Describe how hydrogen and oxygen form water.

2 answers:

7 0

Hydrogen and Oxygen by themselves have 1 and 6 valence electrons, respectively. 8 valence electrons is stable, so the atoms form bonds with each other to achieve 8 valence e-.

1 H atom + 1 H atom + 1 O atom = 8 valence e-

sveta [45]3 years ago

3 0

The oxygen atom requires 2 electrons to complete its valance shell. Hydrogen is special in that it can eighter give or take one electron and still be stable. So hydrogen gives its electron to oxygen, and two hydrogens fill up all of oxygen's valance shell, and it forms water.

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Describe how visible spectroscopy could be used to determine the order of a reaction in which a single colorless reactant produc

d1i1m1o1n [39]

Answer:

Explanation:

In a reaction, where, one of the reactant produces a colored product, visible spectroscopy can be used to determined the order of a reaction, the change in concentration of the reactant which forms the colored product is determined by absorbance measurement over time. The data for the concentration and time are plotted on the y and x axis and If we get a straight line it is a zero-order reaction. If instead, a plot of ln[concentration] versus time gives a straight line, it is a first order reaction. However, If 1/concentration versus time gives a straight line, it is a second order reaction kinetics. The other reactants may be changed while keeping this reactant as constant and change on rate of the reaction is observed to see If the other reactant affects the reaction or not.

8 0

4 years ago

Will mark brainliest

ikadub [295]

Answer:

a. Heterogeneous
b. Homogeneous
c. Homogeneous
d. Heterogeneous
e. Heterogeneous

Explanation:

A heterogeneous mixture is a mixture in which you can see multiple different ingredients in, for example vegetable soup, tea with ice and lemon slices, or fruit salad.

A homogeneous mixture is a mixture in which you can only see one thing, for example tea, seawater, or milk.

8 0

3 years ago

A sample of HI (9.30×10^−3mol) was placed in an empty 2.00 L container at 1000 K. After equilibrium was reached, the concentrati

Sedaia [141]

Answer:

The answer is "29.081"

Explanation:

when the empty 2.00 L container of 1000 kg, a sample of HI (9.30 x 10-3 mol) has also been placed.

$\text{calculating the initial HI}= \frac{mol}{V}$

$=\frac{9.3 \times 10 ^ -3}{2}$

$=0.00465 \ Mol$

$\text{Similarly}\ \ I_2 \ \ \text{follows} \ \ H_2 = 0 }$

Its density of I 2 was 6.29x10-4 M if the balance had been obtained, then we have to get the intensity of equilibrium then:

$HI = 0.00465 - 2x\\\\ I_{2} \ eq = H_2 \ eq = 0 + x \\\\$

It is defined that:

$I_2 = 6.29 \times 10^{-4} \ M \\\\x = I_2 \\\\$

$HI \ eq= 0.00465 - 2x \\$

$=0.00465 -2 \times 6.29 \times 10^{-4} \\\\ = 0.00465 -\frac{25.16 }{10^4} \\\\ = 0.003392\ M$

Now, we calculate the position:

For the reaction $H 2(g) + I 2(g)\rightleftharpoons 2HI(g)$ , you can calculate the value of Kc at 1000 K.

data expression for Kc

$2HI \rightleftharpoons H_2 + I_2 \\\\\to Kc = \frac{H_2 \times I_2}{HI^2}$

$= \frac{6.29\times10^{-4} \times 6.29 \times 10^{-4}}{0.003392^2} \\\\= \frac{6.29\times 6.29 \times 10^{-8}}{0.003392^2} \\\\= \frac{39.564 \times 10^{-8}}{1.150 \times 10-5} \\\\= 0.034386$