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3 years ago

Which change occurs at the anode in an operating electrochemical cell?

Chemistry

1 answer:

murzikaleks [220]3 years ago

6 0

Answer:loss of electrons

Explanation:

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4. A student was not sure that she accurately recorded the final temperature of her melting point sample. Therefore, she decided

Mumz [18]

Answer:

AWF DSF

Explanation:

W AFSDDDDDDDDD

3 0

2 years ago

Do you think it matters if you

vovikov84 [41]

Answer:

yes I do think they mean the same thing

Explanation:

150 grams and 150 grams is the same thing and adding 0 to the end of a decimal does not change its value, you could even put 150.0000000 grams and it would still be equivalent to the other numbers

5 0

2 years ago

What is the percentage of oxygen in Li(NO2)3

BartSMP [9]

Answer:

66.2 % of O

Explanation:

Our compound is the lithium nitrite.

LiNO₂

This salt is ionic and can be dissociated: LiNO₂ → Li⁺ + NO₂⁻

We determine the molar mass:

molar mass of Li + 3 . molar mass of N + 6 . molar mass of O

6.94 g/mol + 3. 14 g/mol + 6 . 16 g/mol = 144.94 g/mol

The mass of oxygen contained in 1 mol of lithium nitrite is:

6 . 16 g/mol = 96 g

So the percentage of oxygen present is:

(96 g / 144.94 g) . 100 = 66.2 %

3 0

2 years ago

Write Democritus in your own words HELP ASAP

harina [27]

Answer:

In explanation.

Explanation:

In the file attached is the answer, couldn´t answer it here.

7 0

2 years ago

Acetylsalicylic acid (aspirin), HC9H7O4, is the most widely used pain reliever and fever reducer. Find the pH of 0.035 M aqueous

KatRina [158]

The pH of 0.035 M aqueous aspirin is 2.48

Explanation:

We are given:

Concentration of aspirin = 0.035 M

The chemical equation for the dissociation of aspirin (acetylsalicylic acid) follows:

$HC_9H_7O_4\rightleftharpoons H^++C_9H_7O_4^-$

Initial: 0.035

At eqllm: 0.035-x x x

The expression of $K_a$ for above equation follows:

$K_a=\frac{[C_9H_7O_4^-][H^+]}{[HC_9H_7O_4]}$

We are given:

$K_a=3.6\times 10^{-4}$

Putting values in above expression, we get:

$3.6\times 10^{-4}=\frac{x\times x}{(0.035-x)}\\\\x=-0.0037,0.0033$

Neglecting the value of x = -0.0037 because concentration cannot be negative

So, concentration of $H^+$ = x = 0.0033 M

To calculate the pH of the solution, we use the equation:

$pH=-\log[H^+]$

We are given:

$H^+$ = 0.0033 M

Putting values in above equation, we get:

$pH=-\log(0.0033)\\\\pH=2.48$

Hence, the pH of 0.035 M aqueous aspirin is 2.48

7 0

3 years ago