What is the positive effect of composed fertilizer?

How does the brønsted-Lowry theory define acids and bases?

Leviafan [203]

They define acids as proton donors, and bases as proton acceptors

If you were to have:

HNO3 + H2O -> H3O+. + NO3-

You can see that the nitric acid (HNO3) gave a hydrogen ion which has 1 proton, 0 neutrons and 0 electrons to the water so we just say that it gave a proton.

Now let's see a base

NH3 + H2O -> NH4+ + OH-

Now, you can see that the ammonia (NH3) gained a hydrogen ion (proton) from the water to become ammonium(NH4). which means it accepted a proton

That's basically it. Feel free to ask if you have any further questions

8 0

3 years ago

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potassium chlorate (kclo3) decomposes into potassium chloride (kcl) and oxygen gas (o2) how many grams of oxygen can be produced

sergeinik [125]

Answer:

$m_{O_2}=354.24gO_2$

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$2KClO_3(s)\rightarrow 2KCl(s)+3O_2(g)$

In such a way, as 7.38 moles of potassium chloride are decomposed, the resulting grams of oxygen are computed considering a 2 to 3 molar relationship in the chemical reaction:

$m_{O_2}=7.38molKClO_3*\frac{3molO_2}{2molKClO_3}*\frac{32gO_2}{1molO_2} \\\\m_{O_2}=354.24gO_2$

Best regards.

7 0

3 years ago

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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

Need little help pls

Julli [10]

Answer:

0.354 molal

Explanation:

The molarity (M) or the concentration of a solution is defined as the number of moles of a compound in the solution per liter of solution (mol/L), whereas molality, is defined as the number of moles of a compound in the solution per kg of the compound (mol/kg).

Given that the density of the solution is 1.202 g/mL, which is equivalent to 1.202 kg/L. Since the prefix mili- denotes a factor of one thousandth ( $10^{-3}$ ) and kilo- denotes a factor of one thousand ( ${10}^3$ ),

$1.202 \ \frac{g}{mL} \ = \ 1.202 \ \frac{g}{(10^{-3}) L} \ = \ 1.202 \ \frac{{10}^3g}{L} \ = \ 1.202 \ \frac{kg}{L}$ .

To calculate the corresponding molality of the solution, the formula

$Molality \ (mol/kg) \ = \ \frac{Molarity \ (mol/L)}{Density \ (kg/L)}$ is used.

Therefore,

$Molality \ = \ \frac{0.426 \ mol/L}{1.202 \ kg/L} \ = \ 0.354 \ mol/kg \ \ (3 \ s.f.)$

4 0

2 years ago

a 143.1 g sample of a compound contains 53.4 g of carbon, 16.9 g of hydrogen, 43 g of nitrogen, and some amount of oxygen. what

Alex Ar [27]

Mass of oxygen = 143,1g - (53,4g + 16,9g + 43g) = 29,8g

If 143,1g ------ is ------ 100%
so
29,8g ------- is ------ %O

%O = (29,8g*100%) / 143,1g = 20,82%

3 0

3 years ago

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