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4 years ago

7

where do most earthquakes occur ? A. within volcanoes where plates interact B. whar plates interact C. whar plates diverge D. at

hot spots

2 answers:

snow_tiger [21]4 years ago

8 0

I think the answer is B.

Naily [24]4 years ago

8 0

Answer:

b

Explanation:

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Turn the word equation into a chemical balanced equation.

yan [13]

(2)KClO3 = (2)KCl + (3)O2

you have 2 Potassium Chlorates for every 2 Potassium Chlorides and 3 Oxygen

7 0

3 years ago

Ka/KbMIXED PRACTICE108.Calculate the [H3O+(aq)], the pH, and the % reaction for a 0.50 mol/L HCN solution. ([H3O+(aq)] = 1.8 x 1

NeTakaya

Answer:

a) [H₃O⁺] = 1.8x10⁻⁵ M

b) pH = 4.75

c) % rxn = 3.5x10⁻³ %

Explanation:

a) The dissociation reaction of HCN is:

HCN(aq) + H₂O(l) ⇄ H₃O⁺(aq) + CN⁻(aq)

0.5 M - x x x

The dissociation constant from the above reactions is given by:

$Ka = \frac{[H_{3}O^{+}][CN^{-}]}{[HCN]} = 6.17 \cdot 10^{-10}$

$6.17 \cdot 10^{-10} = \frac{x*x}{(0.5 - x)}$

$6.17 \cdot 10^{-10}*(0.5 - x) - x^{2} = 0$

By solving the above quadratic equation we have:

x = 1.75x10⁻⁵ M = 1.8x10⁻⁵ M = [H₃O⁺] = [CN⁻]

Hence, the [H₃O⁺] is 1.8x10⁻⁵ M.

b) The pH is equal to:

$pH = -log[H_{3}O^{+}] = -log(1.75 \cdot 10^{-5} M) = 4.75$

Then, the pH of the HCN solution is 4.75.

c) The % reaction is the % ionization:

$\% = \frac{x}{[HCN]} \times 100$

$\% = \frac{1.75 \cdot 10^{-5} M}{0.5 M} \times 100$

$\% = 3.5 \cdot 10^{-3} \%$

Therefore, the % reaction or % ionization is 3.5x10⁻³ %.

I hope it helps you!

6 0

3 years ago

Predict the organic product formed when phosgene reacts with an excess of methanol.

gulaghasi [49]

When phosgene is reacted with an excess of methanol, this produces dimethyl carbonate. Dimethyl carbonate is used in methylating anilines, phenols, and carboxylic acids, and is also being studied for its use as an alternative additive in fuel.

The industrial production of dimethyl carbonate no longer uses phosgene because of the latter's toxicity. Phosgene widely gained infamy when it was used in chemical warfare during WWI.

7 0

4 years ago

Write in exponential form: 34⋅34⋅34⋅34⋅34⋅34⋅34⋅34.

kvv77 [185]

Answer:

34^8

Explanation:

7 0

3 years ago

The compound known as diethyl ether, commonly referred to as ether, contains carbon, hydrogen, and oxygen. A 1.751 g sample of e

fredd [130]

<u>Answer:</u> The empirical formula for the given compound is $C_{4}H_{10}O$

<u>Explanation:</u>

The chemical equation for the combustion of ether follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=4.159g$

Mass of $H_2O=2.128g$

Mass of sample = 1.751 g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

<u>For calculating the mass of carbon:</u>

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 4.159 g of carbon dioxide, $\frac{12}{44}\times 4.159=1.134g$ of carbon will be contained.

<u>For calculating the mass of hydrogen:</u>

In 18 g of water, 2 g of hydrogen is contained.

So, in 2.128 g of water, $\frac{2}{18}\times 2.128=0.236g$ of hydrogen will be contained.

Mass of oxygen in the compound = (1.751) - (1.134 + 0.236) = 0.381 g

To formulate the empirical formula, we need to follow some steps:

<u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{1.134g}{12g/mole}=0.0945moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.236g}{1g/mole}=0.236moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.381g}{16g/mole}=0.0238moles$

<u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0238 moles.

For Carbon = $\frac{0.0945}{0.0238}=3.97\approx 4$

For Hydrogen = $\frac{0.236}{0.0238}=9.91\approx 10$

For Oxygen = $\frac{0.0238}{0.0238}=1$

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 4 : 10 : 1

Hence, the empirical formula for the given compound is $C_{4}H_{10}O$

7 0

3 years ago