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3 years ago

At each step, a dichotomous key involves how many choices

Chemistry

2 answers:

kvasek [131]3 years ago

8 0

Two, hope this helps

GaryK [48]3 years ago

6 0

Two is the correct answer :3

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Cacodyl, which has an intolerable garlicky odor and is used in the manufacture of cacodylic acid, a cotton herbicide, has a mass

Papessa [141]

Answer:

The molecular formula of cacodyl is C₄H₁₂As₂.

Explanation:

<u>Let's assume we have 1 mol of cacodyl</u>, in that case we'd have 209.96 g of cacodyl and the<u> following masses of its components</u>:

209.96 g * 22.88/100 = 48.04 g C

209.96 g * 5.76/100 = 12.09 g H

209.96 g * 71.36/100 = 149.83 g As

Now we convert those masses into moles:

48.04 g C ÷ 12 g/mol = 4.00 mol C

12.09 g H ÷ 1 g/mol = 12.09 mol H

149.83 g As ÷ 74.92 g/mol = 2.00 mol As

Those amounts of moles represent the amount of each component in 1 mol of cacodyl, thus, the molecular formula of cacodyl is C₄H₁₂As₂.

3 0

2 years ago

What is the total number of atoms represented in the formula CuSO4*5H2O

Alenkinab [10]

There are 21 atoms represented in the formula :) hope this helped.

4 0

3 years ago

Find the pH of a 0.010 M HNO2 solution.

lidiya [134]

Data:
Molar Mass of HNO2
H = 1*1 = 1 amu
N = 1*14 = 14 amu
O = 3*16 = 48 amu
------------------------
Molar Mass of HNO2 = 1 + 14 + 48 = 63 g/mol

M (molarity) = 0.010 M (Mol/L)

Now, since the Molarity and ionization constant has been supplied, we will find the degree of ionization, let us see:
M (molarity) = 0.010 M (Mol/L)
Use: Ka (ionization constant) = $5.0*10^{-4}$
$\alpha^2 (degree\:of\:ionization) = ?$

$Ka = M * \alpha^2$
$5.0*10^{-4} = 0.010* \alpha^2$
$0.010\alpha^2 = 5.0*10^{-4}$
$\alpha^2 = \frac{5.0*10^{-4}}{0.010}$
$\alpha^2\approx500*10^{-4}$

$\alpha\approx\sqrt{500*10^{-4}}$
$\alpha \approx 2.23*10^{-3}$

Now, we will calculate the amount of Hydronium [H3O+] in nitrous acid (HNO2), multiply the acid molarity by the degree of ionization, we will have:

$[ H_{3} O^+] = M* \alpha$
$[ H_{3} O^+] = 0.010* 2.23*10^{-3}$
$[ H_{3} O^+] \approx 0.0223*10^{-3}$
$[ H_{3} O^+] \approx 2.23*10^{-5} \:mol/L$

And finally, we will use the data found and put in the logarithmic equation of the PH, thus:

Data:
log10(2.23) ≈ 0.34
pH = ?
$[ H_{3} O^+] = 2.23*10^{-5}$

Formula:
$pH = - log[H_{3} O^+]$

Solving:
$pH = - log[H_{3} O^+]$
$pH = -log2.23*10^{-5}$
$pH = 5 - log2.23$
$pH = 5 - 0.34$
$\boxed{\boxed{pH = 4.66}}\end{array}}\qquad\quad\checkmark$

Note:. The pH <7, then we have an acidic solution.

6 0

3 years ago

Can anyone help me?

fomenos

its C it has soo much PH

8 0

3 years ago

A series stack of alkaline cells produces 6 kW of power. If each cell contributes 14.5 W, how many cells are in the stack? (Roun

liberstina [14]

The answer to this question would be the last option - D) 414

4 0

3 years ago

Read 2 more answers