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3 years ago

what concentration in parts per million in a solution that contains .008 grams of o2 dissolved in 1000 grams of water

1 answer:

7 0

Ppm = mass of solute mg / mass of solvent kg

0.008 * 1000 = 8.0 mg ( solute )

1000 / 1000 = 1.0 kg (solvent )

ppm = 8 / 1

= 8.0 ppm

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Consider the reaction of peroxydisulfate ion (S2O2−8) with iodide ion (I−) in aqueous solution: S2O2−8(aq)+3I−(aq)→2SO2−4(aq)+I−

kolbaska11 [484]

Answer:

r = 3.61x $10^{-6}$ M/s

Explanation:

The rate of disappearance (r) is given by the multiplication of the concentrations of the reagents, each one raised of the coefficient of the reaction.

r = k. $[S2O2^{-8} ]^{x} x [I^{-} ]^{y}$

K is the constant of the reaction, and doesn't depends on the concentrations. First, let's find the coefficients x and y. Let's use the first and the second experiments, and lets divide 1º by 2º :

$\frac{r1}{r2} = \frac{0.018^{x} x0.036^{y} }{0.027^xx0.036^y}$

$\frac{2.6x10^{-6}}{3.9x10^{-6}} = (\frac{0.018}{0.027})^xx(\frac{0.036}{0.036})^y$

$0.67 = 0.67^x$

x = 1

Now, to find the coefficient y let's do the same for the experiments 1 and 3:

$\frac{r1}{r3} = \frac{0.018x0.036^y}{0.036x0.054^y}$

$\frac{2.6x10^{-6}}{7.8x10^{-6}} = (\frac{0.018}{0.036})x(\frac{0.036}{0.054})^y$

$0.33 = 0.5x 0.67^y$

$0.67 = 0.67^y$

y = 1

Now, we need to calculate the constant k in whatever experiment. Using the first :

$2.6x10^{-6} = kx0.018x0.036$ $kx6.48x10^{-4} = 2.6x10^{-6}$

k = 4.01x10^{-3} M^{-1}s^{-1}[/tex]

Using the data given,

r = $4.01x10^{-3}x1.8x10^{-2}x5.0x10^{-2}$

r = 3.61x $10^{-6}$ M/s

7 0

2 years ago

What is the reaction type 2 Mgl2+ Mn(SO3)2 → 2 MgSO3 + Mnl4

Oksana_A [137]

Answer:

The reaction type is double displacement

4 0

2 years ago

Calculate the concentration of hydronium and hydroxide ions in a 0.050 M solution of nitric acid.

Mazyrski [523]

Answer:

[H₃O⁺] = 0.05 M & [OH⁻] = 2.0 x 10⁻¹³.

Explanation:

HNO₃ is completely ionized in water as:

HNO₃ + H₂O → H₃O⁺ + NO₃⁻.

The concentration of hydronium ion is equal to the concentration of HNO₃:

[H₃O⁺] = 0.05 M.

∵ [H₃O⁺][OH⁻] = 10⁻¹⁴.

∴ [OH⁻] = 10⁻¹⁴/[H₃O⁺] = 10⁻¹⁴/0.05 = 2.0 x 10⁻¹³.

3 0

2 years ago

In the laboratory, a volume of 100 mL of sulfuric acid (H2SO4) is recorded. How many g are there of the liquid if its density is

ser-zykov [4K]

Answer:

$\large \boxed{\text{183 g}}$

Explanation:

$\begin{array}{rcl}\text{Density} & = & \dfrac{\text{Mass}}{\text{Volume}}\\\\\rho & = &\dfrac{m}{V}\\\\1.83 \text{ g$\cdot$ cm}^{-3} & = & \dfrac{m}{\text{100 cm}^{3}}\\\\m & = & \text{183 g}\\\end{array}\\\text{There are $\large \boxed{\textbf{183 g}}$ of sulfuric acid.}$

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3 years ago

Enter the correct ground-state (or lowest energy) configuration based on the number of electrons: 1s^4 2s^4 2p^12.

Blababa [14]

Answer:

The ground state configuration is the lowest energy, most stable arrangement. An excited state configuration is a higher energy arrangement (it requires energy input to create an excited state). Valence electrons are the electrons utilised for bonding.

or the

FIGURE 5.9 The arrow shows a second way of remembering the order in which sublevels fill. Table 5.2 shows the electron configurations of the elements with atomic numbers 1 through 18.

Element Atomic number Electron configuration

sulfur 16 1s22s22p63s23p4

chlorine 17 1s22s22p63s23p5

argon 18 1s22s22p63s23p6

or the

Two electrons

Two electrons fill the 1s orbital, and the third electron then fills the 2s orbital. Its electron configuration is 1s22s1.

Explanation:

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6 0

2 years ago