necessarily same as the form in which the compound really exists in the nature. Hence we can say that empirical formula shows the simple ratios in which atoms combined to form a molecule.

For example : Hydrogen peroxide

- Hydrogen peroxide exists in nature as $\sf H_2O_2$ , which is its molecular formula.

but it's empirical formula will be H0 [ simplest ratio of 1 : 1 ]

Therefore, the correct choice is :

C. $\sf H_2O$

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Nadusha1986 [10]

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<h3>Therefore, after long period of time 80kg of salt will remain in tank</h3>

Explanation:

given amount of salt at time t is A(t)

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$dA/dt=2-(A/40)\\\\dA=2dt-(A/40)dt\\\\dA+(A/40)dt=2dt$

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$=e^{\int\limits (1/40) \, dt}$

integrating factor $=e^{(1/40)t}$

multiply on both sides by $=e^{(1/40)t}$

$dAe^{(1/40)t}+(A/40)e^{(1/40)t} dt =2e^{(1/40)t}t\\\\(Ae^{(1/40)t})=2e^{(1/40)t}t$

integrate on both sides

$\int\limits(Ae^{(1/40)t})=\int\limits2e^{(1/40)t}dt\\\\(Ae^{(1/40)t})=2*40e^{(1/40)t}+C\\\\A=80+(C/e^{(1/40)t})\\\\A(0)=0.3\\\\0.3=80+(C/e^{(1/40)t}^*^0)\\\\0.3=80+(C/1)\\\\C=0.3-80\\\\C=-79.7\\\\A(t)=80-(79.7/e^{(1/40)t})$

after long period of time means t - > ∞

${t \to \infty}\\\\ \lim_{t \to \infty} A_t \\\\ \lim_{t \to \infty} (80)-(79/{e^{(1/40)t}}\\\\=80-(0)\\\\=80$

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